Question

The value of c in Lagrange’s theorem for the function

ƒ(x) = log sin x in the interval $\left\lbrack \frac{\pi}{6},\frac{5\pi}{6} \right\rbrack$ is –

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{2\pi}{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi}{2}$

Answer 2

We have, ƒ(x) = log sin x

Ž ƒ¢(x) = $\frac{1}{\sin x}$ cos x = cot x

Clearly ƒ(x) is continuous and differentiable in $\left\lbrack \frac{\pi}{6},\frac{5\pi}{6} \right\rbrack$

Hence mean value theorem is applicable

\ There exist a real number c in $\left( \frac{\pi}{6},\frac{5\pi}{6} \right)$ such that

ƒ¢(3) = $\frac{ƒ\left( \frac{5\pi}{6} \right) - ƒ\left( \frac{\pi}{6} \right)}{\frac{5\pi}{6} - \frac{\pi}{6}}$

Ž cot c = $\frac{{logsin}\frac{5\pi}{6} - {logsin}\frac{\pi}{6}}{\frac{2}{3}\pi}$

Ž cot c = $\frac{\log\frac{1}{2} - \log\frac{1}{2}}{\frac{2\pi}{3}}$ = 0

Ž c = $\frac{\pi}{2}$Ī$\left( \frac{\pi}{6},\frac{5\pi}{6} \right)$.