Question

The least value of $5^{\sin x-1}$+ $5^{-\sin x-1}$ is

• A. $10$
• B. $\frac{5}{2}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{2}{5}$

In order to find the maximum or minimum of a function, we differentiate the function once and equate it to zero.

Given, $5^{\sin x-1}+5^{-\sin x-1}$

upon differentiating and equating to zero we get,

$5^{\sin x-1}\log 5\cos x+5^{-\sin x-1}\log 5(-\cos x)=0$

(since, $\frac{d}{dx}{a}^x=a^x\log a$)

$⟹\log 5\cos x(5^{\sin x-1}-5^{-\sin x-1})=0$

$\cos x=0$ (or) $5^{\sin x-1}-5^{-\sin x-1}=0$

$\cos x=0⟹x=\frac{\pi }{2}$

$5^{\sin x-1}-5^{-\sin x-1}=0⟹5^{\sin x-1}=5^{-\sin x-1}$

when bases are equal, we can equate the powers

$⟹\sin x-1=-\sin x-1⟹2\sin x=0$

$⟹\sin x=0⟹x=0$

Now, substituting $x=\frac{\pi }{2}$ in the function, we get

$5^{1-1}+5^{-1-1}⟹5^0+5^{-}2⟹1+\frac{1}{25}⟹\frac{26}{25}$

Now, substituting $x=0$ in the function, we get

$5^{0-1}+5^{-0-1}⟹5^{-}1+5^{-}1⟹\frac{1}{5}+\frac{1}{5}⟹\frac{2}{5}$

$\frac{2}{5}$ is the least value (since, $\frac{26}{25}>\frac{2}{5}$)