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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

I. Evaluate the following integrals as a limit of a sum.

Question 1.
\(\int_{1}^{3}(3 x-4) \cdot d x\)
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1

Question 2.
\(\int_{0}^{4} x^{2} d x\)
Solution:
Let f(x) = x², for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = \(\frac{4}{n}\) as n → ∞, h → 0
Here, a = 0

Question 3.
\(\int_{0}^{2} e^{x} d x\)
Solution:
Let f(x) = e^x, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0

Question 4.
\(\int_{0}^{2}\left(3 x^{2}-1\right) d x\)
Solution:
Let f(x) = 3x² – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)² – 1
= 3r²h² – 1

Question 5.
\(\int_{1}^{3} x^{3} d x\)
Solution:
Let f(x) = x³, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here a = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Miscellaneous Exercise 3

I. Choose the correct options from the given alternatives:

Question 1.
\(\int \frac{1+x+\sqrt{x+x^{2}}}{\sqrt{x}+\sqrt{1+x}} \cdot d x=\)
(a) \(\frac{1}{2} \sqrt{x+1}+c\)
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)
(c) \(\sqrt{x+1}+c\)
(d) \(2(x+1)^{\frac{3}{2}}+c\)
Answer:
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)

Question 2.
\(\int \frac{1}{x+x^{5}} \cdot d x\) = f(x) + c, then \(\int \frac{x^{4}}{x+x^{5}} \cdot d x=\)
(a) log x – f(x) + c
(b) f(x) + log x + c
(c) f(x) – log x + c
(d) \(\frac{1}{5}\) x⁵ f(x) + c
Answer:
(a) log x – f(x) + c

Question 3.
\(\int \frac{\log (3 x)}{x \log (9 x)} \cdot d x=\)
(a) log(3x) – log(9x) + c
(b) log(x) – (log 3) . log(log 9x) + c
(c) log 9 – (log x) . log(log 3x) + c
(d) log(x) + log(3) . log(log 9x) + c
Answer:
(b) log(x) – (log 3) . log(log 9x) + c

Question 4.
\(\int \frac{\sin ^{m} X}{\cos ^{m+2} X} \cdot d x=\)
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)
(b) (m + 2) tan^m+1 x + c
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{m}+c\)
(d) (m + 1) tan^m+1 x + c
Answer:
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)

Question 5.
∫tan(sin^-1 x) . dx =
(a) \(\left(1-x^{2}\right)^{-\frac{1}{2}}+c\)
(b) \(\left(1-x^{2}\right)^{\frac{1}{2}}+c\)
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{\sqrt{1-x^{2}}}+c\)
(d) \(-\sqrt{1-x^{2}}+c\)
Answer:
(d) \(-\sqrt{1-x^{2}}+c\)

Hint: sin^-1 x = \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\)

Question 6.
\(\int \frac{x-\sin x}{1-\cos x} \cdot d x=\)
(a) x cot(\(\frac{x}{2}\)) + c
(b) -x cot(\(\frac{x}{2}\)) + c
(c) cot(\(\frac{x}{2}\)) + c
(d) x tan(\(\frac{x}{2}\)) + c
Answer:
(b) -x cot(\(\frac{x}{2}\)) + c

Question 7.
If f(x) = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\), g(x) = \(e^{\sin ^{-1} x}\), then ∫f(x) . g(x) . dx =
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)
(b) \(e^{\sin ^{-1} x} \cdot\left(1-\sin ^{-1} x\right)+c\)
(c) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x+1\right)+c\)
(d) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} X-1\right)+c\)
Answer:
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)

Question 8.
If ∫tan³ x . sec³ x . dx = (\(\frac{1}{m}\)) sec^m x – (\(\frac{1}{n}\)) secⁿ x + c, then (m, n) =
(a) (5, 3)
(b) (3, 5)
(c) \(\left(\frac{1}{5}, \frac{1}{3}\right)\)
(d) (4, 4)
Answer:
(a) (5, 3)

Hint: ∫tan³ x . sec³ x dx
= ∫sec² x . tan² x . sec x tan x dx
= ∫sec² x (sec² x – 1) sec x tan x dx
Put sec x = t.

