Question

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...

• A. \(6,-11\)
• B. \(6,11\)
• C. \(-6,11\)
• D. \(-6,-11\)

Hint:

Use f(1) = f(3) and the fact that c is a root of f'(x).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Rolle's theorem needs \(f(1)=f(3)\) and some \(c\) in \((1,3)\) with \(f'(c)=0\).

Step 2: Use f(1) = f(3)
\(1+p+q+5=27+9p+3q+5\), so \(8p+2q=-26\), i.e. \(4p+q=-13\).

Step 3: Use f'(c) = 0
\(f'(x)=3x^2+2px+q\) has \(c=2+\dfrac1{\sqrt3}\) as a root.
Since the polynomial has real coefficients and the other root must be \(2-\dfrac1{\sqrt3}\) (sum of roots \(=4\), product \(=4-\dfrac13=\dfrac{11}{3}\)).

Step 4: Find p and q
Sum of roots \(=-\dfrac{2p}{3}=4\Rightarrow p=-6\).
Product \(=\dfrac q3=\dfrac{11}3\Rightarrow q=11\).
Check: \(4p+q=-24+11=-13\), which is correct.

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