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Integral from 0 to 1 of x f''(2x) dx given f(0), f(2) and f'(2) | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
If f(0)=1f\left(0\right) = 1, f(2)=3f\left(2\right) = 3, f′(2)=5f'\left(2\right) = 5 then the value of the definite integral ∫01x f′′(2x)dx\displaystyle\int_0^1x\,f''\left(2x\right)dx is
Solution
Answer: 2
Step 1: Integrating by parts with
u=x,dv=f′′(2x)dxu = x ,\qquad dv = f''\left(2x\right)dx
 the inner factor 22 has to be compensated when antidifferentiating:
v=f′(2x)2,sinceddx[f′(2x)2]=f′′(2x)v = \frac{f'\left(2x\right)}{2},\qquad\text{since}\qquad \frac{d}{dx}\left[\frac{f'\left(2x\right)}{2}\right] = f''\left(2x\right)
 Step 2:
∫01x f′′(2x)dx=[x f′(2x)2]01−∫01f′(2x)2 dx\int_0^1x\,f''\left(2x\right)dx = \left[\frac{x\,f'\left(2x\right)}{2}\right]_0^1-\int_0^1\frac{f'\left(2x\right)}{2}\,dx
 Halving again,
∫f′(2x)2dx=f(2x)4\int\frac{f'\left(2x\right)}{2}dx = \frac{f\left(2x\right)}{4}
 so
∫01x f′′(2x)dx=[x f′(2x)2−f(2x)4]01\int_0^1x\,f''\left(2x\right)dx = \left[\frac{x\,f'\left(2x\right)}{2}-\frac{f\left(2x\right)}{4}\right]_0^1
 Step 3: At the upper limit,
1⋅f′(2)2−f(2)4=52−34\frac{1\cdot f'\left(2\right)}{2}-\frac{f\left(2\right)}{4} = \frac52-\frac34
 Step 4: At the lower limit,
0⋅f′(0)2−f(0)4=0−14=−14\frac{0\cdot f'\left(0\right)}{2}-\frac{f\left(0\right)}{4} = 0-\frac14 = -\frac14
 f′(0)f'\left(0\right) is multiplied by x=0x = 0, so its value is never needed, which is why the question does not give it. Step 5:
(52−34)−(−14)=52−34+14=52−12=2\left(\frac52-\frac34\right)-\left(-\frac14\right) = \frac52-\frac34+\frac14 = \frac52-\frac12 = 2
 Answer: 22
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