Explore topic-wise InterviewSolutions in .

1. The more the merrier.

2. Sugar is bad for your health.

3. Eggs are sold by the dozen.

4. He plays the guitar so well.

5. I know French.

1. Delhi is our nation’s capital.

2. He was elected chairman of the group.

3. This is the least of his concerns.

4. I prefer to travel by plane.

5. He told me an interesting story.

1. Children like to play.

2. Pride has a fall.

3. I saw the drama in an auditorium.

4. He studies at a university.

5. She asked if I would be there in an hour.

1. The management arrived at a decision.

2. The train went through a tunnel.

3. We congratulate you on your success.

4. He is walking along the road to Bangalore.

5. The athletes ran over the hurdles.

Correct option is D) making

Correct option is C) passing

Correct option is B) otherwise

Correct option is B) degree

\(\int \limits _0 ^{\sqrt3 +2} \left[ \int \limits_0^{\frac \pi 3}(2\, cos \, \theta - 3 \, sin \, 3 \, \theta)d\theta\right]dr\)

= \(\int \limits_0^{\sqrt3 +2}dr \left[ 2\, sin \, \theta + cos \, 3 \theta \right]_0^{\frac \pi 3}\)

= \(\left[r\right]_0^{\sqrt3 +2} \)  ( 2 sin \(\frac \pi 3\) + cos \(\pi\) - 2 sin 0 - cos 0)

= (\(\sqrt3 \) + 2 -0) (  2 x \(\frac {\sqrt3}{2}\) -1 -1)

= (\(\sqrt 3\) +2) ( \(\sqrt 3\) -2)

= 3-4

= -1

Correct option is A) income

Correct option is B) salary

Correct option is C) wages

Correct option is C) grant

(C) have not succeeded

1. plenty

2. agreement

3. knight

4. swift

\(\int\limits_0^{2\pi}\)sin⁴x cos⁶x dx = 4\(\int\limits_0^{\pi/2}\)sin⁴x cos⁶x dx

\(=\cfrac{4\Gamma(\frac{4+1}2)\Gamma(\frac{6+1}2)}{2\Gamma(\frac{4+6+2}2)}\) 

\(=\cfrac{2\Gamma(\frac52)\Gamma(\frac72)}{\Gamma(6)}\)

\(=\frac{2\times3/2\times1/2\,\Gamma(1/2)\times5/2\times3/2\times1/2\,\Gamma(1/2)}{(5\times4\times3\times2\times1)}\) 

\(=\frac{3}{16\times8}\pi.\pi\) (\(\because\) \(\Gamma(1/2)=\sqrt{\pi}\))

\(=\frac{3\pi}{128}\)

Let I = \(\int {\cfrac{sin a \,cosa}{2cos^2a-1}}{da}\)

Put 2cos² a - 1 = t

= - 4 cos a sin a da = dt

= sin a cos a da = \(-\cfrac{dt}4\)

\(\therefore\) I = \(\cfrac{-1}4\int\)  \(\cfrac{dt}t\) = \(\cfrac{-1}4\) lnt + c

= \(\cfrac{-1}4\) ln |2cos² a - 1 | + c (\(\because\) t = 2 cos² a - 1)

Let I = \(\int\frac{(6x^2-6x+5)dx}{\sqrt{2x^3-3x^2+5x-7}}\)

Let 2x³ - 3x² + 5x - 7 = t

⇒ (6x² - 6x + 5)dx = dt

\(\therefore\) I = \(\int\frac{dt}{\sqrt t}=2t^{1/2} + c\)

Hence,  \(\int\frac{(6x^2-6x+5)dx}{\sqrt{2x^3-3x^2+5x-7}}\) 

 = \(2\sqrt{2x^3-3x^2+5x-7}+c\).

