Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.5

Question 1.
Find the second order derivatives of the following:
(i) 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9
Solution:
Let y = 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9

(ii) e^2x . tan x
Solution:
Let y = e^2x . tan x

(iii) e^4x . cos 5x
Solution:
Let y = e^4x . cos 5x

(iv) x³ . log x
Solution:
Let y = x³ . log x

(v) log(log x)
Solution:
Let y = log(log x)

(vi) x^x
Solution:
y = x^x
log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 2.
Find \(\frac{d^{2} y}{d x^{2}}\) of the following:
(i) x = a(θ – sin θ), y = a (1 – cos θ)
Solution:
x = a(θ – sin θ), y = a (1 – cos θ)
Differentiating x and y w.r.t. θ, we get
\(\frac{d x}{d \theta}=a \frac{d}{d \theta}(\theta-\sin \theta)\) = a(1 – cos θ) …….(1)

(ii) x = 2at², y = 4at
Solution:
x = 2at², y = 4at
Differentiating x and y w.r.t. t, we get

(iii) x = sin θ, y = sin³θ at θ = \(\frac{\pi}{2}\)
Solution:
x = sin θ, y = sin³θ
Differentiating x and y w.r.t. θ, we get,

(iv) x = a cos θ, y = b sin θ at θ = \(\frac{\pi}{4}\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

Question 3.
(i) If x = at² and y = 2at, then show that \(x y \frac{d^{2} y}{d x^{2}}+a=0\)
Solution:
x = at², y = 2at ………(1)
Differentiating x and y w.r.t. t, we get

(ii) If y = \(e^{m \tan ^{-1} x}\), show that \(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-m) \frac{d y}{d x}=0\)
Solution:
y = \(e^{m \tan ^{-1} x}\) ……..(1)

(iii) If x = cos t, y = e^mt, show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0\)
Solution:
x = cos t, y = e^mt

(iv) If y = x + tan x, show that \(\cos ^{2} x \cdot \frac{d^{2} y}{d x^{2}}-2 y+2 x=0\)
Solution:
y = x + tan x

(v) If y = e^ax . sin (bx), show that y₂ – 2ay₁ + (a² + b²)y = 0.
Solution:
y = e^ax . sin (bx) ………(1)

(vi) If \(\sec ^{-1}\left(\frac{7 x^{3}-5 y^{3}}{7 x^{3}+5 y^{3}}\right)=m\), show that \(\frac{d^{2} y}{d x^{2}}=0\)
Solution:

(vii) If 2y = \(\sqrt{x+1}+\sqrt{x-1}\), show that 4(x² – 1)y₂ + 4xy₁ – y = 0.
Solution:
2y = \(\sqrt{x+1}+\sqrt{x-1}\) …… (1)
Differentiating both sides w.r.t. x, we get

(viii) If y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\), show that \(\left(x^{2}+a^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=0\)
Solution:
y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\) = \(m \log \left(x+\sqrt{x^{2}+a^{2}}\right)\)

(ix) If y = sin(m cos^-1x), then show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+m^{2} y=0\)
Solution:
y = sin(m cos^-1x)
sin^-1y = m cos^-1x
Differentiating both sides w.r.t. x, we get

(x) If y = log(log 2x), show that xy₂ + y₁(1 + xy₁) = 0.
Solution:
y = log(log 2x)

(xi) If x² + 6xy + y² = 10, show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\)
Solution:
x² + 6xy + y² = 10 …… (1)
Differentiating both sides w.r.t. x, we get

(xii) If x = a sin t – b cos t, y = a cos t + b sin t, Show that \(\frac{d^{2} y}{d x^{2}}=-\frac{x^{2}+y^{2}}{y^{3}}\)
Solution:
x = a sin t – b cos t, y = a cos t + b sin t
Differentiating x and y w.r.t. t, we get

Question 4.
Find the nth derivative of the following:
(i) (ax + b)^m
Solution:
Let y = (ax + b)^m

(ii) \(\frac{1}{x}\)
Solution:
Let y = \(\frac{1}{x}\)

(iii) e^ax+b
Solution:
Let y = e^ax+b

(iv) a^px+q
Solution:
Let y = a^px+q

(v) log(ax + b)
Solution:
Let y = log(ax + b)
Then \(\frac{d y}{d x}=\frac{d}{d x}[\log (a x+b)]\)

(vi) cos x
Solution:
Let y = cos x

(vii) sin(ax + b)
Solution:
Let y = sin(ax + b)

(viii) cos(3 – 2x)
Solution:

(ix) log(2x + 3)
Solution:

(x) \(\frac{1}{3 x-5}\)
Solution:
Let y = \(\frac{1}{3 x-5}\)

(xi) y = e^ax . cos (bx + c)
Solution:
y = e^ax . cos (bx + c)

(xii) y = e^8x . cos (6x + 7)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Question 1.
Find \(\frac{d y}{d x}\) if
(i) x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Solution:
x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Differentiating x and y w.r.t. m, we get
\(\frac{d x}{d m}=\frac{d}{d m}\left(\sqrt{a^{2}+m^{2}}\right)\)

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\), where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\) ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Solution:
x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Put t = tan θ Then θ = tan^-1t

(viii) x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Solution:
x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Put t = cos θ. Then θ = cos^-1t
x = cos^-1(4cos³θ – 3cos θ)

Question 2.
Find \(\frac{d y}{d x}\), if
(i) x = cosec²θ, y = cot³θ at θ = \(\frac{\pi}{6}\)
Solution:
x = cosec²θ, y = cot³θ
Differentiating x and y w.r.t. θ, we get

(ii) x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{3}\)
Solution:
x = a cos³θ, y = a sin³θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\)) at t = 1
Solution:
x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\))
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = \(\frac{\pi}{4}\)
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t = \(\frac{1}{2}\)
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

Question 3.
(i) If x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\), then show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution:
x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\)

(ii) If x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\), then show that \(\frac{d y}{d x}=-\frac{y \log x}{x \log y}\)
Solution:
x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\)
log x = log \(e^{\sin 3 t}\), log y = log \(e^{\cos 3 t}\)
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\), then show that y2 – \(\frac{d y}{d x}\) = 0.
Solution:
x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\)

(iv) If x = a cos³t, y = a sin³t, then show that \(\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}\)
Solution:
x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3), show that \(\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}\)
Solution:
x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3)

(vi) If x = log (1 + t²), y = t – tan^-1t, show that \(\frac{d y}{d x}=\frac{\sqrt{e^{x}-1}}{2}\)
Solution:
x = log (1 + t²), y = t – tan^-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\), show that sin x + \(\frac{d y}{d x}\) = 0
Solution:
x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\)
Differentiating x and y w.r.t. t, we get

(viii) If x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\), show that \(\frac{d x}{d y}=-\frac{b^{2} y}{a^{2} x}\)
Solution:
x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\)

Question 4.
(i) Differentiate x sin x w.r.t tan x.
Solution:
Let u = x sinx and v = tan x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

(ii) Differentiate \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) w.r.t \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)

(iii) Differentiate \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\) w.r.t \(\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\)
Solution:

(iv) Differentiate \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) w.r.t. tan^-1x
Solution:
Let u = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) and v = tan^-1x
Then we want to find \(\frac{d u}{d v}\)
Put x = tan θ. Then θ = tan^-1x.

(v) Differentiate 3x w.r.t. log_x3.
Solution:
Let u = 3x and v = log_x3.
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(3^{x}\right)=3^{x} \cdot \log 3\)

(vi) Differentiate \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) w.r.t. sec^-1x.
Solution:
Let u = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) and v = sec^-1x
Then we want to find \(\frac{d u}{d v}\).
Differentiating u and v w.r.t. x, we get

(vii) Differentiate x^x w.r.t. x^{sin x}.
Solution:
Let u = x^x and v = x^{sin x}
Then we want to find \(\frac{d u}{d x}\).
Take, u = x^x
log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get

(viii) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Solution:
Let u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and v = \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)
u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Put x = tan θ. Then θ = tan^-1x and

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.3

Question 1.
Differentiate the following w.r.t. x:
(i) \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Solution:
Let y = \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Then, log y = log [latex]\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}[/latex]
= log (x + 1)² – log (x + 2)³ – log (x + 3)⁴
= 2 log (x +1) – 3 log (x + 2) – 4 log (x + 3)
Differentiating both sides w.r.t. x, we get

(ii) \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Solution:
Let y = \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Then log y = log [latex]\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}[/latex]

Differentiating both sides w.r.t. x, we get

(iii) \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Solution:
Let y = \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Then log y = log [latex]\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}[/latex]

