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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.2

Question 1.
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f^-1( y) in the following
(i) y = \(\sqrt {x}\)
Solution:
y = \(\sqrt {x}\) … (1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
y² = x ∴ x = y²

(ii) y = \(\sqrt{2-\sqrt{x}}\)
Solution:
y = \(\sqrt{2-\sqrt{x}}\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iii) y = \(\sqrt[3]{x-2}\)
Solution:
y = \(\sqrt[3]{x-2}\) ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(iv) y = log (2x – 1)
Solution:
y = log (2x – 1) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(v) y = 2x + 3
Solution:
y = 2x + 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),

(vi) y = e^x – 3
Solution:
y = e^x – 3 ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
e^x = y + 3
∴ x = log(y + 3)
∴ x = f^-1(y) = log(y + 3)

(vii) y = e^{2x – 3}
Solution:
y = e^{2x – 3} ….(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
2x – 3 = log y ∴ 2x = log y + 3

(viii) y = log₂\(\left(\frac{x}{2}\right)\)
Solution:
y = log₂\(\left(\frac{x}{2}\right)\) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),
\(\frac{x}{2}\) = 2^y ∴ x = 2∙2^y = 2^y+1
∴ x = f^-1(y) = 2^y+1

Question 2.
Find the derivative of the inverse function of
the following
(i) y = x²·e^x
Solution:
y = x²·e^x
Differentiating w.r.t. x, we get

(ii) y = x cos x
Solution:
y = x cos x
Differentiating w.r.t. x, we get

(iii) y = x·7^x
Solution:
y = x·7^x
Differentiating w.r.t. x, we get

(iv) y = x² + logx
Solution:
y = x² + logx
Differentiating w.r.t. x, we get

(v) y = x logx
Solution:
y = x logx
Differentiating w.r.t. x, we get

Question 3.
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them.
(i) y = x⁵ + 2x³ + 3x, at x = 1
Solution:
y = x⁵ + 2x³ + 3x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x⁵ + 2x³ + 3x)
= 5x⁴ + 2 × 3x² + 3 × 1
= 5x⁴ + 6x² + 3
The derivative of inverse function of y = f(x) is given by

(ii) y = e^x + 3x + 2, at x = 0
Solution:
y = e^x + 3x + 2
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(e^x + 3x + 2)
The derivative of inverse function of y = f(x) is given by

(iii) y = 3x² + 2 log x³, at x = 1
Solution:
y = 3x² + 2 log x³
= 3x² + 6 log x
Differentiating w.r.t. x, we get

The derivative of inverse function of y = f(x) is given by

(iv) y = sin (x – 2) + x², at x = 2
Solution:
y = sin (x – 2) + x²
Differentiating w.r.t. x, we get

Question 4.
If f(x) = x³ + x – 2, find (f^-1)’ (0).
Question is modified.
If f(x) = x³ + x – 2, find (f^-1)’ (-2).
Solution:
f(x) = x³ + x – 2 ….(1)
Differentiating w.r.t. x, we get

Question 5.
Using derivative prove
(i) tan^-1x + cot^-1x = \(\frac{\pi}{2}\)
Solution:
let f(x) = tan^-1x + cot^-1x
Differentiating w.r.t. x, we get

Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k
Let x = 0.
Then f(0) = k ….(2)
From (1), f(0) = tan^-1(0) + cot^-1(0)
= 0 + \(\frac{\pi}{2}=\frac{\pi}{2}\)

(ii) sec^-1x + cosec^-1x = \(\frac{\pi}{2}\) . . . [for |x| ≥ 1]
Solution:
Let f(x) = sec^-1x + cosec^-1x for |x| ≥ 1 ….(1)
Differentiating w.r.t. x, we get

Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k, where |x| > 1
Let x = 2.
Then, f(2) = k ……(2)

Question 6.
Diffrentiate the following w. r. t. x.
(i) tan^-1(log x)
Solution:
Let y = tan^-1(log x)
Differentiating w.r.t. x, we get

(ii) cosec^-1(e^-x)
Solution:
Let y = cosec^-1(e^-x)
Differentiating w.r.t. x, we get

(iii) cot^-1(x³)
Solution:
Let y = cot^-1(x³)
Differentiating w.r.t. x, we get

(iv) cot^-1(4^x
Solution:
Let y = cot^-1(4^x
Differentiating w.r.t. x, we get

(v) tan^-1(\(\sqrt {x}\))
Solution:
Let y = tan^-1(\(\sqrt {x}\))
Differentiating w.r.t. x, we get

(vi) sin^-1\(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Solution:
Let y = sin^-1\(\left(\sqrt{\frac{1+x^{2}}{2}}\right)\)
Differentiating w.r.t. x, we get

(vii) cos^-1(1 – x²)
Solution:
Let y = cos^-1(1 – x²)
Differentiating w.r.t. x, we get

(viii) sin^-1\(\left(x^{\frac{3}{2}}\right)\)
Solution:
Let y = sin^-1\(\left(x^{\frac{3}{2}}\right)\)
Differentiating w.r.t. x, we get

(ix) cos³[cos^-1(x³)]
Solution:
Let y = cos³[cos^-1(x³)]
= [cos(cos^-1x³)]³
= (x³)³ = x⁹
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x⁹) = 9x⁸.

(x) sin⁴[sin^-1(\(\sqrt {x}\))]
Solution:
Let y = sin⁴[sin^-1(\(\sqrt {x}\))]
= {sin[sin^-1(\(\sqrt {x}\))]}⁸
= (\(\sqrt {x}\))⁴ = x²
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x²) = 2x.

Question 7.
Diffrentiate the following w. r. t. x.
(i) cot^-1[cot (e^x2)]
Solution:
Let y = cot^-1[cot (e^x2)] = e^x2

(ii) cosec^-1\(\left(\frac{1}{\cos \left(5^{x}\right)}\right)\)
Solution:

(iii) cos^-1\(\left(\sqrt{\frac{1+\cos x}{2}}\right)\)
Solution:

(iv) cos^-1\(\left(\sqrt{\frac{1-\cos \left(x^{2}\right)}{2}}\right)\)
Solution:

(v) tan^-1\(\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)\)
Solution:

(vi) cosec^-1\(\left(\frac{1}{4 \cos ^{3} 2 x-3 \cos 2 x}\right)\)
Solution:

(vii) tan^-1\(\left(\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \left(\frac{x}{3}\right)}\right)\)
Solution:

(viii) cot^-1\(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{\sin 3 x}{1+\cos 3 x}\right)\)

(ix) tan^-1\(\left(\frac{\cos 7 x}{1+\sin 7 x}\right)\)
Solution:

(x) tan^-1\(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)
Solution:
Let y = tan^-1\(\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\)

(xi) tan^-1(cosec x + cot x)
Solution:
Let y = tan^-1(cosec x + cot x)

(xii) cot^-1\(\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)\)
Solution:

Question 8.
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:


= e^x.

(vi)
Solution:

y = sin^-1[sin(2^x)∙cosα – cos(2^x)∙sinα]
= sin^–[sin(2^x – α)]
= 2^x – α, where α is a constant
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(2^x – α)
= \(\frac{d}{d x}\)(2^x) – \(\frac{d}{d x}\)(α)
= 2^x∙log2 – 0
= 2^x∙log2

Question 9.
Diffrentiate the following w. r. t. x.
(i) cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

(ii) tan^-1\(\left(\frac{2 x}{1-x^{2}}\right)\)
Solution:

(iii) sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:

(iv) sin^-1(2x\(\sqrt{1-x^{2}}\))
Solution:

(v) cos^-1(3x – 4x³)
Solution:

(vi) cos^-1\(\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)\)
Solution:

(vii) cos^-1\(\left(\frac{1-9^{x}}{1+9^{x}}\right)\)
Solution:

(viii) sin^-1\(\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\)
Solution:

(ix) sin^-1\(\left(\frac{1-25 x^{2}}{1+25 x^{2}}\right)\)
Solution:

(x) sin^-1\(\left(\frac{1-x^{3}}{1+x^{3}}\right)\)
Solution:

(xi) tan^-1\(\left(\frac{2 x^{\frac{5}{2}}}{1-x^{5}}\right)\)

(xii) cot^-1\(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\)

Question 10.
Diffrentiate the following w. r. t. x.
(i) tan^-1\(\left(\frac{8 x}{1-15 x^{2}}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{8 x}{1-15 x^{2}}\right)\)

(ii) cot^-1\(\left(\frac{1+35 x^{2}}{2 x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{1+35 x^{2}}{2 x}\right)\)

(iii) tan^-1\(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2 \sqrt{x}}{1+3 x}\right)\)

(iv) tan^-1\(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2^{x+2}}{1-3\left(4^{x}\right)}\right)\)

(v) tan^-1\(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{2^{x}}{1+2^{2 x+1}}\right)\)

(vi) cot^-1\(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{a^{2}-6 x^{2}}{5 a x}\right)\)

(vii) tan^-1\(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{a+b \tan x}{b-a \tan x}\right)\)

(viii) tan^-1\(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
Solution:
Let y = tan^-1\(\left(\frac{5-x}{6 x^{2}-5 x-3}\right)\)
= tan^-1\(\left[\frac{5-x}{1+\left(6 x^{2}-5 x-4\right)}\right]\)

(ix) cot^-1\(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)
Solution:
Let y = cot^-1\(\left(\frac{4-x-2 x^{2}}{3 x+2}\right)\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.1

Question 1.
Differentiate the following w.r.t. x :
(i) (x³ – 2x – 1)⁵
Solution:
Method 1:
Let y = (x³ – 2x – 1)⁵
Put u = x³ – 2x – 1. Then y = u⁵

Method 2:
Let y = (x³ – 2x – 1)⁵
Differentiating w.r.t. x, we get

(ii) \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Solution:
Let y = \(\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}\)
Differentiating w.r.t. x, we get

(iii) \(\sqrt{x^{2}+4 x-7}\)
Solution:

(iv) \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Solution:
Let y = \(\sqrt{x^{2}+\sqrt{x^{2}+1}}\)
Differentiating w.r.t. x, we get

(v) \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Solution:
Let y = \(\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}\)
Differentiating w.r.t. x, we get

(vi) \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Solution:
Let y = \(\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}\)
Differentiating w.r.t. x, we get

Question 2.
Diffrentiate the following w.r.t. x
(i) cos(x² + a²)
Solution:
Let y = cos(x² + a²)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[cos(x² + a²)]
= -sin(x² + a²)∙\(\frac{d}{d x}\)x² + a²)
= -sin(x² + a²)∙(2x + 0)
= -2xsin(x² + a²)

(ii) \(\sqrt{e^{(3 x+2)}+5}\)
Solution:
Let y = \(\sqrt{e^{(3 x+2)}+5}\)
Differentiating w.r.t. x, we get

(iii) log[tan(\(\frac{x}{2}\))]
Solution:
Let y = log[tan(\(\frac{x}{2}\))]
Differentiating w.r.t. x, we get

(iv) \(\sqrt{\tan \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\tan \sqrt{x}}\)
Differentiating w.r.t. x, we get

(v) cot³[log (x³)]
Solution:
Let y = cot³[log (x³)]
Differentiating w.r.t. x, we get

(vi) 5^{sin3x+ 3}
Solution:
Let y = 5^{sin3x+ 3}
Differentiating w.r.t. x, we get

(vii) cosec (\(\sqrt{\cos X}\))
Solution:
Let y = cosec (\(\sqrt{\cos X}\))
Differentiating w.r.t. x, we get

(viii) log[cos (x³ – 5)]
Solution:
Let y = log[cos (x³ – 5)]
Differentiating w.r.t. x, we get

(ix) e^{3 sin2x – 2 cos2x}
Solution:
Let y = e^{3 sin2x – 2 cos2x}
Differentiating w.r.t. x, we get

(x) cos²[log (x²+ 7)]
Solution:
Let y = cos²[log (x²+ 7)]
Differentiating w.r.t. x, we get

(xi) tan[cos (sinx)]
Solution:
Let y = tan[cos (sinx)]
Differentiating w.r.t. x, we get

(xii) sec[tan (x⁴ + 4)]
Solution:
Let y = sec[tan (x⁴ + 4)]
Differentiating w.r.t. x, we get

= sec[tan(x⁴ + 4)]∙tan[tan(x⁴ + 4)]∙sec²(x⁴ + 4)(4x³ + 0)
= 4x³sec²(x⁴ + 4)∙sec[tan(x⁴ + 4)]∙tan[tan(x⁴ + 4)].

