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Practice AP Inter 2nd Year Maths Study Material Chapter 7 Integrals MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Integrals MCQ

Indefinite Integrals

Question 1.
The anti derivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) =
1) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)
2) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+C\)
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
4) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Solution:
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
Anti derivative of \(\sqrt{x}+\frac{1}{\sqrt{x}}=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)
⇒ I = \(\int x^{\frac{1}{2}} d x+\int x^{\frac{1}{2}} d x=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c\)

Question 2.
The anti derivative of e^2logcotx
1) cot x – 1
2) tan x – cot x
3) -cot x – x
4) – 1 – cot x
Solution:
3) -cot x – x
I = \(\int e^{2 \log \cot x} d x=\int e^{\log _e \cot ^2 x} d x=\int \cot ^2 x d x=\int\left({cosec}^2 x-1\right) d x\) = -cot x – x + c

Question 3.
If \(\frac{d}{d x}\)f(x) = 4x³ – \(\frac{3}{x^4}\) such that f(2) = 0. Then f(x) is
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
2) \(x^3+\frac{1}{x^4}+\frac{129}{8}\)
3) \(x^4+\frac{1}{x^3}+\frac{129}{8}\)
4) \(x^3+\frac{1}{x^4}-\frac{129}{8}\)
Solution:
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
\(\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4} \Rightarrow f(x)=\int\left(4 x^3-\frac{3}{x^4}\right) d x=\not A \cdot \frac{x^4}{\not A}-\not z\left(\frac{-1}{\not \partial x^3}\right)=x^4+\frac{1}{x^3}+c\) …….(1)
Given, f(x) = 0 ⇒ 0 = 16 + \(\frac{1}{8}+c \Rightarrow c=-\left(\frac{129}{8}\right)(1) \Rightarrow f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}\)

Question 4.
\(\int \frac{10 x^9+10^x \log _e 10}{x^{10}+10^x}\)dx =
1) 10^x – 10¹⁰ + C
2) 10^x + x¹⁰ + C
3) (10^x – x¹⁰)^-1 + C
4) log(10^x + x¹⁰) + C
Solution:
4) log(10^x + x¹⁰) + C
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
I = \(\int \frac{10 x^9+10^x \log _e^{10}}{x^{10}+10^x} d x\) = log(x¹⁰ + x^x) + c

Question 5.
\(\int \frac{x^2}{1+x^3}\) dx =
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
2) \(\frac{2}{3} \log \left|1+x^3\right|+c\)
3) log|1 + x³| + c
4) tan^-1(x^3/2 + c
Solution:
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
Put 1 + x³ = t ⇒ 0 + 3x²dx = dt ⇒ x² dx = \(\frac{1}{3}\)dt
I = \(\int \frac{x^2}{1+x^3} d x=\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+c=\frac{1}{3} \log \left|1+x^3\right|+c\)

Question 6.
\(\int \frac{d x}{\sin ^2 x \cos ^2 x}\) =
1) tan x + cot x + C
2) tan x – cot x + C
3)tan x cot x + C
4) tan x – cot 2x + C
Solution:
2) tan x – cot x + C
I = \(\int \frac{1}{\sin ^2 x \cos ^2 x} d x=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
= \(\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x} d x+\int \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\) = ∫sec² dx + ∫cosec² dx = tan x – cot x + c

Question 7.
\(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx =
1) tanx + cot x + C
2) tan x + cosecx + C
3) -tan x + cot x + C
4) tan x + sec x + C
Solution:
1) tanx + cot x + C
I = \(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x=\int \frac{1}{\cos ^2 x} d x-\int \frac{1}{\sin ^2 x} d x\)
= ∫sec² x dx – ∫cosec²x dx = tan x + cot x + c

Question 8.
\(\int \frac{\cos x+x \sin x}{x(x+\cos x)}\)dx = log|f(x)| + c then f(x) =
1) x(x + cos x)
2) \(\frac{x+\cos x}{x}\)
3) \(\frac{x}{x+\cos x}\)
4) \(\frac{1}{x(x+\cos x)}\)
Solution:
3) \(\frac{x}{x+\cos x}\)

Question 9.
\(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)}\)dx =
1) -cot(e^xx) + C
2) tan(xe^x) + C
3) tan(e^x) + C
4) cot(e^x) + C
Solution:
2) tan(xe^x) + C
Put, x.e^x = t ⇒ (xe^x + e^x)dx = dt ⇒ e^x(x + 1)dx = dt
I = \(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x \Rightarrow I=\int \frac{d x}{\cos ^2 t}\) = ∫sec² dt = tan t + c = tan(xe^x) + c

Question 10.
\(\int \frac{d x}{x^2+2 x+2}\) =
1) x tan^-1(x + 1) + C
2) tan^-1 (x + 1) + C
3) (x + 1)tan^-1x + C
4) tan^-1x + C
Solution:
2) tan^-1 (x + 1) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{x}^2+2 \mathrm{x}+2} \mathrm{dx}=\int \frac{1}{\mathrm{x}^2+2 \mathrm{x}+1+1} \mathrm{dx}=\int \frac{1}{(\mathrm{x}+1)^2+1^2} \mathrm{dx}\) = tan^-1 (x + 1) + C

Question 11.
\(\int \frac{d x}{\sqrt{9 x-4 x^2}}\) =
1) \(\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
3) \(\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
4) \(\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+C\)
Solution:
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
9x – 4x² = \(-4\left[x^2-\frac{9}{4} x\right]=-4\left[x^2-2 \cdot x \cdot \frac{9}{8}+\left(\frac{9}{8}\right)^2-\left(\frac{9}{8}\right)^2\right]=4\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]\)

Question 12.
\(\int \frac{x d x}{(x-1)(x-2)}\) =
1) \(\log \left|\frac{(x-1)^2}{x-2}\right|+C\)
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
3) \(\log \left|\left(\frac{x-1}{x-2}\right)^2\right|+C\)
4) log|(x – 1)(x -2)| + C
Solution:
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
Using partial fractions \(\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\) ⇒ x = A(x – 2) + B(x – 1)
ar x = 1 we get A = -1; at x = 2 we get B = 2
I = \(\int \frac{x}{(x-1)(x-2)} d x=\int \frac{-1}{x-1} d x+\int \frac{2}{x-2} d x\) = -log|x – 1| + 2log|x – 2|
= -log|x – 1| + log|(x – 2)|² = \(\log \left|\frac{(x-2)^2}{(x-1)}\right|+c\)

Question 13.
\(\int \frac{d x}{x\left(x^2+1\right)}\) =
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
2) \(\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
3) \(-\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
4) \(\frac{1}{2} \log |x|+\log \left|x^2+1\right|+C\)
Solution:
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
\(\frac{1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1}\) we get A = 1; B = -1; C = 0
I = \(\int \frac{1}{x\left(x^2+1\right)} d x=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right) d x=\int \frac{1}{x} d x-\frac{1}{2} \int \frac{2 x}{x^2+1} d x=\log |x|-\frac{1}{2} \log \left|x^2+1\right|+c\)

Question 14.
\(\int \frac{x^2}{1-x^4} d x\) =
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)
2) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|+\frac{1}{2} \tan ^{-1} x+C\)
3) \(\frac{1}{4} \log \left|\frac{1+x^2}{1-x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
4) \(\frac{1}{4} \log \left|\frac{1-x^2}{1+x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
Solution:
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)

Question 15.
∫x²e^x dx =
1) \(\frac{1}{3}\)e^x3 + C
2) \(\frac{1}{3}\)e^x2 + C
3) \(\frac{1}{2}\)e^x3 + C
4) \(\frac{1}{2}\)e^x2 + C
Solution:
1) \(\frac{1}{3}\)e^x3 + C
Put, x³ = t ⇒ 3x²dx = dt ⇒ x²dx = \(\frac{1}{3}\)dt
I = ∫x²e^x3 dx = \(\frac{1}{3}\)∫e^tdt = \(\frac{1}{3}\). e^t + c = \(\frac{1}{3}\) e^x3 + c

Question 16.
∫ x sec² x dx =
1) xtanx – log|sec x| + C
2) xtanx – log|cos x| + C
3) xtanx – log|cosec x| + C
4) xtanx – log|sin x| + C
Solution:
1) xtanx – log|sec x| + C
Integration by parts we have
I = x(tan x) – ∫tan x dx = ∫ x sec²x dx = x(tan x) – log|sec| + c

Question 17.
∫e^x sec x( + tan x) dx =
1) e^xcos x + C
2) e^x sec x + C
3) e^x sin x + C
4) e^x tan x + C
Solution:
2) e^x sec x + C
∫ e^x(f(x) + f'(x)) dx = e^xf(x) + c
I = ∫e^xsecx(1 + tan x) dx = ∫e^x[sec x + sec x tan x]dx = e^x sec x + c

Question 18.
\(\int e^x\left(\frac{1+x \log x}{x}\right) d x\) =
1) xe^{log x} + C
2) e^x log x +C
3) e^x log x² + C
4) None
Solution:
2) e^x log x +C
I = ∫e^x[f(x) + f'(x)]dx = e^xf(x) + c
I = \(\int e^x\left(\frac{1+x \log x}{x}\right) d x=\int e^x\left(\frac{1}{x}+\log x\right) d x=\int e^x\left(\log x+\frac{1}{x}\right) d x\) = e^x(log x) + c

Question 19.
\(\int \sqrt{1+x^2} d x\) =
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
2) \(\frac{2}{3}\left(1+x^2\right)^{\frac{3}{2}}+C\)
3) \(\frac{2}{3} x\left(1+x^2\right)^{\frac{3}{2}}+C\)
4) \(\frac{x^2}{2} \sqrt{1+x^2}+\frac{1}{2} x^2 \log \left|x+\sqrt{1+x^2}\right|+C\)
Solution:
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
\(\int \sqrt{a^2+x^2} d x=\frac{x}{2} \sqrt{a^2+x^2}+\frac{a^2}{2} \log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|+c\)
I = \(\int \sqrt{1+\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+1}\right|+\mathrm{c}\)

Question 20.
\(\int \sqrt{x^2-8 x+7} d x\) =
1) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+9 \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
2) \(\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C\)
3) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
Solution:
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)

Question 21.
\(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) =
1) tan^-1(e^x) + C
2) tan^-1(e^-x) + C
3) log(e^x – e^-x) + C
4) log(e^x + e^-x) + C
Solution:
1) tan^-1(e^x) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}=\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}}=\int \frac{\mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}\right)^2+1^2} \mathrm{dx}\). Put e^x = t ⇒ e^x dx = dt
I = \(\int \frac{d t}{t^2+1}\) = tan^-1(t) + c = tan^-1(e^x) + c

Question 22.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x\) =
1) \(\frac{-1}{\sin x+\cos x}+C\)
2) log|sin x – cos x| + C
3) log|sin x – cos x| + C
4) \(\frac{1}{(\sin x+\cos x)^2}\)
Solution:
2) log|sin x – cos x| + C
I = \(\int \frac{\cos 2 x}{(\sin x+\cos x)^2}=d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x\)
= \(\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)(\cos x+\sin x)} d x\) = log|cos x + sin x| + c

Definite Intervals

Question 1.
\(\int_1^{\sqrt{3}} \frac{d x}{1+x^2}\) =
1) \(\frac{\pi}{3}\)
2) \(\frac{2 \pi}{3}\)
3) \(\frac{\pi}{6}\)
4) \(\frac{\pi}{12}\)
Solution:
4) \(\frac{\pi}{12}\)
Textual given key is 1.
I = \(\int \frac{1}{\left(1+x^2\right)} d x=\left(\tan ^{-1}(x)\right)_1^{\sqrt{3}}\) = tan^-1(\(\sqrt{3}\)) – tan^-1(1) = 60 – 45 = 15 = \(\frac{\pi}{12}\)

Question 2.
\(\int_0^{\frac{2}{3}} \frac{d x}{4+9 x^2}\) =
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{12}\)
3) \(\frac{\pi}{24}\)
4) \(\frac{\pi}{4}\)
Solution:
3) \(\frac{\pi}{24}\)
Textual given key is 4.

