Question

Let $f\left(\theta \right)=\frac{1}{2}+\frac{2{\text{cosec}}^2\theta }{3}+\frac{3{\sec }^2\theta }{8}$.The least value of $f\left(\theta \right)$ for all permissible values of $\theta$ is-

• A. $\frac{31}{12}$
• B. $\frac{61}{48}$
• C. $\frac{61}{25}$
• D. $\frac{61}{24}$

Answer 1

Answer: (D)

$\frac{61}{24}$

Given $f\left(\theta \right)=\frac{1}{2}+\frac{2{\text{cosec}}^2\theta }{3}+\frac{3{\sec }^2\theta }{8}$.

For least value of $f\left(\theta \right)$, we must have $f^{\prime }\left(\theta \right)=0$.

Now $f^{\prime }\left(\theta \right)=-\frac{4{\text{cosec}}^2\theta \cot \theta }{3}+\frac{6{\sec }^2\theta \tan \theta }{8}$

Again $f^{\prime \prime }\left(\theta \right)=\frac{8{\text{cosec}}^2\theta {\cot }^2\theta +4{\text{cosec}}^4\theta }{3}+\frac{12{\sec }^2\theta {\tan }^2\theta +6{\sec }^4\theta }{8}$.

Now $f^{\prime }\left(\theta \right)=0$ we get,

$-\frac{4{\text{cosec}}^2\theta \cot \theta }{3}+\frac{6{\sec }^2\theta \tan \theta }{8}=0$

or, ${\cot }^4\theta =\frac{9}{16}$

or, $\cot \theta =\pm \frac{\sqrt{3}}{2}$.

First taking the positive value of $\cot \theta$.

$\cot \theta =\frac{\sqrt{3}}{2}\Rightarrow \text{cosec}\theta =\frac{\sqrt{7}}{2}$ and $\sec \theta =\frac{\sqrt{7}}{\sqrt{3}}$.

Using these values in $f^{\prime \prime }\left(\theta \right)$ then we will get the value of $f^{\prime \prime }\left(\theta \right)>0$.

So the value of $\cot \theta$ gives the least value of $f\left(\theta \right)$.

Then the least value of $f\left(\theta \right)$ will be

$\frac{1}{2}+\frac{7}{6}+\frac{7}{8}=\frac{61}{24}$.

The same value will be obtained when we use $\cot \theta =-\frac{\sqrt{3}}{2}$, and in this case also $f\left(\theta \right)$ will be least.