\(\int\limits_1^2[2x]dx\) = \(\int\limits_1^{3/2}[2x]dx\) = \(\int\limits_{3/2}^2[2x]dx\)

 = \(\int\limits_1^{3/2}2dx+\int\limits_{3/2}^2 3dx\) 

 = \(2[x]_1^{3/2}+3[x]^2_{3/2}\)

 = 2(32/ - 1) + 3(2 - 3/2)

 = 2 x 1/2 + 3 x 1/2 = 5/2

Correct option is D) into

(D) Present continuous

1. \(\int \frac{1 + cos^2x + tan^2x}{sin^2x}dx\)

\(=\int (cosec^2x + cot^2x+sec^2x)dx\)

\(=\int(2cosec^2x-1+sec^2x)dx\,\,\,\,\,\,\,\,\,(\because cot^2x = cosec^2x-1)\)

\(=2\int cosec^2x\,dx-\int dx + \int sec^2x dx\)

= -2 cot x - x + tan x +C

2. \(\int \frac{x^4-8x}{x -2 }dx\)

= \(\int \frac{(x-2)(x^3+2x^2+4x)}{x-2}dx\)

= \(\int (x^3+2x^2+4x)dx\)

= \(\frac{x^4}{4}+\frac{2x^3}{3}+\frac{4x^2}{2}+C\)

= \(\frac{x^4}{4}+\frac23x^3+2x^2 + C\)

3. Let I = \(\int(x+1)\sqrt{2-x}\,\,\,\,\,dx\)

Let 2 - x = f² 

⇒ x = 2 - f²

⇒ -dx = 2tdt

⇒ dx = -2tdt

\(\therefore\) I = \(-\int(2-t^2+1)t.2tdt\)

= \(-2\int(3t^2-t^4)dt\)

= \(-2[3\frac{t^3}3-\frac{t^5}5]+C\)

= \(-2(t^3- \frac{t^5}5)+C\)

= \(-2((2-x)^{\frac32}-\frac15(2-x)^{\frac52})+C\)

= \(-2(2-x)^{\frac32}-\frac25(2-x)^{\frac52}+C\)

\(\therefore\) \(\int(x+1)\sqrt{2-x}\,\,\,\,dx\)  = \(-2(2-x)^{\frac32}-\frac25(2-x)^{\frac52}+C\)

4. Let I = \(\int\frac{cos\theta}{\sqrt{1-sin\theta}}\,\,\,d\theta\)

Put 1 - sinθ = t²

⇒ - cosθ dθ = 2tdt

⇒ cosθ dθ = -2tdt

\(\therefore\) i = \(\int\frac{-2tdt}t= -2\int dt\)

= -2t + C

= \(-2\sqrt{1-sin\theta}+C\)

\(\therefore\) \(\int\frac{cos\theta}{\sqrt{1-sin\theta}}\,\,\,d\theta\) = \(-2\sqrt{1-sin\theta}+C\)

Correct option is D) to

\(\frac{dy}{dx}+\frac{y^2+1}{x^2+1}=0\)

⇒ \(\frac{dy}{y^2+1}+\frac{dx}{x^2+1}=0\)

⇒ tan^-1y + tan^-1x = c (On integrating both sides)

Correct option is D) in

Correct option is A) by

\(\int\frac{tan^{-1}x}{1+\frac1{x^2}}dx\) = \(\int\frac{x^2tan^{-1}x}{1+x^2}dx\)

put tan^-1 x = t

⇒ \(\frac1{1+x^2}dx=dt\)

\(\therefore\) \(\int\)\(\cfrac{tan^{-1}x}{1+\frac1{x^2}}dx\) = \(\int t\,tan^2t dt\) 

= t\(\int tan^2 t dt\) - \(\int 1.(tan t - t)dt\) 

 (\(\because\) \(\int\) tan² t dt = \(\int(sec^2t-1)dt= tan t - t\))

 = t(tan t - t) - log sec t + \(\frac{t^2}2+c\)

