Given f'(x) > 0 and g'(x) < 0 , , x R, then- - Vidyadip

MCQ

Given $f'(x) > 0$ and $g'(x) < 0\,\, \forall x \in R$, then-

A
$g(ƒ(|x| + 1)) > g(ƒ(|x| -1))$
B
$ƒ(ƒ(|x| + 1)) < ƒ(ƒ(|x|-1))$
✓
$g(g(|x| -1)) < g(g(|x|+ 1))$
D
$ƒ(g(|x| -1)) < ƒ(g(|x|+ 1))$

✓

Answer

Correct option: C.

$g(g(|x| -1)) < g(g(|x|+ 1))$

c
$\because|x|-1<|x|+1$

$\Rightarrow \mathrm{g}(|\mathrm{x}|-1)>\mathrm{g}(|\mathrm{x}|+1) $($\because$ $g$ is $ \downarrow $ ing)

$\Rightarrow \mathrm{g}(|\mathrm{x}|-1)<\mathrm{g}(\mathrm{g}|\mathrm{x}|+1)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The real value of $m$ for which the substitution, $y = u^m$ will transform the differential equation,$2x^4y \frac{{dy}}{{dx}} + y^4 = 4x^6$ into a homogeneous equation is :

Let $[\mathrm{t}]$ denote the greatest integer $\leq \mathrm{t}$. Then the value of $8 \cdot \int \limits_{-\frac{1}{2}}^{1}([2 x]+|x|) \,d x$ is .... .

If the line $y = \sqrt 3 x + k$ touches the circle ${x^2} + {y^2} = 16$, then $k =$

The probability that two randomly selected subsets of the set $\{1,2,3,4,5\}$ have exactly two elements in their intersection, is :

Let $Z$ and $W$ be complex numbers such that $\left| Z \right| = \left| W \right|,$ and arg $Z$ denotes the principal argument of $Z.$

Statement $1:$ If arg $Z+$ arg $W = \pi ,$ then $Z = -\overline W $.

Statement $2:$ $\left| Z \right| = \left| W \right|,$ implies arg $Z-$ arg $\overline W = \pi .$

The differential coefficient of ${\tan ^{ - 1}}\left( {{{\sqrt {1 + x} - \sqrt {1 - x} } \over {\sqrt {1 + x} + \sqrt {1 - x} }}} \right)$ is

${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $

Maximum value of sum of arithmetic progression $50, 48, 46, 44 ........$ is :-

If the function $f(x)=\left\{\begin{array}{ll}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi\end{array}\right.$ is twice differentiable, then the ordered pair $\left( k _{1}, k _{2}\right)$ is equal to

$\int_{\,1}^{\,x} {\frac{{\log {x^2}}}{x}\,dx = } $