Question

Verify Lagrange's mean value for the function $f\left(x\right)=\sin x$ in $[0,\frac{\pi }{2}]$

• A. No Lagrange's theorem is not applicable
• B. Yes Lagrange's theorem is applicable and $c={\cos }^{-1}\left(\frac{2}{\pi }\right)$
• C. Yes Lagrange's theorem is applicable and $c={\sin }^{-1}\left(\frac{4}{\pi }\right)$
• D. none of these

Answer 1

Answer: (B)

Yes Lagrange's theorem is applicable and $c={\cos }^{-1}\left(\frac{2}{\pi }\right)$

The function $f\left(x\right)=\sin x$ is continuous and differentiable for all $x\in R$.

In particular it is continuous in the closed interval $[0,\frac{\pi }{2}]$ and differentiable in the open interval $(0,\frac{\pi }{2})$ as is required for the application of lagrange's mean value theorem for it.

By lagrange's mean value theorem, there must exist at least one value of ${}^{\prime }{c}^{\prime }$ of $x$ lying in the open interval $(0,\frac{\pi }{2})$ such that

$\frac{f\left(\frac{\pi }{2}\right)-f\left(0\right)}{\left(\frac{\pi }{2}\right)-0}=f^{\prime }\left(c\right)$ ...(1)

Let us verify it, we have

$f\left(0\right)=\sin \left(0\right)=0;f\left(\frac{\pi }{2}\right)=\sin \left(\frac{\pi }{2}\right)=1$

Also $f^{\prime }\left(x\right)=\cos x$ gives $f^{\prime }\left(c\right)=\cos c$.

Putting these values in (1), we have

$\frac{1-0}{\frac{\pi }{2}}=\cos \left(c\right)\Rightarrow \cos \left(c\right)=\frac{2}{\pi }\Rightarrow c={\cos }^{-1}\left(\frac{2}{\pi }\right)$

Since $0<\frac{2}{\pi }<1,$ therefore the principal value of $c={\cos }^{-1}\left(\frac{2}{\pi }\right)$ lies in the open interval $(0,\frac{\pi }{2})$ and so is the required value of $c$

This verifies lagrange's mean value theorem