Differentiating both sides w.r.t. x, we get

(iv) \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Then log y = log [latex]\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}[/latex]

Differentiating both sides w.r.t. x, we get

(v) \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Solution:
Let y = \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Then log y = log [latex]\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}[/latex]
= log x⁵ + log tan³4x – log sin²3x
= 5 log x+ 3 log (tan 4x) – 2 log (sin 3x)
Differentiating both sides w.r.t. x, we get

(vi) \(x^{\tan ^{-1} x}\)
Solution:
Let y = \(x^{\tan ^{-1} x}\)
Then log y = log (\(x^{\tan ^{-1} x}\)) = (tan^-1 x)(log x)
Differentiating both sides w.r.t. x, we get

(vii) (sin x)^x
Solution:
Let y = (sin x)^x
Then log y = log (sin x)^x = x . log (sin x)
Differentiating both sides w.r.t. x, we get

(viii) sin x^x
Solution:
Let y = (sin x^x)
Then \(\frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin x^{x}\right)\right]\)
\(\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot \frac{d}{d x}\left(x^{x}\right)\) ……. (1)
Let u = x^x
Then log u = log x^x = x . log x
Differentiating both sides w.r.t. x, we get

Question 2.
Differentiate the following w.r.t. x:
(i) x^e + x^x + e^x + e^e
Solution:
Let y = x^e + x^x + e^x + e^e
Let u = x^x
Then log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get
\(\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)\)

(ii) \(x^{x^{x}}+e^{x^{x}}\)
Solution:
Let y = \(x^{x^{x}}+e^{x^{x}}\)
Put u = \(x^{x^{x}}\) and v = \(e^{x^{x}}\)
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Take u = \(x^{x^{x}}\)
log u = log \(x^{x^{x}}\) = x^x . log x
Differentiating both sides w.r.t. x, we get

(iii) (log x)^x – (cos x)^{cot x}
Solution:
Let y = (log x)^x – (cos x)^{cot x}
Put u = (log x)^x and v = (cos x)^{cot x}
Then y = u – v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\) ……..(1)
Take u = (log x)^x
∴ log u = log (log x)^x = x . log (log x)
Differentiating both sides w.r.t. x, we get

(iv) \(x^{e^{x}}+(\log x)^{\sin x}\)
Solution:
Let y = \(x^{e^{x}}+(\log x)^{\sin x}\)
Put u = \(x^{e^{x}}\) and v = (log x)^{sin x}
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….(1)
Take u = \(x^{e^{x}}\)
∴ log u = log \(x^{e^{x}}\) = e^x . log x
Differentiating both sides w.r.t. x, we get

Also, v = (log x)^{sin x}
∴ log v = log (log x)^{sin x} = (sin x) . (log log x)
Differentiating both sides w.r.t. x, we get
\(\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}[(\sin x) \cdot(\log \log x)]\)
= \((\sin x) \cdot \frac{d}{d x}\left[(\log \log x)+(\log \log x) \cdot \frac{d}{d x}(\sin x)\right]\)

(v) \(e^{\tan x}+(\log x)^{\tan x}\)
Solution:
Let y = \(e^{\tan x}+(\log x)^{\tan x}\)
Put u = (log x)^{tan x}
∴ log u =log(log x)^{tan x} = (tan x).(log log x)
Differentiating both sides w.r.t. x, we get

(vi) (sin x)^{tan x} + (cos x)^{cot x}
Solution:
Let y = (sin x)^{tan x} + (cos x)^{cot x}
Put u = (sin x)^{tan x} and v = (cos x)^{cot x}
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………(1)
Take u = (sin x)^{tan x}
∴ log u = log (sin x)^{tan x} = (tan x) . (log sin x)
Differentiating both sides w.r.t. x, we get

(vii) \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Solution:
Let y = \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Put u = \(10^{x^{x}}\), v = \(x^{x^{10}}\) and w = \(x^{10^{x}}\)
Then y = u + v + w
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}\) ………(1)

(viii) \(\left[(\tan x)^{\tan x}\right]^{\tan x}\) at x = \(\frac{\pi}{4}\)
Solution:
Let y = \(\left[(\tan x)^{\tan x}\right]^{\tan x}\)
∴ log y = log [latex]\left[(\tan x)^{\tan x}\right]^{\tan x}[/latex]
= tan x . log(tan x)^{tan x}
= tan x . tan x log (tan x)
= (tan x)² . log (tan x)
Differentiating both sides w.r.t. x, we get

Question 3.
Find \(\frac{d y}{d x}\) if
(i) √x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get

(ii) x√x + y√y = a√a
Solution:
x√x + y√y = a√a
∴ \(x^{\frac{3}{2}}+y^{\frac{3}{2}}=a^{\frac{3}{2}}\)
Differentiating both sides w.r.t. x, we get

(iii) x + √xy + y = 1
Solution:
x + √xy + y = 1
Differentiating both sides w.r.t. x, we get

(iv) x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get

(v) x²y² – tan^-1(\(\sqrt{x^{2}+y^{2}}\)) = cot^-1(\(\sqrt{x^{2}+y^{2}}\))
Solution:
x²y² – tan^-1(\(\sqrt{x^{2}+y^{2}}\)) = cot^-1(\(\sqrt{x^{2}+y^{2}}\))
∴ x²y² = tan^-1(\(\sqrt{x^{2}+y^{2}}\)) + cot^-1(\(\sqrt{x^{2}+y^{2}}\))
∴ x²y² = \(\frac{\pi}{2}\) …….[∵ \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)]
Differentiating both sides w.r.t. x, we get

(vi) xe^y + ye^x = 1
Solution:
xe^y + ye^x = 1
Differentiating both sides w.r.t. x, we get

(vii) e^x+y = cos (x – y)
Solution:
e^x+y = cos (x – y)
Differentiating both sides w.r.t. x, we get

(viii) cos (xy) = x + y
Solution:
cos (xy) = x + y
Differentiating both sides w.r.t. x, we get

(ix) \(e^{e^{x-y}}=\frac{x}{y}\)
Solution:
\(e^{e^{x-y}}=\frac{x}{y}\)
∴ e^x-y = log(\(\frac{x}{y}\)) …….[e^x = y ⇒ x = log y]
∴ e^x-y = log x – log y
Differentiating both sides w.r.t. x, we get

Question 4.
Show that \(\frac{d y}{d x}=\frac{y}{x}\) in the following, where a and p are constants.
(i) x⁷y⁵ = (x + y)¹²
Solution:
x⁷y⁵ = (x + y)¹²
(log x⁷y⁵) = log(x + y)¹²
log x⁷ + log y⁵ = log(x + y)¹²
7 log x + 5 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get

(ii) x^py⁴ = (x + y)^p+4, p∈N
Solution:
x^py⁴ = (x + y)^p+4
Taking log
log (x^py⁴) = log(x + y)^p+4
log x^p + log y⁴ = (p + 4) log(x + y)
p log x + 4 log y = (p + 4) log(x + y)
Differentiating both sides w.r.t. x, we get

(iii) \(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)
Solution:
\(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)

Differentiating both sides w.r.t. x, we get


Differentiating both sides w.r.t. x, we get

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

Differentiating both sides w.r.t. x, we get

(v) \(\cos ^{-1}\left(\frac{7 x^{4}+5 y^{4}}{7 x^{4}-5 y^{4}}\right)=\tan ^{-1} a\)
Solution:

(vi) \(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)
Solution:
\(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)

Differentiating both sides w.r.t. x, we get

(vii) \(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)
Solution:
\(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)

Differentiating both sides w.r.t. x, we get

(viii) \(\sin \left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=a^{3}\)
Solution:

Differentiating both sides w.r.t. x, we get

Question 5.
(i) If log (x + y) = log (xy) + p, where p is a constant, then prove that \(\frac{d y}{d x}=-\frac{y^{2}}{x^{2}}\).
Solution:
log (x + y) = log (xy) + p
∴ log (x + y) = log x + log y + p
Differentiating both sides w.r.t. x, we get

(ii) If \(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{99 x^{2}}{101 y^{2}}\)
Solution:
\(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\)

(iii) If \(\log _{5}\left(\frac{x^{4}+y^{4}}{x^{4}-y^{4}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{12 x^{3}}{13 y^{3}}\)
Solution:

Differentiating both sides w.r.t. x, we get

(iv) If e^x + e^y = e^x+y, then show that \(\frac{d y}{d x}=-e^{y-x}\)
Solution:
e^x + e^y = e^x+y ……(1)
Differentiating both sides w.r.t. x, we get

