Question

If the function \[ f(x)= \begin{cases} \dfrac{\sin x-\tan x}{x^{3}}, & -\dfrac{\pi}{2}[1ex] a, & x=0,[1ex] \dfrac{\sin (b-3)x+\sin bx}{\sin x}, & 0<x<\dfrac{\pi}{2}, \end{cases} \] is continuous in \[ \left(-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right), \] then \(a+2b=\)

• A. \(1\)
• B. \(2\)
• C. \(\dfrac32\)
• D. \(\dfrac54\)

Hint:

For continuity at a point, \[ \boxed{\text{LHL}=\text{Function Value}=\text{RHL}.} \] Useful limits: \[ \lim_{x\to0}\frac{\sin x}{x}=1, \] \[ \sin x=x-\frac{x^3}{6}+\cdots,\qquad \tan x=x+\frac{x^3}{3}+\cdots. \]

Answer 1

The Correct Option is B