Question

Applying Lagrange's Mean Value Theorem for a suitable function $f\left(x\right)$ in $[0,h]$, we have $f\left(h\right)=f\left(0\right)+h{f}^{\prime }\left(\theta h\right),0<\theta <1$. Then for $f\left(x\right)=\cos x$, the value of $\underset{h\to 0^{+}}{lim}\theta$ is

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

We know that in a Lagrange mean value theorem there exist $c\in (a,b)$ such that
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$\therefore f^{\prime }\left(\theta h\right)=\frac{f\left(h\right)-\cos 0}{h-0}$
$\Rightarrow -\sin \left(\theta h\right)=\frac{\cos h-\cos 0}{h-0}$ $[\because f\left(x\right)=\cos x]$
$=\frac{\cos h-1}{h}$
$=\frac{(1-\frac{h^2}{2!})-1}{h}$ [neglecting higher power of $h$]
$\Rightarrow -\sin \left(\theta h\right)=\frac{-\frac{h^2}{2}}{h}$
$\Rightarrow \sin \left(\theta h\right)=\frac{\frac{h}{2}}{1}$
$\Rightarrow \theta h={\sin }^{-1}\left(\frac{h}{2}\right)$
$\Rightarrow \theta =\frac{{\sin }^{-1}\frac{h}{2}\times \frac{1}{2}}{h\times \frac{1}{2}}$
$\therefore \underset{h\to {\theta }^{+}}{lim}\theta =\frac{1}{2}\underset{h\to 0^{+}}{lim}\frac{{\sin }^{-1}\frac{h}{2}}{\frac{h}{2}}$
$=\frac{1}{2}\times 1=\frac{1}{2}$