In the Lagrange's Mean Value Theorem, $f(b)-f(a)=(b-a)f'(c);\ (a\lt c\lt b)$. If $a=2,b=4$ and $f(x)=\sqrt{x^{2}-4}$, then the value of $c$ 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: Substitution check
Test $c=\sqrt{6}$: $f'(c)=\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}$.

Step 2: Compare
The average slope is $\frac{\sqrt{12}-0}{4-2}=\sqrt{3}$, which matches.

Step 3: Other options
For $\sqrt{2}$ the root is not defined, and $\sqrt{7},\sqrt{5}$ give $f'(c)=\sqrt{7/3}$ and $\sqrt{5}$, not $\sqrt{3}$.

Final Answer:
Only $\sqrt{6}$ satisfies the theorem. \[ \boxed{\text{(D) }\sqrt{6}} \]

In the Lagrange's Mean Value Theorem, \(f(b)-f(a)=(b-a)f'(c);\ (a\lt c\lt b)\). If \(a=2,b=4\) and \(f(x)=\sqrt{x^{2}-4}\), then the value of \(c\) is

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The Correct Option is D

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