Question 3.
The value of the integral \(\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^2} d x\) is
1) 6
2) 0
3) 3
4) 4
Solution:
1) 6

Question 4.
If f(x) = \(\int_0^x t\) sin t dt, then f'(x) is
1) cos x + x sin x
2) x sin x
3) x cos x
4) sinx + x cosx
Solution:
2) x sin x
f(x) = \(\int_0^x t \sin t d x\) Diff. w.r.t we get f'(x) = x sin x

Question 5.
The value \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right) d x\) is
1) 0
2) 2
3) π
4) 1
Solution:
3) π
I = \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right)=\int_{-\pi / 2}^{\pi / 2} 1 d x=(x)_{-\pi / 2}^{\pi / 2}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}+\frac{\pi}{2}=\pi\)

Question 6.
The value of \(\int_0^\pi 2 \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\) is
1) 2
2) 3/4
3) 0
4) -2
Solution:
3) 0
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \quad \ldots \ldots \ldots .(1) \quad I=\int_0^{\pi / 2} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x\) …..(2)
I + I = \(\int_0^{\pi / 2}\left[\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right] d x \Rightarrow 2 I=\int_0^{\pi / 2} \log (1) d x \Rightarrow 2 I=0 \Rightarrow I=0\)

Question 7.
If f(a + b – x) = f(x), then \(\int_a^b f(x) d x\) =
1) \(\frac{(a+b)}{2} \int_a^b f(b-x) d x\)
2) \(\frac{(a+b)}{2} \int_a^b f(b+x) d x\)
3) \(\frac{b-a}{2} \int_a^b f(x) d x\)
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
Solution:
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
\(\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x=\int_a^b[(a+b)-x] f(x) d x=\int_a^b(a+b) f(x) d x-\int_a^b x f(x) d x\)
\(\int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x-\int_a^b x f(x) d x\) ⇒ \(2 \int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x\)
⇒ \(\int_a^b x f(x) d x=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x\)

Question 8.
\(\int_0^{\pi / 2} \frac{3 \sin x+5 \cos x}{\sin x+\cos x} d x\) =
1) 2π
2) π
3) 4π
4) 8π
Solution:
1) 2π
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)

Question 9.
\(\int_0^4|2-x| d x\) =
1) 12
2) 4
3) 8
4) 2
Solution:
2) 4
I = \(\int_0^4|2-x| d x\) |2 – x| = 2x if 2 – x ≥ 0; 2 ≥ x; x ≤ 2
= \(\int_0^2|2-x| d x+\int_2^4|2-x| d x=\int_0^2(2-x) d x+\int_2^4-(2-x) d x=\left(2 x-\frac{x^2}{2}\right)_0^2-\left(2 x-\frac{x^2}{2}\right)_2^4\)
= (4 – 2) – 0 – [(8 – 8) – (4 – 2)] = 2 – [0 – 2] = 2 + 2 = 4

Question 10.
\(\int_{-2}^2\left(4-x^2\right)^{\frac{3}{2}} d x\) =
1) 2π
2) 4π
3) 6π
4) 8π
Solution:
3) 6π

Practice AP Inter 2nd Year Maths Study Material Chapter 13 Probability MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Probability MCQ

Question 1.
If P(A) = \(\frac{1}{2}\), P(B) = (), then P(A|B) is
1) 0
2) \(\frac{1}{2}\)
3) not exist
4) 1
Solution:
3) not exist
P(A/B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{0}\) which is not defined

Question 2.
If A and B are events such that P(A|B) = P(B|A), then
1) A ⊂ B but A ≠ B
2) A = B
3) A ∩ B = Φ
4) P(A) = P(B)
Solution:
Given P(A/B) = P(B/A)
⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})} \Rightarrow \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) [∵ A ∩ B = B ∩ A]
⇒ \(\frac{1}{P(B)}=\frac{1}{P(A)}\) ⇒ P(A) = P(B)

Question 3.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
1) 0
2) \(\frac{1}{3}\)
3) \(\frac{1}{12}\)
4) \(\frac{1}{36}\)
Solution:
4) \(\frac{1}{36}\)
When two dice are rolled, the number of outcomes n(S) = 6² = 36.
The only even prime number is 2.
Let E be the event of getting an even prime number on each die. ∴ E = (2, 2) ⇒ P(E) = \(\frac{1}{36}\)

Question 4.
Two events A and B will be independent, if
1) A and B are mutually exclusive
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
3) P(A) = P(B)
4) P(A) + P(B) = 1
Solution:
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
A and B are independent ⇒ A’ and B’ are independent
⇒ P(A’ ∩ B’) = [1 – P(A)] [1 – P(B)] are independent

Question 5.
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
1) P(A|B) = \(\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\)
2) P(A|B) < P(A)
3) P(A|B) ≥ P(A)
4) P(A) = P(B)
Solution:
3) P(A|B) ≥ P(A)
If A ⊂ B, then A ∩ B ⇒ P(A ∩ B) = P(A). Also, P(A) < P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})}\) …..(1)
Since P(B) ≤ 1 ⇒ \(\frac{1}{\mathrm{P}(\mathrm{~B})} \geq 1 \Rightarrow \frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})} \geq \mathrm{P}(\mathrm{~A})\)
From (1), we have P(A|B) ≥ P(A)

Question 6.
If A and B are two events such that P(A) ≠ 0 and P(B | A) = I, then
1) A ⊂ B
2) B ⊂ A
3) B = Φ
4) A = Φ
Solution:
1) A ⊂ B
Given P(A) ≠ 0 and P(B|A) = 1,
∴ \(P(B \mid A)=\frac{P(B \cap A)}{P(A)} \Rightarrow 1=\frac{P(B \cap A)}{P(A)}\) ⇒ P(A) = P(B ∩ A) ⇒ A = A ∩ B ⇒ A⊂ B

Question 7.
If P(A|B) > P(A), then which of the following is correct :
1) P(B|A) < P(B)
2) P(A ∩ B) < P(A) . P(B) 3) P(B|A) > P(B)
4) P(B|A) = P(B)
Solution:
3) P(B|A) > P(B)
Given that
Given, P(A|B) > P(A) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}>\mathrm{P}(\mathrm{~A})\)
⇒ P(A ∩ B) > P(A) × P(B) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) ⇒ P(B) ⇒ P(B|A) > P(B)

Question 8.
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
1) P(B|A) = 1
2) P(A|B) = 1
3) P(B|A) = 0
4)P(A|B) = 0
Solution:
2) P(A|B) = 1
Given that P(A) + P(B) – P(A and B) =P(A),
⇒ P(A) + P(B) – P(A ∩ B) = P(A) ⇒ P(B) – P(A ∩ B) = 0 ⇒ P(A ∩ B) = P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~B})}\) = 1

Question 9.
A bag contains 7 red and 3 white balls. Three balls are drawn one after other without replacement. Then the probability that the first two are red and third one is white is
1) \(\frac{7}{40}\)
2) \(\frac{33}{40}\)
3) \(\frac{23}{40}\)
4) \(\frac{17}{40}\)
Solution:
1) \(\frac{7}{40}\)
7R + 3W = Total 10 balls
P(E) = P(Red and Red and White) = \(\left(\frac{7}{10}\right) \times \frac{6}{9} \times \frac{3}{8}=\frac{7}{40}\) (∵ Drawn ball is not replaced)

Question 10.
A book consists «f 20 pages. If two pages arc drawn (opened) at random, then the probability that both numbers are prime numbers is
1) \(\frac{17}{95}\)
2) \(\frac{16}{95}\)
3) \(\frac{2}{15}\)
4) \(\frac{14}{95}\)
Solution:
4) \(\frac{14}{95}\)
Total no. of pages = 20
Primes up to 20 are 2, 3, 5, 7, 11, 13, 17, 19 & the no. of these primes = 8
∴ P(E) = \(\frac{{ }^8 \mathrm{C}_2}{{ }^{20} \mathrm{C}_2}=\frac{8 \times 7}{20 \times 19}=\frac{2 \times 7}{5 \times 19}=\frac{14}{95}\)

Question 11.
A fair coin is tossed 3 times, then the probability of getting one head and two tails is
1) \(\frac{1}{8}\)
2) \(\frac{1}{4}\)
3) \(\frac{3}{8}\)
4) \(\frac{1}{2}\)
Solution:
3) \(\frac{3}{8}\)
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
E = {HTT, THT, TTH} ⇒ n(E) = 3; n(S) = 8 P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{8}\)

Question 12.
A person appears lor an interview for two posts A and B. The selection of the posts are independent. If P( A) = \(\frac{1}{5}\), P(B) = \(\frac{1}{8}\) then P(A ∪ B) is
1) \(\frac{7}{10}\)
2) \(\frac{3}{10}\)
3) \(\frac{9}{10}\)
4) \(\frac{1}{10}\)
Solution:
2) \(\frac{3}{10}\)
Given A, B are independent event ⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{5}+\frac{1}{8}\) – [P(A).P(B)] = \(\frac{13}{40}-\left(\frac{1}{5} \times \frac{1}{8}\right)=\frac{13}{40}-\frac{1}{40}=\frac{12}{40}=\frac{3}{10}\)

Question 13.
Three events A, B, C are mutually exclussive and exhaustive and P(A) = 0.4, then P(B) + P(C) =
1) 0.4
2) 0.5
3) 0.6
4) 0
Solution:
3) 0.6
A, B, C are mutually exclusive and exhaustive ⇒ A ∪ B ∪ C = S ……..(1)
Given P(A) = 0.4 ……..(2)
(1) ⇒ P(A ∪ B ∪ C) = P(S) ⇒ P(A) + P(B) + P(C) = 1 ⇒ 0.4 + P(B) + P(C) = 1
⇒ P(B) + P(C) = 1 – 0.4 = 0.6

Question 14.
If P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then P(\(\vec{A}\)) + P(\(\vec{B}\)) =3 J
1) 0.8
2) 0.6
3) 1.2
4) 1.4
Solution:
3) 1.2
\(\mathrm{P}(\overline{\mathrm{~A}})+\mathrm{P}(\overline{\mathrm{~B}})\) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)] = 2 – [P(A ∪ B) + P(A ∩ B)]
= 2 – [0.65 + 0.15] = 2 – [0.80] = 1.20 = 1.2

Question 15.
When two dice are rolled, the probability of getting unequal numbers on the faces is
1) \(\frac{1}{6}\)
2) \(\frac{35}{36}\)
3) \(\frac{5}{6}\)
4) \(\frac{1}{3}\)
Solution:
3) \(\frac{5}{6}\)
Two dice are rolled n(S) = 36
Equal number faces = {(1, 1) (2, 2) (3, 3) (4, 4)(5, 5)(6, 6)} ⇒ n(S) = 6
⇒ no.of unequal faces = 36 – 6 = 30 = n(E)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{30}{36}=\frac{5}{6}\)

Question 16.
A fair coin whose faces are marked with I and 2 is thrown for four times. Then the probability of throwing a total of atleast 5 is
1) \(\frac{1}{16}\)
2) \(\frac{5}{16}\)
3) \(\frac{15}{16}\)
4) \(\frac{3}{16}\)
Solution:
3) \(\frac{15}{16}\)
Coin with faces 1. (say H); 2. (say T)
thrown 4 – times
Getting total at least 5 ⇒ total ≥ 5 ⇒ x ≥ 5
Now, P(x ≥ 5) = 1 – P(x < 5) = 1 – P (getting a total 4 faces 4 – times)
= 1 – P(every time a face 1) = 1 – \(\left[\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right]=1-\frac{1}{16}=\frac{16-1}{16}=\frac{15}{16}\)

Question 17.
If A and B are independent events of an experiment then which of the following statements is true
1) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
2) P(A|B) = P(A) and P(B|A) = P(B)
3) P(A ∩ B) = P(A) . P(B)
4) All the above
Solution:
4) All the above
By definition P(A ∩ B) = P(A).P(B)

Question 18.
If A and B are two events of a random experiment of throwing a die given by “A” : throwing an odd face and B : throwing a composite face.
Then which of the following statements is correct. ?
1)A and Bare equally likely
2) A and B are mutually exclusive
3) A and B are mutually exhaustive
4) A and B are linearly independent
Solution:
2) A and B are mutually exclusive
When a die is thrown
A : odd face (1, 3, 5); B : composite face (4, 6). Then P(A) =\(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{2}{6}=\frac{1}{3}\)
1) A, B are likely (✗)
2) A, B are mutually exclusive(✓)
A ∩ B = Φ (or) P(A ∩ B) = 0
3) A, B mutually exhaustive (✗) ∵ A ∪ B ≠ S
4) P(A ∩ B) ≠ P(A).P(B) ⇒ NOT independent (✗)

Question 19.
If A, B and C are independent events of a random experiment such that P(A) = p, P(B) = q, P(C) = r, where p, a, r ∈ (0, 1). Then the probability of the event A only occurs is
1) p . q . r
2) p(1 – q)(1 – r)
3) p. q(1 – r)
4) (1 – p) (1 – q) (1 – r)
Solution:
2) p(1 – q)(1 – r)
A, B, C are independent
P(A only occurs) = \(\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}(\overline{\mathrm{~B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})\) (∵ A, B, C are independent)
= P(A).[1 – P(B)][1 – P(C)] = P[1 – q][1 – r]

Question 20.
A fair die is rolled. Consider the events A = {I, 3, 5} and B = {2, 3}, then P(A|B) is
1) \(\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{2}{3}\)
4) \(\frac{5}{6}\)
Solution:
1) \(\frac{1}{2}\)
S = {1, 2, 3, 4, 8, 6}; A = {1, 3, 5}, B = {2, 3} ⇒ P(B) =\(\frac{2}{6}=\frac{1}{3}\)
∴ (A ∩ B) = {3} ⇒ P(A ∩ B) = \(\frac{1}{6}\)
∴ P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{1 / 6}{1 / 3}=\frac{1}{6} \times \frac{3}{1}=\frac{1}{2}\)

Practice AP Inter 2nd Year Maths Study Material Chapter 12 Linear Programming MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Linear Programming MCQ

Question 1.
Region represented by x > 0, y ≥ 0 is
1) First quadrant
2) Second quadrant
3) Third quadrant
4) Fourth quadrant
Solution:
1) First quadrant
Region x ≥ 0, y ≥ 0 (non-negative) ⇒ Both x and y positive → first quadrant.

