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Question

Let y=f(x) be the solution of dydx=yx+tanyx, y(1)=π2 then f is defined when

A
x[1,)
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B
x(,1]
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C
x[1,1]{0}
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D
x[1,)
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Solution

The correct option is C x[1,1]{0}
Let y=xt
xdtdx+t=t+tant
cottdt=ln|x|+c
log|sint|=ln|x|+c
sinyx=|x| (x0)
y=xsin1|x|
x[1,1]{0}

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