(v) If \(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\), show that \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)
Solution:
\(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\)
\(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}=\sin \frac{\pi}{6}=\frac{1}{2}\)
2x⁵ – 2y⁵ = x⁵ + y⁵
3y⁵ = x⁵
Differentiating both sides w.r.t. x, we get
\(3 \times 5 y^{4} \frac{d y}{d x}=5 x^{4}\)
∴ \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)

(vi) If x^y = e^x-y, then show that \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}\)
Solution:
x^y = e^x-y
log x^y = log e^x-y
y log x = (x – y) log e
y log x = (x – y) ….. [∵ log e = 1]
y + y log x = x – y
y + y log x = x
y(1 + log x) = x
y = \(\frac{x}{1+\log x}\)

(vii) If \(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{\sin x}{1-2 y}\)
Solution:
\(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\)
y² = cos x + \(\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}\)
y² = cos x + y
Differentiating both sides w.r.t. x, we get

(viii) If \(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{1}{x(2 y-1)}\)
Solution:
\(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\)

(ix) If \(y=x^{x^{x^{-\infty}}}\), then show that \(\frac{d y}{d x}=\frac{y^{2}}{x(1-\log y)}\)
Solution:
\(y=x^{x^{x^{-\infty}}}\)

(x) If e^y = y^x, then show that \(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\)
Solution:
e^y = y^x
log e^y = log y^x
y log e = x log y
y = x log y …… [∵log e = 1] ……….(1)
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=x \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.3

Question 1.
Find the vector equation of a plane which is at 42 unit distance 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 is the length of the perpendicular from origin to the plane, then the vector equation of the plane is \(\bar{r} \cdot \hat{n}\) = p

Question 2.
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
Solution:
The equation of the plane is
6x – 2y + 3z – 7 = 0
∴ its vector equation is
\(\bar{r} \cdot(6 \hat{i}-2 \hat{j}+3 \hat{k})\) = 7 ….(1)
where \(\bar{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ \(\bar{n}=6 \hat{i}-2 \hat{j}+3 \hat{k}\) is normal to the plane
\(|\bar{n}|=\sqrt{6^{2}+(-2)^{2}+3^{2}}=\sqrt{49}\) = 7
Unit vector along \(\bar{n}\) is

Comparing with normal form of equation of the plane \(\bar{r} \cdot \hat{n}\) = p, it follows that length of perpendicular from origin is 1 unit.
Alternative Method:
The equation of the plane is 6x – 2y + 3z – 7 = 0 i.e. 6x – 2y + 3z = 7

This is the normal form of the equation of plane.
∴ perpendicular distance of the origin from the plane is p = 1 unit.

Question 3.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63 .
Solution:
The equation of the plane is 2x + 6y – 3z = 63. Dividing each term by \(\sqrt{2^{2}+6^{2}+(-3)^{2}}=\sqrt{49}\) = 7, we get
\(\frac{2}{7} x+\frac{6}{7} y-\frac{3}{7} z=\frac{63}{7}\) = 9
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 = \(\frac{6}{7}\), n = \(-\frac{3}{7}\)
and length of perpendicular from origin to the plane is p = 9.
∴ the coordinates of the foot of the perpendicular from the origin to the plane are (lp, mp, np) i.e. \(\left(\frac{18}{7}, \frac{54}{7},-\frac{27}{7}\right)\).

Question 4.
Reduce the equation \(\bar{r} \cdot(3 \hat{i}+4 \hat{j}+12 \hat{k})\) = 78 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.

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 6.
(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 point having position vector \(\hat{i}+\hat{j}+\hat{k}\) and perpendicular to the vector \(4 \hat{i}+5 \hat{j}+6 \hat{k}\).
Solution:
The vector equation of the plane passing through the point A (\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\)
Here, \(\bar{a}=\hat{i}+\hat{j}+\hat{k}\), \(\bar{n}=4 \hat{i}+5 \hat{j}+6 \hat{k}\)
∴ \(\bar{a} \cdot \bar{n}\) = \((\hat{i}+\hat{j}+\hat{k}) \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})\)
= (1)(4) + (1)(5) + (1)(6)
= 4 + 5 + 6 = 15
∴ the vector equation of the required plane is \(\bar{r} \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})\) = 15.

Question 6.
Find the Cartesian equation of the plane passing through A( -1, 2, 3), the direction ratios of whose normal are 0, 2, 5.
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
0(x +1) + 2(y – 2) + 5(z – 3) = 0
i.e. 0 + 2y – 4 + 5z – 15 = 0
i.e. 2y + 5z = 19.

Question 7.
Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the XY plane.
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 required plane is parallel to XY-plane.
∴ it is perpendicular to Z-axis i.e. Z-axis is normal to the plane. Z-axis has direction ratios 0, 0, 1.
The plane passes through (7, 8, 6).
∴ the cartesian equation of the required plane is
0(x – 7) + 0(y – 8) + 1 (z – 6) = 0
i.e. z = 6.

Question 8.
The foot of the perpendicular drawn from the origin to a plane is M(1, 0, 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, 0, 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 \(\bar{m}\) = \(\hat{i}\)
Normal to the plane is
\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{i}\)
\(\bar{m} \cdot \bar{n}=\hat{i} \cdot \hat{i}\) = 1
∴ the vector equation of the required plane is \(\bar{r} \cdot \hat{i}\) = 1

Question 9.
Find the vector equation of the plane passing through the point A(-2, 7, 5) and parallel to vectors \(\hat{4}-\hat{j}+3 \hat{k}\) and \(\hat{i}+\hat{j}+\hat{k}\).
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
\(\bar{r} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot(\bar{b} \times \bar{c})\) ….(1)

Question 10.
Find the Cartesian equation of the plane \(\bar{r}=(5 \hat{i}-2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})\)
Solution:
The equation \(\) represents a plane passing through a point having position vector \(\bar{a}\) and parallel to vectors \(\bar{b}\) and \(\bar{c}\).


∴ 5x – 2y – 3z = 38.
This is the cartesian equation of the required plane.

Question 11.
Find the vector equation of the plane which makes intercepts 1, 1, 1 on the co-ordinates axes.
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 \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) = \(\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (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, 0, 1)

∴ from (1), the vector equation of the required plane is \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1.

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.
Construct the truth table for each of the following statement patterns:
(i) [(p → q) ∧ q] → p
Solution :
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (p ∧ ~q) ↔ (p → q)
Solution:

(iii) (p ∧ q) ↔ (q ∨ r)
Solution:

(iv) p → [~(q ∧ r)]
Solution:

(v) ~p ∧ [(p ∨ ~q ) ∧ q]
Solution:

(vi) (~p → ~q) ∧ (~q → ~p)
Solution:

(vii) (q → p) ∨ (~p ↔ q)
Solution:

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:

(ix) p → [~(q ∧ r)]
Solution:

(x) (p ∨ ~q) → (r ∧ p)
Solution:

Question 2.
Using truth tables prove the following logical equivalences.
(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:

The entries in the columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:

The entries in the columns 3 and 7 are identical.
∴ ~(p ∨ q) ∧ (~p ∧ q) = ~p.

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]
Solution:

The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

(iv) p → (q → p) ≡ ~p → (p → q)
Solution:

The entries in the columns 4 and 7 are identical.
∴ p → (q → p) ≡ ~p → (p → q).

(v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r)
Solution:

The entries in the columns 5 and 8 are identical.
∴ (p ∨ q ) → r ≡ (p → r) ∧ (q → r).

(vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r)
Solution:

The entries in the columns 5 and 8 are identical.
∴ p → (q ∧ r) ≡ (p → q) ∧ (p → r).

(vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:

The entries in the columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

(viii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r
Solution:

The entries in the columns 3 and 7 are identical.
∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r.

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
Solution:

The entries in the columns 6 and 9 are identical.
∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

Question 3.
Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.
(i) (p ∧ q) → (q ∨ p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → (q ∨ p) is a tautology.

(ii) (p → q) ↔ (~p ∨ q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~p ∨ q) p is a tautology.

(iii) [~(~p ∧ ~q)] ∨ q
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ [~(~p ∧ ~q)] ∨ q is a contingency.

(iv) [(p → q) ∧ q)] → p
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ [(p → q) ∧ q)] → p is a contingency

(v) [(p → q) ∧ ~q] → ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q] → ~p is a tautology.

(vi) (p ↔ q) ∧ (p → ~q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ↔ q) ∧ (p → ~q) is a contingency.

(vii) ~(~q ∧ p) ∧ q
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ ~(~q ∧ p) ∧ q is a contingency.

(viii) (p ∧ ~q) ↔ (p → q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∧ ~q) ↔ (p → q) is a contradiction.

