If the function f(x) = ax^2 + bx + sin x sati | vedclass

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(D) $1$. Rolle's theorem requires $f(0) = f(\pi)$.$2$. $f(0) = a(0)^2 + b(0) + \sin(0) = 0$.$3$. $f(\pi) = a\pi^2 + b\pi + \sin(\pi) = a\pi^2 + b\pi =

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$\frac{\sqrt{2}(\pi + 1)}{\pi}$

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(D) $1$. Rolle's theorem requires $f(0) = f(\pi)$.$2$. $f(0) = a(0)^2 + b(0) + \sin(0) = 0$.$3$. $f(\pi) = a\pi^2 + b\pi + \sin(\pi) = a\pi^2 + b\pi = 0$. Since $\pi 
eq 0$, we have $a\pi + b = 0$, so $b = -a\pi$.$4$. The slope of the tangent is $f'(x) = 2ax + b + \cos x$.$5$. Given $f'(\frac{\pi}{4}) = 0$, so $2a(\frac{\pi}{4}) + b + \cos(\frac{\pi}{4}) = 0$.$6$. $\frac{a\pi}{2} + b + \frac{1}{\sqrt{2}} = 0$.$7$. Substitute $b = -a\pi$: $\frac{a\pi}{2} - a\pi + \frac{1}{\sqrt{2}} = 0 \implies -\frac{a\pi}{2} = -\frac{1}{\sqrt{2}} \implies a = \frac{2}{\pi\sqrt{2}} = \frac{\sqrt{2}}{\pi}$.$8$. Then $b = -a\pi = -\frac{\sqrt{2}}{\pi} \cdot \pi = -\sqrt{2}$.$9$. $a - b = \frac{\sqrt{2}}{\pi} - (-\sqrt{2}) = \frac{\sqrt{2} + \sqrt{2}\pi}{\pi} = \frac{\sqrt{2}(1 + \pi)}{\pi}$.

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 $x = \frac{\pi}{4}$ is zero, then $a - b = \dots$

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