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Question

If y=f(x) satisfies 2exy2+y cos(x2)=4 then value of f(0)5 is

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Solution

at x = 0, y = 2
2exy2(y2+2xyy)+(cos x2)yy sin(x2).2x=0
at x=0 and y=2
2(4+0)+y(0)=0
y(0)=8 f(0)=8
f(0)5=85=1.6

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