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The function f(x), that satisfies the condition $$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} $$, is :

JEE · Math · previous-year question

  1. A.$$x + {2 \over 3}(\pi - 2)\sin x$$
  2. B.$$x + (\pi + 2)\sin x$$
  3. C.$$x + {\pi \over 2}\sin x$$
  4. D.$$x + (\pi - 2)\sin x$$correct

Answer

D. $$x + (\pi - 2)\sin x$$

Explanation

$$f(x) = x + \int\limits_0^{\pi /2} {\sin x\cos y\,f(y)\,dy} $$ $$f(x) = x + sinx\underbrace {\int_0^{\pi /2} {\cos y\,f(y)\,dy} }_K$$ $$ \Rightarrow f(x) = x + K\sin x$$ $$ \Rightarrow f(y) = y + K\sin y$$ Now, $$K = \int_0^{\pi /2} {\mathop {y\cos y\,dy}\limits_{Apply\,IBP} } + K\int_0^{\pi /2} {\mathop {\cos y\sin y\,dy}\limits_{Put\,\sin y = t} } $$ $$K = \left( {y\sin y} \right)_0^{\pi /2} - \int_0^{\pi /2} {\sin y dy + K\int_0^1 {t\,dt} } $$ $$ \Rightarrow K = {\pi \over 2} - 1 + K\left( {{1 \over 2}} \right)$$ $$ \Rightarrow K = \pi - 2$$ So, $$f(x) = x + (\pi - 2)\sin x$$ Option (d)

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