Question 2.
If the objective function Z = ax + by has both a maximum and a minimum value on the region R then R is
1) Bounded
2) Unbounded
3) Concave polygon
4) Infeasible
Solution:
1) Bounded
Both maximum and minimum exist only if region is closed and bounded.

Question 3.
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called
1) constriants
2) Decision variables
3) objective function
4) Linear relations
Solution:
1) constriants
Linear restrictions in LPP are called constraints.

Question 4.
The optimal value of the objective function is attained at the points
1) On X-axis
2) On Y-axis
3) Which are at the corner points of the feasible region
4) Which are at the points of intersection of the inequation with Y-axis
Solution:
3) Which are at the corner points of the feasible region
In LPP, optimum value occurs at vertices of feasible region.

Question 5.
In a linear programming problem the objective function and constraints must be
1) Non-linear
2) Linear
3) Exponential
4) Logarithmic
Solution:
2) Linear
Objective function & constraints must be linear in LPP

Question 6.
The maximum value of Z = 3x + 4y subject to constraints x + y ≤ 4, x ≥ 0, y ≥ 0 is
1) 12
2) 14
3) 16
4) 10
Solution:
3) 16
Corner check points: (0, 0), (4, 0), (0, 4) then Z = 3x + 4y values: 0, 12, 16.
Maximum value is 16.

Question 7.
Maximize Z = 3x + 5y, subject to constriants x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
1) 20 at (1, 0)
2) 30 at (0, 6)
3) 37 at (4, 5)
4) 33 at (6, 3)
Solution:
3) 37 at (4, 5)
Corner check points of feasible region. Z = 3x + 5y
Best point: (4, 5) ⇒ Z = 12 + 25 = 37

Question 8.
The point which does not lie in the half plane 2x- + 3y – 12 < 0 is
1) (2, 1)
2) (1, 2)
3) (-2, 3)
4) (2, 3)
Solution:
4) (2, 3)
Point NOT in 2x + 3y – 12 < 0
Substitute each point → LHS < 0
Check (2, 3): 4 + 9 – 12 = 1 (NOT < 0)

Question 9.
Which of the following is not a component of linear programming problem?
1) Objective function
2) Constriant
3) Decision variable
4) Differential equation
Solution:
4) Differential equation
Not a component of LPP. LPP uses linear equations, not calculus.

Question 10.
The position of points 0(0, 0) and P(2, -3) in the region of graph of inequation 2x – 3y < 5 will be
1) O inside and P outside
2) O and P both inside
3) O and P both outside
4) O outside and P inside
Solution:
1) O inside and P outside
Check points in 2x – 3y < 5
0(0, 0): 0 < 5 → inside P(2, -3): 4 + 9 = 13 > 5 → outside

Practice AP Inter 2nd Year Maths Study Material Chapter 11 Three Dimensional Geometry MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Three Dimensional Geometry MCQ

Question 1.
If α, β, γ are the angles made by the line with positive direction of the coordinate axes, then sin²α + sin²β + sin² γ =
1) 1
2) 2
3) 3
4) \(\frac{3}{2}\)
Solution:
2) 2
α, β, γ are angle made by the line with +ve direction of coordinate axes
1 = cosα, m = cosβ, n = cosγ are dc’s of the line ⇒ l² + m² + n = 1
⇒ cos² α + cos² β + cos² γ = 1 ⇒ (1 – sin² α) + (1 – sin² β) + (1 – sin² γ) = 1
⇒ 3 – 1 = sin² α + sin² β + sin² γ ⇒ sin² a + sin² p + sin² γ = 2

Question 2.
The direction cosines of the median of the triangle formed by A(1, -3, 2) B(3, 1, 2) and C(-1, 3, -3) which passing through the vertex C is
1) \(\left(\frac{3}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
2) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{1}{5 \sqrt{2}}\right)\)
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
4) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)\)
Solution:
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
Mid point of AB = F = \(\left(\frac{1+3}{2}, \frac{-3+1}{2}, \frac{2+2}{2}\right)\) = (2, -1, 2), C = (-1, 3, -3)
d.r’s of Median CF = (a, b, c) = (2 + 1, -1 – 3, 2 + 3) = (3, -4, 5) ⇒ \(\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2}\)
d.c’s = \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}\right)\)

Question 3.
If the line joining the points A(2, 3, 4) and B(3, -2, 2) is parallel to the line joining C(1, -2, z) and D(-1, y, -1), then y + z =
1) 13
2) 3
3) -3
4) -13
Solution:
2) 3
Given AB || CD ⇒ \(\left(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\right)=\frac{-2}{1}=\frac{y+2}{-5}=\frac{-1-z}{-2}\)
⇒ \(\frac{-2}{1}=\frac{y+2}{-5}=\frac{1+z}{2} \Rightarrow-2=\frac{y+2}{-5} \text { and }-2=\frac{1+z}{2}\) ⇒ -4 = 1 + z ⇒ -5 = z
⇒ 10 = y + 2 ⇒ 8 = y ⇒ y + z ⇒ 8 + (-5) = 3

Question 4.
If the two lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\lambda(\mathbf{P} \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \text { and } \overline{\mathbf{r}}=(4 \hat{\mathbf{i}}-\mathbf{P} \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(3 \mathbf{P} \hat{\mathbf{i}}+5 \mathrm{P} \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) are perpendicular then p =
1) 2
2) 3
3) 6
4) 2 or 3
Solution:
4) 2 or 3
Dr’s of line (1) are (p, -3, 6); Dr’s of line (2) are (3p, 5p, 3)
Given lines are perpendicular
⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0 ⇒ 3p(p) + 5p(-3) + 18 = 0 ⇒ 3p² – 15p + 18 = 0
⇒ p² – 5p + 6 = 0 ⇒ (p – 2)(p – 3) = 0 ⇒ p = 2 (or) p = 3

Question 5.
It the two lines \(\frac{x+1}{2 k}=\frac{y-3}{3}=\frac{z-4}{-7}\) and \(\frac{x-1}{1}=\frac{y+1}{-3 k}=\frac{z+2}{2}\) are perpendicular, then k =
1) 2
2) 1
3) -2
4) 14/11
Solution:
3) -2
Given lines are perpendicular ⇒ 2k(1) + 3(-3k) + (-7)(2) = 0 ⇒ 2k – 9k – 14 = 0
⇒ -7k = 14 ⇒ k = -2

Question 6.
The angle between the lines \(\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{1}\) and \(\frac{x+2}{1}=\frac{y+2}{1}=\frac{z-3}{2}\) is
1) \(\frac{\pi}{3}\)
2) \(\frac{\pi}{6}\)
3) \(\cos ^{-1}\left(\frac{5}{6}\right)\)
4) \(\cos ^{-1}\left(\frac{3}{4}\right)\)
Solution:
1) \(\frac{\pi}{3}\)
Dr’s of the lines are (a₁, b₁, c₁) = (2, -1, 1); (a₂, b₂, c₂) = (1, 1, 2)
∴ cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}=\frac{|2-1+2|}{\sqrt{4+1+1} \sqrt{1+1+4}}=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{3}{6}=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=\frac{\pi}{3}\)

Question 7.
If θ is the acute angle between the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\) and \(\vec{r}=(2 \hat{i}-3 \hat{j}-4 \hat{k})+\mu(4 \hat{i}+3 \hat{j}+12 \hat{k})\) then cosθ
1) \(\frac{34}{39}\)
2) \(\frac{22}{39}\)
3) \(\frac{26}{39}\)
4) \(\frac{14}{39}\)
Solution:
4) \(\frac{14}{39}\)
Dr’s of the lines are (a₁, b₁, c₁) = (1, 2, -2); (a₂, b₂, c₂) = (4, 3, 12)
cos θ = \(\frac{|(4+6-24)|}{\sqrt{1+4+4} \sqrt{16+9+144}}=\frac{14}{3 \sqrt{169}}=\frac{14}{3(13)}=\frac{14}{39}\)

Question 8.
Equation of the line passing through (2, 1, -4) and parallel to the line joining the points (1, 0, -1) and (3, 2, 2) is
1) \(\frac{x+1}{2}=\frac{y+1}{2}=\frac{z-4}{3}\)
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
3) \(\frac{x-2}{-2}=\frac{y-1}{-2}=\frac{z+4}{3}\)
4) \(\frac{x+2}{-2}=\frac{y+1}{-2}=\frac{z-4}{3}\)
Solution:
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
Dr’s of line joning points (1, 0, -1) and (3, 2, 2) are (3 – 1, 2 – 0, 2 + 1) = (2, 2, 3)
required line || to given line ⇒ Dr’s of the line = (a, b, c) = (2, 2, 3)
Also (x₁, y₁, z₁) = (2, 1, -4) is a point on the line
∴ Equation of required line = \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \Rightarrow \frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)

Question 9.
Euation of the line passing through the point (1, 2, 3)and parallel to the z axis is
1) \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{0}\)
2) \(\frac{x-1}{0}=\frac{y-2}{1}=\frac{z-3}{1}\)
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
4) \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\)
Solution:
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
Dr’s of z-axis (a, b, c) = (0, 0, 1) . Aslo point on the line is (x₁, y₁, z₁) = (1, 2, 3)
∴ Equation of required line \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)

Question 10.
The direction cossines of the line which is perpendicular to the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\stackrel{\rightharpoonup}{\mathbf{r}}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\boldsymbol{\mu}(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is
1) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
2) \(\left(\frac{-3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
3) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Solution:
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Given lines \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{~b}} \Rightarrow \overline{\mathrm{~b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}\) ………(1) and \(\overline{\mathrm{r}}=\overline{\mathrm{c}}+\mathrm{s} \overline{\mathrm{~d}} \Rightarrow \overline{\mathrm{~d}}=3 \overline{\mathrm{i}}-5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) …………..(2)
Dr’s of the line which is perpendicular to both (1) and (2) and parallel to vector \(\overline{\mathbf{b}} \times \overline{\mathbf{d}}\)
Now \(\overline{\mathrm{b}} \times \overline{\mathrm{d}}=\left|\begin{array}{ccc}
\mathrm{i} & \mathrm{j} & \mathrm{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|=\overline{\mathrm{i}}(3)-\overline{\mathrm{j}}(1)+\overline{\mathrm{k}}(-7)\)
Dr’s of the line = (a, b, c) = (3, -1, -7) = \(\sqrt{3^2+(-1)^2+(-7)^2}=\sqrt{59}\)
∴ d.c’s = \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)

Practice AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Vector Algebra MCQ

Question 1.
In triangle ABC (Fig), which of the following is not true:

1) \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
2) \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\)
3) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}+\overrightarrow{\mathbf{A C}}=\overrightarrow{\mathbf{0}}\)
4) \(\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
Solution:
3) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}+\overrightarrow{\mathbf{A C}}=\overrightarrow{\mathbf{0}}\)
By Triangle Law of Addition of Vectors we have
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}} \text { (or) } \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}} \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)

