# Mark the correct alternative in the following:

Question:

Mark the correct alternative in the following:

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

A. $a \in\left(\frac{1}{2}, \infty\right)$

B. $a \in\left(-\frac{1}{2}, \frac{1}{2}\right)$

C. $a=\frac{1}{2}$

D. $a \in R$

Solution:

Formula:- (i) $a x^{2}+b x+c>0$ for all $x \Rightarrow a>0$ and $b^{2}-4 a c<0$

(ii) $a x^{2}+b x+c<0$ for all $x \Rightarrow a<0$ and $b^{2}-4 a c<0$

(iii) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)>0$ for all $x \in(a, b)$

Given:-

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

$d\left(\frac{f(x)}{d x}\right)=2 \sec ^{2} x+\frac{(2 a+1) \sec x \cdot \tan x}{\sec x}+(a-2)=f^{\prime}(x)$

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

$\Rightarrow f^{\prime}(x)=2\left(\tan ^{2}+1\right)+(2 a+1) \cdot \tan x+(a-2)$

$\Rightarrow f^{\prime}(x)=2 \tan ^{2} x+2 \operatorname{atan} x+\tan x+a$

For increasing function

$f^{\prime}(x)>0$

$\Rightarrow 2 \tan ^{2} x+2 \operatorname{atan} x+\tan x+a>0$

From formula (i)

$(2 a+1)^{2}-8 a<0$

$\Rightarrow 4\left(a-\frac{1}{2}\right)^{2}<0$

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