Q102 / 160 · TS EAMCET 2008 · Mathematics
If $f: R \rightarrow R$ is defined by
$$
f(x)=\left\{\begin{array}{cc}
\frac{\cos 3 x-\cos x}{x^2}, & \text { for } x \neq 0 \\
\lambda & , \text { for } x=0
\end{array}\right.
$$
and if $f$ is continuous at $x=0$, then $\lambda$ is equal to
- A$-2$
- Correct answerB$-4$
- C$-6$
- D$-8$
Solution
Given that,
$$
f(x)=\left\{\begin{array}{cc}
\frac{\cos 3 x-\cos x}{x^2}, & \text { for } x \neq 0 \\
\lambda, & \text { for } x=0
\end{array}\right.
$$
Now,
$$
\begin{aligned}
\text { LHL } & =\lim _{x \rightarrow 0^{-}} f(x) \\
& =\lim _{x \rightarrow 0^{-}} \frac{\cos 3 x-\cos x}{x^2} \\
& =\lim _{h \rightarrow 0} \frac{\cos 3(0-h)-\cos (0-h)}{(0-h)^2} \\
& =\lim _{h \rightarrow 0} \frac{\cos 3 h-\cos h}{h^2} \\
& =\lim _{h \rightarrow 0} \frac{-3 \sin 3 h+\sin h}{2 h} \\
& =\lim _{h \rightarrow 0} \frac{-9 \cos 3 h+\cos h}{2} \\
& =\frac{-9+1}{2}=-4
\end{aligned}
$$
(using L' Hospital's rule)
Since, $f(x)$ is continuous at $x=0$
$$
\begin{array}{ll}
\therefore & \lim _{x \rightarrow 0^{-}} f(x)=f(0) \\
\Rightarrow & -4=\lambda \Rightarrow \lambda=-4
\end{array}
$$
Hence, the correct answer is option: 'B'.