(ix) (~p → q) ∧ (p ∧ r)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (~p → q) ∧ (p ∧ r) is a contingency.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:

All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 1.
State which of the following sentences are statements. Justify your answer. In case of the statement, write down the truth value :
(i) 5 + 4 = 13.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(ii) x – 3 = 14.
Solution:
It is an open sentence, hence it is not a statement.

(iii) Close the door.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) Zero is a complex number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(v) Please get me breakfast.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) Congruent triangles are also similar.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vii) x² = x.
Solution:
It is an open sentence, hence it is not a statement,

(viii) A quadratic equation cannot have more than two roots.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(ix) Do you like Mathematics ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(x) The sun sets in the west.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xi) All real numbers are whole numbers.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(xii) Can you speak in Marathi ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(xiii) x² – 6x – 7 = 0, when x = 7.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xiv) The sum of cuberoots of unity is zero.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xv) It rains heavily.
Solution :
It is an open sentence, hence it is not a statement.

Question 2.
Write the following compound statements symbolically:
(i) Nagpur is in Maharashtra and Chennai is in Tamil Nadu.
Solution:
Let p : Nagpur is in Maharashtra.
q : Chennai is in Tamil Nadu.
Then the symbolic form of the given statement is P∧q.

(ii) Triangle is equilateral or isosceles,
Solution:
Let p : Triangle is equilateral.
q : Triangle is isosceles.
Then the symbolic form of the given statement is P∨q.

(iii) The angle is right angle if and only if it is of measure 90°.
Solution:
Let p : The angle is right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p↔q

(iv) Angle is neither acute nor obtuse.
Solution:
Let p : Angle is acute.
q : Angle is obtuse.
Then the symbolic form of the given statement is
~p ∧ ~q.

(v) If ∆ ABC is right angled at B, then m∠A + m∠C = 90°.
Solution:
Let p : ∆ ABC is right angled at B.
q : m∠A + m∠C = 90°.
Then the symbolic form of the given statement is p → q

(vi) Hima Das wins gold medal if and only if she runs fast.
Solution:
Let p : Hima Das wins gold medal
q : She runs fast.
Then the symbolic form of the given statement is p ↔ q.

(vii) x is not irrational number but it is a square of an integer.
Solution:
Let p : x is not irrational number
q : It is a square of an integer
Then the symbolic form of the given statement is p ∧ q
Note : If p : x is irrational number, then the symbolic form of the given statement is ~p ∧ q.

Question 3.
Write the truth values of the following :
(i) 4 is odd or 1 is prime.
Solution:
Let p : 4 is odd.
q : 1 is prime.
Then the symbolic form of the given statement is p∨q.
The truth values of both p and q are F.
∴ the truth value of p v q is F. … [F ∨ F = F]

(ii) 64 is a perfect square and 46 is a prime number.
Solution:
Let p : 64 is a perfect square.
q : 46 is a prime number.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iii) 5 is a prime number and 7 divides 94.
Solution:
Let p : 5 is a prime number.
q : 7 divides 94.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iv) It is not true that 5 – 3i is a real number.
Solution:
Let p : 5 – 3i is a real number.
Then the symbolic form of the given statement is ~ p.
The truth values of p is F.
∴ the truth values of ~ p is T. … [~ F ≡ T]

(v) If 3 × 5 = 8, then 3 + 5 = 15.
Solution:
Let p : 3 × 5 = 8.
q : 3 + 5 = 15.
Then the symbolic form of the given statement is p → q.
The truth values of both p and q are F.
∴ the truth value of p → q is T. … [F → F ≡ T]

(vi) Milk is white if and only if sky is blue.
Solution:
Let p : Milk is white.
q : Sky is blue
Then the symbolic form of the given statement is p ↔ q.
The truth values of both p and q are T.
∴ the truth value of p ↔ q is T. … [T ↔ T ≡ T]

(vii) 24 is a composite number or 17 is a prime number.
Solution :
Let p : 24 is a composite number.
q : 17 is a prime number.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are T.
∴ the truth value of p ∨ q is T. … [T ∨ T ≡ T]

Question 4.
If the statements p, q are true statements and r, s are false statements, then determine the truth values of the following:
(i) p ∨ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∨ (q ∧ r) ≡ T ∨ (T ∧ F)
≡ T ∧ F ≡ T
Hence the truth value of the given statement is true.

(ii) (p → q) ∨ (r → s)
Solution:
(p → q) ∨ (r → s) ≡ (T → T) ∨ (F → F)
≡ T ∨ T ≡ T
Hence the truth value of the given statement is true.

(iii) (q ∧ r) ∨ (~p ∧ s)
Solution:
(q ∧ r) ∨ (~p ∧ s) ≡ (T ∧ F) ∨ (~T ∧ F)
≡ F ∨ (F ∧ F)
≡ F ∨ F ≡ F
Hence the truth value of the given statement is false.

(iv) (p → q) ∧ (~ r)
Solution:
(p → q) ∧ (~ r) ≡ (T → T) ∧ (~ F)
≡ T ∧ T ≡ T
Hence the truth value of the given statement is true.

(v) (~r ↔ p) → (~q)
Solution:
(~r ↔ p) → (~q) ≡ (~F ↔ T) → (~T)
≡ (T ↔ T) → F
≡ T → F ≡ F
Hence the truth value of the given statement is false.

(vi) [~p ∧ (~q ∧ r) ∨ (q ∧ r) ∨ (p ∧ r)]
Solution:
[~p ∧ (~q ∧ r)∨(q ∧ r)∨(p ∧ r)]
≡ [~T ∧ (~T ∧ F)] ∨ [(T ∧ F) V (T ∧ F)]
≡ [F ∧ (F ∧ F)] ∨ [F V F]
≡ (F ∧ F) ∨ F
≡ F ∨ F ≡ F
Hence the truth value of the given statement is false.

(vii) [(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
Solution:
[(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
≡ [(~T ∧ T) ∧ (~F)] ∨ [(T → T) → (~F ∨ F)]
≡ [(F ∧ T) ∧ T] ∨ [T → (T ∨ F)]
≡ (F ∧ T) ∨ (T → T)
≡ F ∨ T ≡ T
Hence the truth value of the given statement is true.

(viii) ~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
Solution :
~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
≡ ~ [(~T ∧ F) ∨ (F → ~T)] ↔ (T ∧ F)
≡ ~ [(F ∧ F) ∨ (F → F)] ↔ F
≡ ~ (F ∨ T) ↔ F
≡ ~T ↔ F
≡ F ↔ F ≡ T
Hence the truth value of the given statement is true.

Question 5.
Write the negations of the following :
(i) Tirupati is in Andhra Pradesh.
Solution:
The negations of the given statements are :
Tirupati is not in Andhra Pradesh.

(ii) 3 is not a root of the equation x² + 3x – 18 = 0.
Solution:
3 is a root of the equation x² + 3x – 18 = 0.

(iii) \(\sqrt {2}\) is a rational number.
Solution:
\(\sqrt {2}\) is not a rational number.

(iv) Polygon ABCDE is a pentagon.
Solution:
Polygon ABCDE is not a pentagon.

(v) 7 + 3 > 5.
Solution :
7 + 3 > 5.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 8 Binomial Distribution Miscellaneous Exercise 8

(I) Choose the correct option from the given alternatives:

Question 1.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
(a) √50
(b) 5
(c) 25
(d) 10
Answer:
(b) 5

Question 2.
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probablity of 2 successes is
(a) \(\frac{128}{256}\)
(b) \(\frac{219}{256}\)
(c) \(\frac{37}{256}\)
(d) \(\frac{28}{256}\)
Answer:
(d) \(\frac{28}{256}\)

Question 3.
For a binomial distribution, n = 5. If P(X = 4) = P(X = 3) then p = ___________
(a) \(\frac{1}{3}\)
(b) \(\frac{3}{4}\)
(c) 1
(d) \(\frac{2}{3}\)
Answer:
(d) \(\frac{2}{3}\)

Question 4.
In a binomial distribution, n = 4. If 2 P(X = 3) = 3 P(X = 2) then p = ___________
(a) \(\frac{4}{13}\)
(b) \(\frac{5}{13}\)
(c) \(\frac{9}{13}\)
(d) \(\frac{6}{13}\)
Answer:
(c) \(\frac{9}{13}\)

Question 5.
If X ~ B (4, p) and P (X = 0) = \(\frac{16}{81}\), then P (X = 4) = ___________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{81}\)
(c) \(\frac{1}{27}\)
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{1}{81}\)

Question 6.
The probability of a shooter hitting a target is \(\frac{3}{4}\). How many minimum numbers of times must he fie so that the probability of hitting the target at least once is more than 0·99?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 7.
If the mean and variance of a binomial distribution are 18 and 12 respectively, then n = ___________
(a) 36
(b) 54
(c) 18
(d) 27
Answer:
(b) 54

