Question

Observe the following lists

List - I	List - II
(A) $f\left(x\right)=x+\frac{1}{x}$ is decreasing in	1) R
(B) $f\left(x\right)=1-x^3$ decreasing in is	2) $(0,\frac{1}{e})$
(C) $f\left(x\right)=x{e}^{-x}$ is	3) $(-1,0)\cup (0,1)$
(D) $f\left(x\right)=x^x$ is decreasing in	4) $(1,\infty )$
	5) $(0,\infty )$

• A. A -3, B - 4, C - 2, D - 5.
• B. A -2, B - 3, C - 5, D - 4.
• C. A -3, B - 1, C - 4, D - 2.
• D. A -1, B - 2, C - 3, D - 4.

Answer 1

The correct option is C A -3, B - 1, C - 4, D - 2.

(A) $f^{\prime }\left(x\right)=1-\frac{1}{x^2}$

$=\frac{x^2-1}{x^2}$

$f^{\prime }\left(x\right)<0$ If $\mid x\mid <1-0$

i.e. $|x|=x,$ if $1>x>0$ or $|x|=-x$ if $-1<x<0$

i.e. if $x\in (-1,0)\cup (0,1)$

So, A - 3

(B) $f^{\prime }\left(x\right)=-3x^2$

$f^{\prime }\left(x\right)<0$ $\forall x\in R$ as $x^2\ge 0\forall x\in R$

So, B - 1

(C) $f^{\prime }\left(x\right)=e^{-x}-x{e}^{-x}$

$=e^{-x}(1-x)$

$f^{\prime }\left(x\right)<0$ when $x>1$ i.e. $x\in (1,\infty )$

So, C - 4

(D) $f^{\prime }\left(x\right)=x^x(\log x+1)$

$0=\log x+1$

$x=\frac{1}{e}$

$f^{\prime }\left(x\right)<0$ when $0<x<\frac{1}{e}$

Thus, D - 2