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

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: Check the Options
Test each option in $f'(x)\propto -x^{2}+x+6=-(x-3)(x+2)$.

Step 2: Zeros
The zeros are $x=3$ and $x=-2$. Of the options $0,-1,-2,-3$, only $-2$ is a zero.

Step 3: Conclusion
So $c=-2$, option (C).

Final Answer:
The point guaranteed by Rolle's theorem is $c=-2$. \[ \boxed{\text{(C) }-2} \]

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

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The Correct Option is C

Solution and Explanation

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