JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 14)
The function f(x), that satisfies the condition
$$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} $$, is :
$$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} $$, is :
$$x + {2 \over 3}(\pi - 2)\sin x$$
$$x + (\pi + 2)\sin x$$
$$x + {\pi \over 2}\sin x$$
$$x + (\pi - 2)\sin x$$
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)
$$f(x) = x + sinx\underbrace {\int_0^{\pi /2} {\cos y\,f(y)\,dy} }_K$$
$$ \Rightarrow f(x) = x + K\sin x$$
$$ \Rightarrow f(y) = y + K\sin y$$
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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