Question 9.
\(\int \frac{1}{\cos x-\cos ^{2} x} \cdot d x=\)
(a) log(cosec x – cot x) + tan(\(\frac{x}{2}\)) + c
(b) sin 2x – cos x + c
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c
(d) cos 2x – sin x + c
Answer:
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c

Question 10.
\(\int \frac{\sqrt{\cot x}}{\sin x \cdot \cos x} \cdot d x=\)
(a) \(2 \sqrt{\cot x}+c\)
(b) \(-2 \sqrt{\cot x}+c\)
(c) \(\frac{1}{2} \sqrt{\cot x}+c\)
(d) \(\sqrt{\cot X}+c\)
Answer:
(b) \(-2 \sqrt{\cot x}+c\)

Question 11.
\(\int \frac{e^{x}(x-1)}{x^{2}} \cdot d x=\)
(a) \(\frac{e^{x}}{x}+c\)
(b) \(\frac{e^{x}}{x^{2}}+c\)
(c) \(\left(x-\frac{1}{x}\right) e^{x}+c\)
(d) x e^-x + c
Answer:
(a) \(\frac{e^{x}}{x}+c\)

Question 12.
∫sin(log x) . dx =
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c
(b) \(\frac{x}{2}\) [sin(log x) + cos(log x)] + c
(c) \(\frac{x}{2}\) [cos(log x) – sin(log x)] + c
(d) \(\frac{x}{4}\) [cos(log x) – sin(log x)] + c
Answer:
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c

Question 13.
∫x^x (1 + log x) . dx =
(a) \(\frac{1}{2}\) (1 + log x)² + c
(b) x^2x + c
(c) x^x log x + c
(d) x^x + c
Answer:
(d) x^x + c

Hint: \(\frac{d}{d x}\)(xx) = x^x (1 + log x)

Question 14.
\(\int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \cdot d x=\)
(a) \(\log \left(\sin ^{-\frac{4}{7}} x\right)+c\)
(b) \(\frac{4}{7} \tan ^{\frac{4}{7}} x+c\)
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)
(d) \(\log \left(\cos ^{\frac{3}{7}} x\right)+c\)
Answer:
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)

Hint: \(\int \cos ^{-\frac{3}{7}} x \sin ^{-\frac{11}{7}} x d x\)
= \(\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x \cdot \cos ^{2} x} d x\)
= \(\int \tan ^{-\frac{11}{7}} x \sec ^{2} x d x\)
Put tan x = t.

Question 15.
\(2 \int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x+\sin ^{2} x} \cdot d x=\)
(a) sin 2x + c
(b) cos 2x + c
(c) tan 2x + c
(d) 2 sin 2x + c
Answer:
(a) sin 2x + c

Question 16.
\(\int \frac{d x}{\cos x \sqrt{\sin ^{2} x-\cos ^{2} x}} \cdot d x=\)
(a) log(tan x – \(\sqrt{\tan ^{2} x-1}\)) + c
(b) sin^-1 (tan x) + c
(c) 1 + sin^-1 (cot x) + c
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c
Answer:
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c

Question 17.
\(\int \frac{\log x}{(\log e x)^{2}} \cdot d x=\)
(a) \(\frac{x}{1+\log x}+c\)
(b) x(1 + log x) + c
(c) \(\frac{1}{1+\log x}+c\)
(d) \(\frac{1}{1-\log x}+c\)
Answer:
(a) \(\frac{x}{1+\log x}+c\)

Question 18.
∫[sin(log x) + cos(log x)] . dx =
(a) x cos(log x) + c
(b) sin(log x) + c
(c) cos(log x) + c
(d) x sin(log x) + c
Answer:
(d) x sin(log x) + c

