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If f(x)={sin⁡(a−b)xx,x<02a+b,x=0−tan⁡12xbx,x>0f(x)=\begin{cases}\frac{\sin(a-b)x}{x}, & x<0\\ 2a+b, & x=0\\ \frac{-\tan 12x}{bx}, & x>0\end{cases}, (b>0)(b>0) is continuous at x=0x=0, then

Solution

✅ Correct Option: 4

Left limit: lim⁡x→0−sin⁡(a−b)xx=a−b\lim_{x\to0^-}\frac{\sin(a-b)x}{x}=a-b.

Right limit: lim⁡x→0+−tan⁡12xbx=−12b\lim_{x\to0^+}\frac{-\tan 12x}{bx}=-\frac{12}{b}.

Continuity needs a−b=2a+b=−12ba-b=2a+b=-\frac{12}{b} with b>0b>0.

For a=−4, b=2a=-4,\,b=2: a−b=−6a-b=-6, 2a+b=−62a+b=-6, −122=−6-\frac{12}{2}=-6, all equal.

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