Question

If the function \[ f(x)=x^3+bx^2+ax \] satisfies the conditions of Rolle's theorem in \([1,3]\) with \[ c=2+\frac{1}{\sqrt3}, \] then \((a,b)=\)

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

Hint:

For Rolle's theorem problems, always use the two conditions: \[ f(a)=f(b) \] and \[ f'(c)=0. \] These usually provide enough equations to determine the unknown constants.

Answer 1

The Correct Option is B

Concept: If a function satisfies Rolle's theorem on \([1,3]\), then \[ f(1)=f(3) \] and there exists \[ c\in(1,3) \] such that \[ f'(c)=0. \] Use these two conditions to determine \(a\) and \(b\).

Step 1: Apply the condition \(f(1)=f(3)\). Given \[ f(x)=x^3+bx^2+ax. \] Now, \[ f(1)=1+b+a, \] \[ f(3)=27+9b+3a. \] Since \[ f(1)=f(3), \] \[ 1+b+a=27+9b+3a. \] \[ 26+8b+2a=0. \] \[ a+4b+13=0. \] \[ a=-4b-13. \]

Step 2: Use the condition \(f'(c)=0\). Differentiating, \[ f'(x)=3x^2+2bx+a. \] Given \[ c=2+\frac1{\sqrt3}. \] Therefore, \[ 3c^2+2bc+a=0. \] Now, \[ c^2 = \left(2+\frac1{\sqrt3}\right)^2 = 4+\frac{4}{\sqrt3}+\frac13 = \frac{13}{3}+\frac{4}{\sqrt3}. \] Hence, \[ 3c^2 = 13+4\sqrt3. \] Substituting, \[ 13+4\sqrt3 + 2b\left(2+\frac1{\sqrt3}\right) +a = 0. \] Using \[ a=-4b-13, \] \[ 13+4\sqrt3 + 4b+\frac{2b}{\sqrt3} -4b-13 = 0. \] \[ 4\sqrt3+\frac{2b}{\sqrt3}=0. \] Multiplying by \(\sqrt3\), \[ 12+2b=0. \] \[ b=-6. \]

Step 3: Find \(a\). Using \[ a=-4b-13, \] \[ a=-4(-6)-13. \] \[ a=24-13. \] \[ a=11. \] Therefore, \[ (a,b)=(11,-6). \] \[ \boxed{(a,b)=(11,-6)} \] \[ \boxed{\text{Answer = (B)}} \]