Question

Let (x) is a derivable function which is increasing for all $x\in R$ (having no critical point), then

	List I		List II
P.	$f^{\prime }\left(x\right)$ is	1.	decreasing function for all $x$
Q.	$f(3-4x)$ is	2.	increasing for $x>\frac{1}{2}$
R.	$f(x^2-x)$ is	3.	increasing function for all $x$
S.	$\left(f\right(x\left)\right\}{4}^3$ is	4.	cant say about increasing or decreasing

• A. P - 4, Q - 1, R - 3, S - 2
• B. P - 1, Q - 2, R - 3, S - 4
• C. P - 4, Q - 1, R - 2, S - 3
• D. P - 4, Q - 3, R - 2, S - 1

Answer 1

The correct option is C P - 4, Q - 1, R - 2, S - 3

$f\left(x\right)$ is a derivable function which is increasing for all $xϵR$

(Image)

P. $f^{\prime }\left(x\right)$ will give slope of $f\left(x\right)$ and $f^{\prime }\left(x\right)$ will be different at different points and slope can be either increasing or decreasing.

So cant say about increasing or decreasing.

Q. $f(3-4x)\Rightarrow$ Let $y=3-4x,x=\frac{3-y}{4}$ (a decreasing value)

Thus $f\left(y\right)=f(3-4x)$ decreasing for all x

R. $f(x^2-x)\Rightarrow$ Let $y=x^2-x,\frac{dy}{dx}=(2x-1)$ (increasing for $x>\frac{1}{2}$)

Thus $f(x^2-x)$ is increasing for $x>\frac{1}{2}$

S. $\left\{f\left(x\right)\right\}{4}^3\Rightarrow$ If $f\left(x\right)$ is increasing $k\left\{f\left(x\right)\right\}$ is also increasing

Therefore, $\left\{f\left(x\right)\right\}{4}^3$ is increasing for all x.