Question

Let $f\left(x\right)$ be an even function in $R$. If $f\left(x\right)$ is monotonic increasing in $[2,6]$, then

• A. $f\left(3\right)>f(-5)$
• B. $f(-2)<f\left(2\right)$
• C. $f(-2)>f\left(2\right)$
• D. $f(-3)<f\left(5\right)$

Answer 1

The correct option is D $f(-3)<f\left(5\right)$

$f\left(x\right)$ is an even function
$\therefore f(-x)=f\left(x\right)⟶1$
Also, $f\left(x\right)$ is monotonically increasing in [2,6]
$\Rightarrow f\left(x_1\right)<f\left(x_2\right)$ Whenever $x_1<x_2$
$\left(A\right)f\left(3\right)>f(-5)$
$f(-5)=f\left(5\right)\therefore f\left(3\right)>f\left(5\right)$
But $3<5\therefore f\left(3\right)\ngtr f\left(5\right)$
$\left(B\right)f(-2)<f\left(2\right)$
$f(-2)=f\left(2\right)\therefore f\left(2\right)<f\left(2\right)$
But $2\nless 2\therefore f\left(2\right)\nless f\left(2\right)$
$\left(C\right)f(-2)>f\left(2\right)$ wrong
$\left(D\right)f(-3)<f\left(5\right)$
$f(-3)=f\left(3\right)$
$\therefore f\left(3\right)<f\left(5\right)$ which is true $\because 3<5$
$\therefore$ Option $\left(D\right)$ is correct