(xiii) e^{log[(logx)2 – logx2]}
Solution:
Let y = e^{log[(logx)2 – logx2]}
= (log x)² – log x² …[∵ e^{log x} = x]
Differentiating w.r.t. x, we get

(xiv) sin\(\sqrt{\sin \sqrt{x}}\)
Solution:
Let y = sin\(\sqrt{\sin \sqrt{x}}\)
Differentiating w.r.t. x, we get

(xv) log[sec(e^x2)]
Solution:
Let y = log[sec(e^x2)]
Differentiating w.r.t. x, we get

(xvi) log_e²(logx)
Solution:
Let y = log_e²(logx) = \(\frac{\log (\log x)}{\log e^{2}}\)

(xvii) [log{log(logx)}]²
Solution:
let y = [log{log(logx)}]²
Differentiating w.r.t. x, we get

(xviii) sin²x² – cos²x²
Solution:
Let y = sin²x² – cos²x²
Differentiating w.r.t. x, we get

= 2sinx²∙cosx² × 2x + 2sinx²∙cosx² × 2x
= 4x(2sinx²∙cosx²)
= 4xsin(2x²).

Question 3.
Diffrentiate the following w.r.t. x
(i) (x² + 4x + 1)³ + (x³ – 5x – 2)⁴
Solution:
Let y = (x² + 4x + 1)³ + (x³ – 5x – 2)⁴
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[(x² + 4x + 1)³ + (x³ – 5x – 2)⁴]
= \(\frac{d}{d x}\) = (x² + 4x + 1)³ + \(\frac{d}{d x}\)(x³ – 5x – 2)⁴
= 3(x² + 4x + 1)²∙\(\frac{d}{d x}\)(x² + 4x + 1) + 4(x³ – 5x – 2)⁴∙\(\frac{d}{d x}\)(x³ – 5x – 2)
= 3(x² + 4x + 1)³∙(2x + 4 × 1 + 0) + 4(x³ – 5x – 2)³∙(3x² – 5 × 1 – 0)
= 6 (x + 2)(x² + 4x + 1)² + 4 (3x² – 5)(x³ – 5x – 2)³.
(ii) (1 + 4x)⁵(3 + x − x²)⁸
Solution:
Let y = (1 + 4x)⁵(3 + x − x²)⁸
Differentiating w.r.t. x, we get

= 8 (1 + 4x)⁵ (3 + x – x²)⁷∙(0 + 1 – 2x) + 5 (1 + 4x)⁴ (3 + x – x²)⁸∙(0 + 4 × 1)
= 8 (1 – 2x)(1 + 4x)⁵(3 + x – x²)⁷ + 20(1 + 4x)⁴(3 + x – x²)⁸.

(iii) \(\frac{x}{\sqrt{7-3 x}}\)
Solution:
Let y = \(\frac{x}{\sqrt{7-3 x}}\)
Differentiating w.r.t. x, we get

(iv) \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Solution:
Let y = \(\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\left[\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\right]\)

(v) (1 + sin²x)²(1 + cos²x)³
Solution:
Let y = (1 + sin²x)²(1 + cos²x)³
Differentiating w.r.t. x, we get

= 3(1 + sin²x)² (1 + cos²x)²∙[2cosx(-sinx)] + 2 (1 + sin²x)(1 + cos²x)³∙[2sinx-cosx]
= 3 (1 + sin²x)² (1 + cos²x)² (-sin 2x) + 2(1 + sin²x)(1 + cos²x)³(sin 2x)
= sin2x (1 + sin²x) (1 + cos²x)² [-3(1 + sin²x) + 2(1 + cos²x)]
= sin2x (1 + sin²x)(1 + cos²x)²(-3 – 3sin²x + 2 + 2cos²x)
= sin2x (1 + sin²x)(1 + cos²x)² [-1 – 3 sin²x + 2 (1 – sin²x)]
= sin 2x(1 + sin²x)(1 + cos²x)² (-1 – 3 sin²x + 2 – 2 sin²x)
= sin2x (1 + sin²x)(1 + cos²x)²(1 – 5 sin²x).

(vi) \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Solution:
Let y = \(\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}]\)

(vii) log(sec 3x+ tan 3x)
Solution:
Let y = log(sec 3x+ tan 3x)
Differentiating w.r.t. x, we get

(viii) \(\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}\)
Solution:

(ix) cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Solution:
Let y = cot\(\left(\frac{\log x}{2}\right)\) – log\(\left(\frac{\cot x}{2}\right)\)
Differentiating w.r.t. x, we get

(x) \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution:

(xi) \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Solution:
let y = \(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\)
Differentiating w.r.t. x, we get

(xii) log[tan³x·sin⁴x·(x² + 7)⁷]
Solution:
Let y = log [tan³x·sin⁴x·(x² + 7)⁷]
= log tan³x + log sin⁴x + log (x² + 7)⁷
= 3 log tan x + 4 log sin x + 7 log (x² + 7)
Differentiating w.r.t. x, we get

= 6cosec2x + 4 cotx + \(\frac{14 x}{x^{2}+7}\)

(xiii) log\(\left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)\)
Solution:

(xiv) log\(\left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.\)
Solution:
Using log\(\left(\frac{a}{b}\right)\) = log a – log b
log a^b = b log a


\(-\frac{5}{2}\)cosec\(\left(\frac{5 x}{2}\right)\)

(xv) log\(\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\)
Solution:

(xvi) log\(\left[4^{2 x}\left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{\frac{3}{2}}\right]\)
Solution:

(xvii) log\(\left[\frac{e^{x^{2}}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]\)
Solution:

(xviii) log\(\left[\frac{a^{\cos x}}{\left(x^{2}-3\right)^{3} \log x}\right]\)
Solution:

(xix) y= (25)^log₅(secx) − (16)^log₄(tanx)
Solution:
y = (25)^log₅(secx) − (16)^log₄(tanx)
= 5^{2log₅(secx)} – 4^{2log₄(tanx)}
= 5^{log₅(sec5x)} – 4^{log₄(tan2x)}
= sec²x – tan²x … [∵ = x]
∴ y = 1
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(1) = 0

(xx) \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}\)
Differentiating w.r.t. x, we get

Question 4.
A table of values of f, g, f ‘ and g’ is given

(i) If r(x) = f [g(x)] find r’ (2).
Solution:
r(x) = f[g(x)]
∴ r'(x) = \(\frac{d}{d x}\)f[g(x)]
= f'[g(x)]∙\(\frac{d}{d x}\)[g(x)]
= f'[g(x)∙[g'(x)]
∴ r'(2) = f'[g(2)]∙g'(2)
= f'(6)∙g'(2) … [∵ g(x) = 6, when x = 2]
= -4 × 4 … [From the table]
= -16.

(ii) If R(x) = g[3 + f(x)] find R’ (4).
Solution:
R(x) = g[3 + f(x)]
∴ R'(x) = \(\frac{d}{d x}\){g[3+f(x)]}
= g'[3 + f(x)]∙\(\frac{d}{d x}\)[3 + f(x)]
= g'[3 +f(x)]∙[0 + f'(x)]
= g'[3 + f(x)]∙f'(x)
∴ R'(4) = g'[3 + f(4)]∙f'(4)
= g'[3 + 3]∙f'(4) … [∵ f(x) = 3, when x = 4]
= g'(6)∙f'(4)
= 7 × 5 … [From the table]
= 35.

(iii) If s(x) = f[9− f(x)] find s’ (4).
Solution:
s(x) = f[9− f(x)]
∴ s'(x) = \(\frac{d}{d x}\){f[9 – f(x)]}
= f'[9 – f(x)]∙\(\frac{d}{d x}\)[0 – f(x)]
= f'[9 – f(x)]∙[0 – f'(x)]
= -f'[9 – f(x)] – f'(x)
∴ s'(4) = -f'[9 – f(4)] – f'(4)
= -f'[9 – 3] – f'(4) … [∵ f(x) = 3, when x = 4]
= -f'(6) – f'(4)
= -(-4)(5) … [From the table]
= 20.

(iv) If S(x) = g[g(x)] find S’ (6)
Solution:
S(x) = g[g(x)]
∴ S'(x) = \(\frac{d}{d x}\)g[g(x)]
= g'[g(x)]∙\(\frac{d}{d x}\)[g(x)]
= g'[g(x)]∙g'(x)
∴ S ‘(6) = g'[g'(6)]∙g'(6)
= g'(2)∙g'(6) … [∵ g (x) = 2, when x = 6]
= 4 × 7 … [From the table]
= 28.

Question 5.
Assume that f ‘(3) = -1, g'(2) = 5, g(2) = 3 and y = f[g(x)] then \(\left[\frac{d y}{d x}\right]_{x=2}\) = ?
Solution:
y = f[g(x)]
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\){[g(x)]}

Question 6.
If h(x) = \(\sqrt{4 f(x)+3 g(x)}\), f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 find h'(1).
Solution:
Given f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 …..(1)

Question 7.
Find the x co-ordinates of all the points on the curve y = sin 2x – 2 sin x, 0 ≤ x < 2π where \(\frac{d y}{d x}\) = 0.
Solution:
y = sin 2x – 2 sin x, 0 ≤ x < 2π

= cos2x × 2 – 2cosx
= 2 (2 cos²x – 1) – 2 cosx
= 4 cos²x – 2 – 2 cos x
= 4 cos²x – 2 cos x – 2
If \(\frac{d y}{d x}\) = 0, then 4 cos²x – 2 cos x – 2 = 0
∴ 4cos²x – 4cosx + 2cosx – 2 = 0
∴ 4 cosx (cosx – 1) + 2 (cosx – 1) = 0
∴ (cosx – 1)(4cosx + 2) = 0
∴ cosx – 1 = 0 or 4cosx + 2 = 0
∴ cos x = 1 or cos x = \(-\frac{1}{2}\)
∴ cos x = cos 0

Question 8.
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity]
“Let f (x) = x² + 5 and g(x) = e^x + 3 then
f [g(x)] = _ _ _ _ _ _ _ _ and g [f(x)] =_ _ _ _ _ _ _ _.
Now f ‘(x) = _ _ _ _ _ _ _ _ and g'(x) = _ _ _ _ _ _ _ _.
The derivative off [g (x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ _.
Therefore \(\frac{d}{d x}\)[f[g(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[f[g(x)]]]_{x = 0} = _ _ _ _ _ _ _ _ _ _ _.
The derivative of g[f(x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ __ _ _ _ _.
Therefore \(\frac{d}{d x}\)[g[f(x)]] = _ _ _ _ _ _ _ _ _ and [\(\frac{d}{d x}\)[g[f(x)]]]_{x = 1} = _ _ _ _ _ _ _ _ _ _ _.”
Hint basket : { f ‘[g(x)]·g'(x), 2e^2x + 6e^x, 8, g'[f(x)]·f ‘(x), 2xe^{x2 + 5}, -2e⁶, e^2x + 6e^x + 14, e^{x2 + 5} + 3, 2x, e^x}
Solution:
f[g(x)] = e^2x + 6e^x + 14
g[f(x)] = e^{x2 + 5} + 3
f'(x) = 2x, g’f(x) = e^x
The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]∙g'(x).
∴ \(\frac{d}{d x}\){f[g(x)]} = 2e^2x + 6e^x and \(\frac{d}{d x}\){f[g(x)]}_{x = 0} = 8
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]∙f'(x).
∴ \(\frac{d}{d x}\){g[(f(x)]} = 2xe^{x2 + 5} and
\(\frac{d}{d x}\){g[(f(x)]}_{x = -1} = -2e⁶.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Miscellaneous Exercise 7

I) Select the appropriate alternatives for each of the following :
Question 1.
The value of objective function is maximum under linear constraints _______.
(A) at the centre of the feasible region
(B) at (0, 0)
(C) at a vertex of the feasible region
(D) the vertex which is of maximum distance from (0, 0)
Solution:
(C) at a vertex of the feasible region

Question 2.
Which of the following is correct _______.
(A) every L.P.P. has an optimal solution
(B) a L.P.P. has unique optimal solution
(C) if L.P.P. has two optimal solutions then it has an infinite number of optimal solutions
(D) the set of all feasible solutions of L.P.P. may not be a convex set
Solution:
(C) if L.P.P. has two optimal solutions then it has an infinite number of optimal solutions

Question 3.
Objective function of L.P.P. is _______.
(A) a constraint
(B) a function to be maximized or minimized
(C) a relation between the decision variables
(D) equation of a straight line
Solution:
(B) a function to be maximized or minimized