Question 2.
If \(\vec{a} \text { and } \vec{b}\) are two collinear vectors, then which of the following is correct
1) \(\overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}\) for some scalar λ ≠ 0.
2) \(\vec{a}= \pm \vec{b}\)
3) the respective components of \(\vec{a} \text { and } \vec{b}\) are not proportional
4) both the vectors \(\vec{a} \text { and } \vec{b}\) have same direction, but different magnitudes.
Solution:
1) \(\overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}\) for some scalar λ ≠ 0.
If \(\vec{a} \text { and } \vec{b}\) are two collinear vectors, then \(\vec{b}\) = λ\(\vec{a}\).
The other options (2) & (4) are only true for particular values of λ

Question 3.
If \(\overrightarrow{\mathbf{a}}\) is a nonzero vector of magnitude ‘a’ and λ. a nonzero scalar, then \(\lambda \overrightarrow{\mathbf{a}}\) is unit vector if
1) λ = 1
2) λ = – 1
3) a = |λ|
4) a = 1/| λ|
Solution:
4) a = 1/| λ|
\(|\lambda \bar{a}|=1 \Rightarrow|\lambda \| \vec{a}|=1 \Rightarrow|\vec{a}|=\frac{1}{|\lambda|} \Rightarrow a=\frac{1}{|\lambda|}\)

Question 4.
Let the vectors \(\vec{a} \text { and } \vec{b}\) be such that \(|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{~b}}|=\frac{\sqrt{2}}{3}\), then \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\) is a unit vector, if the angle between \(\vec{a} \text { and } \vec{b}\) is
1) π/6
2) π/4
3) π/3
4) π/2
Solution:
2) π/4
Given that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\) and \(\vec{a} \text { and } \vec{b}\) is a unit vector. ⇒ \(|\vec{a} \times \vec{b}|=1 \Rightarrow|\vec{a} \| \vec{b}| \sin \theta=1\)
⇒ \(3\left(\frac{\sqrt{2}}{3}\right) \sin \theta=1 \Rightarrow \sqrt{2} \sin \theta=1 \Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4}\)

Question 5.
Area of a rectangle having vertices A, B, C and D with position vectors \(-\hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \hat{\mathbf{i}}-\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { and }-\hat{\mathbf{i}}-\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\), respectively is
1) 1/2
2) 1
3) 2
4) 4
Solution:
3) 2
Given ABCD is a rectangle

Area of rectangle ABCD = Length × Breadth = (AB) × (AD) = 2(1) = 2 sq. units

Question 6.
If θ is the angle between two vectors \(\vec{a} \text { and } \vec{b}\), then \(\vec{a} \text.\vec{b}\) > 0 only when
1) 0 < θ < \(\frac{\pi}{2}\)
2) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
3) 0 < θ < π
4) 0 ≤ θ ≤ π
Solution:
2) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
We have \(\vec{a} \cdot \vec{b} \geq 0 \Rightarrow|\vec{a} \| \vec{b}| \cos \theta \geq 0 \Rightarrow \cos \theta \geq 0\) [∵ \(|\overrightarrow{\mathrm{a}}| \geq 0 \text { and }|\overrightarrow{\mathrm{b}}| \geq 0\)]
⇒ 0 ≤ θ ≤ \(\frac{\pi}{2}\) Hence \(\vec{a}\).\(\vec{b}\) ≥ 0 of 0 ≤ θ ≤ \(\frac{\pi}{2}\)

Question 7.
Let \(\vec{a} \text { and } \vec{b}\) be two unit vectors and θ is the angle between them. Then \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\) is a unit vector if
1) θ = \(\frac{\pi}{4}\)
2) θ = \(\frac{\pi}{3}\)
3) θ = \(\frac{\pi}{2}\)
4) θ = \(\frac{2\pi}{3}\)
Solution:
4) θ = \(\frac{2\pi}{3}\)
We have \(\vec{a} \text { and } \vec{b}\) two unit vectors and θ is the angle between them. Then, |\(\vec{a}\)|=|\(\vec{b}\)|= 1
Now \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\) is a unit vector if \(|\vec{a}+\vec{b}|=1 \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1 \Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=1\)
⇒ \(|\vec{a}|^2+2 \vec{a} \vec{b}+|\vec{b}|^2=1 \Rightarrow 1^2+2|\vec{a}| \vec{b} \cos \theta+1^2=1\)
⇒ 1 + 2(1)(1) cosθ + 1 = 1 ⇒ cos θ = \(-\frac{1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\)

Question 8.
The value of \(\hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\hat{\mathbf{j}} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{k}})+\hat{\mathbf{k}} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})\) is
1) 0
2) -1
3) 1
4) 3
Solution:
3) 1
\(\hat{\mathrm{i}} \cdot \hat{\mathrm{j}} \times \hat{\mathrm{k}})+\hat{\mathrm{j}} \cdot(\hat{\mathrm{i}} \times \hat{\mathrm{k}})+\hat{\mathrm{k}} .(\hat{\mathrm{i}} \times \hat{\mathrm{j}})=\hat{\mathrm{i}} . \hat{\mathrm{i}}+\hat{\mathrm{j}} .(-\hat{\mathrm{j}})+\hat{\mathrm{k}} . \hat{\mathrm{k}}=1-1+1=1\)

Question 9.
If θ is the angle between any two vectors \(\vec{a} \text { and } \vec{b}\), then \(|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\) when θ is equal to 10.
1) 0
2) \(\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) π
Solution:
2) \(\frac{\pi}{4}\)
\(|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta \Rightarrow \cos \theta=\sin \theta \Rightarrow \tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}\)

Question 10.
The value of the dot product of \(\vec{a}-\vec{b} \text { and } \vec{a}+\vec{b}/latex] is
1) a² – b²
2) [latex](\vec{a} \times \vec{b})\)
3) \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\)
4) \(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}\)
Solution:
1) a² – b²
\((\bar{a}-\bar{b}) \cdot(\bar{a}+\bar{b})=\bar{a} \cdot \bar{a}+\bar{a}-\bar{b}-\bar{b} \cdot \bar{a}-\bar{b} \cdot \bar{b}=|\bar{a}|^2-|\bar{b}|^2=a^2-b^2 \text { where }|\bar{a}|=a ;|\bar{b}|=b\)

Question 11.
The position vector of the point (1, 2, 0)is
1) \(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}\)
2) \(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{2j}}+\overrightarrow{\mathrm{k}}\)
3) \(\vec{i}+2 \vec{j}\)
4) \(2 \vec{j}+\vec{k}\)
Solution:
3) \(\vec{i}+2 \vec{j}\)
PV of P = (1, 2, 0) is \(\overline{\mathrm{OP}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+0 \overline{\mathrm{k}}\)

Question 12.
If \(|(\vec{a} \times \vec{b})|=4 \text { and }|\vec{a} \cdot \vec{b}|=2\) then \(\left.\overrightarrow{\mathbf{a}}\right|^2|\overrightarrow{\mathbf{b}}|^2\) is equal to
1) 4
2) 2
3) 20
4) 2
Solution:
3) 20
Relation between \(\vec{a} \text { and } \vec{b}\) and \(\bar{a} \cdot \bar{b} \text { is }|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=(\bar{a})^2(\bar{b})^2\)
⇒ (4)² + (2)² = \((\bar{a})^2(\bar{b})^2 \Rightarrow(\bar{a})^2 \cdot(\bar{b})^2\) = 16 + 4 = 20

Question 13.
The points with position vectors \(10 \bar{i}+3 \bar{j}, 12 i-5 \vec{j} \text { and } a \dot{i}+11 j\) are collinear, if a is
1) 2
2) -8
3) 4
4) 8
Solution:
4) 8

Question 14.
The vector cos α cosβ \(\vec{i}\) + cosα sinβ \(\vec{j}\) + sinα \(\vec{k}\) is
1) null vector
2) unit vector
3) constant vector
4) vector with magnitude > 1
Solution:
2) unit vector
Consider \(|(\cos \alpha \cdot \cos \beta) \overline{\mathrm{i}}+(\cos \alpha \cdot \sin \beta) \overline{\mathrm{j}}+(\sin \alpha) \overline{\mathrm{k}}|\)
= \(\sqrt{\cos ^2 \alpha \cos ^2 \beta+\cos ^2 \alpha \sin ^2 \beta+\sin ^2 \alpha}=\sqrt{\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right)+\sin ^2 \alpha}\)
= \(\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}=\sqrt{1}=1\). Hence a Unit vector.

Question 15.
If \(\vec{a}\), \(\vec{b}\), \(\vec{b}\) are mutually perpendicular unit vectors, then the value of |\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| is
1) 1
2) \(\sqrt{2}\)
3) \(\sqrt{3}\)
4) 2
Solution:
3) \(\sqrt{3}\)
Given \(|\bar{a}|=|\bar{b}|=|\bar{c}|=1 \text { and } \bar{a} \cdot \bar{b}=\bar{b}-\bar{c}=\bar{c} \cdot \bar{a}=0\)
⇒ \(|\bar{a}+\bar{b}+\bar{c}|^2=(\bar{a})^2+(\bar{b})^2+(\bar{c})^2+2(\bar{a} \cdot \bar{b}+\bar{b}-\bar{c}+\bar{c}-\bar{a})\) = 1 + 1 + 1 + 0 = 3
⇒ \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=\sqrt{3}\)

Question 16.
If \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\), |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7 then the angle between \(\vec{a} \text { and } \vec{b}\) is
1) \(\frac{\pi}{6}\)
2) \(\frac{2\pi}{3}\)
3) \(\frac{5\pi}{3}\)
4) \(\frac{\pi}{3}\)
Solution:
4) \(\frac{\pi}{3}\)
\(\bar{a}+\bar{b}+\bar{c}=0 \Rightarrow \bar{a}+\bar{b}=-\bar{c} \quad \Rightarrow|\bar{a}+\bar{b}|=\bar{c}\). Squaring on both sides, we get
⇒ \((\bar{a})^2+(\bar{b})^2+2 \bar{a} \cdot \bar{b}=(\bar{c})^2 \Rightarrow 9+25+2 \bar{a} \cdot \bar{b}=49 \Rightarrow 2 \bar{a} \cdot \bar{b}=15 \Rightarrow 2(\bar{a})(\bar{b}) \cos (\bar{a} \bar{b})=15\)
⇒ 2(3)(5) cos θ = 15 ⇒ 2 cos θ = 1 ⇒ cos θ = \(\frac{\pi}{2}\) = cos 60°
∴ θ = 60° = π/3

Question 17.
If \(\vec{a} \text { and } \vec{b}\) are two unit vectors inclined atan angle θ then the Value of |\(\vec{a}\) – \(\vec{b}\)| is
1) 2sin\(\frac{\theta}{2}\)
2) 2sinθ
3) 2cos\(\frac{\theta}{2}\)
4) 2cosθ
Solution:
1) 2sin\(\frac{\theta}{2}\)
Given \(\bar{a}=|\bar{b}|=1\langle\bar{a}, \bar{b}\rangle\) = θ
consider \(|\bar{a}-\bar{b}|^2=(\bar{a})^2+(\bar{b})^2-2 \bar{a}-\bar{b}=1+1-2(\bar{a})(\bar{b}) \cos \theta\) = 2 – 2 cosθ
= 2(1 – cosθ) = \(2 \sin ^2 \theta / 2 \Rightarrow|\bar{a}-\bar{b}|=\sqrt{4 \sin ^2(\theta / 2)}=2 \sin (\theta / 2)\)

Question 18.
If |\(\vec{a}\)|= 3 and -1 ≤ k ≤ 2 then | k\(\vec{a}\) |lies in the internal
1) [0, 6]
2) [-3, 6]
3) [3, 6]
4) [1, 2]
Solution:
1) [0, 6]
|k\(\vec{a}\)| ⇒ |k||\(\vec{a}\)| ⇒ 3|k|
-1 ≤ k ≤ 2
0 ≤ |k| ≤ 2
0 × 3 ≤ 3 |k| ≤ 2 × 3
0 ≤ 3|k| ≤ 6 ⇒ |k\(\vec{a}\)| ∈ [0, 6]

Practice AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Differential Equations MCQ

I. Select the correct option from the given choices.

Question 1.
The degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0\) is
1) 3
2) 2
3) 1
4) not defined
Solution:
4) not defined
Given D.E is not a polynomial equation in its derivatives. Its degree is not defined.

