Step 1: Understanding the Concept:
We need to find the derivative of the given function $y$ with respect to $x$, evaluate it at $x = 1/2$, and set it equal to the given value $2\sqrt{3}$ to solve for the unknown constant $k$.
Step 2: Key Formula or Approach:
Use logarithm property: $\log_e(x^n) = n \log_e x$.
Derivatives:
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$
$\frac{d}{dx}(kx^2) = 2kx$
Step 3: Detailed Explanation:
The given function is:
\[ y = \log_e(x^3) + 3 \sin^{-1} x + kx^2 \]
First, simplify the logarithmic term using exponent properties:
\[ y = 3 \ln x + 3 \sin^{-1} x + kx^2 \]
Now, differentiate $y$ with respect to $x$:
\[ y' = \frac{dy}{dx} = 3\left(\frac{1}{x}\right) + 3\left(\frac{1}{\sqrt{1 - x^2}}\right) + k(2x) \]
\[ y' = \frac{3}{x} + \frac{3}{\sqrt{1 - x^2}} + 2kx \]
We are given that $y'\left(\frac{1}{2}\right) = 2\sqrt{3}$. Substitute $x = \frac{1}{2}$ into the derivative:
\[ y'\left(\frac{1}{2}\right) = \frac{3}{(1/2)} + \frac{3}{\sqrt{1 - (1/2)^2}} + 2k\left(\frac{1}{2}\right) \]
\[ 2\sqrt{3} = 6 + \frac{3}{\sqrt{1 - 1/4}} + k \]
\[ 2\sqrt{3} = 6 + \frac{3}{\sqrt{3/4}} + k \]
\[ 2\sqrt{3} = 6 + \frac{3}{\sqrt{3}/2} + k \]
\[ 2\sqrt{3} = 6 + \frac{6}{\sqrt{3}} + k \]
Rationalize the term $\frac{6}{\sqrt{3}}$:
\[ \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \]
Substitute this back:
\[ 2\sqrt{3} = 6 + 2\sqrt{3} + k \]
Subtract $2\sqrt{3}$ from both sides:
\[ 0 = 6 + k \]
\[ k = -6 \]
Step 4: Final Answer:
The value of $k$ is $-6$.