Question 4.
The maximum value of z = 5x + 3y subjected to the constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y≥ 0 is _______.
(A) 235
(B) \(\frac{235}{9}\)
(C) \(\frac{235}{19}\)
(D) \(\frac{235}{3}\)
Solution:
(C) \(\frac{235}{19}\)

Question 5.
The maximum value of z = 10x + 6y subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y≥ 0. _______.
(A) 56
(B) 65
(C) 55
(D) 66
Solution:
(A) 56

Question 6.
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained at _______.
(A) (30, 25)
(B) (20, 35)
(C) (35, 20)
(D) (40, 15)
Solution:
(D) (40, 15)

Question 7.
Of all the points of the feasible region, the optimal value of obtained at the point lies _______.
(A) inside the feasible region
(B) at the boundary of the feasible region
(C) at the vertex of the feasible region
(D) outside the feasible region
Solution:
(C) at the vertex of the feasible region

Question 8.
The feasible region is the set of points that satisfy _______.
(A) the objective function
(B) all of the given constraints
(C) some of the given constraints
(D) only one constraint
Solution:
(B) all of the given constraints

Question 9.
Solution of L.P.P. to minimize z = 2x + 3y such that x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is _______.
(A) x = 0, y = \(\frac{1}{2}\)
(B) x = \(\frac{1}{2}\), y = 0
(C) x = 1, y = 2
(D) x = \(\frac{1}{2}\), y = \(\frac{1}{2}\)
Solution:
(A) x = 0, y = \(\frac{1}{2}\)

Question 10.
The corner points of the feasible solution given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0 are _______.
(A) (0, 0), (4, 0), (7, 1), (0, 4)
(B) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)
(C) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 7)
(D) (0, 0), (4, 0), (3, 1), (0, 7)
Solution:
(B) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)

Question 11.
The corner points of the feasible solution are (0, 0), (2, 0), (\(\frac{12}{7}\), \(\frac{3}{7}\)), (0, 1). Then z = 7x + y is maximum at _______.
(A) (0, 0)
(B) (2, 0)
(C) (\(\frac{12}{7}\), \(\frac{3}{7}\))
(D) (0, 1)
Solution:
(B) (2, 0)

Question 12.
If the corner points of the feasible solution are (0, 0), (3, 0), (2, 1) and (0, \(\frac{7}{3}\) ), the maximum value of z = 4x + 5y is _______.
(A) 12
(B) 13
(C) 35
(D) 0
Solution:
(B) 13

Question 13.
If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0) then the point of minimum z = 3x + 2y is _______.
(A) (2, 2)
(B) (0, 10)
(C) (4, 0)
(D) (3 ,4)
Solution:
(A) (2, 2)

Question 14.
The half plane represented by 3x + 2y < 8 contains the point _______.
(A) (1, \(\frac{5}{2}\))
(B) (2, 1)
(C) (0, 0)
(D) (5, 1)
Solution:
(C) (0, 0)

Question 15.
The half plane represented by 4x + 3y > 14 contains the point _______.
(A) (0, 0)
(B) (2, 2)
(C) (3, 4)
(D) (1, 1)
Solution:
(C) (3, 4)

II) Solve the following :
Question 1.
Solve each of the following inequations graphically using X Y plane.
(i) 4x – 18 ≥ 0
Solution:
Consider the line whose equation is 4x – 18 ≥ 0 i.e. x = \(\frac{18}{4}=\frac{9}{2}\) = 4.5
This represents a line parallel to Y-axis passing3through the point (4.5, 0)
Draw the line x = 4.5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, 4x – 18 = 4 × 0 – 18 = -18 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = 4.5 and the non-origin side of the line which is shaded in the graph.

(ii) -11x – 55 ≤ 0
Solution:
Consider the line whose equation is -11x – 55 ≤ 0 i.e. x = -5
This represents a line parallel to Y-axis passing3through the point (-5, 0)
Draw the line x = – 5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, -11x – 55 = – 11(0) – 55 = -55 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = -5 and the non-origin side of the line which is shaded in the graph.

(iii) 5y – 12 ≥ 0
Solution:
Consider the line whose equation is 5y – 12 ≥ 0 i.e. y = \(\frac{12}{5}\)
This represents a line parallel to X-axis passing through the point (o, \(\frac{12}{5}\))
Draw the line y = \(\frac{12}{5}\)
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y – 12 = 5(0) – 12 = -12 > 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{12}{5}\) and the non-origin side of the line which is shaded in the graph.

(iv) y ≤ -3.5
Solution:
Consider the line whose equation is y ≤ – 3.5 i.e. y = – 3.5
This represents a line parallel to X-axis passing3through the point (0, -3.5)
Draw the line y = – 3.5
To find the solution set, we have to check the position of the origin (0, 0).
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = – 3.5 and the non-origin side of the line which is shaded in the graph.

Question 2.
Sketch the graph of each of following inequations in XOY co-ordinate system.
(i) x ≥ 5y
Solution:

(ii) x + y≤ 0
Solution:

(iii) 2y – 5x ≥ 0
Solution:

(iv) |x + 5| ≤ y
Solution:
|x + 5| ≤ y
∴ -y ≤ x + 5 ≤ y
∴ -y ≤ x + 5 and x + 5 ≤ y
∴ x + y ≥ -5 and x – y ≤ -5
First we draw the lines AB and AC whose equations are
x + y= -5 and x – y = -5 respectively.

The graph of |x + 5| ≤ y is as below:

Question 3.
Find graphical solution for each of the following system of linear inequation.
(i) 2x + y ≥ 2, x – y ≤ 1
Solution:
First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(ii) x + 2y ≥ 4, 2x – y ≤ 6
Solution:

(iii) 3x + 4y ≤ 12, x – 2y ≥ 2, y ≥ -1
Solution:
First we draw the lines AB, CD and ED whose equations are 3x + 4y = 12, x – 2y = 2 and y = -1 respectively.


The solution set of given system of inequation is shaded in the graph.

Question 4.
Find feasible solution for each of the following system of linear inequations graphically.
(i) 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0
Solution:

The feasible solution is OCPBO.

(ii) 3x + 4y ≥ 12, 4x + 7y ≤ 28, x ≥ 0, y ≥ 0
Solution:

The feasible solution is ACDBA.

Question 5.
Solve each of the following L.P.P.
(i) Maximize z = 5x₁ + 6x₂ subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x₁ ≥ 0, x₂ ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 2x₁ + 3x₂ = 18 and 2x₁ + x₂ = 12 respectively.


The feasible region is OCPBO which is shaded in the graph. The vertices of the feasible region are O(0, 0), C(6, 0), P and B (0,6).
P is the point of intersection of the lines
2x₁ + 3x₂ = 18 ….(1)
and 2x₁ + x₂ = 12
On subtracting, we get
2x₂ = 6 ∴ x₂ = 3
Substituting x₂ = 3 in (2), we get
2x₁ + 3 = 12 ∴ x₂ = 9
∴ P is (\(\frac{9}{2}\), 3)
The values of objective function z = 5x₁ + 6x₂ at these vertices are
z(O) = 5(0) + 6(0) = 0 + 0 = 0
z(C) = 5(6) + 6(0) = 30 + 0 = 30
z(P) = 5(\(\frac{9}{2}\)) + 6(3) = \(\frac{45}{2}\) + 18 = \(\frac{45+36}{2}=\frac{81}{2}\) = 40.5
z(B) = 5(0) + 6(3) = 0 + 18 = 18
Maximum value of z is 40.5 when x₁ = 9/2, y = 3.

(ii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21
Question is modified.
Maximize z = 4x + 2y subject to 3x + y ≤ 27, x + y ≤ 21, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + y = 27 and x + y = 21 respectively.


The feasible region is OAPDO which is shaded region in the graph. The vertices of the feasible region are 0(0, 0), A (9, 0), P and D(0, 21). P is the point of intersection of lines
3x + y = 27 … (1)
and x + y = 21 … (2)
On substracting, we get 2x = 6 ∴ x = 3
Substituting x = 3 in equation (1), we get
9 + y = 27 ∴ y = 18
∴ P = (3, 18)
The values of the objective function z = 4x + 2y at these vertices are
z(O) = 4(0) + 2(0) = 0 + 0 = 0
z(a) = 4(9) + 2(0) = 36 + 0 = 36
z(P) = 4(3) + 2(18) = 12 + 36 = 48
z (D) = 4(0) + 2(21) = 0 + 42 = 42
∴ 2 has minimum value 48 when x = 3, y = 18.

(iii) Maximize z = 6x + 10y subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 10 and 5x + 3y = 15 respectively.


The feasible region is OCPBD which is shaded in the graph.
The vertices of the feasible region are 0(0, 0), C(3, 0), P and B (0, 2).
P is the point of intersection of the lines
3x + 5y = 10 … (1)
and 5x + 3y = 15 … (2)
Multiplying equation (1) by 5 and equation (2) by 3, we get
15x + 25y = 50
15x + 9y = 45
On subtracting, we get
16y = 5 ∴ y = \(\frac{5}{16}\)
Substituting y = \(\frac{5}{16}\) in equation (1), we get
3x + \(\frac{25}{16}\) = 10 ∴ 3x = 10 – \(\frac{25}{16}=\frac{135}{16}\)
∴ x = \(\frac{45}{16}\) ∴ P ≡ \(\left(\frac{45}{16}, \frac{5}{16}\right)\)
The values of objective function z = 6x + 10y at these vertices are
z(O) = 6(0) + 10(0) = 0 + 0 = 0
z(C) = 6(3) + 10(0) = 18 + 0 = 18
z(P) = 6\(\left(\frac{45}{16}\right)\) + 10\(\left(\frac{5}{10}\right)\) = \(\frac{270}{16}+\frac{50}{16}=\frac{320}{16}\) = 20
z(B) = 6(0) + 10(2) = 0 + 20 = 20
The maximum value of z is 20 at P\(\left(\frac{45}{16}, \frac{5}{16}\right)\) and B (0, 2) two consecutive vertices.
∴ z has maximum value 20 at each point of line segment PB where B is (0, 2) and P is \(\left(\frac{45}{16}, \frac{5}{16}\right)\).
Hence, there are infinite number of optimum solutions.

(iv) Maximize z = 2x + 3y subject to x – y ≥ 3, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB whose equation is x – y = 3.


The feasible region is shaded which is unbounded.
Therefore, the value of objective function can be in- j creased indefinitely. Hence, this LPP has unbounded solution.

Question 6.
Solve each of the following L.P.P.
(i) Maximize z = 4x₁ + 3x₂ subject to 3x₁ + x₂ ≤ 15, 3x₁ + 4x₂ ≤ 24, x₁ ≥ 0, x₂ ≥ 0
Solution:
We first draw the lines AB and CD whose equations are 3x₁ + x₂ = 15 and 3x₁ + 4x₂ = 24 respectively.


The feasible region is OAPDO which is shaded in the graph.
The Vertices of the feasible region are 0(0, 0), A(5, 0), P and D(0, 6).
P is the point of intersection of lines.
3x₁ + 4x₂ = 24 … (1)
and 3x₁ + x₂ = 15 … (2)
On subtracting, we get
3x₂ = 9 ∴ x₂ = 3
Substituting x₂ = 3 in (2), we get
3x₁ + 3 = 15
∴ 3x₁ = 12 ∴ x₁ = 4 ∴ P is (4, 3)
The values of objective function z = 4x₁ + 3x₂ at these vertices are
z(O) = 4(0) + 3(0) = 0 + 0 = 0
z(a) = 4(5) + 3(0) = 20 + 0 = 20
z(P) = 4(4) + 3(3) = 16 + 9 = 25
z(D) = 4(0) + 3(6) = 0 + 18 = 18
∴ z has maximum value 25 when x = 4 and y = 3.

(ii) Maximize z = 60x + 50y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (20, 0), P and B (0, 20).
P is the point of intersection of the lines.
3x + 2y = 60 … (1)
and x + 2y = 40 … (2)
On subtracting, we get
2x = 20 ∴ x = 10
Substituting x = 10 in (2), we get
10 + 2y = 40
∴ 2y = 30 ∴ y = 15 ∴ P is (10, 15)
The values of the objective function z = 60x + 50y at these vertices are
z(O) = 60(0) + 50(0) = 0 + 0 = 0
z(C) = 60(20) + 50(0) = 1200 + 0 = 1200
z(P) = 60(10) + 50(15) = 600 + 750 = 1350
z(B) = 60(0) + 50(20) = 0 + 1000 = 1000 .
∴ z has maximum value 1350 at x = 10, y = 15.

