Question

If the function \(f(x) = ax^2+bx+sinx\) satisfies all the conditions of Rolle's theorem on \([0,π]\) and the slope of the tangent to the curve \(y = f(x)\) at \(x = \frac{π}{4}\) is zero, then \(a-b =\)

• A. \(\frac{\sqrt{2}(1-π)}{π}\)
• B. \(\frac{\sqrt{2}(2+π)}{π}\)
• C. \(\frac{\sqrt{2}(π-1)}{π}\)
• D. \(\frac{\sqrt{2}(π+1)}{π}\)

Hint:

Rolle's theorem needs f(0) = f(pi). Combine that with f'(pi/4) = 0 to find a and b.

Answer 1

The Correct Option is D

Step 1: Two equations, two unknowns:
Rolle: $a\pi^2 + b\pi = 0$, so $a\pi + b = 0$ (dividing by $\pi$).
Tangent: $\frac{a\pi}{2} + b + \frac{1}{\sqrt2} = 0$.

Step 2: Eliminate b:
Subtract the first from the second: $\frac{a\pi}{2} - a\pi + \frac{1}{\sqrt2} = 0$, so $a = \frac{2}{\sqrt2\,\pi} = \frac{\sqrt2}{\pi}$.
Then $b = -a\pi = -\sqrt2$.

Step 3: Compute:
$a - b = \frac{\sqrt2}{\pi} + \sqrt2 = \sqrt2\cdot\frac{1+\pi}{\pi}$.

Step 4: Match:
This is option (D). The others have $(1-\pi)$ or $(2+\pi)$, which do not follow from the two equations.

Final Answer:
Option (D). \[ \boxed{\frac{\sqrt2(\pi+1)}{\pi} \text{ (D)}} \]