Question

If $f: R \rightarrow R$ is defined by $f(x)=\begin{cases}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\\ a, & \text { if } x=0\end{cases}$ then the value of a so that f is continuous at $0$ is

• A. 2
• B. 1
• C. -1
• D. 0

Answer 1

Answer: (D)

0

Answer 2

$Given, f(x)=\begin{cases}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\\ a, & \text { if } x=0\end{cases}$

and f is continuous at x=0

$\therefore$ The left hand limit (LHL)

$= \displaystyle \lim_{x \to 0}f(x)$

$= \displaystyle \lim_{x \to 0} \frac{\cos3x-\cos x}{x^2}$

$= \displaystyle \lim_{h \to 0}\frac{\cos3(0-h)-\cos(0-h)}{0-h^2}$

$= \displaystyle \lim_{h \to 0}\frac{\cosh-\cosh}{h^2} (\frac00 form)$

$= \displaystyle \lim_{h \to 0}-\frac{3\sin3h+\sinh}{2h}$

(using L hospital's rule)
= $\frac{-9+1}{2}= -4$

As per the question f(x) is continuous at x=0

i,e, $= \displaystyle \lim_{x \to 0}f(x)$ = f(o)

⇒ -4 = λ

λ = -4