JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 9)
Let y = y(x) satisfies the equation $${{dy} \over {dx}} - |A| = 0$$, for all x > 0, where $$A = \left[ {\matrix{
y & {\sin x} & 1 \cr
0 & { - 1} & 1 \cr
2 & 0 & {{1 \over x}} \cr
} } \right]$$. If $$y(\pi ) = \pi + 2$$, then the value of $$y\left( {{\pi \over 2}} \right)$$ is :
$${\pi \over 2} + {4 \over \pi }$$
$${\pi \over 2} - {1 \over \pi }$$
$${{3\pi } \over 2} - {1 \over \pi }$$
$${\pi \over 2} - {4 \over \pi }$$
Explanation
$$|A| = - {y \over x} + 2\sin x + 2$$
$${{dy} \over {dx}} = |A|$$
$${{dy} \over {dx}} = - {y \over x} + 2\sin x + 2$$
$${{dy} \over {dx}} + {y \over x} = 2\sin x + 2$$
$$I.F. = {e^{\int {{1 \over x}dx} }} = x$$
$$ \Rightarrow yx = \int {x(2\sin x + 2)dx} $$
$$xy = {x^2} - 2x\cos x + 2\sin x + c$$ ..... (i)
Now, x = $$\pi$$, y = $$\pi$$ + 2
Use in (i)
c = 0
Now, (i) becomes
$$xy = {x^2} - 2x\cos x + 2\sin x$$
put $$x = \pi /2$$
$${\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}$$
$$y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2$$
$${{dy} \over {dx}} = |A|$$
$${{dy} \over {dx}} = - {y \over x} + 2\sin x + 2$$
$${{dy} \over {dx}} + {y \over x} = 2\sin x + 2$$
$$I.F. = {e^{\int {{1 \over x}dx} }} = x$$
$$ \Rightarrow yx = \int {x(2\sin x + 2)dx} $$
$$xy = {x^2} - 2x\cos x + 2\sin x + c$$ ..... (i)
Now, x = $$\pi$$, y = $$\pi$$ + 2
Use in (i)
c = 0
Now, (i) becomes
$$xy = {x^2} - 2x\cos x + 2\sin x$$
put $$x = \pi /2$$
$${\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}$$
$$y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2$$
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