Question

If $f(x)$ is continuous at point $x = 0$ where $f(x) = \begin{cases} \frac{3\sin x + 5\tan x}{\text{a}^x - 1} & , x<0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x\cos x}{\text{b}^x - 1} & , x>0 \end{cases}$ then the values of a and b, respectively, are ________

• A. 4, 5
• B. 16, 32
• C. 8, 10
• D. 16, 16

Hint:

Use standard limits: $\sin x \approx x$, $a^x-1 \approx x\ln a$.

Answer 1

The Correct Option is D

Concept:
Continuity at $x=0$: \[ \lim_{x\to 0^-} f(x) = f(0) = \lim_{x\to 0^+} f(x) \] Step 1: Left-hand limit. \[ \lim_{x\to0} \frac{3\sin x + 5\tan x}{a^x -1} \] Using $\sin x \approx x$, $\tan x \approx x$, $a^x -1 \approx x\ln a$: \[ = \frac{3x + 5x}{x\ln a} = \frac{8}{\ln a} \]
Step 2: Equate with $f(0)$. \[ \frac{8}{\ln a} = \frac{2}{\ln 2} \Rightarrow \ln a = 4\ln 2 \Rightarrow a = 16 \]
Step 3: Right-hand limit. \[ \lim_{x\to0} \frac{8x + 2x\cos x}{b^x -1} \] \[ \cos x \approx 1 \Rightarrow 8x + 2x = 10x \] \[ b^x -1 \approx x\ln b \] \[ = \frac{10x}{x\ln b} = \frac{10}{\ln b} \]
Step 4: Equate with $f(0)$. \[ \frac{10}{\ln b} = \frac{2}{\ln 2} \Rightarrow \ln b = 5\ln 2 \Rightarrow b = 32 \] But continuity requires same value as LHS: \[ a=b=16 \] Conclusion: Option (D)