Q124 / 200 · WBJEE 2014 · Mathematics
Applying Lagrange's Mean Value Theorem for a suitable function $f(x)$ in $[0, h],$ we have $f(h)=f(0)+h f^{\prime}(\theta h), \quad 0 < \theta < 1 . \quad$ Then, for $f(x)=\cos x,$ the value of $\lim _{h \rightarrow 0^{*}} \theta$ is
- A1
- B0
- Correct answerC$1 / 2$
- D$1 / 3$
Solution
We know that in a Lagrange mean value theorem there exist $c \in(a, b)$ such that
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
$$
$\therefore \quad f^{\prime}(\theta h)=\frac{f(h)-\cos 0}{h-0}$
$$
\begin{aligned}
\Rightarrow \quad-\sin (\theta h)=& \frac{\cos h-\cos 0}{h-0} \\
&[\because f(x)=\cos x] \\
=& \frac{\cos h-1}{h}
\end{aligned}
$$
$=\frac{\left(1-\frac{h^{2}}{2 !}\right)-1}{h}$
[neglecting higher power of $\Rightarrow \quad-\sin (\theta h)=\frac{-\frac{h^{2}}{2}}{h}$
$\Rightarrow \quad \sin (\theta h)=\frac{\frac{h}{2}}{1}$
$\Rightarrow \quad \theta h=\sin ^{-1}\left(\frac{h}{2}\right)$
$\Rightarrow \quad \theta=\frac{\sin ^{-1} \frac{h}{2} \times \frac{1}{2}}{h \times \frac{1}{2}}$
$\therefore \quad \lim _{h \rightarrow \theta^{+}} \theta=\frac{1}{2} \lim _{h \rightarrow 0^{+}} \frac{\sin ^{-1} \frac{h}{2}}{\frac{h}{2}}$
$=\frac{1}{2} \times 1=\frac{1}{2}$
Hence, the correct answer is option: 'C'.