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

Step 1: Use continuity at \(x=0\). Since \(f(x)\) is continuous at \(x=0\), \[ \lim_{x\to0^-}f(x) = a = \lim_{x\to0^+}f(x). \]

Step 2: Evaluate the left-hand limit. Using the standard expansions, \[ \sin x=x-\frac{x^3}{6}+O(x^5), \] and \[ \tan x=x+\frac{x^3}{3}+O(x^5), \] we get \[ \sin x-\tan x = -\frac{x^3}{2}+O(x^5). \] Hence, \[ \lim_{x\to0^-}\frac{\sin x-\tan x}{x^3} = -\frac12. \] Therefore, \[ a=-\frac12. \]

Step 3: Evaluate the right-hand limit. As \(x\to0\), \[ \sin((b-3)x)\sim(b-3)x, \] \[ \sin(bx)\sim bx, \] and \[ \sin x\sim x. \] Hence, \[ \lim_{x\to0^+} \frac{\sin((b-3)x)+\sin(bx)}{\sin x} = (b-3)+b = 2b-3. \] By continuity, \[ 2b-3=a=-\frac12. \] Thus, \[ 2b=\frac52, \] \[ b=\frac54. \]

Step 4: Find \(a+2b\). \[ a+2b = -\frac12+\frac52 = 2. \] Hence, \[ \boxed{a+2b=2.} \] Therefore, the correct option is \(\boxed{(B)}\).