If Rolle's theorem holds for the function $f(x)=x^{3}+px^{2}+qx+5,x\in [1,3]$ with $c=2+\frac{1}{\sqrt{3}}$, then the values of $p$ and $q$ are respectively...

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: Condition from the endpoints
$f(1)=f(3)$ gives $4p+q=-13$.

Step 2: Condition from c
The roots of $3x^2+2px+q=0$ are $2\pm\dfrac1{\sqrt3}$, so the equation is $3(x-2)^2-1=0$, i.e. $3x^2-12x+11=0$.

Step 3: Compare
Comparing $2p=-12$ and $q=11$ gives $p=-6$, $q=11$, which also satisfies $4p+q=-13$.

Final Answer:
The values are $p=-6$ and $q=11$. \[ \boxed{\text{(C) }-6,\,11} \]

If Rolle's theorem holds for the function \(f(x)=x^{3}+px^{2}+qx+5,x\in [1,3]\) with \(c=2+\frac{1}{\sqrt{3}}\), then the values of \(p\) and \(q\) are respectively...

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The Correct Option is C

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