(iii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30; x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB, CD and EF whose equations are 3x + y = 27, x + y = 21, x + 2y = 30 respectively.


The feasible region is XEPQBY which is shaded in the graph.
The vertices of the feasible region are E (30,0), P, Q and B (0,27).
P is the point of intersection of the lines
x + 2y = 30 … (1)
and x + y = 21 … (2)
On subtracting, we get
y = 9
Substituting y = 9 in (2), we get
x + 9 = 21 ∴ x = 12
∴ P is (12, 9)
Q is the point of intersection of the lines
x + y = 21 … (2)
and 3x + y = 27 … (3)
On subtracting, we get
2x = 6 ∴ x = 3
Substituting x = 3 in (2), we get
3 + y = 21 ∴ y = 18
∴ Q is (3, 18).
The values of the objective function z = 4x + 2y at these vertices are
z(E) = 4(30) + 2(0) = 120 + 0 = 120
z(P) = 4(12) + 2(9) = 48 + 18 = 66
z(Q) = 4(3) + 2(18) = 12 + 36 = 48
z(B) = 4(0) + 2(27) = 0 + 54 = 54
∴ z has minimum value 48, when x = 3 and y = 18.

Question 7.
A carpenter makes chairs and tables. Profits are ₹140/- per chair and ₹ 210/- per table. Both products are processed on three machines : Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by following table:

Formulate the above problem as L.P.P. Solve it graphically to get maximum profit.
Solution:
Let the number of chairs and tables made by the carpenter be x and y respectively.
The profits are ₹ 140 per chair and ₹ 210 per table.
∴ total profit z = ₹ (140x + 210y) This is the objective function which is to be maximized.
The constraints are as per the following table :

From the table, the constraints are
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60.
The number of chairs and tables cannot be negative.
∴ x ≥ 0, y ≥ 0
Hence, the mathematical formulation of given LPP is :
Maximize z = 140x + 210y, subject to
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.
We first draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.


The feasible region is OCPQFO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (10, 0), P, Q and F (0, 10).
P is the point of intersection of the lines
5x + 2y = 50 … (1)
and 3x + 3y = 36 … (2)
Multiplying equation (1) by 3 and equation (2) by 2, we get
15x + 6y = 150
6x + 6y = 72
On subtracting, we get 26
9x = 78 ∴ x = \(\frac{26}{3}\)
Substituting x = \(\frac{26}{3}\) in (2), we get
3\(\left(\frac{26}{3}\right)\) + 3y = 36
3y = 1o y = \(\frac{10}{3}\)
Q is the point of intersection of the lines
3x + 3y = 36 … (2)
and 2x + 6y = 60 … (3)
Multiplying equation (2) by 2, we get
6x + 6 y = 72
Subtracting equation (3) from this equation, we get
4x = 12 ∴ x = 3
Substituting x = 3 in (2), we get
3(3) + 3y = 36
∴ 3y = 27 ∴ y = 9
∴ Q is (3, 9).
Hence, the vertices of the feasible region are O (0, 0),
C(10, 0), P\(\left(\frac{26}{3}, \frac{10}{3}\right)\), Q(3, 9) and F(0, 10).
The values of the objective function z = 140x + 210y at these vertices are
z(O) = 140(0) + 210(0) = 0 + 0 = 0
z(C) = 140 (10) + 210(0) = 1400 + 0 = 1400
z(P) = 140\(\left(\frac{26}{3}\right)\) + 210\(\left(\frac{10}{3}\right)\) = \(\frac{3640+2100}{3}=\frac{5740}{3}\) = 1913.33
z(Q) = 140(3) + 210(9) = 420 + 1890 = 2310
z(F) = 140(0) + 210(10) = 0 + 2100 = 2100
∴ z has maximum value 2310 when x = 3 and y = 9 Hence, the carpenter should make 3 chairs and 9 tables to get the maximum profit of ₹ 2310.

Question 8.
A company manufactures bicycles and tricycles, each of which must be processed through two machines A and B. Maximum availability of Machine A and B is respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on Machine A and 3 hours on Machine B. Manufacturing a tricycles requires 4 hours on Machine A and 10 hours on Machine B. If profits are ₹180/- for a bicycle and ₹220/- for a tricycle. Determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
Solution:
Let x bicycles and y tricycles are to be manu¬factured. Then the total profit is z = ₹ (180x + 220y)
This is a linear function which is to be maximized. Hence, it is the objective function. The constraints are as per the following table :

From the table, the constraints are
6x + 4y ≤ 120, 3x +10y ≤ 180
Also, the number of bicycles and tricycles cannot be i negative.
∴ x ≥ 0, y ≥ 0.
Hence, the mathematical formulation of given LPP is :
Maximize z = 180x + 220y, subject to
6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are 6x + 4y = 120 and 3x + 10y = 180 respectively.


The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), A(20, 0) P and D(0, 18).
P is the point of intersection of the lines
3x + 10y = 180 … (1)
and 6x + 4y = 120 … (2)
Multiplying equation (1) by 2, we get
6x + 20y = 360
Subtracting equation (2) from this equation, we get
16y = 240 ∴ y = 15
∴ from (1), 3x + 10(15) = 180
∴ 3x = 30 ∴ x = 10
∴ P = (10, 15)
The values of the objective function z = 180x + 220y at these vertices are
z(O) = 180(0) + 220(0) = 0 + 0 = 0
z(a) = 180(20) + 220(0) = 3600 + 0 = 3600
z(P) = 180(10) + 220(15) = 1800 + 3300 = 5100
z(D) = 180(0) +220(18) = 3960
∴ the maximum value of z is 5100 at the point (10, 15).
Hence, 10 bicycles and 15 tricycles should be manufactured in order to have the maximum profit of ₹ 5100.

Question 9.
A factory produced two types of chemicals A and B. The following table gives the units of ingredients P and Q (per kg) of chemicals A and B as well as minimum requirements of P and Q and also cost per kg. chemicals A and B :

Find the number of units of chemicals A and B should be produced so as to minimize the cost.
Solution:
Let the factory produce x units of chemical A and y units of chemical B. Then the total cost is z = ₹ (4x + 6y). This is the objective function which is to be minimized.
From the given table, the constraints are
x + 2y ≥ 80, 3x + y ≥ 75.
Also, the number of units x and y of chemicals A and B cannot be negative.
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Minimize z = 4x + 6y, subject to
x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + 2y = 80 and 3x + y = 75 respectively.


The feasible region is shaded in the graph.
The vertices of the feasible region are A (80, 0), P and D (0, 75).
P is the point of intersection of the lines
x + 2y = 80 … (1)
and 3x + y = 75 … (2)
Multiplying equation (2) by 2, we get
6x + 2 y = 150
Subtracting equation (1) from this equation, we get
5x = 70 ∴ x = 14
∴ from (2), 3(14) + y = 75
∴ 42 + y = 75 ∴ y = 33
∴ P = (14, 33)
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(80)+ 6(0) =320 + 0 = 320
z(P) = 4(14)+ 6(33) = 56+ 198 = 254
z(D) = 4(0) + 6(75) = 0 + 450 = 450
∴ the minimum value of z is 254 at the point (14, 33).
Hence, 14 units of chemical A and 33 units of chemical B are to be produced in order to have the j minimum cost of ₹ 254.

Question 10.
A company produces mixers and food processors. Profit on selling one mixer and one food processor is ₹ 2,000/- and ₹ 3,000/- respectively. Both the products are processed through three Machines A, B, C. The time required in hours by each product and total time available in hours per week on each machine are as follows :

How many mixers and food processors should be produced to maximize the profit?
Solution:
Let the company produce x mixers and y food processors.
Then the total profit is z = ₹ (2000x + 3000y)
This is the objective function which is to be maximized. From the given table in the problem, the constraints are 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60
Also, the number of mixers and food processors cannot be negative,
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Maximize z = 2000x + 3000y, subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤60, x ≥ 0, y ≥ 0.
First we draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.


The feasible region is OCPQFO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), C(10, 0), P, Q and F(0,10).
P is the point of intersection of the lines
3x + 3y = 36 … (1)
and 5x + 2y = 50 … (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
6x + 6y = 72
15x + 6y = 150
On subtracting, we get
9x = 78
∴ x = \(\frac{26}{3}\)
∴ from (1), 3\(\left(\frac{26}{3}\right)\) + 3y = 36
∴ 3y = 10
∴ y = \(\frac{10}{3}\)
∴ P = \(\left(\frac{26}{3}, \frac{10}{3}\right)\)
Q is the point of intersection of the lines
3x + 3y = 36 … (1)
and 2x + 6y = 60 … (3)
Multiplying equation (1) by 2, we get
6x + 6y = 72
Subtracting equation (3), from this equation, we get
4x = 12
∴ x = 3
∴ from (1), 3(3) + 3y = 36
∴ 3y = 27
∴ y = 9
∴ Q = (3, 9)
The values of the objective function z = 2000x + 3000y at these vertices are
z(O) = 2000(0) + 3000(0) = 0 + 0 = 0
z(C) = 2000(10) + 3000(0) = 20000 + 0 = 20000
z(P) = 2000\(\left(\frac{26}{3}\right)\) + 3000\(\left(\frac{10}{3}\right)\) = \(\frac{52000}{3}+\frac{30000}{3}=\frac{82000}{3}\)
z(Q) = 2000(3) + 3000(9) = 6000 + 27000 = 33000
z(F) = 2000(0) + 3000(10) = 30000 + 0 = 30000
∴ the maximum value of z is 33000 at the point (3, 9).
Hence, 3 mixers and 9 food processors should be produced in order to get the maximum profit of ₹ 33,000.

Question 11.
A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound I is ₹ 800/- and that of compound II is ₹ 640/-. Formulate the problem as L.P.P. and solve it to minimize the cost.
Solution:
Let the company buy x units of compound I and y units of compound II.
Then the total cost is z = ₹(800x + 640y).
This is the objective function which is to be minimized.
The constraints are as per the following table :

From the table, the constraints are
4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18.
Also, the number of units of compound I and compound II cannot be negative.
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Minimize z = 800x + 640y, subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x ≥ 0, y ≥ 0.
First we draw the lines AB, CD and EF whose equations are 4x + 2y = 16, 12x + 2y = 24 and 2x + 6y = 18


The feasible region is shaded in the graph.
The vertices of the feasible region are E(9, 0), P, Q, and D(0, 12).
P is the point of intersection of the lines
2x + 6y = 18 … (1)
and 4x + 2y = 16 … (2)
Multiplying equation (1) by 2, we get 4x + 12y = 36
Subtracting equation (2) from this equation, we get
10y = 20
∴ y = 2
∴ from (1), 2x + 6(2) = 18
∴ 2x = 6
∴ x = 3
∴ P = (3, 2)
Q is the point of intersection of the lines
12x + 2y = 24 … (3)
and 4x + 2y = 16 … (2)
On subtracting, we get
8x = 8 ∴ x = 1
∴ from (2), 4(1) + 2y = 16
∴ 2y = 12 ∴ y = 6
∴ Q = (1, 6)
The values of the objective function z = 800x + 640y at these vertices are
z(E) = 800(9)+ 640(0) =7200 + 0 = 7200
z(P) = 800(3) + 640(2) = 2400 + 1280 = 3680
z(Q) = 800(1) + 640(6) =800 + 3840 =4640
z(D) = 800(0) + 640(12) = 0 + 7680 = 7680
∴ the minimum value of z is 3680 at the point (3, 2).
Hence, the company should buy 3 units of compound I and 2 units of compound II to have the minimum cost of ₹ 3680.

Question 12.
A person makes two types of gift items A and B requires the services of a cutter and a finisher. Gift item A requires 4 hours of cutter’s time and 2 hours of finisher’s time. B requires 2 hours of cutter’s time and 4 hours of finisher’s time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75/- and on gift item B is ₹ 125/-. Assuming that the person can sell all the gift items produced, determine how many gift items of each type should he make every month to obtain the best returns?
Solution:
Let x: number of gift item A
y: number of gift item B
As numbers of the items are never negative
x ≥ 0; y ≥ 0

Total time required for the cutter = 4x + 2y
Maximum available time 208 hours
∴ 4x+ 2y ≤ 208
Total time required for the finisher 2x +4y
Maximum available time 152 hours
2x + 4y ≤ 152
Total Profit is 75x + 125y
∴ L.P.P. of the above problem is
Minimize z = 75x + 125y
Subject to 4x+ 2y ≤ 208
2x + 4y ≤ 152
x ≥ 0; y ≥ 0
Graphical solution

Corner points
Now, Z at
x = (75x + 125y)
O(0, 0) = 75 × 0 + 125 × 0 = 0
A(52,0) = 75 × 52 + 125 × 0 = 3900
B(44, 16) = 75 × 44 + 125 × 16 = 5300
C(0, 38) = 75 × 0 + 125 × 38 = 4750
A person should make 44 items of type A and 16 Uems of type Band his returns are ₹ 5,300.