(II) Solve the following:

Question 1.
Let X ~ B(10, 0.2). Find
(i) P(X = 1)
(ii) P(X ≥ 1)
(iii) P(X ≤ 8).
Solution:
X ~ B(10, 0.2)
∴ n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
The p,m.f. of X is given by

Question 2.
Let X ~ B(n, p).
(i) If n = 10, E(X) = 5, find p and Var(X).
(ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) Given: n = 10 and E(X) = 5
But E(X) = np
∴ np = 5.
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Var(X) = npq = 10(\(\frac{1}{2}\))(\(\frac{1}{2}\)) = 2.5.
Hence, p = \(\frac{1}{2}\) and Var(X) = 2.5

(ii) Given: E(X) = 5 and Var(X) = 2.5
∴ np = 5 and npq = 2.5
∴ \(\frac{n p q}{n p}=\frac{2.5}{5}\)
∴ q = 0.5 = \(\frac{5}{10}=\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Substituting p = \(\frac{1}{2}\) in np = 5, we get
n(\(\frac{1}{2}\)) = 5
∴ n = 10
Hence, n = 10 and p = \(\frac{1}{2}\)

Question 3.
If a fair coin is tossed 10 times and the probability that it shows heads (i) 5 times (ii) in the first four tosses and tail in the last six tosses.
Solution:
Let X = number of heads.
p = probability that coin tossed shows a head
∴ p = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 10
∴ X ~ B(10, \(\frac{1}{2}\))
The p.m.f. of X is given by
P(X = x) = \({ }^{n} C_{x} P^{x} q^{n-x}\)

(i) P(coin shows heads 5 times) = P[X = 5]

Hence, the probability that can shows heads exactly 5 times = \(\frac{63}{256}\)

(ii) P(getting heads in first four tosses and tails in last six tosses) = P(X = 4)

Hence, the probability that getting heads in first four tosses and tails in last six tosses = \(\frac{105}{512}\).

Question 4.
The probability that a bomb will hit a target is 0.8. Find the probability that out of 10 bombs dropped, exactly 2 will miss the target.
Solution:
Let X = the number of bombs hitting the target.
p = probability that bomb will hit the target
∴ p = 0.8 = \(\frac{8}{10}=\frac{4}{5}\)
∴ q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Given: n = 10
∴ X ~ B(10, \(\frac{4}{5}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e.p(x) = \({ }^{10} \mathrm{C}_{x}\left(\frac{4}{5}\right)^{x}\left(\frac{1}{5}\right)^{10-x}\)
P(exactly 2 bombs will miss the target) = P(exactly 8 bombs will hit the target)
= P[X = 8]
= p(8)

Hence, the probability that exactly 2 bombs will miss the target = 45\(\left(\frac{2^{16}}{5^{10}}\right)\)

Question 5.
The probability that a mountain bike travelling along a certain track will have a tire burst is 0.05. Find the probability that among 17 riders:
(i) exactly one has a burst tyre
(ii) at most three have a burst tyre
(iii) two or more have burst tyres.
Solution:
Let X = number of burst tyres.
p = probability that a mountain bike travelling along a certain track will have a tyre burst.
∴ p = 0.05
∴ q = 1 – p = 1 – 0.05 = 0.95
Given: n = 17
∴ X ~ B(17, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} P^{x} q^{n-x}\)
i.e.(x) = \({ }^{17} \mathrm{C}_{x}(0.05)^{x}(0.95)^{17-x}\), x = 0, 1, 2, ……, 17
(i) P(exactly one has a burst tyre)
P(X = 1) = p(1) = \({ }^{17} \mathrm{C}_{1}\) (0.05)¹ (0.95)^17-1
= 17(0.05) (0.95)¹⁶
= 0.85(0.95)¹⁶
Hence, the probability that riders has exactly one burst tyre = (0.85)(0.95)¹⁶

(ii) P(at most three have a burst tyre) = P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= p(0) + p(1) + p(2) + p(3)


Hence, the probability that at most three riders have burst tyre = (2.0325)(0.95)¹⁴.

(iii) P(two or more have tyre burst) = P(X ≥ 2)
= 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – [p(0) + p(1)]
= 1 – [\({ }^{17} \mathrm{C}_{0}\) (0.05)⁰ (0.95)¹⁷ + \({ }^{17} \mathrm{C}_{1}\) (0.05)(0.95)¹⁶]
= 1 – [1(1)(0.95)¹⁷ + 17(0.05)(0.95)¹⁶]
= 1 – (0.95)¹⁶[0.95 + 0.85]
= 1 – (1.80)(0.95)¹⁶
= 1 – (1.8)(0.95)¹⁶
Hence, the probability that two or more riders have tyre burst = 1 – (1.8)(0.95)¹⁶.

Question 6.
The probability that a lamp in a classroom will be burnt out is 0.3. Six such lamps are fitted in the classroom. If it is known that the classroom is unusable if the number of lamps burning in it is less than four, find the probability that the classroom cannot be used on a random occasion.
Solution:
Let X = number of lamps burnt out in the classroom.
p = probability of a lamp in a classroom will be burnt
∴ p = 0.3 = \(\frac{3}{10}\)
∴ q = 1 – p = 1 – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Given: n = 6
∴ X ~ B(6, \(\frac{3}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{6} \mathrm{C}_{x}\left(\frac{3}{10}\right)^{x}\left(\frac{7}{10}\right)^{6-x}\)
Since the classroom is unusable if the number of lamps burning in it is less than four, therefore
P(classroom cannot be used) = P[X < 4]
= P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]
= p(0) + p(1) + p(2) + p(3)

Hence, the probability that the classroom cannot be used on a random occasion is 0.92953.

Question 7.
A lot of 100 items contain 10 defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective items.
p = probability that item is defective
∴ p = \(\frac{10}{100}=\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{10}\))
The p.m.f. of X is given as:
P[X = x] = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{5} C_{x}\left(\frac{1}{10}\right)^{x}\left(\frac{9}{10}\right)^{5-x}\)
P (store will receive at most one defective item) = P[X ≤ 1]
=P[X = 0] + P[X = 1]
= p(0) + p(1)

Hence, the probability that the store will receive at most one defective item is (1.4)(0.9)⁴.

Question 8.
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. The inspector of the retailer picks 20 items from a shipment. What is the probability that the store will receive at most one defective item?
Solution:
Let X = number of defective electronic devices.
p = probability that device is defective
∴ p = 3% = \(\frac{3}{100}\)
∴ q = 1 – p = 1 – \(\frac{3}{100}\) = \(\frac{97}{100}\)
Given: n = 20
∴ X ~ B(20, \(\frac{3}{100}\))
The p.m.f. of X is given as:

Hence, the probability that the store will receive at most one defective item = (1.57)(0.97)¹⁹.

Question 9.
The probability that a certain kind of component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 tested components tested survive.
Solution:
Let X = number of tested components survive.
p = probability that the component survives the check test

Hence, the probability that exactly 2 of the 4 tested components survive is 0.3456.

Question 10.
An examination consists of 10 multiple choice questions, in each of which a candidate has to deduce which one of five suggested answers is correct. A completely unprepared student guesses each answer completely randomly. What is the probability that this student gets 8 or more questions correct? Draw the appropriate moral.
Solution:
Let X = number of correct answers.
p = probability that student gets correct answer
∴ p = \(\frac{1}{5}\)
∴ q = 1 – p = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Given: n = 10 (number of total questions)
∴ X ~ B(10, \(\frac{1}{5}\))
The p.m.f. of X is given by

Hence, the probability that student gets 8 or more questions correct = \(\frac{30.44}{5^{8}}\)

Question 11.
The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that (i) all 8 machines (ii) 7 or 8 machines (iii) at most 6 machines will produce all bolts within specification.
Solution:
Let X = number of machines which produce the bolts within specification.
p = probability that a machine produce bolts within specification
p = 0.998 and q = 1 – p = 1 – 0.998 = 0.002
Given: n = 8
∴ X ~ B(8, 0.998)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{8} \mathrm{C}_{x}(0.998)^{x}(0.002)^{8-x}\), x = 0, 1, 2, …, 8
(i) P(all 8 machines will produce all bolts within specification) = P[X = 8]
= p(8)
= \({ }^{8} \mathrm{C}_{8}\) (0.998)⁸ (0.002)^8-8
= 1(0.998)⁸ . (1)
= (0.998)⁸
Hence, the probability that all 8 machines produce all bolts with specification = (0.998)⁸.