Question 19.
\(\int \frac{\cos 2 x-1}{\cos 2 x+1} \cdot d x=\)
(a) tan x – x + c
(b) x + tan x + c
(c) x – tan x + c
(d) -x – cot x + c
Answer:
(c) x – tan x + c

Question 20.
\(\int \frac{e^{2 x}+e^{-2 x}}{e^{x}} \cdot d x=\)
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)
(b) \(e^{x}+\frac{1}{3 e^{3 x}}+c\)
(c) \(e^{-x}+\frac{1}{3 e^{3 x}}+c\)
(d) \(e^{-x}-\frac{1}{3 e^{3 x}}+c\)
Answer:
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)

II. Integrate the following with respect to the respective variable:

Question 1.
(x – 2)² √x
Solution:

Question 2.
\(\frac{x^{7}}{x+1}\)
Solution:

Question 3.
\((6 x+5)^{\frac{3}{2}}\)
Solution:

Question 4.
\(\frac{t^{3}}{(t+1)^{2}}\)
Solution:

Question 5.
\(\frac{3-2 \sin x}{\cos ^{2} x}\)
Solution:

Question 6.
\(\frac{\sin ^{6} \theta+\cos ^{6} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}\)
Solution:

Question 7.
cos 3x cos 2x cos x
Solution:

Question 8.
\(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\)
Solution:

Question 9.
\(\cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right)\)
Solution:
Let I = \(\int \cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x\)

III. Integrate the following w.r.t. x:

Question 1.
\(\frac{(1+\log x)^{3}}{x}\)
Solution:

Question 2.
cot^-1 (1 – x + x²)
Solution:

Question 3.
\(\frac{1}{x \sin ^{2}(\log x)}\)
Solution:

Question 4.
\(\sqrt{x} \sec \left(x^{\frac{3}{2}}\right) \tan \left(x^{\frac{3}{2}}\right)\)
Solution:

Question 5.
log(1 + cos x) – x tan(\(\frac{x}{2}\))
Solution:

Question 6.
\(\frac{x^{2}}{\sqrt{1-x^{6}}}\)
Solution:

Question 7.
\(\frac{1}{(1-\cos 4 x)(3-\cot 2 x)}\)
Solution:

Question 8.
log(log x) + (log x)^-2
Solution:

Question 9.
\(\frac{1}{2 \cos x+3 \sin x}\)
Solution:

Question 10.
\(\frac{1}{x^{3} \sqrt{x^{2}-1}}\)
Solution:

Question 11.
\(\frac{3 x+1}{\sqrt{-2 x^{2}+x+3}}\)
Solution:

Question 12.
log(x² + 1)
Solution:

Question 13.
e^2x sin x cos x
Solution:

Question 14.
\(\frac{x^{2}}{(x-1)(3 x-1)(3 x-2)}\)
Solution:

Question 15.
\(\frac{1}{\sin x+\sin 2 x}\)
Solution:

Question 16.
\(\sec ^{2} x \sqrt{7+2 \tan x-\tan ^{2} x}\)
Solution:

Question 17.
\(\frac{x+5}{x^{3}+3 x^{2}-x-3}\)
Solution:

Question 18.
\(\frac{1}{x\left(x^{5}+1\right)}\)
Solution:

Question 19.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:

Question 20.
sec⁴ x cosec² x
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.4

I. Integrate the following w. r. t. x:

Question 1.
\(\frac{x^{2}+2}{(x-1)(x+2)(x+3)}\)
Solution:

Question 2.
\(\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}-2\right)\left(x^{2}+3\right)}\)
Solution:

Question 3.
\(\frac{12 x+3}{6 x^{2}+13 x-63}\)
Solution:

Question 4.
\(\frac{2 x}{4-3 x-x^{2}}\)
Solution:

Question 5.
\(\frac{x^{2}+x-1}{x^{2}+x-6}\)
Solution:

Question 6.
\(\frac{6 x^{3}+5 x^{2}-7}{3 x^{2}-2 x-1}\)
Solution:

Question 7.
\(\frac{12 x^{2}-2 x-9}{\left(4 x^{2}-1\right)(x+3)}\)
Solution:

Question 8.
\(\frac{1}{x\left(x^{5}+1\right)}\)
Solution:

Question 9.
\(\frac{2 x^{2}-1}{x^{4}+9 x^{2}+20}\)
Solution:

Question 10.
\(\frac{x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}-2\right)}\)
Solution:

Question 11.
\(\frac{2 x}{\left(2+x^{2}\right)\left(3+x^{2}\right)}\)
Solution:

Question 12.
\(\frac{2^{x}}{4^{x}-3 \cdot 2^{x}-4}\)
Solution:

Question 13.
\(\frac{3 x-2}{(x+1)^{2}(x+3)}\)
Solution:

Question 14.
\(\frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}\)
Solution:
Let I = ∫\(\frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}\) dx

Question 15.
\(\frac{1}{x\left(1+4 x^{3}+3 x^{6}\right)}\)
Solution:

Question 16.
\(\frac{1}{x^{3}-1}\)
Solution:

Question 17.
\(\frac{(3 \sin x-2) \cdot \cos x}{5-4 \sin x-\cos ^{2} x}\)
Solution:

Question 18.
\(\frac{1}{\sin x+\sin 2 x}\)
Solution:

Question 19.
\(\frac{1}{2 \sin x+\sin 2 x}\)
Solution:

Question 20.
\(\frac{1}{\sin 2 x+\cos x}\)
Solution:

Question 21.
\(\frac{1}{\sin x \cdot(3+2 \cos x)}\)
Solution:

Question 22.
\(\frac{5 \cdot e^{x}}{\left(e^{x}+1\right)\left(e^{2 x}+9\right)}\)
Solution:

Question 23.
\(\frac{2 \log x+3}{x(3 \log x+2)\left[(\log x)^{2}+1\right]}\)
Solution:

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

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

Question 1.
Let f(1) = 3, f'(1) = \(-\frac{1}{3}\), g(1) = -4 and g'(1) = \(-\frac{8}{3}\). The derivative of \(\sqrt{[f(x)]^{2}+[g(x)]^{2}}\) w.r.t. x at x = 1 is
(a) \(-\frac{29}{15}\)
(b) \(\frac{7}{3}\)
(c) \(\frac{31}{15}\)
(d) \(\frac{29}{15}\)
Answer:
(d) \(\frac{29}{15}\)

Question 2.
If y = sec(tan^-1 x), then \(\frac{d y}{d x}\) at x = 1, is equal to
(a) \(\frac{1}{2}\)
(b) 1
(c) \(\frac{1}{\sqrt{2}}\)
(d) 2
Answer:
(c) \(\frac{1}{\sqrt{2}}\)

Question 3.
If f(x) = \(\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\), which of the following is not the derivative of f(x)?
(a) \(\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}\)
(b) \(\frac{4^{x+1} \log 2}{1+4^{2 x}}\)
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)
(d) \(\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}\)
Answer:
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)

Question 4.
If x^y = y^x, then\(\frac{d y}{d x}\) = _______
(a) \(\frac{x(x \log y-y)}{y(y \log x-x)}\)
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)
(c) \(\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}\)
(d) \(\frac{y(1-\log x)}{x(1-\log y)}\)
Answer:
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)

Question 5.
If y = sin (2 sin^-1 x), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)
(b) \(\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}\)
(c) \(\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)

Question 6.
If y = \(\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{1-x^{2}}}\)
(b) \(\frac{1-2 x}{\sqrt{1-x^{2}}}\)
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)

Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
Answer:
(d) 1

Question 8.
If g is the inverse of function f and f'(x) = \(\frac{1}{1+x^{7}}\), then the value of g'(x) is equal to:
(a) 1 + x⁷
(b) \(\frac{1}{1+[g(x)]^{7}}\)
(c) 1 + [g(x)]⁷
(d) 7x⁶
Answer:
(c) 1 + [g(x)]⁷

Question 9.
If \(x \sqrt{y+1}+y \sqrt{x+1}=0\) and x ≠ y, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{1}{(1+x)^{2}}\)
(b) \(-\frac{1}{(1+x)^{2}}\)
(c) (1 + x)²
(d) \(-\frac{x}{x+1}\)
Answer:
(b) \(-\frac{1}{(1+x)^{2}}\)

Question 10.
If y = \(\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)\), where -a < x < a, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{a^{2}-x^{2}}}\)
(b) \(\frac{a}{\sqrt{a^{2}-x^{2}}}\)
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
(d) \(\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
Answer:
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
[Hint: Put x = a cos θ]

Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then \(\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}\) = _______
(a) \(\frac{8 \sqrt{2}}{a \pi}\)
(b) \(-\frac{8 \sqrt{2}}{a \pi}\)
(c) \(\frac{a \pi}{8 \sqrt{2}}\)
(d) \(\frac{4 \sqrt{2}}{a \pi}\)
Answer:
(a) \(\frac{8 \sqrt{2}}{a \pi}\)

Question 12.
If y = a cos (log x) and \(A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0\), then the values of A, B, C are _______
(a) x², -x, -y
(b) x², x, y
(c) x², x, -y
(d) x², -x, y
Answer:
(b) x², x, y

(II) Solve the following:

Question 1.


Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:

Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:

Match the following:

Solution:

Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.

(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of \(\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}\) is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:

Question 4.
Differentiate the following w.r.t. x:
(i) \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Solution:
Let y = \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Put x = cos θ, Then θ = cos^-1 x and

(ii) \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Solution:
Let y = \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Put x = cos θ. Then θ = cos^-1 x and

(iii) \(\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]\)
Solution:

(iv) \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Solution:
Let y = \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Put x = cos θ. Then θ = cos^-1 x and

(v) \(\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)\)
Solution:

(vi) \(\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]\)
Solution:

Question 5.
(i) If \(\sqrt{y+x}+\sqrt{y-x}=c\), show that \(\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}\)
Solution:
\(\sqrt{y+x}+\sqrt{y-x}=c\)
Differentiating both sides w.r.t. x, we get

(ii) If \(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\), then show that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Solution:
\(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\)
\(y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1\)
Differentiating both sides w.r.t. x, we get

(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get

(iv) If sin y = x sin(a + y), then show that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:

(v) If x = \(e^{\frac{x}{y}}\), then show that \(\frac{d y}{d x}=\frac{x-y}{x \log x}\)
Solution:
x = \(e^{\frac{x}{y}}\)
\(\frac{x}{y}\) = log x …..(1)
y = \(\frac{x}{\log x}\)

(vi) If y = f(x) is a differentiable function of x, then show that \(\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}\)
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f^-1(y) exists,

Question 6.
(i) 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:

(ii) Differentiate \(\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]\) w.r.t. cos(log x)
Solution:

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

Question 7.
(i) If y² = a² cos²x + b² sin²x, show that \(y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}\)
Solution:
y² = a² cos²x + b² sin²x …… (1)
Differentiating both sides w.r.t. x, we get

(ii) If log y = log(sin x) – x², show that \(\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0\)
Solution:

(iii) If x = a cos θ, y = b sin θ, show that \(a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

(iv) If y = A cos(log x) + B sin(log x), show that x²y₂ + xy₁ + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(v) If y = A e^mx + B e^nx, show that y₂ – (m + n) y₁ + (mn) y = 0.
Solution:
y = A e^mx + B e^nx
Differentiating w.r.t. x, we get

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.