Question 2.
The order of the differential equation \(2 x^2 \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+y=0\) is
1) 2
2) 1
3) 0
4) not defined
Solution:
1) 2
Highest order derivative present in the given D.E is \(\frac{d^2 y}{d x^2}\). Its order is two.

Question 3.
The number of arbitrary constants in the general solution of a differential equation of fourth order is
1) 0
2) 2
3) 3
4) 4
Solution:
4) 4
Number of constants in the GS= Order
Number of constants in the general solution of D.E of order n is equal to its order.
The number of constants in fourth order differential equation is 4.

Question 4.
The number of arbitrary constants in the particular solution of a differential equation of third order is
1) 3
2) 2
3) 1
4) 0
Solution:
4) 0
In a particular solution of a differential equation, there are no arbitrary constants.

Question 5.
The general solution of the differential equation \(\frac{d y}{d x}=e^{x+y}\) is
1) e^x + e^-y = C
2) e^x + e^y = C
3) e^-x + e^y = C
4) e^-x + e^-x = C
Solution:
1) e^x + e^-y = C
Given D.E is \(\frac{d y}{d x}\) = e^x+y = e^x.e^y ⇒ \(\frac{d y}{e^y}\) = e^x dx ⇒ e^-y dy = e^x dx
x ∫e^-y dy = ∫e^x dx ⇒ -e^-y = e^x + k ⇒ e^x + e^-y = -k ⇒ e^x + e^-y = C

Question 6.
A homogeneous differential equation of the from \(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\) can be solved by making the substitution.
1) y = vx
2) v = yx
3) x = vy
4) x = v
Solution:
3) x = vy
For solving homogeneous equation of form \(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\), we need to make substitution as x = vy
Thus, the correct option is C.

Question 7.
Which of the following is a homogeneous differential equation?
1) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
2) (xy) dx – (x³ + y³) dy = 0
3) (x³ + 2y²) dx + 2xy dy = 0
4) y² dx + (x² – xy – y²) dy = 0
Solution:
4) y² dx + (x² – xy – y²) dy = 0
F(x,y) is homogeneous function of degree n, if F (λx, λy) = λF\(x, y)
Consider D.E in (D) y²dx + (x² – xy² – y²)dy = 0 ⇒ \(\frac{d y}{d x}=\frac{y^2}{y^2+x y^2-x^2}\) F(x, y) = \(\frac{y^2}{y^2+x y^2-x^2}\)
F(λx, λy) = \(\frac{(\lambda y)^2}{(\lambda y)^2+(\lambda x)(\lambda y)^2-(\lambda x)^2}=\frac{\lambda^2 y^2}{\lambda^2\left(y^2+x y^2-x^2\right)}=\lambda^2\left(\frac{y^2}{y^2+x y^2-x^2}\right)\) = λ°F(x, y)
Differential equation given in D is a homogeneous equation

Question 8.
The Integrating Factor of the differential equation \(\frac{d y}{d x}-y=2 x^2\) is
1) e^-x
2) e^-y
3) \(\frac{1}{\mathrm{x}}\)
4) x
Solution:
3) \(\frac{1}{\mathrm{x}}\)
Given D.E is \(x \frac{d y}{d x}-y=2 x^2 \Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x\) This is in the \(\frac{d y}{d x}+P y=Q\) form
where, P = \(-\frac{1}{x}\) and Q = 2x ∴ IF = \(e^{-\int \frac{1}{x} d x}=e^{-\log x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x}\)

Question 9.
The Integrating Factor of the D.E (1 – y²)\(\frac{d x}{d y}\) + yx = ay, (-1 < y < 1) is
1) \(\frac{1}{y^2-1}\)
2) \(\frac{1}{\sqrt{y^2-1}}\)
3) \(\frac{1}{1-y^2}\)
4) \(\frac{1}{\sqrt{1-y^2}}\)
Solution:
4) \(\frac{1}{\sqrt{1-y^2}}\)
Given D.E is (1 – y²)\(\frac{d x}{d y}\) + yx = ay ⇒ \(\frac{d x}{d y}+\frac{y x}{1-y^2}=\frac{a y}{1-y^2}\) This is in the \(\frac{d y}{d x}+P y=Q\) form

Question 10.
The general solution of the differential equation \(\frac{y d x-x d y}{y}=0\) is
1) xy = C
2) x = Cy²
3) y = Cx
4) y = Cx²
Solution:
3) y = Cx
Given D.E. is \(\frac{y d x-x d y}{y}=0 \Rightarrow \frac{y d x-x d y}{x y}=0 \Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0\)
⇒ log |x| = log |y| = log k ⇒ \(\log \left|\frac{x}{y}\right|=\log k \Rightarrow \frac{x}{y}=k \Rightarrow y=\frac{1}{k} x \Rightarrow y=C x\) (where, C = \(\frac{1}{k}\))

Question 11.
The general solution of a D.E of the type \(\frac{d x}{d y}+P_1 x=Q_1\) (P₁, Q₁ are functions of y) is
1) \(y e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
2) \(y . e^{\int P_1 d x}=\int\left(Q_1 e^{\int P_1 d x}\right) d x+C .\)
3) \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
4) \(x e^{\int P_1 d x}=\int\left(\mathbf{Q}_1 e^{\int \mathbf{P}_1 d \mathrm{x}}\right) \mathrm{dx}+\mathbf{C}\)
Solution:
3) \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
IF for \(\frac{d x}{d y}+P_1 x=Q_1^{\prime} \text { is } e^{\int P_1 d y} \Rightarrow x(\text { I.F. })=\left(\int Q_1 \times \text { IF }\right) d y+C \Rightarrow x . e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)

Question 12.
The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is
1) x e^y + x² = C
2) x e^y + y² = C
3) y e^x + x² = C
4) y e^y + x² = C
Solution:
3) y e^x + x² = C
Given D.E is e^x dy + (ye^x + 2x)dx = 0 ⇒ e^x\(\frac{d y}{d x}\) + ye^x + 2x = 0 ⇒ \(\frac{d y}{d x}\) + y = \(\frac{2 x}{e^x}\) = 0
⇒ \(\frac{d y}{d x}\) + y = 2xe^-x = 0 ⇒ \(\frac{d y}{d x}\) + y = -2xe^-x
This is a Linear D.E form \(\frac{d y}{d x}\) + Py = Q where, P = I and Q = -2xe^-x
Now, IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \mathrm{dx}}=\mathrm{e}^{\mathrm{x}} \Rightarrow \overline{\mathrm{y}}(\mathrm{IF})=\int(\mathrm{Q} \times \mathrm{IF}) \mathrm{dx}+\mathrm{C}\)
∴ ye^x = \(\int\left(-2 x e^{-x} \cdot e^x\right) d x+C \Rightarrow y e^x=-\int 2 x d x+C\) ⇒ ye^x = -x² + C ⇒ ye^x + x² = C

Question 13.
General solution of the differential equation \(\log \left(\frac{d y}{d x}\right)\) = 2x + y is
1) \(e^{-y}=\frac{1}{2} e^{2 x}+C\)
2) \(\frac{1}{e^y}+\frac{1}{2} e^{2 x}=C\)
3) \(-e^{-y}=\frac{1}{2} e^{2 x}+C\)
4) \(e^y=\frac{1}{2} e^{2 x}+C\)
Solution:
3) \(-e^{-y}=\frac{1}{2} e^{2 x}+C\)
\(\log _{\mathrm{e}}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]\) = 2x + y \(\frac{d y}{d x}\) = e^2x+y ⇒ \(\frac{d y}{d x}\) = e^2x.e^y \(\frac{1}{e^y}\)dy = e^2x dx
Integrating \(\int e^{-y} d y=\int e^{2 x} d x \Rightarrow-e^{-y}=\frac{e^{2 x}}{2}+c\)

Question 14.
General solution of differential equation \(\frac{d y}{d x}=\frac{y}{x}\) is
1) log y = Cx
2) y = Cx
3) xy = C
4) y = C log x
Solution:
2) y = Cx
\(\frac{d y}{d x}=\frac{y}{x} \Rightarrow \frac{d y}{y}=\frac{d x}{x} \Rightarrow \int \frac{1}{y} d y=\int \frac{1}{x} d x\) ⇒ log |y| = log |x| + log |c| ⇒ log |y| = log |cx| ⇒ y = cx.

Question 15.
The degree of the differential equation \(\left(1+\frac{d y}{d x}\right)^3=\left(\frac{d y}{d x}\right)^2\) is
1) 1
2) 2
3) 3
4) 4
Solution:
3) 3
order = 1; degree = 3

Question 16.
The degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\) is
1) 1
2) 2
3) 4
4) not defined
Solution:
4) not defined
The given equation is not a polynomial equation in \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\).
Here, its degree is not defined. Hence, degree not defined

Question 17.
Order of differential equation corresponding to family of curves y = Ae^2x + Be^-2x is
1) 2
2) 1
3) 3
4) 4
Solution:
1) 2
y = Ae^2x + Be^-2x arbitary constants = 2
∴ Order of D.E is ‘2’

Question 18.
The general solution of differential equation \(\frac{d y}{d x}=e^{x-y}\) is
1) e^y = e^x + C
2) e^x + e^y = C
3) e^x+y = C
4) e^x-y = C
Solution:
1) e^y = e^x + C
\(\frac{d y}{d x}=e^x \cdot e^{-y} \Rightarrow \frac{1}{e^{-y}} d y=e^x d x \Rightarrow e^y d y=e^x d x \Rightarrow \int e^y d y=\int e^x d x \Rightarrow e^y=e^x+c\)

Question 19.
The order and degree of the differential equation \(\frac{d y}{d x}=\left(\frac{d^2 y}{d x^2}+2\right)^{1 / 2}+\frac{d^2 y}{d x^2}+5\) are respectively
1) 2, 1
2) 2, 4
3) 2, 2
4) 2, 3
Solution:
3) 2, 2
Transposing the terms properly and squaring on both sides we get \(\left[\left(\frac{d y}{d x}\right)-\left(\frac{d^2 y}{d x^2}\right)-5\right]^2=\frac{d^2 y}{d x^2}+2\)
∴ order = 2 ; degree = 2

Question 20.
The differential equation for which ax + by = 1 is general solution (a, b are arbitrary constants) is
1) \(\frac{d y}{d x}=x+C\)
2) \(y \frac{d^2 y}{d x^2}+x=1\)
3) \(\frac{d^2 y}{d x^2}=0\)
4) \(\frac{d^3 y}{d x^3}=0\)
Solution:
3) \(\frac{d^2 y}{d x^2}=0\)
Given ax + by = 1 ⇒ a(1) + b\(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) = 0 Again diff w.r.t ‘x’, 0 + b\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)=0 \Rightarrow \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}=0\)

Practice AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Application of Integrals MCQ

I. Select the correct option from the given choices.

Question 1.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
1) π
2) \(\frac{\pi}{2}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{4}\)
Solution:
1) π

Area(OAB) = \(\int_0^2 \mathrm{ydx}=\int_0^2 \sqrt{4-\mathrm{x}^2} \mathrm{dx}=\left[\frac{\mathrm{x}}{2} \sqrt{4-\mathrm{x}^2}+\frac{4}{2} \sin ^{-1} \frac{\mathrm{x}}{2}\right]_0^2=2\left(\frac{\pi}{2}\right)\) = π sq. units

Question 2.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
1) 2
2) \(\frac{9}{4}\)
3) \(\frac{9}{3}\)
4) \(\frac{9}{2}\)
Solution:
2) \(\frac{9}{4}\)

Area (OAM) = \(\int_0^3 x d y=\int_0^3 \frac{y^2}{4} d y=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3=\frac{1}{12}(27)=\frac{9}{4} \text { sq.units }\)

Question 3.
Area bounded by the curve y = x³, the x-axis and the ordinates x = – 2 and x = 1 is
1) -9
2) \(\frac{-15}{4}\)
3) \(\frac{15}{4}\)
4) \(\frac{17}{4}\)
Solution:
4) \(\frac{17}{4}\)
Required area = \(-\int_{-2}^0 y d x+\int_0^1 y d x\)
= \(-\int_{-2}^0 x^3 d x+\int_0^1 x^3 d x=-\left[\frac{x^4}{4}\right]_{-2}^0+\left[\frac{x^4}{4}\right]_0^1=-\left[0-\frac{(-2)^4}{4}\right]+\left[\frac{1}{4}-0\right]=\left(4+\frac{1}{4}\right)=\frac{17}{4} \text { sq.units }\)

