Mathematics · 2021
JEE Main · 27 July 2021, Shift 2 · Q61
Let f: R → R be defined as f(x+y)+f(x-y)=2 f(x) f(y), f(1/2)=-1. Then, the value of Σ_k =1^20 1/(sin ( k ) sin ( k +f( k ))) is equal to:
Let $\displaystyle f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as
$\displaystyle f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1$. Then, the value of
$\displaystyle \sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+f(\mathrm{k}))}$ is equal to :
Official answer
(3)
$\displaystyle \operatorname{cosec}^2(1) \operatorname{cosec}(21) \sin (20)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.