Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

Match List-I woth List-II

List-I List-II
A $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}sin^7x\, dx$ I $\pi $
B $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(x\, cos \, x+1)dx$ II $\frac{\pi}{12}$
C $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{sinx}}{\sqrt{sin\, x}+\sqrt{cos\, x}}dx$ III 0
D $\int\limits^{\sqrt{3}}_{1}\frac{dx}{1+x^2}$ IV $\frac{\pi}{4}$

Choose the correct answer from the options given below :

Options:

A-II, B-III, C-IV, D-I

A-III, B-I, C-IV, D-II

A-I, B-II, C-III, D-IV

A-IV, B-I, C-II, D-III

Correct Answer:

A-III, B-I, C-IV, D-II

Explanation:

The correct answer is option (2) → A-III, B-I, C-IV, D-II

(A) $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}sin^7x\, dx=0$ (III)

(B) $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(x\, \cos x+1)dx$

(C) $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(1)

$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(2)

$⇒2I=\int\limits^{\frac{\pi}{2}}_{0}1dx⇒I=\frac{\pi}{4}$ (IV)

(D) $\int\limits^{\sqrt{3}}_{1}\frac{dx}{1+x^2}=\left[\tan^{-1}x\right]^{\sqrt{3}}_{1}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$ (II)