Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Match List-I with List-II.

List-I (Function)

List-II (Interval in which function is increasing)

(A) $\frac{x}{\log_ex}$

(I) $(-∞,-2) (2,∞)$

(B) $\frac{x}{2}+\frac{2}{x},x≠0$

(II) $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$

(C) $x^x, x>0$

(III) $\left(\frac{1}{e},∞\right)$

(D) $\sin x – \cos t$

(IV) $(e, ∞)$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

(A) $f(x)=\frac{x}{\log_ex}$

$f'(x)=\frac{\log_ex-1}{(\log_ex)^2}$

for $f(x)$ to be increasing, $f'(x)>0$

$⇒\log_ex-1>0$

$⇒\log_ex>1$

$⇒x>e$

$⇒x∈(e, ∞)$ (IV)

(B) $f(x)=\frac{x}{2}+\frac{2}{x}=\frac{x^2+4}{2x}$

$f'(x)=\frac{(2x)(2x)-(x^2+4)(2)}{4x^2}$

$=\frac{4x^2-2x^2-8}{4x^2}=\frac{2x^2-8}{4x^2}$

$=\frac{x^2-4}{2x^2}$

$f'(x)>0$

$⇒x^2-4>0$

$⇒x>2,x<-2$

$⇒x∈(-∞,-2) (2,∞)$ (I)

(C) $f(x)=x^x$

$f'(x)=x^x+x^x.\log(x)$

$=x^x(1+\log (x))$

$⇒1+\log (x)>0$

$⇒\log (x)>-1$

$⇒x>\frac{1}{e}$

$⇒x∈\left(\frac{1}{e},∞\right)$ (III)

(D) $f(x)=\sin x-\cos t$

$f'(x)=\cos x$

$⇒\cos x>0$

$⇒x∈\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ (II)