Mathematics · 2025
JEE Main · 24 January 2025, Shift 1 · Q1
Let f(x)=(2^x+2+16)/(2^2 x+1+2^x+4+32). Then the value of 8(f(1/15)+f(2/15)+…+f(59/15)) is equal to
Let $\displaystyle f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $\displaystyle 8\left(f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)\right)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 118$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.