Question 13.
A firm manufactures two products A and B on which profit earned per unit ₹3/- and ₹4/- respectively. Each product is processed on two machines M₁ and M₂. The product A requires one minute of processing time on M₁ and two minute of processing time on M₂, B requires one minute of processing time on M₁ and one minute of processing time on M₂. Machine M₁ is available for use for 450 minutes while M₂ is available for 600 minutes during any working day. Find the number of units of product A and B to be manufactured to get the maximum profit.
Solution:
Let the firm manufactures x units of product
A and y units of product B.
The profit earned per unit of A is ₹3 and B is ₹ 4.
Hence, the total profit is z = ₹ (3x + 4y).
This is the linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table :

From the table, the constraints are
x + y ≤ 450, 2x + y ≤ 600
Since, the number of gift items cannot be negative, x ≥ 0, y ≥ o.
∴ the mathematical formulation of LPP is,
Maximize z = 3x + 4y, subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0.
Now, we draw the lines AB and CD whose equations are x + y = 450 and 2x + y — 600 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O(0, 0), C(300, 0), P and B (0, 450).
P is the point of intersection of the lines
2x + y = 600 … (1)
and x + y = 450 … (2)
On subtracting, we get
∴ x = 150
Substituting x = 150 in equation (2), we get
150 + y = 450
∴ y = 300
∴ P = (150, 300)
The values of the objective function z = 3x + 4y at these vertices are
z(O) = 3(0) + 4(0) = 0 + 0 = 0
z(C) = 3(300) + 4(0) = 900 + 0 = 900
z(P) = 3(150) + 4(300) = 450 + 1200 = 1650
z(B) = 3(0) + 4(450) = 0 + 1800 = 1800
∴ z has the maximum value 1800 when x = 0 and y = 450 Hence, the firm gets maximum profit of ₹ 1800 if it manufactures 450 units of product B and no unit product A.

Question 14.
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20/- per unit of A and ₹ 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each units of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should the manufacture per month to maximize profit? How much is the maximum profit?
Solution:
Let the firm manufactures x units of item A and y units of item B.
Firm can make profit of ₹ 20 per unit of A and ₹ 30 per unit of B.
Hence, the total profit is z = ₹ (20x + 30y).
This is the objective function which is to be maximized. The constraints are as per the following table :

From the table, the constraints are
3x + 2y ≤ 210, 2x + 4y ≤ 300
Since, number of items cannot be negative, x ≥ 0, y ≥ 0.
Hence, the mathematical formulation of given LPP is :
Maximize z = 20x + 30y, subject to 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0.
We draw the lines AB and CD whose equations are 3x + 2y = 210 and 2x + 4y = 300 respectively.

The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), A (70, 0), P and D (0, 75).
P is the point of intersection of the lines
2x + 4y = 300 … (1)
and 3x + 2y = 210 … (2)
Multiplying equation (2) by 2, we get
6x + 4y = 420
Subtracting equation (1) from this equation, we get
∴ 4x = 120 ∴ x = 30
Substituting x = 30 in (1), we get
2(30) + 4y = 300
∴ 4y = 240 ∴ y = 60
∴ P is (30, 60)
The values of the objective function z = 20x + 30y at these vertices are
z(O) = 20(0) + 30(0) = 0 + 0 = 0
z(A) = 20(70) + 30(0) = 1400 + 0 = 1400
z(P) = 20(30) + 30(60) = 600 + 1800 = 2400
z(D) = 20(0) + 30(75) = 0 + 2250 = 2250
∴ z has the maximum value 2400 when x = 30 and y = 60. Hence, the firm should manufactured 30 units of item A and 60 units of item B to get the maximum profit of ₹ 2400.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Question 1.
Maximize : z = 11x + 8y subject to x ≤ 4, y ≤ 6,
x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.


The feasible region is shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2).
∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6).
The values of the objective function z = 11x + 8y at these vertices are
z (O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z (P) = 11(4) + 8(2) = 44 + 16 = 60
z (D) = 11(0) + 8(2) = 0 + 16 = 16
∴ z has maximum value 60, when x = 4 and y = 2.

Question 2.
Maximize : z = 4x + 6y subject to 3x + 2y ≤ 12,
x + y ≥ 4, x, y ≥ 0.
Solution:
First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.


The feasible region is the ∆ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0)+ 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ has maximum value 36, when x = 0, y = 6.

Question 3.
Maximize : z = 7x + 11y subject to 3x + 5y ≤ 26
5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 26 and 5x + 3y = 30 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (6, 0), p and B(0, \(\frac{26}{5}\))
The vertex P is the point of intersection of the lines
3x + 5y = 26 … (1)
and 5x + 3y = 30 … (2)
Multiplying equation (1) by 3 and equation (2) by 5, we get
9x + 15y = 78
and 25x + 15y = 150
On subtracting, we get
16x = 72 ∴ x = \(\frac{72}{16}=\frac{9}{2}\) = 4.5
Substituting x = 4.5 in equation (2), we get
5(4.5) + 3y = 30
22.5 + 3y = 30
∴ 3y = 7.5 ∴ y = 2.5
∴ P is (4.5, 2.5)
The values of the objective function z = 7x + 11y at these corner points are
z (O) = 7(0) + 11(0) = 0 + 0 = 0
z (C) = 7(6) + 11(0) = 42 + 0 = 42
z (P) = 7(4.5) + 11 (2.5) = 31.5 + 27.5 = 59.0 = 59
z(B) = 7(0) + 11\(\left(\frac{26}{5}\right)=\frac{286}{5}\) = 57.2
∴ z has maximum value 59, when x = 4.5 and y = 2.5.

Question 4.
Maximize : z = 10x + 25y subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3, x + y ≤ 5 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.


The feasible region is OAPQDO which is shaded in the i graph.
The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3).
t P is the point of intersection of the lines x + y = 5 and x = 3.
Substituting x = 3 in x + y = 5, we get
3 + y = 5 ∴ y = 2
∴ P is (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get
x + 3 = 5 ∴ x = 2
∴ Q is (2, 3)
The values of the objective function z = 10x + 25y at these vertices are
z(O) = 10(0) + 25(0) = 0 + 0 = 0
z(a) = 10(3) + 25(0) = 30 + 0 = 30
z(P) = 10(3) + 25(2) = 30 + 50 = 80
z(Q) = 10(2) + 25(3) = 20 + 75 = 95
z(D) = 10(0)+ 25(3) = 0 + 75 = 75
∴ z has maximum value 95, when x = 2 and y = 3.

Question 5.
Maximize : z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21,
x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.


The feasible region is OCPQBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6).
P is the point of intersection of the lines
3x + y = 21 … (1)
and x + y = 9 … (2)
On subtracting, we get 2x = 12 ∴ x = 6
Substituting x = 6 in equation (2), we get
6 + y = 9 ∴ y = 3
∴ P = (6, 3)
Q is the point of intersection of the lines
x + 4y = 24 … (3)
and x + y = 9 … (2)
On subtracting, we get
3y = 15 ∴ y = 5
Substituting y = 5 in equation (2), we get
x + 5= 9 ∴ x = 4
∴ Q = (4, 5)
∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6).
The values of the objective function 2 = 3x + 5y at these corner points are
z(O) = 3(0)+ 5(0) = 0 + 0 = 0
z(C) = 3(7) + 5(0) = 21 + 0 = 21
z(P) = 3(6) + 5(3) = 18 + 15 = 33
z(Q) = 3(4) + 5(5) = 12 + 25 = 37
z(B) = 3(0)+ 5(6) = 0 + 30 = 30
∴ z has maximum value 37, when x = 4 and y = 5.

Question 6.
Minimize : z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3,
x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 5x + y = 5 and x + y = 3 respectively.


The feasible region is XCPBY which is shaded in the graph.
The vertices of the feasible region are C (3, 0), P and B (0, 5).
P is the point of the intersection of the lines
5x + y = 5
and x + y = 3
On subtracting, we get
4x = 2 ∴ x = \(\frac{1}{2}\)
Substituting x = \(\frac{1}{2}\) in x + y = 3, we get
\(\frac{1}{2}\) + y = 3
∴ y = \(\frac{5}{2}\) ∴ P = \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
The values of the objective function z = 7x + y at these vertices are
z(C) = 7(3) + 0 = 21
z(B) = 7(0) + 5 = 5
∴ z has minimum value 5, when x = 0 and y = 5.

Question 7.
Minimize : z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15,
y ≥ 2, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.


The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15
∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 … (1)
and 2x + y = 7 … (2)
On subtracting, we get
2y = 8 ∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3 ∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) +10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4

Question 8.
Minimize : z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4,
3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 2y = 3, x + 4y = 4 and 3x + y = 3 respectively.


The feasible region is XCPQFY which is shaded in the graph.
The vertices of the feasible region are C (4, 0), P, Q and F(0, 3).
P is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
On subtracting, we get
2y = 1 ∴ y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in x + 2y = 3, we get
x + 2\(\left(\frac{1}{2}\right)\) = 3
∴ x = 2
∴ P = (2, \(\frac{1}{2}\))
Q is the point of intersection of the lines
x + 2y = 3 … (1)
and 3x + y = 3 ….(2)
Multiplying equation (1) by 3, we get 3x + 6y = 9
Subtracting equation (2) from this equation, we get
5y = 6
∴ y = \(\frac{6}{5}\)
∴ from (1), x + 2\(\left(\frac{6}{5}\right)\) = 3
∴ x = 3 – \(\frac{12}{5}=\frac{3}{5}\)
Q ≡ \(\left(\frac{3}{5}, \frac{6}{5}\right)\)
The values of the objective function z = 6x + 21y at these vertices are
z(C) = 6(4) + 21(0) = 24
z(P) = 6(2) + 21\(\left(\frac{1}{2}\right)\)
= 12 + 10.5 = 22.5
z(Q)= 6\(\left(\frac{3}{5}\right)\) + 21\(\left(\frac{6}{5}\right)\)
= \(\frac{18}{5}+\frac{126}{5}=\frac{144}{5}\) = 28.8
2 (F) = 6(0) + 21(3) = 63
∴ z has minimum value 22.5, when x = 2 and y = \(\frac{1}{2}\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.3

Question 1.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to a machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B per unit and the number of man-hours available for the firm is as follows:

Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.
Solution:
Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y.
The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit.
∴ total profit is z = 30x + 20y.
This is a linear function which is to be maximized. Hence it is the objective function.
The constraints are as per the following table :

From the table total man hours of labour required for x units of gadget A and y units of gadget B in foundry is (10x + 6y) hours and total man hours of labour required in machine shop is (5x + 4y) hours.
Since, maximum time avilable in foundry and machine shops are 60 hours and 35 hours respectively.
Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35. Since, x and y cannot be negative, we have x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize z = 30x + 20y, subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.

Question 2.
In a cattle breading firm, it is prescribed that the food ration for one animal must contain 14, 22 and 1 units of nutrients A, B and C respectively. Two different kinds of fodder are available. Each unit of these two contains the following amounts of these three nutrients :

The cost of fodder 1 is ₹3 per unit and that of fodder ₹ 2, Formulate the L.P.P. to minimize the cost.
Solution:
Let x units of fodder 1 and y units of fodder 2 be prescribed.
The cost of fodder 1 is ₹ 3 per unit and cost of fodder 2 is ₹ 2 per unit.
∴ total cost is z = 3x + 2y
This is the linear function which is to be minimized. Hence it is the objective function. The constraints are as per the following table :

From table fodder contains (2x + y) units of nutrients A, (2x + 3y) units of nutrients B and (x + y) units of nutrients C. The minimum requirements of these nutrients are 14 units, 22 units and 1 unit respectively.
Therefore, the constraints are
2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1
Since, number of units (i.e. x and y) cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as
Minimize z = 3x + 2y, subject to
2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Question 3.
A company manufactures two types of chemicals A and B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B and the total availability of P and Q.

