Q87 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics

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

  • A
    $a = -1, b = \dfrac{1}{2}$
  • Correct answer
    B
    $a = 1, b = -6$
  • C
    $a = -1, b = 6$
  • D
    $a = -2, b = 1$

Solution

Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$, we must have $f(1) = f(3)$.

$f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5$

$f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9b + 27$

Equating $f(1)$ and $f(3)$:

$a + b + 5 = 27a + 9b + 27$

$26a + 8b + 22 = 0 \Rightarrow 13a + 4b + 11 = 0$

By Rolle's theorem, $f'(c) = 0$ for $c = 2 + \dfrac{1}{\sqrt{3}}$.

$f'(x) = 3ax^2 + 2bx + 11$

$f'\left(2 + \dfrac{1}{\sqrt{3}}\right) = 3a\left(2 + \dfrac{1}{\sqrt{3}}\right)^2 + 2b\left(2 + \dfrac{1}{\sqrt{3}}\right) + 11 = 0$

$3a\left(4 + \dfrac{1}{3} + \dfrac{4}{\sqrt{3}}\right) + 4b + \dfrac{2b}{\sqrt{3}} + 11 = 0$

$13a + 4\sqrt{3}a + 4b + \dfrac{2b}{\sqrt{3}} + 11 = 0$

$(13a + 4b + 11) + \dfrac{1}{\sqrt{3}}(12a + 2b) = 0$

Substituting $13a + 4b + 11 = 0$ into the above equation:

$\dfrac{1}{\sqrt{3}}(12a + 2b) = 0 \Rightarrow 12a + 2b = 0 \Rightarrow b = -6a$

Substituting $b = -6a$ into $13a + 4b + 11 = 0$:

$13a + 4(-6a) + 11 = 0$

$-11a + 11 = 0 \Rightarrow a = 1$

$b = -6(1) = -6$

Answer: $a = 1, b = -6$

Hence, the correct answer is option: 'B'.

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