 = t tan t - \(\frac{t^2}2\) - log sec t + c

 = t tan t - \(\frac{t^2}2\) - log (\(\sqrt{1+tan^2t}+c\))

 = x tan^-1x - \(\frac12(tan^{-1}x)^2\) - log\(\sqrt {1+x^2}\) + c

Hence, \(\int\cfrac{tan^{-1}x}{1+\frac1{x^2}}dx\) = x tan^-1x - \(\frac12\) (tan^-1x)² - log\(\sqrt{1+x^2}+c\)

Correct option is C) is

Correct option is A) been

Correct option is C) would have to

a) Stress

Stress leads to Physical damage, illness, quality of life, mental and social problems.

Correct option is C) a tin of fruit

Correct option is B) Why

Correct option is C) Where

Correct option is C) about

\(\int\frac{\sqrt x}{\sqrt{1-x^2}}dx\)

Let 

1 - x² = t²

-2xdx = 2tdt

⇒ xdx = -tdt

∴ \(\int\frac{\sqrt x\,dx}{\sqrt{1-x^2}}=-\int\frac{tdt}{t\sqrt{1-t^2}}\)

\(= -\int\frac{dt}{\sqrt{1-t^2}}\)

= - sin^-1 t + C

= - sin^-1 \(\sqrt{1-x^2}\) + C

Correct option is B) about

Correct option is B) about 

integration dt|{a^2+(a-z)^2}^3/2 

let a-z=t ,-dz=dt 

integration -dt/(a^2+t^2)^3/2

integration -dt/t^3(a^2/t^2+1)^3/2

let a^2/t^2 =k , -2a^2)t^3 ×dt =dk 

dt/t^3 =dk/ -2a^2 

1/2a^2 integration dk/(k+1)^3/2

1/2a^2 { (k+1)^-3/2+1/-3/2+1}

1/2a^2 {(k+1)^-1/2 /-1/2 } +C 

1/2a^2 ×-2{1/(k+1)^1/2 } +C 

-1/a^2 {1/{a^2/(a-z)^2+1}1/2}+C 

-1/a^2{1/{a^2+(a-z)^2/(a-z)^2}1/2}

-1/a^2{ a-z/(a^2+(a-z)^2}1/2 }  +C answr 

Let I = \(\int\frac{dz}{[a^2+(a-z)^2]^{3/2}}\)

Put a - z = a tan θ

⇒ dz = -asec² θ dθ

\(\therefore\) I = \(-\int\frac{asec^2\theta d\theta}{(a^2+a^2tan^2\theta)^{3/2}}\) 

 = \(-\int\frac{asec^2\theta d\theta}{a^3(1+tan^2\theta)^{3/2}}\) 

 = \(-\frac1{a^2}\int\frac{sec^2\theta d\theta}{sec^3\theta}\)

( ∵ 1 + tan² \(\theta\) = sec² \(\theta\)) 

= -\(\frac1{a^2}\int\frac1{sec\theta}d\theta\)

= - \(\frac1{a^2}\int cos\theta d\theta\) 

= \(-\frac{sin\theta}{a^2}\) ( ∵ \(\int cos\theta=sin\theta\))

 = \(-\frac1{a^2}\frac{tan\theta}{\sqrt{1+tan^2\theta}}\) (∵ sin \(\theta\) = \(\frac{tan\theta}{\sqrt{1+tan^2\theta}}\))

 =  \(-\frac1{a^2}\cfrac{\frac{a-z}a}{\sqrt{1+\frac{(a-z)^2}{a^2}}}\) \((\because tan\theta=\frac{a-z}a)\)

\(=-\frac1{a^2}\frac{a-z}{\sqrt{a^2+(a-z)^2}}\)

Hence, \(\int\frac{dz}{[a^2+(a-z)^2]^{3/2}}\) = \(-\frac1{a^2}\frac{a-z}{\sqrt{a^2+(a-z)^2}}\)