The company gets profits of ₹350 and ₹400 by selling one unit of A and one unit of B respectively. (Assume that the entire production of A and B can be sold). How many units of the chemicals A and B should be manufactured so that the company get maximum profit? Formulate the problem as L.P.P. to maximize the profit.
Solution:
Let the company manufactures x units of chemical A and y units of chemical B. Then the total profit f to the company is p = ₹ (350x + 400y).
This is a linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table:

The raw material P required for x units of chemical A and y units of chemical B is 3x + 2y. Since, the maximum availability of P is 120, we have the first constraint as 3x + 2y ≤ 120.
Similarly, considering the raw material Q, we have : 2x + 5y ≤ 160.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as :
Maximize p = 350x + 400y, subject to
3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.

Question 4.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II and 2 hours on Machine III. Magazine B requires 3 hours on Machine I, 2 hours on Machine II and 6 hours on Machine III. Machines I, II, III are available for 36, 50, 60 hours per week respectively. Formulate the L.P.P. to determine weekly production of A and B, so that the total profit is maximum.
Solution:
Let the company prints x magazine of type A and y magazine of type B.
Profit on sale of magazine A is ₹ 10 per copy and magazine B is ₹ 15 per copy.
Therefore, the total earning z of the company is
z = ₹ (10x + 15y).
This is a linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table:

From the table, the total time required for Machine I is (2x + 3y) hours, for Machine II is (5x + 2y) hours and for Machine III is (2x + 6y) hours. The machines I, II, III are available for 36,50 and 60 hours per week. Therefore, the constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60.
Since x and y cannot be negative. We have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize z = 10x + 15y, subject to
2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

Question 5.
A manufacture produces bulbs and tubes. Each of these must be processed through two machines M₁ and M₂. A package of bulbs require 1 hour of work on Machine M₁ and 3 hours of work on M₂. A package of tubes require 2 hours on Machine M₁ and 4 hours on Machine M₂. He earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. Formulate the LLP to maximize the profit, if he operates the machine M₁, for atmost 10 hours a day and machine M₂ for atmost 12 hours a day.
Solution:
Let the number of packages of bulbs produced by manufacturer be x and packages of tubes be y. The manufacturer earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes.
Therefore, his total profit is p = ₹ (13.5x + 55y)
This is a linear function which is to be maximized.
Hence, it is the objective function.
The constraints are as per the following table :

From the table, the total time required for Machine M₁ is (x + 2y) hours and for Machine M₂ is (3x + 4y) hours.
Given Machine M₁ and M₂ are available for atmost 10 hours and 12 hours a day respectively.
Therefore, the constraints are x + 2y ≤ 10, 3x + 4y ≤ 12. Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as :
Maximize p = 13.5x + 55y, subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.

Question 6.
A company manufactures two types of fertilizers F₁ and F₂. Each type of fertilizer requires
two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F₁ and F₂ and availability of the raw materials A and B per day are given in the
table below :

By selling one unit of F₁ and one unit of F₂, company gets a profit of ₹ 500 and ₹ 750
respectively. Formulate the problem as L.P.P. to maximize the profit.
Solution:
Let the company manufactures x units of fertilizers F₁ and y units of fertilizers F₁. Then the total profit to the company is
z = ₹(500x + 750y).
This is a linear function that is to be maximized. Hence, it is an objective function.

The raw material A required for x units of Fertilizers F₁ and y units of Fertilizers F₂ is 2x + Since the maximum availability of A is 40, we have the first constraint as 2x + 3y ≤ 40.
Similarly, considering the raw material B, we have x + 4y ≤ 70.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as:
Maximize z = 500x + 750y, subject to
2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.

Question 7.
A doctor has prescribed two different units of foods A and B to form a weekly diet for a sick person. The minimum requirements of fats, carbohydrates and proteins are 18, 28, 14 units respectively. One unit of food A has 4 units of fats. 14 units of carbohydrates and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is ₹ 4.5 per unit and that of food B is ₹ 3.5 per unit. Form the L.P.P. so that the sick person’s diet meets the requirements at a minimum cost.
Solution:
Let the diet of sick person include x units of food A and y units of food B.
Then x ≥ 0, y ≥ 0.
The prices of food A and B are ₹ 4.5 and ₹ 3.5 per unit respectively.
Therefore, the total cost is z = ₹ (4.5x + 3.5y)
This is the linear function which is to be minimized.
Hence, it is objective function.
The constraints are as per the following table :

From the table, the sick person’s diet will include (4x + 6y) units of fats, (14x + 12y) units of carbohydrates and (8x + 8y) units of proteins. The minimum requirements of these ingredients are 18 units, 28 units and 14 units respectively.
Therefore, the constraints are
4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14.
Hence, the given LPP can be formulated as
Minimize z = 4.5x + 3.5y, subject to
4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.

Question 8.
If John drives a car at a speed of 60 kms/hour he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 kms/hour, the cost of petrol increases to ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as L.P.P.
Solution:
Let John travel xl km at a speed of 60 km/ hour and x₁ km at a speed of 90 km/hour.
Therefore, time required to travel a distance of x₁ km is \(\frac{x_{1}}{60}\) hours and the time required to travel a distance of
x₂ km is \(\frac{x_{2}}{90}\) hours.
∴ total time required to travel is \(\left(\frac{x_{1}}{60}+\frac{x_{2}}{90}\right)\) hours.
Since he wishes to travel the maximum distance within an hour,
\(\frac{x_{1}}{60}+\frac{x_{2}}{90}\) ≤ 1
He has to spend ₹ 5 per km on petrol at a speed of 60 km/hour and ₹ 8 per km at a speed of 90 km/hour.
∴ the total cost of travelling is ₹ (5x₁ + 8x₂)
Since he has ₹ 600 to spend on petrol,
5x₁ + 8x₂ ≤ 600
Since distance is never negative, x₁ ≥ 0, x₂ ≥ 0.
Total distance travelled by John is z. = (x₁ + x₂) km.
This is the linear function which is to be maximized.
Hence, it is objective function.
Hence, the given LPP can be formulated as :
Maximize z = x₁ + x₂, subject to
\(\frac{x_{1}}{60}+\frac{x_{2}}{90}\) ≤ 1, 5x₁ + 8x₂ ≤ 600, x₁ ≥ 0, x₂ ≥ 0.

Question 9.
The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be least 5 kg. Cement costs ₹ 20 per kg. and sand costs of ₹ 6 per kg. strength consideration dictate that a concrete brick should contain minimum 4 kg. of cement and not more than 2 kg. of sand. Form the L.P.P. for the cost to be minimum.
Solution:
Let the company use x₁ kg of cement and x₂ kg of sand to make concrete bricks.
Cement costs ₹ 20 per kg and sand costs ₹ 6 per kg.
∴ the total cost c = ₹ (20x₁ + 6x₂)
This is a linear function which is to be minimized.
Hence, it is the objective function.
Total weight of brick = (x₁ + x₂) kg
Since the weight of concrete brick has to be at least 5 kg,
∴ x₁ + x₂ ≥ 5.
Since concrete brick should contain minimum 4 kg of cement and not more than 2 kg of sand,
x₁ ≥ 4 and 0 ≤ x₂ ≤ 2
Hence, the given LPP can be formulated as :
Minimize c = 20x₁ + 6x₂, subject to
x₁ + x₂ ≥ 5, x₁ ≥ 4, 0 ≤ x₂ ≤ 2.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.2

I) Find the feasible solution of the following inequations graphically.
Question 1.
3x + 2y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 2y = 18 and 2x + y = 10 respectively.


The feasible solution is OCPBO which is shaded in the graph.

Question 2.
2x + 3y ≤ 6, x + y ≥ 2, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CB whose equations are 2x + 3y = 6 and x + y = 2 respectively.


The feasible solution is ∆ABC which is shaded in the graph.

Question 3.
3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are 3x + 4 y = 12, 4x + 7y = 28 and y = 1 respectively.


The feasible solution is PQDBP. which is shaded in the graph.

Question 4.
x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.


The feasible solution is OCPQBO. which is shaded in the graph.

Question 5.
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4
Solution:
First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x = 3 and y = 3 respectively.


The feasible solution is CEPQRC. which is shaded in the graph.

Question 6.
x – 2y ≤ 2, x + y ≥ 3, -2x + y ≤ 4, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are x – 2y = 2, x + y = 3 and -2x + y = 4 respectively.


The feasible solution is shaded in the graph.

Question 7.
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find the feasible solution graphically.
Solution:
Let the company produces x units of article A and y units of article B.
The given data can be tabulated as:

Inequations are :
x + 2y ≤ 4 and 3x + 2y ≤ 6
x and y are number of items, x ≥ 0, y ≥ 0
First we draw the lines AB and CD whose equations are x + 2y = 4 and 3x + 2y = 6 respectively.


The feasible solution is OCPBO. which is shaded in the graph.

Question 8.
A furniture dealer deals in tables and chairs. He has Rs.1,50,000 to invest and a space to store at most 60 pieces. A table costs him Rs.1500 and a chair Rs.750. Construct the inequations and find the feasible solution.
Question is modified
A furniture dealer deals in tables and chairs. He has ₹ 15,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 150 and a chair ₹ 750. Construct the inequations and find the feasible solution.
Solution:
Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0.
The dealer has a space to store at most 60 pieces.
∴ x + y ≤ 60
Since, the cost of each table is ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15,000
Hence the system of inequations are
x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + y = 60 and 150x + 750y = 15,000 respectively.


The feasible solution is OAPDO. which is shaded in the graph.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1

Question 1.
Solve graphically :
(i) x ≥ 0
Solution:
Consider the line whose equation is x = 0. This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1)
When x = 1, x ≥ 0
∴ (1, 1) lies in the required region
Therefore, the solution set is the Y-axis and the right
side of the Y-axis which is shaded in the graph.

(ii) x ≤ 0
Solution:
Consider the line whose equation is x = 0.
This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When x = 1, x ≰ 0
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the Y-axis and the left side of the Y-axis which is shaded in the graph.

(iii) y ≥ 0
Solution:
Consider the line whose equation is y = 0. This represents the X-axis. To find the solution set, we have to check any point other than origin. Let us check the point (1, 1).
When y = 1, y ≥ 0
∴ (1, 1) lies in the required region.
Therefore, the solution set is the X-axis and above the X-axis which is shaded in the graph.

(iv) y ≤ 0
Solution:
(iv) Consider the line whose equation is y = 0. This represents the X-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When y = 1, y ≰ 0.
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the X-axis and below the X-axis which is shaded in the graph.

Question 2.
Solve graphically :
(i) x ≥ 0 and y ≥ 0
Solution:
Consider the lines whose equations are x = 0, y = 0.
These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
(i) Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.

(ii) x ≤ 0 and y ≥ 0
Solution:
Since x ≤ 0, y ≥ 0, the solution set is in the second quadrant which is shaded in the graph.

(iii) x ≤ 0 and y ≤ 0
Solution:
Since x ≤ 0, y ≤ 0, the solution set is in the third quadrant which is shaded in the graph.

(iv) x ≥ 0 and y ≤ 0
Solution:
Since x ≥ 0, y ≤ 0, the solution set is in the fourth ! quadrant which is shaded in the graph.

Question 3.
Solve graphically :
(i) 2x – 3 ≥ 0
Solution:
Consider the line whose equation is 2x – 3 = 0,
i.e. x = \(\frac{3}{2}\)
This represents a line parallel to Y-axis passing through the point (\(\frac{3}{2}\), 0)
Draw the line x =\(\frac{3}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 2x – 3 = 2 × 0 – 3 = -3 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(\frac{3}{2}\) and the non-origin side of the line which is shaded in the graph.

(ii) 2y – 5 ≥ 0
Solution:
Consider the line whose equation is 2y – 5 = 0, i.e. y = \(\frac{5}{2}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{5}{2}\)).
Draw the line y = \(\frac{5}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 2y – 5 = 2 × 0 – 5 = -5 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{5}{2}\) and the non-origin side of the line which is shaded in the graph.

(iii) 3x + 4 ≤ 0
Solution:
(iii) Consider the line whose equation is 3x + 4 = 0,
i.e. x = \(-\frac{4}{3}\)
This represents a line parallel to Y-axis passing through the point (\(-\frac{4}{3}\), 0).
Draw the line x = \(-\frac{4}{3}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 3x + 4 = 3 × 0 + 4= 4 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(-\frac{4}{3}\) and the non-origin side of the line which is shaded in the graph.

(iv) 5y + 3 ≤ 0
Solution:
(iv) Consider the line whose equation is 5y + 3 = 0,
i.e. y = \(\frac{-3}{5}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{-3}{5}\))
Draw the line y = \(\frac{-3}{5}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y + 3 = 5 × 0 + 3 = 3 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{-3}{5}\) and the non-origin side of the line which is shaded in the graph.

