Match List-I with List-II.
Choose the correct answer from the options given below: |
(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) |
(A)-(IV), (B)-(I), (C)-(III), (D)-(II) |
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) |