Question

\( f(x)= \begin{cases} \frac{\sin^3(\sqrt{3}) \cdot \log(1+3x)}{(\tan^{-1}\sqrt{x})^2 (e^{5\sqrt{x}}-1)x}, & x\neq 0 \\ a, & x=0 \end{cases} \) is continuous in \( [0,1] \), then \( a \) equals to

• A. $0$
• B. $\frac{3}{5}$
• C. $2$
• D. $\frac{5}{3}$

Hint:

For continuity at 0, always use standard approximations like $\log(1+x)\approx x$.

Answer 1

The Correct Option is B

Concept: Continuity at $x=0$ requires: \[ \lim_{x \to 0} f(x) = f(0) = a \]

Step 1: Use small value approximations.
\[ \log(1+3x) \approx 3x \] \[ \tan^{-1}\sqrt{x} \approx \sqrt{x} \] \[ e^{5\sqrt{x}} -1 \approx 5\sqrt{x} \]

Step 2: Substitute approximations.
\[ f(x) \approx \frac{\sin^3(\sqrt{3}) \cdot 3x}{(x)(5\sqrt{x})x} \]

Step 3: Simplify expression.
\[ = \frac{3\sin^3(\sqrt{3})}{5} \]

Step 4: Use known value.
\[ \sin(\sqrt{3}) \approx 1 \Rightarrow \sin^3(\sqrt{3}) = 1 \]

Step 5: Final value.
\[ a = \frac{3}{5} \] Conclusion:
$a = \frac{3}{5}$