JEE Main 2015 (Online) 11th AprilQuestion 1
Let $f:\mathbb R\to\mathbb R$ be a function such that $f(2-x)=f(2+x)$ and $f(4-x)=f(4+x)$, for all $x\in\mathbb R$ and $\int_0^2f(x)\,dx=5$. Then the value of $\int_{10}^{50}f(x)\,dx$ is
Definite Integration
Solution
Concept. Symmetry about $x=2$ and about $x=4$ together make $f$ periodic with period $4$. Formulas used. (1) $f(2-x)=f(2+x)$ and $f(4-x)=f(4+x)$ $\Rightarrow$ $f(x+4)=f(x)$(2) For period $T$: $\int_a^{a+nT}f=n\int_0^Tf$Step 1 - Period. $f(x+4)=f(4-x)=f\big(2+(2-x)\big)=f\big(2-(2-x)\big)=f(x)$. So $f$ has period $4$. Step 2 - One period. By symmetry about $x=2$, $\int_2^4f=\int_0^2f=5$, so $\int_0^4f=10$. Step 3 - Count periods. $[10,50]$ has length $40=10\times4$. By (2), $$\int_{10}^{50}f(x)\,dx=10\cdot10=100.$$ Answer: (A)