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

Answers (1)

Correct option (c)

Hint: Iff(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} 

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