Q241 / 270 · NDA 2022 (Phase 2) · Mathematics
Under which one of the following conditions does the function $f(x)=(p \sec x)^2+(q \operatorname{cosec} x)^2$ attain minimum value?
- Correct answerA$\tan ^2 x=\frac{q}{p}$
- B$\cot ^2 x=\frac{q}{p}$
- C$\tan ^2 x=p q$
- D$\cot ^2 x=p q$
Solution
$\begin{aligned} & f(x)=(P \sec x)^2+(q \operatorname{cosec} x)^2 \\ & f^{\prime}(x)=2 P^2 \sec ^2 x \tan x-2 q^2 \operatorname{cosec}^2 x \cdot \cot x=0 \\ & \Rightarrow \frac{p^2}{q^2}=\frac{\operatorname{cosec}^2 x \cdot \cot x}{\sec ^2 x \cdot \tan x}=\cot ^4 x \\ & \Rightarrow \cot ^2 x=\frac{p}{q} \Rightarrow \tan ^2 x=\frac{q}{p} \\ & f^{\prime}(x)=2 p^2\left(\tan x+\tan ^3 x\right)-2 q^2\left(\cot x+\cot ^3 x\right) \\ & f^{\prime \prime}(x)=2 p^2 \sec ^2 x\left(1+3 \tan ^2 x\right)+2 q^2 \operatorname{cosec}^2 x \\ & \left(1+3 \cot ^2 x\right) \\ & \text {For } \tan ^2 x=\frac{q}{p} \\ & f^{\prime \prime}(x) \gt 0 \\ & \therefore \quad f(x) \text { is minimum when } \tan ^2 x=\frac{q}{p}\end{aligned}$
Hence, the correct answer is option: 'A'.