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∫fgh over [0,2a] with h(x)+h(2a−x)=3 | JEE

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

Question
Let f(x), g(x), h(x)f(x),\,g(x),\,h(x) be continuous on [0,2a][0,2a] with f(2a−x)=f(x)f(2a-x)=f(x), g(2a−x)=g(x)g(2a-x)=g(x) and h(x)+h(2a−x)=3h(x)+h(2a-x)=3. Then ∫02af(x) g(x) h(x) dx\displaystyle\int_{0}^{2a}f(x)\,g(x)\,h(x)\,dx equals
A∫02af(x) g(x) dx\displaystyle\int_{0}^{2a}f(x)\,g(x)\,dx
B3∫0af(x) g(x) dx3\displaystyle\int_{0}^{a}f(x)\,g(x)\,dxcorrect
C2∫0af(x) g(x) dx2\displaystyle\int_{0}^{a}f(x)\,g(x)\,dx
D∫0af(x) g(x) dx\displaystyle\int_{0}^{a}f(x)\,g(x)\,dx
Solution
Step 1: Apply the reflection x↦2a−xx\mapsto 2a-x to I=∫02af(x)g(x)h(x) dxI=\displaystyle\int_{0}^{2a}f(x)g(x)h(x)\,dx:
I=∫02af(2a−x) g(2a−x) h(2a−x) dx.I=\int_{0}^{2a}f(2a-x)\,g(2a-x)\,h(2a-x)\,dx.
 Step 2: Use the given symmetries f(2a−x)=f(x)f(2a-x)=f(x), g(2a−x)=g(x)g(2a-x)=g(x), and h(2a−x)=3−h(x)h(2a-x)=3-h(x):
I=∫02af(x) g(x) (3−h(x)) dx=3∫02af(x)g(x) dx−I.I=\int_{0}^{2a}f(x)\,g(x)\,\big(3-h(x)\big)\,dx=3\int_{0}^{2a}f(x)g(x)\,dx-I.
 Step 3: Solve for II:
2I=3∫02af(x)g(x) dx ⇒ I=32∫02af(x)g(x) dx.2I=3\int_{0}^{2a}f(x)g(x)\,dx\ \Rightarrow\ I=\dfrac32\int_{0}^{2a}f(x)g(x)\,dx.
 Step 4: Since ff and gg are each symmetric about x=ax=a, so is fgfg, giving ∫02afg dx=2∫0afg dx\displaystyle\int_{0}^{2a}f g\,dx=2\int_{0}^{a}fg\,dx. Therefore
I=32⋅2∫0af(x)g(x) dx=3∫0af(x)g(x) dx.I=\dfrac32\cdot 2\int_{0}^{a}f(x)g(x)\,dx=3\int_{0}^{a}f(x)g(x)\,dx.
 Correct answer: (2)
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