#### please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise Multiple choice question , question 8 maths textbook solution

Correct option (c)

Hint: If$f(x)$  is increasing function${f}'(x)>0$

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

Explanation: It is given that

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

Case i:

If $\sec x>0$

Now,Differentiate (i) w.r.t  x

\begin{aligned} &f^{\prime}(x)=2 \sec ^{2} x+(2 a+1) \frac{1}{\sec x} \sec x \cdot \tan x+(a-2) \\ &f^{\prime}(x)=2 \sec ^{2} x+(2 a+1) \tan x+(a-2) \end{aligned}

$\because f(x)$ is increasing, ${f}'(x)>0$

\begin{aligned} &\Rightarrow 2 \sec ^{2} x+(2 a+1) \tan x+(a-2)>0 \\ &\Rightarrow 2\left(\tan ^{2} x+1\right)+(2 a+1) \tan x+(a-2)>0 \end{aligned}

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

The above equation is quadratic in $\tan x$

Its discriminant is

$(2 a+1)^{2}-4 \times 2 a<0$

$\Rightarrow(2 a-1)^{2}<0$ , which is impossible.

Thus, if $\sec x <0$ then $\left | \sec x \right |=-\sec x$

$\therefore 2 \sec ^{2} x-(2 a+1) \tan x+(a-2) \geq 0$

$\Rightarrow(2 a-1)^{2} \leq 0$ , which is not possible

\begin{aligned} &\therefore(2 a-1)^{2}=0 \\ &\therefore a=\frac{1}{2} \end{aligned}