JEE Mains · Maths · STD 12 - 1. relation and function
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be defined as \(f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1 .\) Then, the value of \(\sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+\mathrm{f}(\mathrm{k}))}\) is equal to:
- A \(\operatorname{cosec}^{2}(1) \operatorname{cosec}(21) \sin (20)\)
- B \(\sec ^{2}(1) \sec (21) \cos (20)\)
- C \(\operatorname{cosec}^{2}(21) \cos (20) \cos (2)\)
- D \(\sec ^{2}(21) \sin (20) \sin (2)\)
Answer & Solution
Correct Answer
(A) \(\operatorname{cosec}^{2}(1) \operatorname{cosec}(21) \sin (20)\)
Step-by-step Solution
Detailed explanation
\(f(x)=\cos a x\) \(\because f\left(\frac{1}{2}\right)=-1\) So, \(-1=\cos \frac{a}{2}\) \(\Rightarrow a=2(2 n+1) \pi\) Thus \(f(x)=\cos 2(2 n+1) \pi x\) Now k is natural number Thus \(f(k)=1\)…
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