Question

Match the following list

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

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

Answer 1

The correct option is A A - 5, B - 1, C - 4, D - 2

A) $f\left(x\right)=x^2-2x+5$
$\Rightarrow f^{\prime }\left(x\right)=2x-2$
For f(x) to be increasing f'(x)>0
$\Rightarrow x>1$
$\Rightarrow x\in (1,\infty )$
B) $f\left(x\right)=e^{-x}$
$\Rightarrow f^{\prime }\left(x\right)=-e^{-x}$
For f(x) to be increasing f'(x)>0
$\Rightarrow -e^{-x}>0$
$\Rightarrow e^{-x}<0$
There is no value of x satisfying the above inequality.
C)$f\left(x\right)={\log }_{e}x$
Clearly , f(x) is defined for x>0
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}$
For f(x) to be increasing f'(x)>0
$\Rightarrow \frac{1}{x}>0$
$\Rightarrow x>0$
$\Rightarrow x\in (0,\infty )$
D) $f\left(x\right)=\frac{x^3}{3}-3\frac{x^2}{2}+2x+5$
$\Rightarrow f^{\prime }\left(x\right)=x^2-3x+2$
For f(x) to be increasing f'(x)>0
$\Rightarrow (x-2)(x-1)>0$
$\Rightarrow x\in (-\infty ,1)\cup (2,\infty )$