\(\int\frac{3x+4}{\sqrt{5x^2+8}}dx\)

 = \(\int(\frac{3x}{\sqrt{5x^2+8}}+\frac4{\sqrt{5x^2+8}})dx\) 

 = \(\frac3{10}\int\frac{10x}{\sqrt{5x^2+8}}dx+\frac4{\sqrt 5}\int\frac{dx}{\sqrt{x^2+(\sqrt{\frac85})^2}}\)

 = \(\frac3{10}(2\sqrt{5x^2+8})\) + \(\frac4{\sqrt 5}log|x + \sqrt{x^2+\frac85}|+k\) 

 = \(\frac35\sqrt{{5x^2+8}}\)  + \(\frac4{\sqrt 5}log(\frac{\sqrt5x+\sqrt{5x^2+8}}{\sqrt5})+k\) 

 = \(\frac35\sqrt{{5x^2+8}}\) + \(\frac4{\sqrt 5}log(\sqrt 5x+\sqrt{5x^2+8})\) - \(\frac4{\sqrt5}log\sqrt5+k\) 

 = \(\frac35\sqrt{{5x^2+8}}\) + \(\frac4{\sqrt 5}log(\sqrt5x + \sqrt{5x^2+8})+c\)

\( I = \int\frac{2x}{(x^2 + 1)(x^2 + 2)^2}dx\)

Let x² = t

2x dx = dt

\(\therefore I = \int \frac{dt}{(t+1)(t+2)^2}\)

Let \(\frac{1}{(t + 1)(t + 2)^2} = \frac{A}{t+1}+ \frac{B}{t+2}+\frac{C}{(t+2)^2}\)

⇒ I = A (t + 2)² + B(t +1) (t + 2) + c(t +1)

Put t = -1, we get 

A = 1

Put t = -2, we get 

C = -1

Put t = 0, we get

4A + 2B + C = 1

⇒ 2B + 4 - 1 = 1

⇒ 2B = -2

⇒ B = -1

\(\therefore \frac{1}{(t +1)(t +2)^2}= \frac1{t+1}-\frac1{t+2}-\frac1{(t+2)^2}\)

\(\therefore I = \int \left(\frac{1}{t +1}- \frac1{t+ 2}-\frac1{(t + 2)^2}\right)dt = log(t +1) - log (t +2) + \frac1{t+2}+C\)

\(= log \left|\frac{t+1}{t+2}\right|+ \frac1 {t+2}+C \)

\(= \log\left|\frac{x^2 +1}{x^2 +2 }\right|+\frac{1}{x^2+2}+C\)

\(f^1(x) = \frac{e^x(x -1)}{x^2}\)

\(f(x) = \int\frac{xe^x - e^x}{x^2}dx +c\)

\(= \int e^x(\frac1x - \frac1{x^2})dx +c\)

\(=\frac{e^x}{x}+C\) 

\(\left(\therefore \int e^x (f(x) + f^1(x))dx = e^xf(x) + c \right)\) 

\(\left(Here\,f(x) = \frac1x⇒f^1(x) = \frac{-1}{x^2}\right)\)

\(\because f (1) = e +3\)

⇒ \(e + 3 = e + c\)

⇒ \(c =3\)

\(\therefore f(x) = \frac{e^x}{x} +3\)

\(\int\frac{2x+1}{\sqrt{x^2+10x+9}}\)dx

 = \(\int\frac{2x+10-9}{\sqrt{x^2+10x+9}}dx\)

 = \(\int\frac{2x+10}{\sqrt{x^2+10x+9}}dx\) - 9\(\int\frac{1}{\sqrt{(x+5)^2-16}}dx\)

 = 2\(\sqrt{x^2+10x+9}\) - 9log|x + 5 + \(\sqrt{x^2+10x+9}\)| + c

(C) It is the stored form of iron

(D) All of these

Correct option is C) at 

Correct option is E) conscious