Mathematics · 2015
JEE Main · 11 April 2015, Shift 1 · Q75
Let f: R → R be a function such that f(2-x)=f(2+x) and f(4-x)=f(4+x), for all x ∈ R and ∫_0^2 f(x) d x=5. Then the value of ∫_10^50 f(x) d x is:
Let $\displaystyle f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that $\displaystyle f(2-x)=f(2+x)$ and $\displaystyle f(4-x)=f(4+x)$, for all $\displaystyle x \in \mathbf{R}$ and $\displaystyle \int_0^2 f(x) \mathrm{d} x=5$. Then the value of $\displaystyle \int_{10}^{50} f(x) \mathrm{d} x$ is :
Official answer
(2)
$\displaystyle 100$
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JEE Main 2015 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.