Q86 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If the function $f(x) = ax^2 + bx + \sin x$ satisfies all the conditions of Rolle's theorem on $[0, \pi]$ and the slope of the tangent to the curve $y = f(x)$ at $x = \dfrac{\pi}{4}$ is zero, then $a - b = $
- A$\dfrac{\sqrt{2}(1-\pi)}{\pi}$
- B$\dfrac{\sqrt{2}(2+\pi)}{\pi}$
- C$\dfrac{\sqrt{2}(\pi-1)}{\pi}$
- Correct answerD$\dfrac{\sqrt{2}(\pi+1)}{\pi}$
Solution
Since $f(x)$ satisfies Rolle's theorem on $[0, \pi]$, we have $f(0) = f(\pi)$.
$a(0)^2 + b(0) + \sin 0 = a\pi^2 + b\pi + \sin \pi$
$0 = a\pi^2 + b\pi$
$b = -a\pi$
The slope of the tangent at $x = \dfrac{\pi}{4}$ is zero, so $f'\left(\dfrac{\pi}{4}\right) = 0$.
$f'(x) = 2ax + b + \cos x$
$f'\left(\dfrac{\pi}{4}\right) = 2a\left(\dfrac{\pi}{4}\right) + b + \cos\left(\dfrac{\pi}{4}\right) = 0$
$\dfrac{a\pi}{2} + b + \dfrac{1}{\sqrt{2}} = 0$
Substituting $b = -a\pi$:
$\dfrac{a\pi}{2} - a\pi + \dfrac{1}{\sqrt{2}} = 0$
$-\dfrac{a\pi}{2} + \dfrac{1}{\sqrt{2}} = 0$
$a = \dfrac{\sqrt{2}}{\pi}$
Using $b = -a\pi$, we get $b = -\sqrt{2}$.
Therefore, $a - b = \dfrac{\sqrt{2}}{\pi} - (-\sqrt{2}) = \dfrac{\sqrt{2}(\pi+1)}{\pi}$.
Answer: $\dfrac{\sqrt{2}(\pi+1)}{\pi}$
$a(0)^2 + b(0) + \sin 0 = a\pi^2 + b\pi + \sin \pi$
$0 = a\pi^2 + b\pi$
$b = -a\pi$
The slope of the tangent at $x = \dfrac{\pi}{4}$ is zero, so $f'\left(\dfrac{\pi}{4}\right) = 0$.
$f'(x) = 2ax + b + \cos x$
$f'\left(\dfrac{\pi}{4}\right) = 2a\left(\dfrac{\pi}{4}\right) + b + \cos\left(\dfrac{\pi}{4}\right) = 0$
$\dfrac{a\pi}{2} + b + \dfrac{1}{\sqrt{2}} = 0$
Substituting $b = -a\pi$:
$\dfrac{a\pi}{2} - a\pi + \dfrac{1}{\sqrt{2}} = 0$
$-\dfrac{a\pi}{2} + \dfrac{1}{\sqrt{2}} = 0$
$a = \dfrac{\sqrt{2}}{\pi}$
Using $b = -a\pi$, we get $b = -\sqrt{2}$.
Therefore, $a - b = \dfrac{\sqrt{2}}{\pi} - (-\sqrt{2}) = \dfrac{\sqrt{2}(\pi+1)}{\pi}$.
Answer: $\dfrac{\sqrt{2}(\pi+1)}{\pi}$
Hence, the correct answer is option: 'D'.