Q87 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If the function $f(x) = ax^3 - bx^2 - 8x - 4$ satisfies Roll's theorem in $[1,3]$, if $f'(2) = 0$ then $a - b$ is equal to...
- Correct answerA$2$
- B$-2$
- C$0$
- D$1$
Solution
Since $f(x)$ satisfies Rolle's theorem in $[1, 3]$, we have $f(1) = f(3)$.
$a(1)^3 - b(1)^2 - 8(1) - 4 = a(3)^3 - b(3)^2 - 8(3) - 4$
$a - b - 12 = 27a - 9b - 28$
$26a - 8b = 16 \Rightarrow 13a - 4b = 8$
Also, $f'(x) = 3ax^2 - 2bx - 8$.
Given $f'(2) = 0$, we have:
$3a(2)^2 - 2b(2) - 8 = 0$
$12a - 4b = 8$
Subtracting the second equation from the first:
$(13a - 4b) - (12a - 4b) = 8 - 8$
$a = 0$
Substituting $a = 0$ into $12a - 4b = 8$:
$-4b = 8 \Rightarrow b = -2$
Therefore, $a - b = 0 - (-2) = 2$.
$a(1)^3 - b(1)^2 - 8(1) - 4 = a(3)^3 - b(3)^2 - 8(3) - 4$
$a - b - 12 = 27a - 9b - 28$
$26a - 8b = 16 \Rightarrow 13a - 4b = 8$
Also, $f'(x) = 3ax^2 - 2bx - 8$.
Given $f'(2) = 0$, we have:
$3a(2)^2 - 2b(2) - 8 = 0$
$12a - 4b = 8$
Subtracting the second equation from the first:
$(13a - 4b) - (12a - 4b) = 8 - 8$
$a = 0$
Substituting $a = 0$ into $12a - 4b = 8$:
$-4b = 8 \Rightarrow b = -2$
Therefore, $a - b = 0 - (-2) = 2$.
Hence, the correct answer is option: 'A'.