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

The Correct Option is D

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

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