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: Condition 1, f(1) = f(3):
\(f(1)=a+b+11-6=a+b+5\). \(f(3)=27a+9b+33-6=27a+9b+27\). Setting them equal gives \(26a+8b+22=0\), so \(13a+4b=-11\).

Step 2: Condition 2, f prime of c is 0:
\(f'(x)=3ax^2+2bx+11\). \(f'(c)=0\) at \(c=2+\tfrac{1}{\sqrt3}\), where \(c^2=4+\tfrac{4}{\sqrt3}+\tfrac13\).

Step 3: Test the options:
Option B: \(a=1,\ b=-6\). Check \(13-24=-11\) which is true. Then \(f'(x)=3x^2-12x+11\) with roots \(x=\dfrac{12\pm\sqrt{144-132}}{6}=2\pm\dfrac{1}{\sqrt3}\). So \(c=2+\tfrac{1}{\sqrt3}\) works and lies in \((1,3)\).

Step 4: Why the other options are wrong.
Option A satisfies \(13a+4b=-11\), but \(f'(x)=-3x^2+x+11\) has roots \(\dfrac{1\pm\sqrt{133}}{6}\), not \(2\pm\tfrac1{\sqrt3}\). Option C gives \(13a+4b=11\) and option D gives \(-22\), so both fail \(f(1)=f(3)\).

Final Answer:
a = 1 and b = -6. \[ \boxed{\text{(B) }a=1,\ b=-6} \]