Question

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

• A. \(2\)
• B. \(1\)
• C. \(-1\)
• D. \(0\)

Hint:

To make piecewise function continuous at 0, set \(a\) equal to \(\lim_{x\to 0}f(x)\). Use identities like \(\sin2x=2\sin x\cos x\).

Answer 1

The Correct Option is D

Step 1: Condition for continuity at \(0\).
We need:
\[ a=\lim_{x\to 0}\frac{2\sin x-\sin 2x}{2x\cos x} \]
Step 2: Simplify numerator using \(\sin2x=2\sin x\cos x\).
\[ 2\sin x-\sin2x=2\sin x-2\sin x\cos x \]
\[ =2\sin x(1-\cos x) \]
So limit becomes:
\[ \lim_{x\to 0}\frac{2\sin x(1-\cos x)}{2x\cos x} =\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x\cos x} \]
Step 3: Split into standard limits.
\[ =\lim_{x\to 0}\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{\cos x}\right) \]
\[ \frac{\sin x}{x}\to 1 \]
\[ 1-\cos x \to 0,\;\cos x\to 1 \Rightarrow \frac{1-\cos x}{\cos x}\to 0 \]
Thus:
\[ a=1\cdot 0=0 \]
Final Answer:
\[ \boxed{0} \]