If the function f(x)=2 +(2a+1) _e| |+(a-2)x is increasing on R, t… - Vidyadip

MCQ

If the function $\text{f}(\text{x})=2\tan\text{x}+(2\text{a}+1)\log_\text{e}|\sec\text{x}|+(\text{a}-2)\text{x}$ is increasing on R, then:

A
$\text{a}\in\Big(\frac{1}{2},\infty\Big)$
B
$\text{a}\in\Big(-\frac{1}{2},\frac{1}{2}\Big)$
✓
$\text{a}=\frac{1}{2}$
D
$\text{a}\in\text{R}$

✓

Answer

Correct option: C.

$\text{a}=\frac{1}{2}$

$\text{f}(\text{x})=2\tan\text{x}+(2\text{a}+1)\log_\text{e}|\sec\text{x}|+(\text{a}-2)\text{x}$

If $\sec\text{x}>0$

$\Rightarrow\text{f}'(\text{x})=2\sec^2\text{x}+(2\text{a}+1)\frac{1}{\sec\text{x}}\sec\text{x}\tan\text{x}+(\text{a}-2)$

$\Rightarrow\text{f}'(\text{x})=2\sec^2\text{x}+(2\text{a}+1)\tan\text{x}+(\text{a}-2)$

Function is increasing

$=2\sec^2\text{x}+(2\text{a}+1)\tan\text{x}+(\text{a}-2)>0$

$\Rightarrow2\big(1+\tan^2\text{x}\big)+(2\text{a}+1)\tan\text{x}+(\text{a}-2)>0$

$\Rightarrow2\tan^2\text{x}+(2\text{a}+1)\tan\text{x}+\text{a}>0$

$\Rightarrow(2\text{a}+1)^2-4\times2\text{a}<0$

$\Rightarrow(2\text{a}-1)^2<0$ which is not possible.

$\Rightarrow(2\text{a}-1)^2=0$

$\Rightarrow\text{a}=\frac{1}{2}$

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