(ii) P(7 or 8 machines will produce all bolts within i specification) = P (X = 7) + P (X = 8)

Hence, the probability that 7 or 8 machines produce all bolts within specification = (1.014)(0.998)⁷.

(iii) P(at most 6 machines will produce all bolts with specification) = P[X ≤ 6]
= 1 – P[x > 6]
= 1 – [P(X = 7) + P(X = 8)]
= 1 – [P(7) + P(8)]
= 1 – (1.014)(0.998)⁷
Hence, the probability that at most 6 machines will produce all bolts with specification = 1 – (1.014)(0.998)⁷.

Question 12.
The probability that a machine develops a fault within the first 3 years of use is 0.003. If 40 machines are selected at random, calculate the probability that 38 or more will develop any faults within the first 3 years of use.
Solution:
Let X = the number of machines who develop a fault.
p = probability that a machine develops a fait within the first 3 years of use
∴ p = 0.003 and q = 1 – p = 1 – 0.003 = 0.997
Given: n = 40
∴ X ~ B(40, 0.003)
The p.m.f. of X is given by

Hence, the probability that 38 or more machines will develop the fault within 3 years of use = (775.44)(0.003)³⁸.

Question 13.
A computer installation has 10 terminals. Independently, the probability that anyone terminal will require attention during a week is 0.1. Find the probabilities that (i) 0 (ii) 1 (iii) 2 (iv) 3 or more, terminals will require attention during the next week.
Solution:
Let X = number of terminals which required attention during a week.
p = probability that any terminal will require attention during a week
∴ p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9
Given: n = 10
∴ X ~ B(10, 0.1)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{10} C_{x}(0.1)^{x}(0.9)^{10-x}\), x = 0, 1, 2, …, 10
(i) P(no terminal will require attention) = P(X = 0)

Hence, the probability that no terminal requires attention = (0.9)¹⁰

(ii) P(1 terminal will require attention)

Hence, the probability that 1 terminal requires attention = (0.9)⁹.

(iii) P(2 terminals will require attention)

Hence, the probability that 2 terminals require attention = (0.45)(0.9)⁸.

(iv) P(3 or more terminals will require attention)

Hence, the probability that 3 or more terminals require attention = 1 – (2.16) × (0.9)⁸.

Question 14.
In a large school, 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics.
(i) Calculate the probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of the pupils.
(ii) Find the probability that the visitor obtains answer yes from at least 2 pupils:
(a) when the number of pupils questioned remains at 4.
(b) when the number of pupils questioned is increased to 8.
Solution:
Let X = number of pupils like Mathematics.
p = probability that pupils like Mathematics

(i) The probabilities of obtaining an answer yes from 0, 1, 2, 3, 4 of pupils are P(X = 0), P(X = 1), P(X = 2), P(X = 3) and P(X = 4) respectively

(ii) (a) P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)

(b) P(the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8)

Question 15.
It is observed that it rains 12 days out of 30 days. Find the probability that
(i) it rains exactly 3 days of the week.
(ii) it will rain at least 2 days of a given week.
Solution:
Let X = the number of days it rains in a week.
p = probability that it rains

(i) P(it rains exactly 3 days of week) = P(X = 3)

Hence, the probability that it rains exactly 3 days of week = 0.2903.

(ii) P(it will rain at least 2 days of the given week)

Hence, the probability that it rains at least 2 days of a given week = 0.8414

Question 16.
If the probability of success in a single trial is 0.01. How many trials are required in order to have a probability greater than 0.5 of getting at least one success?
Solution:
Let X = number of successes.
p = probability of success in a single trial
∴ p = 0.01
and q = 1 – p = 1 – 0.01 = 0.99
∴ X ~ B(n, 0.01)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)

Hence, the number of trials required in order to have a probability greater than 0.5 of getting at least one success is \(\frac{\log 0.5}{\log 0.99}\) or 68.

Question 17.
In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.
Solution:
Given: X ~ B(n = 5, p)
The probability of X success is


Hence, the probability of success is \(\frac{1}{5}\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 8 Binomial Distribution Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 8 Binomial Distribution Ex 8.1

Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of
(i) 5 successes
(ii) at least 5 successes
(iii) at most 5 successes.
Solution:
Let X = number of successes, i.e. number of odd numbers.
p = probability of getting an odd number in a single throw of a die
∴ p = \(\frac{3}{6}=\frac{1}{2}\) and q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{2}\))
The p.m.f. of X is given by

Hence, the probability of 5 successes is \(\frac{3}{32}\).

(ii) P(at least 5 successes) = P[X ≥ 5]
= p(5) + p(6)

Hence, the probability of at least 5 successes is \(\frac{7}{64}\).

(iii) P(at most 5 successes) = P[X ≤ 5]
= 1 – P[X > 5]

Hence, the probability of at most 5 successes is \(\frac{63}{64}\).

Question 2.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
Let X = number of doublets.
p = probability of getting a doublet when a pair of dice is thrown
∴ p = \(\frac{6}{36}=\frac{1}{6}\) and
q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 4
∴ X ~ B(4, \(\frac{1}{6}\))
The p.m.f. of X is given by

Hence, the probability of two successes is \(\frac{25}{216}\).

Question 3.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?
Solution:
Let X = number of defective items.
p = probability of defective item
∴ p = 5% = \(\frac{5}{100}=\frac{1}{20}\)
and q = 1 – p = 1 – \(\frac{1}{20}\) = \(\frac{19}{20}\)
∴ X ~ B(10, \(\frac{1}{20}\))
The p.m.f. of X is given by

P(sample of 10 items will include not more than one defective item) = P[X ≤ 1]

Hence, the probability that a sample of 10 items will include not more than one defective item = 29\(\left(\frac{19^{9}}{20^{10}}\right)\).

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that
(i) all the five cards are spades
(ii) only 3 cards are spades
(iii) none is a spade.
Solution:
Let X = number of spade cards.
p = probability of drawing a spade card from a pack of 52 cards.
Since there are 13 spade cards in the pack of 52 cards.
∴ p = \(\frac{13}{52}=\frac{1}{4}\) and
q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{4}\))
The p.m.f. of X is given by

(i) P(all five cards are spade)

Hence, the probability of all the five cards are spades = \(\frac{1}{1024}\)

(ii) P(only 3 cards are spade) = P[X = 3]

Hence, the probability of only 3 cards are spades = \(\frac{45}{512}\)

(iii) P(none of cards is spade) = P[X = 0]

Hence, the probability of none of the cards is a spade = \(\frac{243}{1024}\)

Question 5.
The probability of a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.
Solution:
Let X = number of fuse bulbs.
p = probability of a bulb produced by a factory will fuse after 150 days of use.
∴ p = 0.05
and q = 1 – p = 1 – 0.05 = 0.95
Given: n = 5
∴ X ~ B(5, 0.05)
The p.m.f. of X is given by
P(X = x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
i.e. p(x) = \({ }^{5} C_{x}(0.05)^{x}(0.95)^{5-x}\), x = 0, 1, 2, 3, 4, 5
(i) P(none of a bulb produced by a factory will fuse after 150 days of use) = P[X = 0]
= p(0)
= \({ }^{5} \mathrm{C}_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 × 1 × (0.95)⁵
= (0.95)⁵
Hence, the probability that none of the bulbs will fuse after 150 days = (0.95)⁵.

(ii) P(not more than one bulb will fuse after 150 days of j use) = P[X ≤ 1]
= p(0) + p(1)
= \({ }^{5} \mathrm{C}_{0} \cdot(0.05)^{0}(0.95)^{5-0}+{ }^{5} \mathrm{C}_{1}(0.05)^{1}(0.95)^{4}\)
= 1 × 1 × (0.95)⁵ + 5 × (0.05) × (0.95)⁴
= (0.95)⁴ [0.95 + 5(0.05)]
= (0.95)⁴ (0.95 + 0.25)
= (0.95)⁴ (1.20)
= (1.2) (0.95)⁴
Hence, the probability that not more than one bulb will fuse after 150 days = (1.2)(0.95)⁴.

(iii) P(more than one bulb fuse after 150 days)
= P[X > 1]
= 1 – P[X ≤ 1]
= 1 – (1.2)(0.95)⁴
Hence, the probability that more than one bulb fuse after 150 days = 1 – (1.2)(0.95)⁴.

(iv) P(at least one bulb fuse after 150 days)
= P[X ≥ 1]
= 1 – P[X = 0]
= 1 – p(0)
= 1 – \({ }^{5} C_{0}(0.05)^{0}(0.95)^{5-0}\)
= 1 – 1 × 1 × (0.95)⁵
= 1 – (0.95)⁵
Hence, the probability that at least one bulb fuses after 150 days = 1 – (0.95)⁵.

Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
Let X = number of balls marked with digit 0.
p = probability of drawing a ball from 10 balls marked with the digit 0.
∴ p = \(\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
The p.m.f. of X is given by

P(none of the ball marked with digit 0) = P(X = 0)

Hence, the probability that none of the bulb marked with digit 0 is \(\left(\frac{9}{10}\right)^{4}\)

Question 7.
On a multiple-choice examination with three possible answers for each of the five questions. What is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
Let X = number of correct answers.
p = probability that a candidate gets a correct answer from three possible answers.
∴ p = \(\frac{1}{3}\) and q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Given: n = 5
∴ X ~ B(5, \(\frac{1}{3}\))
The p.m.f. of X is given by

P(four or more correct answers) = P[X ≥ 4]
= p(4) + p(5)

Hence, the probability of getting four or more correct answers = \(\frac{11}{243}\).

Question 8.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is \(\frac{1}{100}\), find the probability that he will win a prize
(i) at least once
(ii) exactly once
(iii) at least twice.
Solution:
Let X = number of winning prizes.
p = probability of winning a prize
∴ p = \(\frac{1}{100}\)
and q = 1 – p = 1 – \(\frac{1}{100}\) = \(\frac{99}{100}\)
Given: n = 50
∴ X ~ B(50, \(\frac{1}{100}\))
The p.m.f. of X is given by
\(P(X=x)={ }^{n} C_{x} p^{x} q^{n-x}\)
i.e., p(x) = \({ }^{50} \mathrm{C}_{x}\left(\frac{1}{100}\right)^{x}\left(\frac{99}{100}\right)^{50-x}\), x = 0, 1, 2,… 50
(i) P(a person wins a prize at least once)

Hence, probability of winning a prize at least once = 1 – \(\left(\frac{99}{100}\right)^{50}\)

(ii) P(a person wins exactly one prize) = P[X = 1] = p(1)

Hence, probability of winning a prize exactly once = \(\frac{1}{2}\left(\frac{99}{100}\right)^{49}\)

(iii) P(a persons wins the prize at least twice) = P[X ≥ 2]
= 1 – P[X < 2]
= 1 – [p(0) + p(1)]

Hence, the probability of winning the prize at least twice = 1 – 149\(\left(\frac{99^{49}}{100^{50}}\right)\).

Question 9.
In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that (i) none (ii) 1 (iii) 2 (iv) all 3 of the sample will work.
Solution:
Let X = number of working discs.
p = probability that a floppy disc works
∴ p = 95% = \(\frac{95}{100}=\frac{19}{20}\)
and q = 1 – p = 1 – \(\frac{19}{20}\) = \(\frac{1}{20}\)
Given: n = 3
∴ X ~ B(3, \(\frac{19}{20}\))
The p.m.f. of X is given by

(i) P(none of the floppy discs work) = P(X = 0)

Hence, the probability that none of the floppy disc will work = \(\frac{1}{20^{3}}\).

(ii) P(exactly one floppy disc works) = P(X = 1)

Hence, the probability that exactly one floppy disc works = 3\(\left(\frac{19}{20^{3}}\right)\)

(iii) P(exactly two floppy discs work) = P(X = 2)

Hence, the probability that exactly 2 floppy discs work = 3\(\left(\frac{19^{2}}{20^{3}}\right)\)

(iv) P(all 3 floppy discs work) = P(X = 3)

Hence, the probability that all 3 floppy discs work = \(\left(\frac{19}{20}\right)^{3}\).

Question 10.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
Let X = number of sixes.
p = probability that a die shows six in a single throw
∴ p = \(\frac{1}{6}\)
and q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Given: n = 6
∴ X ~ B(6, \(\frac{1}{6}\))
The p.m.f. of X is given by


Hence, probability of throwing at most 2 sixes = \(\frac{7}{3}\left(\frac{5}{6}\right)^{5}\).

Question 11.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?
Solution:
Let X = number of defective articles.
p = probability of defective articles.
∴ p = 10% = \(\frac{10}{100}=\frac{1}{10}\)
and q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
Given: n = 12
∴ X ~ B(12, \(\frac{1}{10}\))
The p.m.f. of X is given by

Hence, the probability of getting 9 defective articles = \(22\left(\frac{9^{3}}{10^{11}}\right)\)

Question 12.
Given X ~ B(n, P)
(i) If n = 10 and p = 0.4, find E(x) and Var(X).
(ii) If p = 0.6 and E(X) = 6, find n and Var(X).
(iii) If n = 25, E(X) = 10, find p and SD(X).
(iv) If n = 10, E(X) = 8, find Var(X).
Solution:
(i) Given: n = 10 and p = 0.4
∴ q = 1 – p = 1 – 0.4 = 0.6
∴ E(X) = np = 10(0.4) = 4
Var(X) = npq = 10(0.4)(0.6) = 2.4
Hence, E(X) = 4, Var(X) = 2.4.

(ii) Given: p = 0.6, E (X) = 6
E(X) = np
6 = n(0.6)
n = \(\frac{6}{0.6}\) = 10
Now, q = 1 – p = 1 – 0.6 = 0.4
∴ Var(X) = npq = 10(0.6)(0.4) = 2.4
Hence, n = 10 and Var(X) = 2.4.

(iii) Given: n = 25, E(X) = 10
E(X) = np
10 = 25p
p = \(\frac{10}{25}=\frac{2}{5}\)
∴ q = 1 – p = 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Var(X) = npq = \(25 \times \frac{2}{5} \times \frac{3}{5}\) = 6
∴ SD(X) = √Var(X) = √6
Hence, p = \(\frac{2}{5}\) and S.D.(X) = √6.

(iv) Given: n = 10, E(X) = 8
E(X) = np
8 = 10p
p = \(\frac{8}{10}=\frac{4}{5}\)
q = 1 – p = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Var(X) = npq = \(10\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)=\frac{8}{5}\)
Hence, Var(X) = \(\frac{8}{5}\).

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

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

(I) Choose the correct option from the given alternatives:

Question 1.
P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 2.
If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x²), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) =
(a) 2x – 3x²
(b) 3x – 4x³
(c) 3x – 2x³
(d) 2x³ – 3x
Answer:
(c) 3x – 2x³

Question 3.
If the p.d.f. of a c.r.v. X is f(x) = \(\frac{x^{2}}{18}\), for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) =
(a) \(\frac{1}{27}\)
(b) \(\frac{1}{28}\)
(c) \(\frac{1}{29}\)
(d) \(\frac{1}{26}\)
Answer:
(a) \(\frac{1}{27}\)

Question 4.
If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1) . 5-x, where k is a constant, then P(X = 0) =
(a) \(\frac{7}{25}\)
(b) \(\frac{16}{25}\)
(c) \(\frac{18}{25}\)
(d) \(\frac{19}{25}\)
Answer:
(b) \(\frac{16}{25}\)

Question 5.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{\left({ }^{5} \mathrm{C}_{x}\right)}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then
(a) a < b
(b) a > b
(c) a = b
(d) a + b
Answer:
(c) a = b

Question 6.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{x}{n(n+1)}\), for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) =
(a) \(\frac{n}{1}+\frac{1}{2}\)
(b) \(\frac{n}{3}+\frac{1}{6}\)
(c) \(\frac{n}{2}+\frac{1}{5}\)
(d) \(\frac{n}{1}+\frac{1}{3}\)
Answer:
(b) \(\frac{n}{3}+\frac{1}{6}\)

Question 7.
If p.m.f. of a d.r.v. X is P(x) = \(\frac{c}{x^{3}}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) =
(a) \(\frac{343}{297}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 8.
If the d.r.v. X has the following probability distribution:

then P(X = -1) =
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{4}{10}\)
Answer:
(a) \(\frac{1}{10}\)

Question 9.
If the d.r.v. X has the following probability distribution:

then k =
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Question 10.
Find the expected value of X for the following p.m.f.

(a) 0.85
(b) -0.35
(c) 0.15
(d) -0.15
Answer:
(b) -0.35

(II) Solve the following:

Question 1.
Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:
(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.
(ii) Amount of syrup prescribed by a physician.
(iii) The person on a high protein diet is interesting to gain weight in a week.
(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.
(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.
Solution:
(i) Let X = number of unemployed graduates in a town.
Since the population of the town is 1 lakh, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.
Then X takes uncountable infinite values.
∴ random variable X is continuous.

(iii) Let X = gain of weight in a week
Then X takes uncountable infinite values
∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.
Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.
Then X takes uncountable infinite values
∴ random variable X is continuous.

