JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 8)
Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is :
$${1 \over 8}(\pi - 1)$$
$${1 \over {12}}(\pi - 3)$$
$${1 \over 4}(\pi - 2)$$
$${1 \over 6}(\pi - 1)$$
Explanation
We have,
$${{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}$$
$$\therefore$$ $${{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)$$
Put $${y \over x} = v$$
$$ \Rightarrow y = vn$$
$$\therefore$$ $${{dy} \over {dx}} = v + {{ndv} \over {dx}}$$
Now, we get
$$v + n{{dv} \over {dx}} = v - \cot (v)$$
$$ \Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}} $$
$$\therefore$$ $$\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c$$
As, $$\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0$$
$$\therefore$$ $$\sec \left( {{y \over x}} \right) = {1 \over x}$$
$$ \Rightarrow \cos \left( {{y \over x}} \right) = x$$
$$\therefore$$ $$y = x{\cos ^{ - 1}}(x)$$
So, required bounded area
$$ = \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)} $$ (I. B. P.)
$$\therefore$$ Option (1) is correct.
$${{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}$$
$$\therefore$$ $${{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)$$
Put $${y \over x} = v$$
$$ \Rightarrow y = vn$$
$$\therefore$$ $${{dy} \over {dx}} = v + {{ndv} \over {dx}}$$
Now, we get
$$v + n{{dv} \over {dx}} = v - \cot (v)$$
$$ \Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}} $$
$$\therefore$$ $$\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c$$
As, $$\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0$$
$$\therefore$$ $$\sec \left( {{y \over x}} \right) = {1 \over x}$$
$$ \Rightarrow \cos \left( {{y \over x}} \right) = x$$
$$\therefore$$ $$y = x{\cos ^{ - 1}}(x)$$
So, required bounded area
$$ = \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)} $$ (I. B. P.)
$$\therefore$$ Option (1) is correct.
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