Question 4.
The area bounded by the curve y = x |x| , x-axis and the ordinates x = – 1 and x = 1 is given by [Hint: y = x² if x > 0 and y = -x² if x < 0|
1) 0
2) \(\frac{1}{3}\)
3) \(\frac{2}{3}\)
4) \(\frac{4}{3}\)
Solution:
3) \(\frac{2}{3}\)
Required area = \(\int_{-1}^1 y d x=\int_{-1}^1 x|x| d x=-\int_{-1}^0 x^2 d x+\int_0^1 x^2 d x\)
= \(\left[\frac{x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1=-\left(-\frac{1}{3}\right)+\frac{1}{3}=\frac{2}{3} \text { sq.units }\)

Question 5.
Area under the curve y = \(\sqrt{a^2-x^2}\) included between the lines x = 0 and x = a is
1) \(\frac{\pi \mathrm{a}^2}{2}\)
2) \(\frac{\pi \mathrm{a}^2}{4}\)
3) \(\frac{\pi \mathrm{a}}{2}\)
4) \(\frac{\pi \mathrm{a}}{4}\)
Solution:
1) \(\frac{\pi \mathrm{a}^2}{2}\)
Area \(\int_0^a \sqrt{a^2-x^2} d x\) = Area of the circle x² + y² = a² in 1st quadrant = \(\frac{1}{4}\)(πa²)

Question 6.
The area bounded by y = sin2x the x – axis and the lines x = \(\frac{\pi}{2}\) and x = \(\frac{3\pi}{4}\) is
1) 1sq units
2) 2sq. units
3) 4sq. units
4) \(\frac{3}{2}\)sq. units
Solution:
1) 1sq units
y = sin2x ⇒ y > 0 if x < 2x < π; i.e., 0 < x <\(\frac{\pi}{2}\) and y < 0 if π < 2x < 2π; i.e., \(\frac{\pi}{2}\) < x < π
A = \(\int_{\pi / 4}^{3 \pi / 4} \sin (2 x) d x=\int_{\pi / 4}^{\pi / 2} \sin (2 x) d x-\int_{\pi / 2}^{3 \pi / 4} \sin (2 x) d x=-\left[\frac{\cos (2 x)}{2}\right]_{\pi / 4}^{\pi / 2}-\left[-\frac{\cos (2 x)}{2}\right]_{\pi / 2}^{3 \pi / 4}\)
= \(-\frac{1}{2}\left[\cos \pi-\cos \frac{\pi}{2}\right]+\frac{1}{2}\left[\cos \left(\frac{3 \pi}{2}\right)-\cos \pi\right]=-\frac{1}{2}[-1-0]+\frac{1}{2}[0-(-11)]=\frac{1}{2}+\frac{1}{2}(1)=1 \text { sq.units }\)

Question 7.
The area bounded by the curve y = x² – 4, and the lines y = 0 and y = 5 is
1) \(\frac{38}{3}\)
2) \(\frac{76}{3}\)
3) \(\frac{16}{3}\)
4) \(\frac{8}{3}\)
Solution:
2) \(\frac{76}{3}\)
Given y = x² – 4 ⇒ x² = y + 4 ⇒ x = \(\sqrt{y+4}\)
Required Area A = \(2\left[\int_0^5 \mathrm{xdx}\right]=2\left[\int_0^5 \sqrt{\mathrm{y}+4} \mathrm{dy}\right]=2\left[\frac{2}{3}(\mathrm{y}+4) \sqrt{\mathrm{y}+4}\right]_0^5\)
= \(\frac{4}{3}[9 \sqrt{9}-(4 \sqrt{4})]=\frac{4}{3}[27-8]=\frac{4 \times 19}{3}=\frac{76}{3} \text { sq.units }\)

Question 8.
The area of the region bounded by parabola y² = 8x and latus rectum is
1) \(\frac{4}{3}\)
2) \(\frac{16}{3}\)
3) \(\frac{32}{3}\)
4) \(\frac{8}{3}\)
Solution:
3) \(\frac{32}{3}\)
y² = 8x y = \(\sqrt{8 x}=2 \sqrt{2 x}\)
Area = \(2 \int_0^2(y) d x=2\left[\int_0^2 2 \times 2 \sqrt{x} d x\right]=2 \times 2 \sqrt{2}\left(\frac{2}{3} x \sqrt{x}\right)_0^2=\frac{8 \sqrt{2}}{3}(2 \sqrt{2}-0)=\frac{16 \times 2}{3}=\frac{32}{3} \text { sq.units }\)

Question 9.
The area bounded by the curve y = 2x – x² and the line y = -x is
1) \(\frac{7}{2}\)
2) 7
3) \(\frac{9}{2}\)
4) 9
Solution:
3) \(\frac{9}{2}\)
Given y = 2x – x² ………..(1) (Upper curve) y = -x …….(2) (Lower curve)
Solving (1) and (2)
-x = 2x – x² ⇒ x² – x – 2x = 0 ⇒ x² – 3x = 0 ⇒ x(x – 3) = 0 ⇒ x = 0 x = 3
Area = \(\int_0^3\left(2 x-x^2\right)-(-x) d x=\int_0^3\left(3 x-x^2\right) d x=\left(3 \frac{x^2}{2}-\frac{x^3}{3}\right)_0^3\)
= \(\frac{3}{2} \times 9-\frac{27}{3}-(0)=\frac{27}{2}-\frac{27}{3}=27\left(\frac{1}{6}\right)=\frac{9}{2} \text { sq.units }\)

Question 10.
The area enclosed between the graph of y = x³ and the lines x = 0, y = 1, y = 8 is
1) 7
2) 14
3) \(\frac{45}{4}\)
4) \(\frac{54}{4}\)
Solution:
3) \(\frac{45}{4}\)
y = x³
x = 0 (y-axis), y = 1 , y =8
A = \(\int_1^8(x) d x=\int_1^8 y^{\frac{1}{3}} d x=\left(\frac{y^{\frac{1}{3}}+1}{\frac{1}{3}+1}\right)_1^8=\frac{3}{4}\left(y^{\frac{4}{3}}\right)_1^8=\frac{3}{4}\left[2^4-1\right]=\frac{3 \times 15}{4}=\frac{45}{4} \text { sq.units }\)

Question 11.
The area of the region bounded by the curve y² = x, the Y-axis and between y = 2 and y = 12.
1) \(\frac{52}{2}\)
2) \(\frac{54}{3}\)
3) \(\frac{56}{3}\)
4) \(\frac{58}{3}\)
Solution:
3) \(\frac{56}{3}\)
y² = x; y-axis(x = 0), y = 2, y = 4
Area = \(\int_2^4(x) d x=\int_2^4 y^2 d x=\left[\frac{y^3}{3}\right]_2^4=\frac{1}{3}[64-8]=\frac{1}{3}[56]=\frac{56}{3} \text { sq.units }\)

Question 12.
Area of the region bounded by the curve y = cos x between x – 0 and x = π and the X-axis is
1) 1
2) 2
3) 3
4) 4
Solution:
2) 2
y = cos x, x = 0 (y-axis), x = π, x-axis (y = 0)
Required Area = \(2 \int_0^{\pi / 2}(y) d x=2 \int_0^{\pi / 2} \cos x d x=2[\sin x]_0^{\pi / 2}=2\left[\sin 90^{\circ}-\sin 0^{\circ}\right]=2[1-0]=2\)

Question 13.
Area of the region bounded by the curve x = 2y + 3, the Y-axis and between y = -1 and y = 1 is
1) 6
2) 4
3) 8
4) 3/2
Solution:
1) 6
Given x = 2y + 3
Area A = \(\int_{-1}^1(x) d y=\int_{-1}^1(2 y+3) d y=\left(\frac{2 y^2}{2}+3 y\right)_{-1}^1\) = 1 + 3 – [1 – 3] = 4 – (-2) = 6 sq. units

Question 14.
The area bounded by the curve y = x³, X-axis and two ordinates x = 1 and x = 2 is
1) \(\frac{15}{2}\)
2) \(\frac{15}{4}\)
3) \(\frac{17}{2}\)
4) \(\frac{17}{4}\)
Solution:
2) \(\frac{15}{4}\)
y = x³ x – axis (y = 0) x = 1, x = 2
Area = \(\int_1^2(y) d x=\int_1^2 x^3 d x=\left(\frac{x^4}{4}\right)_1^2=\frac{16}{4}-\frac{1}{4}=\frac{15}{4} \text { sq.units }\)

Question 15.
The area bounded by the curves y² = 4x and y = x is equal to
1) \(\frac{1}{3}\)
2) \(\frac{8}{3}\)
3) \(\frac{35}{6}\)
4) \(\frac{7}{3}\)
Solution:
2) \(\frac{8}{3}\)
y² = 4x ⇒ y = 2\(\sqrt{x}\) …(1) (Upper curve) y = x ……(2) (Lower curve)
Solving (1) and (2) y² = 4y y(y – 4) = 0 y = 0; y = 4
Area = \(\int_1^4(2 \sqrt{x}-x) d x=2 \int_1^4 \sqrt{x} d x=\int_1^4 x d x=2 \frac{2}{3}(x \sqrt{x})_0^4-\left(\frac{x^4}{2}\right)_0^4=\frac{4}{3}[4 \sqrt{4}]-\frac{1}{2}\)
= \(\frac{32}{3}-\frac{16}{2}=\frac{32}{3}-8=\frac{32-24}{3}=\frac{8}{3} \text { sq.units }\)

Practice AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Application of Derivatives MCQ

I. Select the correct option from the given choices.

Question 1.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm
1) 10π
2) 12π
3) 8π
4) 11π
Solution:
2) 12π
Area of a circle A = πr²; Diff w.r.t ‘r’
Rate of change of Area = \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = ²(2r); Now, = \(\frac{\mathrm{dA}}{(\mathrm{dr})}\) = ²(2)(6) = 12² at r = 6

Question 2.
The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
1)116
2) 96
3) 90
4) 126
Solution:
4) 126
Revenue = R(x) = 3x² + 36x + 5; Marginal Revenue = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) = 3(2x) + 36
\(\begin{aligned}
&\frac{\mathrm{dR}}{\mathrm{dx}}\\
&\text { at } x=15
\end{aligned}\) = 6(15) + 36 = 90 + 36 = 126

Question 3.
Which of the following functions are decreasing on (0, \(\frac{\pi}{2}\))?
1) cos x
2) cos2x
3) cos3x
4) tanx
Solution:
1) cos x
Let f(x) = cosx. For decreasing interval f'(x) < 0 ⇒ -sin x < 0 ⇒ sin x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))

Question 4.
On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x – 1 is decreasing ?
1) (0, 1)
2) (\(\frac{\pi}{2}\), π)
3) (0, \(\frac{\pi}{2}\))
4) (-π, \(\frac{\pi}{2}\))
Solution:
4) (-π, \(\frac{\pi}{2}\))
Give f(x) = x¹⁰⁰ + sinx – 1 ⇒ f’ (x) = 100x⁹⁹ + cos x. For decreasing interval f'(x) < 0
check option
1) In (0, 1) = (0, radian) = (0,57°) f'(x) = 100x⁹⁹ + cos x > 0 (+ve)
2) In (\(\frac{\pi}{2}\), π), f'(x) = 100x⁹⁹ + cosx – a large+ve value + ve (∵ -1 ≥ cos + ve)
3) In (0, \(\frac{\pi}{2}\)), f'(x) = +ve + +ve (+ve);
4) In (-π, \(\frac{-\pi}{2}\)), f'(x) = -ve- = -ve < 0

Question 5.
In which interval y = x² e^-x is increases
1) (-∞, ∞)
2) (-2, 0)
3) (2, ∞)
4) (0, 2)
Solution:
4) (0, 2)
f(x) = x².e^-x ⇒ f'(x) = x²(-e^=x) + e^-x(2x)
For increasing interval f'(x) > 0 ⇒ e^-x(2x – x²) > 0
⇒ 2x – x² > 0 ∵ e^-x > 0 ∀x ∈ R ⇒ x² – 2x < 0 ⇒ x(x – 2) < 0 ⇒ x ∈ (0, 2)

