Question

If $y = \log_e x^3 + 3 \sin^{-1} x + kx^2$ and $y'\left(\frac{1}{2}\right) = 2\sqrt{3}$, then $k =$}

• A. $6$
• B. $-6$
• C. $2\sqrt{3}$
• D. $1$

Hint:

Use log properties ($ \log x^n = n \log x $) to simplify the expression before differentiating to avoid the chain rule.

Answer 1

The Correct Option is B

Step 1: Concept
Differentiate the function term by term. Note that $\log_e x^3 = 3 \log_e x$.

Step 2: Meaning
$y' = \frac{3}{x} + \frac{3}{\sqrt{1-x^2}} + 2kx$.

Step 3: Analysis
Substitute $x = 1/2$ and $y' = 2\sqrt{3}$: $2\sqrt{3} = \frac{3}{1/2} + \frac{3}{\sqrt{1-(1/2)^2}} + 2k(1/2)$. $2\sqrt{3} = 6 + \frac{3}{\sqrt{3/4}} + k$. $2\sqrt{3} = 6 + \frac{3 \cdot 2}{\sqrt{3}} + k \implies 2\sqrt{3} = 6 + 2\sqrt{3} + k$. $2\sqrt{3} - 2\sqrt{3} - 6 = k \implies k = -6$.

Step 4: Conclusion
The value of $k$ is $-6$. Final Answer: (B)