Question 4.
Solve graphically :
(i) x + 2y ≤ 6
Solution:
Consider the line whose equation is x + 2y = 6.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get x = 6.
∴ A = (6, 0) is a point on the line.
Put x = 0, we get 2y = 6, i.e. y = 3
∴ B = (0, 3) is another point on the line.

Draw the line AB joining these points. This line divide the line into two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
When x = 0, y = 0, then x + 2y = 0 which is less than 6.
∴ x + 2y ≤ 6 in this case.
Hence, origin lies in the required region. Therefore, the given inequality is the origin side which is
shaded in the graph.
This is the solution set of x + 2y ≤ 6.

(ii) 2x – 5y ≥ 10
Solution:
Consider the line whose equation is 2x – 5y = 10.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get 2x = 10, i.e. x = 5.
∴ A = (5, 0) is a point on the line.
Put x = 0, we get -5y = 10, i.e. y = -2
∴ B = (0, -2) is another point on the line.

Draw the line AB joining these points. This line J divide the plane in two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then 2x – 5y = 0 which is neither greater nor equal to 10.
∴ 2x – 5y ≱ 10 in this case.
Hence (0, 0) will not lie in the required region.
Therefore, the given inequality is the non-origin side, which is shaded in the graph.
This is the solution set of 2x – 5y ≥ 10.

(iii) 3x + 2y ≥ 0
Solution:
Consider the line whose equation is 3x + 2y = 0.
The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is O ≡ (0, 0).
To find the another point, we can give any value of x and get the corresponding value of y.
Put x = 2, we get 6 + 2y = 0 i.e. y = – 3
∴ A = (2, -3) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check (0, 0) as it is already on the line.
We can check any other point which is not on the line.
Let us check the point (1, 1)

When x = 1, y = 1, then 3x + 2y = 3 + 2 = 5 which is greater than zero.
∴ 3x + 2y > 0 in this case.
Hence (1, 1) lies in the required region. Therefore, the required region is the upper side which is shaded in the graph.
This is the solution set of 3x + 2y ≥ 0.

(iv) 5x – 3y ≤ 0
Solution:
Consider the line whose equation is 5x – 3y = 0. The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is the origin O = (0, 0).
To find the other point, we can give any value of x and get the corresponding value of y.
Put x = 3, we get 15 – 3y = 0, i.e. y = 5
∴ A ≡ (3, 5) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check 0(0, 0), as it is already on the line. We can check any other point which is not on the line.
Let us check the point (1, -1).
When x = 1, y = -1 then 5x – 3y = 5 + 3 = 8
which is neither less nor equal to zero.
∴ 5x – 3y ≰ 0 in this case.
Hence (1, -1) will not lie in the required region. Therefore, the required region is the upper side which is shaded in the graph.

This is the solution set of 5x – 3y ≤ 0.

Question 5.
Solve graphically :
(i) 2x + y ≥ 2 and x – y ≤ 1
Solution:
First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(ii) x – y ≤ 2 and x + 2y ≤ 8
Solution:
First we draw the lines AB and CD whose equations are x – y = 2 and x + 2y = 8 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(iii) x + y ≥ 6 and x + 2y ≤ 10
Solution:
First we draw the lines AB and CD whose equations are x + y = 6 and x + 2y = 10 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(iv) 2x + 3y ≤ 6 and x + 4y ≥ 4
Solution:
First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.

The solution set of the given system of inequalities is shaded in the graph.

(v) 2x + y ≥ 5 and x – y ≤ 1
Solution:
First we draw the lines AB and CD whose equations are 2x + y = 5 and x – y = 1 respectively.


The solution set of the given system of inequations is shaded in the graph.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6B

Question 1.
If the line \(\frac{x}{3}=\frac{y}{4}\) = z is perpendicular to the line \(\frac{x-1}{k}=\frac{y+2}{3}=\frac{z-3}{k-1}\) then the value of k is:

Solution:
(b) \(-\frac{11}{4}\)

Question 2.
The vector equation of line 2x – 1 = 3y + 2 = z – 2 is

Solution:
(a) \(\bar{r}=\left(\frac{1}{2} \hat{i}-\frac{2}{3} \hat{j}+2 \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})\)

Question 3.
The direction ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}\) are
(A) 4, 5, 7
(B) 4, -5, 7
(C) 4, -5, -7
(D) -4, 5, 8
Solution:
(A) 4, 5, 7

Question 4.
The length of the perpendicular from (1, 6, 3) to the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\)
(A) 3
(B) \(\sqrt {11}\)
(C) \(\sqrt {13}\)
(D) 5
Solution:
(C ) \(\sqrt {13}\)

Question 5.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is
Question is modified.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is

Solution:
(c) \(\frac{3}{\sqrt{2}}\)

Question 6.
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\). and coplanar if
(A) k = 1 or -1
(B) k = 0 or -3
(C) k = + 3
(D) k = 0 or -1
Solution:
(B ) k = 0 or -3

Question 7.
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{6}\) and are
(A) perpendicular
(B) inrersecting
(C) skew
(D) coincident
Solution:
(B) inrersecting

Question 8.
Equation of X-axis is
(A) x = y = z
(B) y = z
(C) y = 0, z = 0
(D) x = 0, y = 0
Solution:
(C) y = 0, z = 0

Question 9.
The angle between the lines 2x = 3y = -z and 6x = -y = -4z is
(A ) 45º
(B ) 30º
(C ) 0º
(D ) 90º
Solution:
(D ) 90º

Question 10.
The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are
(A ) 2, 1, 6
(B ) 2, 1, -6
(C ) 2, -1, 6
(D ) -2, 1, 6
Solution:
(B ) 2, 1, -6

Question 11.
The perpendicular distance of the plane 2x + 3y – z = k from the origin is \(\sqrt {14}\) units, the value
of k is
(A ) 14
(B ) 196
(C ) \(2\sqrt {14}\)
(D ) \(\frac{\sqrt{14}}{2}\)
Solution:
(A ) 14

Question 12.
The angle between the planes and \(\bar{r} \cdot(\bar{i}-2 \bar{j}+3 \bar{k})+4=0\) and \(\bar{r} \cdot(2 \bar{i}+\bar{j}-3 \bar{k})+7=0\) is

Solution:
(d) cos^-1\(\left(\frac{9}{14}\right)\)

Question 13.
If the planes \(\bar{r} \cdot(2 \bar{i}-\lambda \bar{j}+\bar{k})=3\) and \(\bar{r} \cdot(4 \bar{i}-\bar{j}+\mu \bar{k})=5\) are parallel, then the values of λ and μ are respectively.

Solution:
(d) \(\frac{1}{2}\), 2

Question 14.
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
(A ) x + y + z =1
(B ) x + y + z = 2
(C ) x + y + z = 3
(D ) x + y + z = 4
Solution:
(D ) x + y + z = 4

Question 15.
Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is
(A ) 0º
(B ) 30º
(C ) 45º
(D ) 90º
Solution:
(A ) 0º

Question 16.
The direction cosines of the normal to the plane 2x – y + 2z = 3 are

Solution:
(a) \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)

Question 17.
The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is :
(A ) 3x + 2z – 1 = 0
(B ) 3x – 2z = 1
(C ) 3x + 2z + 1 = 0
(D ) 3x + 2z = 2
Solution:
(B ) 3x – 2z = 1

Question 18.
The equation of the plane in which the line \(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-4}{1}=\frac{z+5}{3}\) lie, is
(A ) 17x – 47y – 24z + 172 = 0
(B ) 17x + 47y – 24z + 172 = 0
(C ) 17x + 47y + 24z +172 = 0
(D ) 17x – 47y + 24z + 172 = 0
Solution:
(A ) 17x – 47y – 24z + 172 = 0

Question 19.
If the line \(\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is:
(A ) 5
(B ) 3
(C ) 2
(D ) -5
Solution:
(A ) 5

Question 20.
The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is
(A ) 4x + y + 5z = 14
(B ) 4x – 2y – 5z = 45
(C ) x – 2y – 5z = 10
(D ) 4x + y + 6z = 11
Solution:
(B ) 4x – 2y – 5z = 45

II. Solve the following :
Question 1.
Find the vector equation of the plane which is at a distance of 5 unit from the origin and which is normal to the vector \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
If \(\hat{n}\) is a unit vector along the normal and p i the length of the perpendicular from origin to the plane, then the vector equation of the plane \(\bar{r} \cdot \hat{n}\) = p
Here, \(\overline{\mathrm{n}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and p = 5

Question 2.
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is

= 1units.

Question 3.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
Solution:
The equation of the plane is 2x + 3y + 6z = 49
Dividing each term by
\(\sqrt{2^{2}+3^{2}+(-6)^{2}}\)
= \(\sqrt{49}\)
= 7
we get
\(\frac{2}{7}\)x + \(\frac{3}{7}\)y – \(\frac{6}{7}\)z = \(\frac{49}{7}\) = 7
This is the normal form of the equation of plane.
∴ the direction cosines of the perpendicular drawn from the origin to the plane are
l = \(\frac{2}{7}\), m = \(\frac{3}{7}\), n = \(\frac{6}{7}\)
and length of perpendicular from origin to the plane is p = 7.
the coordinates of the foot of the perpendicular from the origin to the plane are
(lp, ∓, np)i.e.(2, 3, 6)

Question 4.
Reduce the equation \(\bar{r} \cdot(\hat{i}+8 \hat{j}+24 \hat{k})=13\) to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Solution:
The normal form of equation of a plane is \(\bar{r} \cdot \hat{n}\) = p where \(\hat{n}\) is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane.
Given pane is \(\text { r. }(6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+24 \hat{\mathrm{k}})=13\) …(1)

This is the normal form of the equation of plane.
Comparing with \(\bar{r} \cdot \hat{n}\) = p,
(i) the length of the perpendicular from the origin to plane is \(\frac{1}{2}\).
(ii) direction cosines of the normal are \(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\)

Question 5.
Find the vector equation of the plane passing through the points A(1, -2, 1), B (2, -1, -3) and C (0, 1, 5).
Solution:
The vector equation of the plane passing through three non-collinear points A(\(\bar{a}\)), B(\(\bar{b}\)) and C(\(\bar{c}\)) is \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (1)

Question 6.
Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.
Solution:
The Cartesian equation of the plane passing through (x₁, y₁, z₁), the direction ratios of whose normal are a, b, c, is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
∴ the cartesian equation of the required plane is
o(x + 1) + 2(y + 2) + 5(z – 3) = 0
i.e. 0 + 2y – 4 + 10z – 15 = 0
i.e. y + 2 = 0.

Question 7.
Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\)
Solution:
The cartesian equation of the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\) is 6x + 8y + 7z = 0 The required plane is parallel to it
∴ its cartesian equation is
6x + 8y + 7z = p …(1)
A (7, 8, 6) lies on it and hence satisfies its equation
∴ (6)(7) + (8)(8) + (7)(6) = p
i.e., p = 42 + 64 + 42 = 148.
∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

Question 8.
The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). Find the vector equation of the plane.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\).
M(1, 2, 0) is the foot of the perpendicular drawn from origin to the plane. Then the plane is passing through M and is
perpendicular to OM.
If \(\bar{m}\) is the position vector of M, then \(\bar{m}\) = \(\hat{\mathrm{i}}\).
Normal to the plane is
\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{\mathrm{i}}\)
\(\overline{\mathrm{m}} \cdot \overline{\mathrm{n}}\) = \(\hat{\mathrm{i}}, \hat{i}\) = 5
∴ the vector equation of the required plane is
\(\bar{r} \cdot(\hat{i}+2 \hat{j})\) = 5

Question 9.
A plane makes non zero intercepts a, b, c on the co-ordinates axes. Show that the vector equation of the plane is \(\bar{r} \cdot(b c \hat{i}+c a \hat{j}+a b \hat{k})\) = abc
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)), B(\(\bar{b}\)).. C(\(\bar{c}\)), where A, B, C are non collinear is
\(\overline{\mathrm{r}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) …(1)
The required plane makes intercepts 1, 1, 1 on the coordinate axes.
∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0, , 1)

Question 10.
Find the vector equation of the plane passing through the pointA(-2, 3, 5) and parallel to vectors \(4 \hat{i}+3 \hat{k}\) and \(\hat{i}+\hat{j}\)
Solution:
The vector equation of the plane passing through the point A(\(\bar{a}\)) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is


= (-2)(-4) + (7)(-1) + (5)(4)
= 8 – 7 + 8
= 35
∴ From (1), the vector equation of the required plane is \(\overline{\mathrm{r}} \cdot(-3 \hat{\mathrm{i}}-3 a t \mathrm{j}+4 \hat{\mathrm{k}})\) = 35.