Question 6.
On the curve x² = 2y which is nearest to the pojnt (0, 5) is
1) (\(2 \sqrt{2}\), 4)
2) (\(2 \sqrt{2}\), 0)
3) (0, 0)
4) (2, 2)
Solution:
1) (\(2 \sqrt{2}\), 4)
Let P(t, \(\frac{t^2}{2}\)) is a point on x² = 2y and A = (0, 5)
consider PA² = (t – 0)² + (\(\frac{t^2}{2}\) – 5)² …..(1) ⇒ PA² = f(x) = t² + (\(\frac{t^2}{2}\) – 5)²
For maxima (or) minimum f'(x) = 0 ⇒ 2t + 2(\(\frac{t^2}{2}\) – 5)\(\left[\frac{2 \mathrm{t}}{2}\right]\) = 0 ⇒ 2t + (t² – 10)t = 0
⇒ 2t + t³ – 10t = 0 ⇒ t³ – 8f = 0 ⇒ f(t² – 8) = 0 ⇒ t = 0 (or) t = \(\sqrt{8}=2 \sqrt{2}\)
From (1) at t = 0 ⇒ PA² = 0 + (-5)² = 25; at t = \(\sqrt{8}\) ⇒ PA² = 8 + (-1)² = 9 minimum
∴ at t = \(\sqrt{8}\) ⇒ P = (\(\sqrt{8}\), 4) = (2\(\sqrt{2}\), 4) is nearest

Question 7.
For all real values of x, the minimum value of \(\frac{1-x+x^2}{1+x+x^2}\) is
1) 0
2) 1
3) 3
4) 1/3
Solution:
4) 1/3
f(x) = \(\frac{1-x+x^2}{1+x+x^2} \Rightarrow f^{\prime}(x)=\frac{\left(1+x+x^2\right)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2}=\frac{2\left(x^2-1\right)}{\left(1+x+x^2\right)^2}\)
∴ f'(x) = 0 ⇒ 2(x² – 1) = 0 ⇒ x² = 1 ⇒ x = ±1
By second derivative test, f is the minimum at x = 1 and f(1) = \(\frac{1-1+1}{1+1+1}=\frac{1}{3}\)

Question 8.
The maximum value of |x(x – 1) + 1|^{\(\frac{1}{3}\)}, 0 ≤ x ≤ 1 is
1) \(\left(\frac{1}{3}\right)^{\frac{1}{3}}\)
2) \(\frac{1}{2}\)
3) 1
4) 0
Solution:
3) 1
y = f(x) = \([x(x-1)+1]^{\frac{1}{3}}=\left(x^2-x+1\right)^{\frac{1}{3}}=\left(\left(x-\frac{1}{2}\right)+\frac{3}{4}\right)^{\frac{1}{3}}\)
Since, extreme values (maximum (or) minimum) occurs at critical points (or) at the end of the interval. Solving, f'(x) = 0 we get x = \(\frac{1}{2}\) (critical calue)
∴ f_max = Max of {(f(0), f(1), f\(\left(\frac{1}{2}\right)\)} = Max of {1, 1, \(\left(\frac{3}{4}\right), \frac{1}{3}\)} ⇒ f_max = 1

Question 9.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
1) 1 m/h
2) 0.1 m/h
3) 1.1 m/h
4) 0.5 m/h
Solution:
1) 1 m/h
Given r = 10 = radius, \(\frac{d v}{d t}\) = 314, h = depth, \(\frac{d h}{d t}\) = ?
Volume = V = πr²h ⇒ V = π(100)h ⇒ V = (3.14)100h ⇒ V = (314)h
Diff w.r.t ‘f’ \(\frac{d v}{d t}\) = (314)\(\frac{d h}{d t}\) ⇒ (314) = (314)\(\frac{d h}{d t}\) ⇒ \(\frac{d h}{d t}\) = 1 ∴ \(\frac{d h}{d t}\) = 1 m/h

Question 10.
The function f(x) = x³ + 3x is increasing in interval
1) (-∞, 0)
2) (0, ∞)
3) R
4) (0, 1)
Solution:
3) R
f(x) = x³ + 3x
For increasing interval f'(x) > 0 ⇒ 3x² + 3 > 0 ∀x ∈ R

Question 11.
The interval in which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing
1) (-1, ∞)
2) (-2, -1)
3) (-0, -2)
4) (-1, 1)
Solution:
2) (-2, -1)
f(x) = 2x³ + 9x² + 12x – 1
For decreasing interval f'(x) < 0 ⇒ 2(3x²) + 9(2x) + 12 < 0
⇒ x² + 3x + 2 < 0 ⇒ (x + 1)(x + 2) < 0 x ∈ (-2, -1)

Question 12.
At which point the function f(x) = |x – 3| attains minimum value
1) x = 1
2) x < 3 3) x = 3 4) x > 3
Solution:
3) x = 3

y = f(x) = |x – 3| graph
clearly f(x) is maximum at x = 3

Question 13.
Minimum value of the function f(x) = |x – 2| + |x – 5| is
1) 1
2) 2
3) 3
4) 4
Solution:
3) 3
f(x) = |x – 2| + |x – 5| = |x – a| + |x – b|
Range of f(x) is [|a – b|, ∞) ⇒ f_Minimum = |a – b|
f_min = |2 – 5| = |3| = 3

Question 14.
The maximum value of is \(\frac{\log x}{x}\) is 0 < x < ∞ is
1) ∞
2) e
3) 1
4) e^-1
Solution:
4) e^-1
f(x) = \(\frac{\log x}{x} \Rightarrow f^{\prime}(x)=\frac{x\left(\frac{1}{x}\right)-\log x(1)}{x^2}=\frac{1-\log x}{x^2}\)
For maxima (or) Minima f'(x) = 0 ⇒ 1 – log x = 0 ⇒ log_e x = 1 ⇒ x = e
f_max at x = e = \(\frac{\log _{\mathrm{e}}}{\mathrm{e}}=\frac{1}{\mathrm{e}}=\mathrm{e}^{-1}\)

Question 15.
The minimum value of (x – α) (x – β) is
1) 0
2) αβ
3) \(\frac{1}{4}(\alpha-\beta)^2\)
4) \(\frac{-1}{4}(\alpha-\beta)^2\)
Solution:
4) \(\frac{-1}{4}(\alpha-\beta)^2\)
f(x) = (x – α)(x – β) = x²(α + β)x + αβ = ax² + bx + c
⇒ A = 1 (+ve)
⇒ f_min = \(\frac{4 a c-b^2}{4 a}=\frac{4(1)(\alpha \beta)-(\alpha+\beta)^2}{4}=\frac{-\left[(\alpha+\beta)^2+4 \alpha \beta\right]}{4}=-\left(\frac{(\alpha-\beta)^2}{4}\right) .\)

Practice AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Continuity and Differentiability MCQ

I. Select the correct option from the given choices.

Question 1.
Which of the following statement is true?
1) Every polynomial function is continuous
2) The function f(x) = 5x + 3 is continuous at x = 0
3) The function f(x) = |x| is continuous at x = 0
4) All of the options are correct
Solution:
4) All of the options are correct
By definition, all are correct

Question 2.
If f(x) = \(\begin{cases}3 a x-2 b, & x>1 \\ a x+b+1, & x<1\end{cases}\) and \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) f(x) exists.
Then the relation between a and b is
1) 3a – 2b = 1
2) 2a – 3b = 1
3) 2a + 3b = 1
4) 2a + 3b = 1
Solution:
2) 2a – 3b = 1
\(\underset{x \rightarrow 1}{\mathrm{Lim}}\) f(x) exists ⇒ LHL = RHL ⇒ \(\underset{{x \rightarrow 1-\\(x<1)}}{{Lim}}\) f(x) = \(\underset{{x \rightarrow 1+\\(x>1)}}{{Lim}}\) f(x)
⇒ a + b + 1 = 3a – 2b ⇒ 2a – 3b = 1

Question 3.
The function f(x) = \(\begin{cases}\frac{2}{5-x}, & x<3 \\ 5-x, & x \geq 3\end{cases}\) is
1) Left discontinuous at x = 3
2) Left continuous at x = 3
3) Right discontinuous at x = 5
4) Discontinuous at x = 5
Solution:
1) Left discontinuous at x = 3
At x = 3, f(3) = 5 – 3 = 2

LHL ≠ f(3) ⇒ f(x) is Left discontinuous at x = 3

Question 4.
If the function f(x) = \(\frac{\sqrt{1+x}-1}{x}\) is continuous at x = 0. Then f(0) =
1) \(-\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{1}{2}\)
4) \(-\frac{1}{3}\)
Solution:
3) \(\frac{1}{2}\)
f(x) is continuous at x = 0

Question 5.
If a function f(x) defined on [a, b] is discontinuous at x = α ∈ [a, b] . Then

Solution:
f(x) is discontinuous at x = α ⇒ \(\underset{x \rightarrow \alpha}{\mathrm{Lim}}\) f(x) ≠ f(x) [by definition]

Question 6.
If the function f defined by f(x) = \(\begin{cases}\cos x, & \text { if } x \leq 0 \\ 3 x+\alpha, & \text { if } 0<x<2 \\ \beta x+3, & \text { if } 2 \leq x \leq 4 \\ 11, & \text { if } x>4\end{cases}\)
where α,β are real constants is continuous on R. Then α² + β² =
1) 3
2) 9
3) 5
4) 4
Solution:
3) 5
Given f is continuouson R f is continuous at every real number.
Consider continuity of f(x) ar x = 0

⇒ cos 0° = 3(0) + α ⇒ 1 = 0 + α ⇒ α = 1
Now, consider continuity at x = 4

⇒ β(4) + 3 = 11 ⇒ 4β = 8 ⇒ β = 2 Now, α² + β² = 1² + 2² = 5

Question 7.
In the interval [0, 3]. The function f(x) = |x – 1| + |x – 2| is
1) discontinuous
2) differentiable
3) continuous but not differentiable at x = 2 only
4) continuous but not differentiable at x = 1 and x = 2.
Solution:
4) continuous but not differentiable at x = 1 and x = 2.

Graph f(x) = |x – 1| + |x – 2| is
f(x) is continuous on [0, 3]
bot not differentiable at x = 1 and x = 2 (turning points)

Question 8.
If y = \(\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\sqrt{\mathbf{x}+\ldots . . \infty}}}\). Then \(\frac{d y}{d x}\) is equal to
1) \(\frac{1}{y}\)
2) \(\frac{1}{x}\)
3) \(\frac{1}{2x-1}\)
4) \(\frac{1}{2y-1}\)
Solution:
4) \(\frac{1}{2y-1}\)
Formula: If y = \(\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+\ldots}}}\) ∞, then \(\frac{d y}{d x}=\frac{f^{\prime}(x)}{2 y-1}\)
Given f(x) = x ⇒ f'(x) = 1 ∴ \(\frac{d y}{d x}=\frac{1}{2 y-1}\)

Question 9.
The set of all points where the function f(x) = 2x|x| is differentiable is
1) (-∞, ∞)
2) (-∞, 0) ∪ (0, ∞)
3) (0, ∞)
4) (-∞, 0)
Solution:
1) (-∞, ∞)

f(x) = 2x |x| = \(\begin{cases}-2 x^2 & \forall x \leq 0 \\ 2 x^2 & \forall x>0\end{cases}\)
f'(x) = \(\left\{\begin{aligned}
-4 \mathrm{x} & \forall \mathrm{x} \leq 0 \\
4 \mathrm{x} & \forall \mathrm{x}>0
\end{aligned}\right.\) exists ∀x ∈ R ⇒ f(x) is differentiable ∀x ∈ R ⇒ x ∈ (-∞, ∞)

Question 10.
Differentiation of (x² – 5x + 8) (x³ + 7x + 9) can be done
1) only by using product rule
2) only by obtaining a single polynomial expanding it
3) only by using logarithmic differentiation
4) All of the options are correct
Solution:
4) All of the options are correct
All are correct.

