The function $f(x) = x(x+3)e^{-(\frac{1}{2})x}$ satisfies all the conditions of Rolle's theorem in $[-3,0]$, then $c =$

Question

If $f(x)$ and $g(x)$ are differentiable functions for $0 \leq x \leq 1$ such that, $f(1)-f(0) = k(g(1)-g(0))$, $k \neq 0$, and there exists a 'c' satisfying $0<c<1$. Then, the value of $\frac{f'(c)}{g'(c)}$ is equal to

Answer 1

Step 1: Approach
Use the product rule in a different arrangement, with the logarithmic derivative.

Step 2: Log derivative
$\dfrac{f'}{f}=\dfrac1x+\dfrac1{x+3}-\dfrac12$ for $f\ne0$. Setting this to zero:
\[ \frac{2x+3}{x(x+3)}=\frac12\Rightarrow4x+6=x^2+3x \]
\[ x^2-x-6=0 \]

Step 3: Roots
$x=-2$ or $x=3$. Only $-2$ is in $(-3,0)$. At $x=-2$, $f=(-2)(1)e^{1}\ne0$, so the division was valid. So $c=-2$, option (B).

Final Answer:
Setting f'(x) = 0 gives x = -2 or 3, and only -2 lies in the interval, option (B). \[ \boxed{-2} \]

The function \(f(x) = x(x+3)e^{-(\frac{1}{2})x}\) satisfies all the conditions of Rolle's theorem in \([-3,0]\), then \(c =\)

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The Correct Option is B

Solution and Explanation

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