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Functions question

2021 · 27 Jul · Shift 2 · Q25
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  5. /2021 · 27 Jul · Shift 2 · Q25

Functions question

2021 · 27 Jul · Shift 2 · Q25

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f : R →\to→ R be defined as f(x+y)+f(x−y)=2f(x)f(y),f(12)=−1f(x + y) + f(x - y) = 2f(x)f(y),f\left( {{1 \over 2}} \right) = - 1f(x+y)+f(x−y)=2f(x)f(y),f(21​)=−1. Then, the value of ∑k=1201sin⁡(k)sin⁡(k+f(k))\sum\limits_{k = 1}^{20} {{1 \over {\sin (k)\sin (k + f(k))}}}k=1∑20​sin(k)sin(k+f(k))1​ is equal to :
  1. A
    cosec2(21) cos(20) cos(2)
  2. B
    sec2(1) sec(21) cos(20)
  3. C
    cosec2(1) cosec(21) sin(20)
  4. D
    sec2(21) sin(20) sin(2)
View written solutionFree

Correct answer: C

  1. Use the functional equation to identify fff.

We are given f(x+y)+f(x−y)=2f(x)f(y),f(x+y)+f(x-y)=2f(x)f(y),f(x+y)+f(x−y)=2f(x)f(y), with f(12)=−1.f\left(\tfrac12\right)=-1.f(21​)=−1.

This is the standard d'Alembert-type functional equation satisfied by cosine. Let us derive the needed values.

  1. Find f(0)f(0)f(0).

Put y=0y=0y=0: f(x)+f(x)=2f(x)f(0).f(x)+f(x)=2f(x)f(0).f(x)+f(x)=2f(x)f(0). So, 2f(x)=2f(x)f(0).2f(x)=2f(x)f(0).2f(x)=2f(x)f(0). Since f(12)=−1≠0f\left(\tfrac12\right)=-1\neq 0f(21​)=−1=0, we get f(0)=1.f(0)=1.f(0)=1.

  1. Find f(1)f(1)f(1).

Put x=y=12x=y=\tfrac12x=y=21​: f(1)+f(0)=2f(12)f(12)=2(−1)(−1)=2.f(1)+f(0)=2f\left(\tfrac12\right)f\left(\tfrac12\right)=2(-1)(-1)=2.f(1)+f(0)=2f(21​)f(21​)=2(−1)(−1)=2. Since f(0)=1f(0)=1f(0)=1, f(1)+1=2  ⟹  f(1)=1.f(1)+1=2 \implies f(1)=1.f(1)+1=2⟹f(1)=1.

  1. Find f(x+12)f\left(x+\tfrac12\right)f(x+21​) and deduce periodicity.

Take y=12y=\tfrac12y=21​ in the functional equation: f(x+12)+f(x−12)=2f(x)f(12)=−2f(x).f\left(x+\tfrac12\right)+f\left(x-\tfrac12\right)=2f(x)f\left(\tfrac12\right)=-2f(x).f(x+21​)+f(x−21​)=2f(x)f(21​)=−2f(x).

Now replace xxx by x+12x+\tfrac12x+21​: f(x+1)+f(x)=−2f(x+12).f(x+1)+f(x)=-2f\left(x+\tfrac12\right).f(x+1)+f(x)=−2f(x+21​). But f(1)=1f(1)=1f(1)=1 strongly suggests cosine-type behavior. In fact, the cosine family f(x)=cos⁡(2nπx)f(x)=\cos(2n\pi x)f(x)=cos(2nπx) satisfies the equation, and the condition f(12)=cos⁡(nπ)=−1f\left(\tfrac12\right)=\cos(n\pi)=-1f(21​)=cos(nπ)=−1 forces nnn to be odd. Hence one valid form is f(x)=cos⁡(2(2m+1)πx).f(x)=\cos(2(2m+1)\pi x).f(x)=cos(2(2m+1)πx). For every integer kkk, f(k)=cos⁡(2(2m+1)πk)=1.f(k)=\cos(2(2m+1)\pi k)=1.f(k)=cos(2(2m+1)πk)=1. So for the given sum, we only need f(k)=1for all integers k.f(k)=1\quad \text{for all integers }k.f(k)=1for all integers k.

  1. Simplify the summand.

Since f(k)=1f(k)=1f(k)=1, 1sin⁡k sin⁡(k+f(k))=1sin⁡k sin⁡(k+1).\frac{1}{\sin k\,\sin(k+f(k))}=\frac{1}{\sin k\,\sin(k+1)}.sinksin(k+f(k))1​=sinksin(k+1)1​. Therefore, S=∑k=1201sin⁡ksin⁡(k+1).S=\sum_{k=1}^{20}\frac{1}{\sin k\sin(k+1)}.S=∑k=120​sinksin(k+1)1​.

  1. Use the identity for telescoping.

Recall: cot⁡k−cot⁡(k+1)=sin⁡((k+1)−k)sin⁡ksin⁡(k+1)=sin⁡1sin⁡ksin⁡(k+1).\cot k-\cot(k+1)=\frac{\sin((k+1)-k)}{\sin k\sin(k+1)}=\frac{\sin 1}{\sin k\sin(k+1)}.cotk−cot(k+1)=sinksin(k+1)sin((k+1)−k)​=sinksin(k+1)sin1​. Hence, 1sin⁡ksin⁡(k+1)=csc⁡1 [cot⁡k−cot⁡(k+1)].\frac{1}{\sin k\sin(k+1)}=\csc 1\,[\cot k-\cot(k+1)].sinksin(k+1)1​=csc1[cotk−cot(k+1)].

So, S=csc⁡1∑k=120[cot⁡k−cot⁡(k+1)].S=\csc 1\sum_{k=1}^{20}[\cot k-\cot(k+1)].S=csc1∑k=120​[cotk−cot(k+1)]. This telescopes: S=csc⁡1 [cot⁡1−cot⁡21].S=\csc 1\,[\cot 1-\cot 21].S=csc1[cot1−cot21].

  1. Rewrite in the required trigonometric form.
=\frac{\cos1\sin21-\sin1\cos21}{\sin^2 1\sin21}.$$ Using $$\sin A\cos B-\cos A\sin B=\sin(A-B),$$ we get $$\cos1\sin21-\sin1\cos21=\sin(21-1)=\sin20.$$ Thus, $$S=\frac{\sin20}{\sin^2 1\sin21}=\csc^2(1)\csc(21)\sin(20).$$ --- 8. **Match with the options.** This is exactly $$\boxed{\cosec^2(1)\,\cosec(21)\,\sin(20)}.$$ So the correct option is **C**. --- 9. **Compare with stored correct answer.** Stored correct answer: **C** Our derived answer: **C** They agree.
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