Question 11.
If y = cos^-1(cos x) the find \(\frac{d y}{d x}\) at x = \(\frac{5 \pi}{4}\)
1) 1
2) -1
3) 0
4) \(-\frac{1}{\sqrt{2}}\)
Solution:
2) -1
\(\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}\)
y = \(\cos ^{-1}(\cos x) \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-(\cos x)^2}} \cdot \frac{d}{d x}(\cos x)=\frac{(-1)(-\sin x)}{\sqrt{-1(\cos x)^2}}=\frac{\sin x}{\sqrt{1-(\cos x)^2}}\)

Question 12.
If f(x) = x⁴ – x³ + 7x² + 14, then what is the value of f¹(5)?
1) 594
2) 549
3) 954
4) 495
Solution:
4) 495
f(x) = x⁴ – x³ + 7x² + 14 ⇒ f'(x) = 4x³ – 3x² + 14x
at x = 5; f'(5) = 4(5)³ – 3(5)² + 14(5) = 4(125) – 3 × 25 + 70 = 500 – 75 + 70 = 495

Question 13.
If y = x + \(\frac{1}{\mathbf{x}}\) then which among the following holds?
1) x²y¹ + xy = 0
2) x²y¹ + xy + 2 = 0
3) x²y¹ – xy + 2 = 0
4) x²y¹ + xy – 2 = 0
Solution:
3) x²y¹ – xy + 2 = 0
Given y = x + \(\frac{1}{x}\) ..(1); y = 1 – \(\frac{1}{x^2}\)
⇒ x²y¹ = x² – 1 …….(2) ⇒ x²y¹ – ⇒ x² + 1 = 0 x²y¹ – [xy – 1] + 1 = 0
⇒ x²y¹ – xy + 1 + 1 = 0 ⇒ x²y¹ – xy + 2 = 0

Question 14.
\(\frac{d}{d x}\left(e^{\log _e \sqrt{1+\tan ^2 x}}\right)\) when x ∈ Q₁
1) sec²(x) tan x
2) sec x tan²(x)
3) sec x tan x
4) tan² (x)
Solution:
3) sec x tan x
Given y = \(e^{\log _e \sqrt{1+\tan ^2 x}}\) [∵ e^logN_e = N]
y = \(\sqrt{1+\tan ^2 x}\) = sec x ∴ \(\frac{d y}{d x}=\frac{d}{d x}(\sec x)\)= sec x tan x

Question 15.
If y = log(cosh x) then \(\frac{d^2 y}{d x^2}\) =
1) sech² x
2) -sech² x
3) sinh x
4) -sinh x
Solution:
1) sech² x
y = log(cosh x)
⇒ \(\frac{d y}{d x}=\frac{1}{\cosh x}(\sinh x)=\tanh x \Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tanh x) \Rightarrow \frac{d^2 y}{d x^2}=\operatorname{sech}^2 x\)

Question 16.
If f(x) = \(\begin{cases}\frac{\sin ^2(a x)}{x^2} ; & x \neq 0 \\ 1 ; & x=0\end{cases}\) is continuous at x = 0, then the value of ‘a’ is
1) -1
2) 1
3) 0
4) ±1
Solution:
4) ±1

Question 17.
If y = sinh^-1\(\left[\frac{1-\mathbf{x}}{1+\mathbf{x}}\right]\). Then \(\frac{d y}{d x}\) is equal to
1) \(\frac{-\sqrt{2}}{(1+x) \sqrt{1+x^2}}\)
2) \(\frac{-1}{(1+x) \sqrt{x}}\)
3) \(\frac{1}{\left(1+x^2\right) \sqrt{1+x}}\)
4) \(\frac{\sqrt{2}}{1+x \sqrt{1-x^2}}\)
Solution:
1) \(\frac{-\sqrt{2}}{(1+x) \sqrt{1+x^2}}\)
Formula: \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=\frac{1}{\sqrt{x^2+1}}\)
Given y = \(\sinh ^{-1}\left[\frac{1-x}{1+x}\right] \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{\left(\frac{1-x}{1+x}\right)^2+\frac{1}{1}}} \cdot \frac{d}{d x}\left(\frac{1-x}{1+x}\right)\)
= \(\frac{1+x}{\sqrt{(1-x)^2+(1+x)^2}}\left[\frac{(1+x)[-1]-[(1-x)(1)]}{(1+x)^2}\right]\)
= \(\frac{-1-x-1+x}{\sqrt{2\left(1^2+x^2\right)}(1+x)}=\frac{-2}{\sqrt{2} \sqrt{1+x^2}(1+x)}=\frac{-2}{\sqrt{1+x^2}(1+x)}\)

Question 18.
[x] represents the greatest integer function of x. At x = \(-1 \frac{\mathrm{~d}}{\mathrm{dx}}(\sin \pi|\mathrm{x}|)\) =
1) 0
2) 2
3) -2
4) 1/2
Solution:
1) 0
Let y = sin π[x] \(\frac{\mathrm{d}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\sin \pi[\mathrm{x}]]=\frac{\mathrm{d}}{\mathrm{dx}}[\sin (\mathrm{n} \pi)]=\frac{\mathrm{d}}{\mathrm{dx}}(0)=0\)
where n = [x] = An integer ∈ Z ∀x ∈ R
G.S of θ = nπ ∀n ∈ Z ⇒ sin (nπ) = 0 ∀n ∈ Z

Question 19.
If 3.sin(xy) + 4.cos(xy) = 5, then \(\frac{d y}{d x}\) is equal to
1) \(\frac{3 \sin x y+4 \cos x y}{3 \cos x y-4 \sin x y}\)
2) \(\frac{3 \cos x y+4 \sin x y}{4 \cos x y-3 \sin x y}\)
3) \(\frac{-y}{x}\)
4) \(\frac{x}{y}\)
Solution:
3) \(\frac{-y}{x}\)

Given 3 sin(xy) + 4 cos(xy) = 5
⇒ \(\frac{3}{5}\)sin (xy) + \(\frac{4}{5}\)cos(xy) = \(\frac{5}{5}\) ⇒ sin (xy)cos α + cos (xy)sin α = 1
⇒ sin(xy + α) = sin 90° ⇒ xy = \(\frac{\pi}{2}\) – α = A constant
Diff. w.r.t x
⇒ \(x \frac{d y}{d x}+y(1)=0 \Rightarrow x \frac{d y}{d x}=-y\)
\(\frac{d y}{d x}=-\frac{y}{x}\)

Question 20.
If y = log^x_y then \(\frac{d y}{d x}\) is equal to
1) \(\frac{1}{x \log y}\)
2) \(\frac{\log y}{x(1+\log y)}\)
3) \(\frac{1}{x(1+\log y)}\)
4) \(\frac{1}{1+\log y}\)
Solution:
3) \(\frac{1}{x(1+\log y)}\)
Formula: \(\log _{\mathrm{b}}^{\mathrm{a}}=\frac{\log \mathrm{a}}{\log \mathrm{~b}}, \frac{\mathrm{~d}}{\mathrm{dx}}(\mathrm{U} \cdot \mathrm{~V})=\mathrm{U} \cdot \frac{\mathrm{dU}}{\mathrm{dx}}+\mathrm{V} \frac{\mathrm{dV}}{\mathrm{dx}}\)
Given y = \(\log _y^x \Rightarrow y=\frac{\log x}{\log y} \Rightarrow y \cdot(\log y)=\log x\)
⇒ \(y\left(\frac{1}{y} ; \frac{d y}{d x}\right)+(\log y) \frac{d y}{d x}=\frac{1}{x} \Rightarrow(1+\log y) \frac{d y}{d x}=\frac{1}{x} \Rightarrow \frac{d y}{d x}=\frac{1}{x(1+\log y)}\)

Practice AP Inter 2nd Year Maths Study Material Chapter 4 Determinants MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Determinants MCQ

Question 1.
If \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\), then x is equal to
1) 6
2) ±6
3) -6
4) 0
Solution:
2) ±6
G.E = x² – 36 = 36 – 36 ⇒ x² – 36 = 0 ⇒ x² = 36 ⇒ x = ±6

Question 2.
If A is 3 × 3 matrix and det (3A) = k (deta A), then k =
1) 9
2) 6
3) 1
4) 27
Solution:
4) 27
Given A_{3 × 3} ∴ |3A| = 3³|A| = 27 (det A) = k (det A) ⇒ k = 27

Question 3.
Value of k, which \(\) is singular
1) 4
2) -4
3) ±4
4) 0
Solution:
3) ±4
If A is singular |A| = 0 ⇒ \(\left|\begin{array}{ll}
\mathrm{k} & 2 \\
8 & \mathrm{k}
\end{array}\right|\) = 0 ⇒ k² – 16 = 0 ⇒ k = ±4

Question 4.
The area of a triangle with vertices (-3, 0), (0, 3) and (0, k) is 9 sq. units the value of k will be
1) 9
2) 3
3) -9
4) 6
Solution:
1) 9
Area of the triangle formed by (x₁, y₁) (x₂, y₂) (x₃, y₃) is
= \(\frac{1}{2}\left|\begin{array}{lll}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{array}\right|=9 \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|=9 \Rightarrow\left|\begin{array}{ccc}
1 & -3 & 0 \\
1 & 0 & 3 \\
1 & 0 & k
\end{array}\right|\) = 18 ⇒ 3|k – 3| = 18 ⇒ |k – 3| = 6
k – 3 = 6 ⇒ k = 9; k – 3 = -6 ⇒ k = -3

Question 5.
If A is square matrix of order 3 and |A| = -4, then |adj A| is equal to
1) -4
2) 4
3) -16
4) 16
Solution:
4) 16
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = (|A|)² = (-4)² = 16

Question 6.
If area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
1) 12
2) -2
3) -12, -2
4) 12, -2
Solution:
4) 12, -2
Area of triangle = 35
⇒ \(\frac{1}{2}\left|\begin{array}{ccc}
1 & 2 & -6 \\
1 & 5 & 4 \\
1 & \mathrm{k} & 4
\end{array}\right|\) = 35 ⇒ |1(20 – 4k) – 2(4 – 4) – 6(k – 5)| = 70
⇒ 20 – 4k + 0 – 6k + 30 = ±70 ⇒ 50 – 10k = ±70 ⇒ 5 – k = ±7
5 – k = 7 ⇒ k = -2; 5 – k = -7 ⇒ k = 12

Question 7.
If A = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and A_ij is Cofactors of a_ij then value of ∆ is given by
1) a₁₁A₃₁ + a₁₁A₃₂ + a₁₃A₃₃
2) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
3) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Solution:
4) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
∆ = Determinant of a mathix
= sum of the products of the elements of a row (or) column with the corresponding co-factors = (a₁₁)A₁₁ + (a₁₂)A₁₂ + (a₃₁)A₃₁ [using 1^st row]

Question 8.
Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
1) |A|
2) |A|²
3) |A|³
4) 3|A|
Solution:
2) |A|²
If A_n×n, then |AdjA| = |A|^n-1
We have A_3×3 ∴ |AdjA| = |A|²

Question 9.
If A is an invertible matrix of order 2, then det (A^-1) is equal to
1) det(A)
2) \(\frac{1}{{det}(\mathrm{~A})}\)
3) 1
4) 0
Solution:
2) \(\frac{1}{{det}(\mathrm{~A})}\)
A_2×2, and A, exists ∵ AA^-1 = A^-1A = I
Consider AA^-1 = I
|AA^-1| = |I| ⇒ |A||A^-1| = I ⇒ |A^-1| = \(\frac{1}{|\mathrm{~A}|}\) ⇒ det(A^-1| = \(\frac{1}{{det} A}\)

Question 10.
If x, y, z are nonzero real numbers, then the inverse of matrix A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
2) \(x y z\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
3) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}
\mathrm{x} & 0 & 0 \\
0 & \mathrm{y} & 0 \\
0 & 0 & \mathrm{z}
\end{array}\right]\)
4) \(\frac{1}{\mathrm{xyz}}\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
1) \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
If A = diag[a b c] then A^-1 = \(\frac{1}{|\mathrm{~A}|}\) AdjA = \(\left[\begin{array}{lll}
a^{-1} & b^{-1} & c^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
1 / a & 0 & 0 \\
0 & 1 / b & 0 \\
0 & 0 & 1 / c
\end{array}\right]\)
∴ A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) = diag [x y z]
⇒ A^-1 = diag\(\left[\begin{array}{lll}
x^{-1} & y^{-1} & z^{-1}
\end{array}\right]=\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)

Question 11.
Let A = \(\left[\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right]\), where 0 ≤ θ ≤ 2π. Then
1) Det(A) = 0
2) Det(A) ∈ (2, ∞)
3) Det(A) ∈ (2, 4)
4) Det(A) ∈ [2, 4]
Solution:
4) Det(A) ∈ [2, 4]
|A| = 1(1 + sin²θ) – sinθ[-sin θ + sin θ] + 1[sin² θ + 1]
= 1 + sin² θ – 0 + sin² θ + 1 = 2 + 2sin² θ
⇒ det A = 2 + 2sin² θ
∵ 0 ≤ sin² θ ≤ 1 ⇒ 0 ≤ 2sin² θ ≤ 2 ⇒ 0 + 2 ≤ (2 + 2sin² θ) ≤ 2 + 2 ⇒ 2 ≤ |A| ≤ 4
⇒ |A| ∈ [2, 4]