Question 2.
The probability distribution of discrete r.v. X is as follows:

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:

Question 3.
The following is the probability distribution of X:

Find the probability that
(i) X is positive
(ii) X is non-negative
(iii) X is odd
(iv) X is even.
Solution:
(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.25 + 0.15 + 0.1
= 0.50

(ii) P(X is non-negative)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0.25 + 0.15 + 0.1
= 0.70

(iii) P(X is odd)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.1
= 0.55

(iv) P(X is even)
= P(X = -2) + P(X = 0) + P(X = 2)
= 0.10 + 0.20 + 0.15
= 0.45.

Question 4.
The p.m.f. of a r.v. X is given by P(X = x) = x = \(\frac{{ }^{5} \mathrm{C}_{\mathrm{x}}}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}\)
= \(\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}\) ………[latex]{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}[/latex]
= P(X = 5) + P(X = 4) + P(X = 3)
= P(X ≥ 3)
∴ P(X ≤ 2) = P(X ≥ 3).

Question 5.
In the p.m.f. of r.v. X

Find a and obtain c.d.f. of X.
Solution:
For p.m.f. of a r.v. X
\(\sum_{i=1}^{5} P(X=x)=1\)
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{20}\)
F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2)
\(=\frac{1}{20}+\frac{3}{20}=\frac{4}{20}=\frac{1}{5}\)
P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}=\frac{9}{20}\)
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}=\frac{19}{20}\)
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}+\frac{1}{20}=\frac{20}{20}=1\)
Hence, the c.d.f. of the random variable X is as follows:

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.
Solution:
When a fair coin is tossed 4 times then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n (X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{0}}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{1}}{16}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{{ }^{4} \mathrm{C}_{2}}{16}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{3}}{16}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{4}}{16}\)
∴ the probability distribution of X is as follows:

Also, the formula for p.m.f. of X is
P(x) = \(\frac{{ }^{4} \mathrm{C}_{x}}{16}\), x = 0, 1, 2, 3, 4 and = 0, otherwise.

Question 7.
Find the probability distribution of the number of successes in two tosses of a die, where success is defined as
(i) number greater than 4
(ii) six appear on at least one die.
Solution:
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P(X = 0) = P(number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{16}{36}=\frac{4}{9}\)
P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}\)
= \(\frac{8}{36}+\frac{8}{36}\)
= \(\frac{16}{36}\)
= \(\frac{4}{9}\)
P(X = 2) = P(number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
Thus, the probability distribution is as follows:

(ii) Here, success means six appears on at least one die.
P(Y = 0) = P(six appears on none of the dice) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(Y = 1) = P(six appears on none of the dice x six appears on at least one of the dice ) + P(six appears on none of the dice x six appears on at least one of the dice)
= \(\frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\)
P(Y = 2) = P(six appears on at least one of the dice) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
Thus, the required probability distribution is as follows:

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X > 6)
(iii) P(0 < X < 3).

Question 9.
The following is the c.d.f. of a r.v. X:

Find
(i) p.m.f. of X
(ii) P( -1 ≤ X ≤ 2)
(iii) P(X ≤ X > 0).
Solution:
(i) From the given table
F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 1
P(X = -3) = F(-3) = 0.1
P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2
P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ the p.m.f of X is as follows:

(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55

(iii) (X ≤ 3) ∩ (X > 0)
= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}
= {1, 2, 3}

Question 10.
Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

Solution:
(i) We construct the following table to find the expected value, variance, and standard deviation:

(ii) We construct the following table to find the expected value, variance, and standard deviation:



(iii) We construct the following table to find the expected value, variance, and standard deviation:



(iv) We construct the following table to find the expected value, variance, and standard deviation:

Question 11.
A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.
Solution:
When a coin is tossed twice, the sample space is
S = {HH, HT, TH, HH}
Let X denote the amount he wins.
Then X takes values 10, 5, 2.
P(X = 10) = P(2 heads appear) = \(\frac{1}{4}\)
P(X = 5) = P(1 head appears) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P(no head appears) = \(\frac{1}{4}\)
We construct the following table to calculate the mean and the variance of X:

From the table Σx_i . P(x_i) = 5.5, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 38.5
E(X) = Σx_i . P(x_i) = 5.5
Var(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 38.5 – (5.5)²
= 38.5 – 30.25
= 8.25
∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

Question 12.
Let the p.m.f. of r.v. X be P(x) = \(\frac{3-x}{10}\), for x = -1, 0, 1, 2 and = 0, otherwise.
Calculate E(X) and Var(X).
Solution:
P(X) = \(\frac{3-x}{10}\)
X takes values -1, 0, 1, 2
P(X = -1) = P(-1) = \(\frac{3+1}{10}=\frac{4}{10}\)
P(X = 0) = P(0) = \(\frac{3-0}{10}=\frac{3}{10}\)
P(X = 1) = P(1) = \(\frac{3-1}{10}=\frac{2}{10}\)
P(X = 2) = P(2) = \(\frac{3-2}{10}=\frac{1}{10}\)
We construct the following table to calculate the mean and variance of X:

From the table
Σx_iP(x_i) = 0 and \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) = 1
E(X) = Σx_iP(x_i) = 0
Var(X) = \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 1 – 0
= 1
Hence, E(X) = 0, Var (X) = 1.

Question 13.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and = 0 otherwise.
Compute
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(X < -0.5 or X > 0.5).
Solution:
(i) P(X > 0)


(ii) P(-1 < X < 1)


(iii) P(X < -0.5 or X > 0.5)

Question 14.
The p.d.f. of a continuous r.v. X is given by f(x) = \(\frac{1}{2 a}\), for 0 < x < 2a and = 0, otherwise. Show that P( X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
The p.d.f. of r.v. X is given by f(x) = \(\frac{k}{\sqrt{x}}\), for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).
Solution:
Since f is p.d.f. of the r.v. X,

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

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and

Hence, f(x) is the p.d.f. of X.

(ii) f(x) = x ≥ 0 if 0 ≤ x ≤ 1
For 1 < x < 2, -2 < -x < -1
-2 – 2 < -2 – x < -2 – 1
i.e. -4 < f(x) < -3 if 1 < x < 2
Hence, f(x) is not p.d.f. of X.

(iii) (a) f(x) = 2 ≥ 0 for 0 ≤ x ≤ 1

Hence, f(x) is not p.d.f. of X.

Question 2.
The following is the p.d.f. of r.v. X:
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
Find
(a) P(x < 1.5)
(b) P(1 < x < 2) (c) P(x > 2).
Solution:

Question 3.
It is known that error in measurement of reaction temperature (in 0°C) in a certain experiment is continuous r.v. given by
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0. otherwise.
(i) Verify whether f(x) is p.d.f. of r.v. X
(ii) Find P(0 < x ≤ 1)
(iii) Find the probability that X is negative.
Solution:

Question 4.
Find k if the following function represents p.d.f. of r.v. X
(i) f(x) = kx. for 0 < x < 2 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{3}{2}\)).
(ii) f(x) = kx(1 – x), for 0 < x < 1 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{1}{2}\)), P(x < \(\frac{1}{2}\)).
Solution:
(i) Since, the function f is p.d.f. of X

(ii) Since, the function f is the p.d.f. of X,

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected students and suppose X has p.d.f.
f(x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate:
(i) P(x ≤ 1)
(ii) P(0.5 ≤ x ≤ 1.5)
(iii) P(x ≥ 1.5).
Solution:
(i) P(x ≤ 1)

(ii) P(0.5 ≤ x ≤ 1.5)

(iii) P(x ≥ 1.5)

Question 6.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = \(\frac{1}{5}\), for 0 ≤ x ≤ 5 and = 0 otherwise. Find the probability that
(i) waiting time is between 1 and 3
(ii) waiting time is more than 4 minutes.
Solution:
(i) Required probability = P(1 < X < 3)

(ii) Required probability = P(X > 4)

Question 7.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and 0 otherwise.
Compute:
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(-0.5 < X or X > 0.5).
Solution:
Since, f is the p.d.f. of X,

Question 8.
The following is the p.d.f. of continuous r.v. X
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
(i) Find expression for c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7 and 5.
Solution:
(i) Let F(x) be the c.d.f. of X

Question 9.
Given the p.d.f. of a continuous random r.v. X, f(x) = \(\frac{x^{2}}{3}\), for -1 < x < 2 and = 0 otherwise. Determine c.d.f. of X and hence find P(X < 1); P(X < -2), P(X > 0), P(1 < X < 2).
Solution:

Question 10.
If a r.v. X has p.d.f.
f(x) = \(\frac{c}{x}\) for 1 < x < 3, c > 0. Find c, E(X), Var (X).
Solution:
Since f(x) is p.d.f of r.v. X