Question 11.
Find the Cartesian equation of the plane \(\bar{r}=\lambda(\hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+2 \hat{j}+3 \hat{k})\)
Solution:
The equation \(\bar{r}=\bar{a}+\lambda \bar{b}+\mu \bar{c}\) represents a plane passing through a point having position vector \(\overline{\mathrm{a}}\) and parallel to vectors \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
Here,

Question 12.
Find the vector equations of planes which pass through A(1, 2, 3), B (3, 2, 1) and make equal intercepts on the co-ordinates axes.
Question is modified
Find the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the cartesian equation of the plane is
ax + by + cz = 0 …..(1)
(1, 2, 3) and (3, 2, 1) lie on the plane.
∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ a, b, c are proportional to 1, -2, 1
∴ from (1), the required cartesian equation is x – 2y + z = 0.
Case 2 : Let the plane make non zero intercept p on each axis.
then its equation is \(\frac{x}{p}+\frac{y}{p}+\frac{z}{p}\) = 1
i.e. x + y + z = p …(2)
Since this plane pass through (1, 2, 3) and (3, 2, 1)
∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p
∴ p = 6
∴ from (2), the required cartesian equation is
x + y + z = 6
Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

Question 13.
Find the vector equation of the plane which makes equal non-zero intercepts on the co-ordinates axes and passes through (1, 1, 1).
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the vector equation of the plane is ax + by + cz …(1)
(1, 1, 1) lie on the plane.
∴ 1a + 1b + 1c = 0

∴ from (1), the required cartesian equation is x – y + z = 0
Case 2 : Let he plane make non zero intercept p on each axis.
then its equation is \(\frac{\hat{\mathrm{i}}}{p}+\frac{\hat{\mathrm{j}}}{p}+\frac{\hat{\mathrm{k}}}{p}=1\) = 1
i.e. \(\hat{i}+\hat{j}+\hat{k}=p\) = p ….(2)
Since this plane pass through (1, 1, 1)
∴ 1 + 1 + 1 = p
∴ p = 3
∴ from (2), the required cartesian equation is \(\hat{i}+\hat{j}+\hat{k}\) = 3
Hence, the cartesian equations of required planes are \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=3\)

Question 14.
Find the angle between planes \(\bar{r} \cdot(-2 \hat{i}+\hat{j}+2 \hat{k})=17\) and \(\bar{r} \cdot(2 \hat{i}+2 \hat{j}+\hat{k})=71\).
Solution:
The acute angle between the planes

= (1)(2) + (1)(1) + (2)(1)
= 2 + 1 + 2
= 5
Also,

Question 15.
Find the acute angle between the line \(\bar{r}=\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=23\)
Solution:
The acute angle θ between the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) and the plane \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}\) = d is given by

= (2)(2) + (3)(-1) + (-6)(1)
= 4 – 3 – 6
= -5

Question 16.
Show that lines \(\bar{r}=(\hat{i}+4 \hat{j})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(3 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
Solution:

Question 17.
Find the distance of the point \(3 \hat{i}+3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}+6 \hat{k})=21\)
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}\) = p is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}\) ……(1)

= (3)(2) + (3)(3) + (1)(-6)
= 6 + 9 – 6
= 9

Question 18.
Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.
Solution:

= 19units.

Question 19.
Find the vector equation of the plane passing through the origln and containing the line \(\bar{r}=(\hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})\).
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) … (1)
We can take \(\bar{a}\) = \(\bar{0}\) since the plane passes through the origin.
The point M with position vector \(\bar{m}\) =\(\hat{i}+4 \hat{j}+\hat{k}\) lies on the line and hence it lies on the plane.
.’. \(\overline{\mathrm{OM}}=\bar{m}=\hat{i}+4 \hat{j}+\hat{k}\) lies on the plane.
The plane contains the given line which is parallel to \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
Let \(\bar{n}\) be normal to the plane. Then \(\bar{n}\) is perpendicular to \(\overline{\mathrm{OM}}\) as well as \(\bar{b}\)

Question 20.
Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B( 4, 3, -2) at right angle.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) ….(1)
The position vectors \(\bar{a}\) and \(\bar{b}\) of the given points A and B are \(\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) and \(\bar{b}=4 \hat{i}+3 \hat{j}-2 \hat{k}\)
If M is the midpoint of segment AB, the position vector \(\bar{m}\) of M is given by

Question 21.
Show thatlines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.
Solution:
Given lines are x = y, z = 0 and x + y = 0, z = 0.
It is clear that (0, 0, 0) satisfies both the equations.
∴ the lines intersect at O whose position vector is \(\overline{0}\)
Since z = 0 for both the lines, both the lines lie in XY- plane.
Hence, we have to find equation of XY-plane.
Z-axis is perpendicular to XY-plane.
∴ normal to XY plane is \(\hat{k}\).
0(\(\overline{0}\)) lies on the plane.
By using \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\), the vector equation of the required plane is \(\bar{r} \cdot \hat{k}=\overline{0} \cdot \bar{k}\)
i.e. \(\bar{r} \cdot \hat{k}=0\).
Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \cdot \hat{k}=0\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6A Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6A

Question 1.
Find the vector equation of the line passing through the point having position vector \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector
\(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to the vector \(6 \hat{i}-\hat{j}+\hat{k}\) is
\(\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(\hat{6 \hat{\mathrm{i}}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\).

Question 2.
Find the vector equation of the line which passes through the point (3, 2, 1) and is parallel to the vector \(2 \hat{i}+2 \hat{j}-3 \hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and parallel to the vector
\(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \text { is } \overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)

Question 3.
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\)
Solution:
The line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\) has direction ratios 3, 5, 6. The required line has direction ratios 3, 5, 6 as it is parallel to the given line.
It passes through the point (-2, 4, -5).
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

Question 4.
Obtain the vector equation of the line \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
Solution:
The cartesian equations of the line are \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6.
Let \(\bar{a}\) be the position vector of the point A w.r.t. the origin and \(\bar{b}\) be the vector parallel to the line.
Then \(\bar{a}=-5 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\bar{b}=3 \hat{i}+5 \hat{j}+6 \hat{k}\).
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=(-5 \hat{i}-4 \hat{j}-6 \hat{k})+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

Question 5.
Find the vector equation of the line which passes through the origin and the point (5, -2, 3).
Solution:
Let \(\bar{b}\) be the position vector of the point B(5, -2, 3).
Then \(\bar{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
Origin has position vector \(\overline{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\).
The vector equation the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=\overline{0}+\lambda(\bar{b}-\overline{0})=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)

Question 6.
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).
Solution:
Let A = (3, -2, -5), B = (3, -2, 6)
The direction ratios of the line AB are
3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are
x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ
i.e. x = 3, y = -2, z = 11λ – 5
∴ the cartesian equations of the line are
x = 3, y = -2, z = 11λ – 5, λ is a scalar.

Question 7.
Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).
Solution:
The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are
x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ)
x – 3 = 2λ, y – 2 = -λ, z = 1
∴ \(\frac{x-3}{2}=\frac{y-2}{-1}\) = λ, z = 1
∴ the cartesian equations of the line are
\(\frac{x-3}{2}=\frac{y-2}{-1}\), z = 1.

Question 8.
Find the Cartesian equations of the line passing through the point A(1, 1, 2) and perpendicular to vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
Solution:
Let the required line have direction ratios p, q, r. ,
It is perpendicular to the vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3p + 2q – r = 0

∴ the required line has direction ratios -1, 1, -1.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are
\(\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-2}{-1}\)

Question 9.
Find the Cartesian equations of the line which passes through the point (2, 1, 3) and perpendicular
to lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}\).
Solution:
Let the required line have direction ratios p, q, r.
It is perpendicular to the vector \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3 + 2q – r = 0

∴ the required line has direction ratios 2, -7, 4.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x=x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are
\(\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-2}{4}\)

Question 10.
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
Solution:
The given line is \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\) = λ … (Say)
∴ coordinates of any point on the line are
x = λ + 1, y = λ + 2, z = λ + 3
∴ position vector of any point on the line is
(λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … (1)
If \(\bar{b}\) is parallel to the given line whose direction ratios are 1, 1, 1, then \(\bar{b}=\hat{i}+\hat{j}+\hat{k}\).
Let the required line passing through O meet the given line at M.
∴ position vector of M
= \(\bar{m}\) = (λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … [By (1)]
The required line is perpendicular to given line

The vector equation of the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\), λ is a scalar.
∴ the vector equation of the line passing through o(\(\bar{o}\)) and M(\(\bar{m}\)) is
\(\bar{r}=\overline{0}+\lambda(\bar{m}-\overline{0})=\lambda \bar{m}=\lambda(-\hat{i}+\hat{k})\) where λ is a scalar.
Hence, vector equation of the required line is \(\).

Question 11.
Find the value of λ so that lines \(\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angle.
Solution:
The equations of the given lines are

Question 12.
Find the acute angle between lines \(\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{2}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}\).
Solution:

Question 13.
Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.
Solution:
The equations x = y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0.
∴ its directiton ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0.

If θ is the acute angle between the lines, then

Question 14.
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.
Solution:
The equations x = -y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0.
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.
∴ its direction ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0

Question 15.
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\).
Solution:
Let P = (0, 2, 3)
Let M be the foot of the perpendicular drawn from P to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) = λ ……(Say)
The coordinates of any point on the line are given by
x = 5λ – 3, y = 2λ + 1, z = 3λ – 4
Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1)
The direction ratios of PM are
5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7
Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3,
5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0
25λ – 15 + 4λ – 2 + 9λ – 21 =0
38λ – 38 = 0 ∴ λ = 1
Substituting λ = 1 in (1), we get.
M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1).
Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

Question 16.
By computing the shortest distance determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+2 \hat{j}-3 \hat{k})+\mu(\hat{i}+\hat{j}-2 \hat{k})\)
Solution:
The shortest distance between the lines


Shortest distance between the lines is 0.
∴ the lines intersect each other.

(ii) \(\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}\) and x – 6 = y – 8 = z + 2.
Solution:
The shortest distance between the lines

∴ x₁ = 5, y₁ = 7, z₁ = 3, x₂ = 6, y₂ = 8, z₂ = 2,
l₁ = 4, m₁ = 5, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines

or
Shortest distance between the lines is 0.
∴ the lines intersect each other.

Question 17.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}\) intersect each other then find m.
Solution:

Here, (x₁, y₁, z₁) ≡ (1, -1, 1),
(x₂, y₂, z₂) ≡ (2, -m, 2),
a₁ = 2, b₁ = 3, c₁ = 4,
a₂ = 1, b₂ = 2, c₂ = 1
Substituting these values in (1), we get

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0
∴ -5 + 2 – 2m + 1 = 0
∴ -2m = 2
∴ m = -1.

Question 18.
Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.
Solution:
Let \(\bar{a}\) be the position vector of the point A (-1, -1, 2) w.r.t. the origin.
Then \(\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}\)
The equation of given line is
x – 2 = 3y + 1 = 6z – 2.

The direction ratios of this line are

Question 19.
Find the direction cosines of the line \(\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})\).
Solution:

Question 20.
Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersects the \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\).
Solution:
Let the required line have direction ratios a, b, c
Since the line passes through the origin, its cartesian equations are
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) …(1)
This line is perpendicular to the line
x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0
∴ 4b – 3c + 4a – 2c + 3a – 2b = 0
∴ 7a + 2b – 5c = 0
From (2) and (3), we get

Question 21.
Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution:
4x – 3z + 5 = 0 can be written as

This line passes through the point A(0, 2, \(\frac{5}{3}\)) position vector is \(\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}\)
Also the line has direction ratio 3, 0, 4.
If \(\bar{b}\) is a vector parallel to the line, then \(\bar{b}=3 \hat{i}+4 \hat{k}\)
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is \(\bar{a}\) scalar,
∴ the vector equation of the required line is
\(\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})\).

Question 22.
Find the co-ordinates of points on the line \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) which are at the distance 3 unit from the base point A(1, 2, 3).
Solution:
The cartesian equations of the line are \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) = λ
The coordinates of any point on this line are given by
x = λ + 1, y = -2λ + 2, z = 2λ + 3
Let M(λ + 1, -2λ + 2, 2λ + 3) … (1)
be the point on the line whose distance from A(1, 2, 3) is 3 units.

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)]
i. e. M = (2, 0, 5)
When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)]
i. e. M = (0, 4, 1)
Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).