Question

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

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

Answer 1

Answer: (C)

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