Question

If the function \(f(x) = ax^3+bx^2+11x-6\), defined on \([1,3]\), satisfies all the conditions of Rolle's theorem for \(c = 2+\frac{1}{\sqrt{3}}\), then

• A. \(a = -1,b = \frac{1}{2}\)
• B. \(a = 1,b = -6\)
• C. \(a = -1,b = 6\)
• D. \(a = -2,b = 1\)

Hint:

Use f(1) = f(3) and f'(c) = 0 for c = 2 + 1/sqrt3.

Answer 1

The Correct Option is B

Step 1: Use roots of f prime:
If $c=2+\tfrac1{\sqrt3}$ is a root of the quadratic $f'(x)=3ax^2+2bx+11$ with real coefficients, and the options give integers, the conjugate $2-\tfrac1{\sqrt3}$ is also a root.

Step 2: Use sum and product of roots:
Sum $=4=-\dfrac{2b}{3a}$ and product $=4-\dfrac13=\dfrac{11}{3}=\dfrac{11}{3a}$, so $a=1$. Then $b=-6$.

Step 3: Check f(1) = f(3):
$f(1)=1-6+11-6=0$ and $f(3)=27-54+33-6=0$. Both are 0, so all Rolle conditions hold. Option B.

Final Answer:
Roots of f prime are 2 plus or minus 1/sqrt3, which gives a = 1, b = -6. \[ \boxed{\text{(B) }a=1,\ b=-6} \]