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Question

The slope of the tangent at (x, y) to a curve passing through (1,π4) is given by yxcos2(yx), then the equation of the curve is

A
y=tan1[log(ex)]
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B
y=x tan1[log(xe)]
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C
y=x tan1[log(ex)]
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D
None of these
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Solution

The correct option is C y=x tan1[log(ex)]

We have dydx=yxcos2(yx)
Putting y =vx so that dydx=v+xdvdx, we get
v+xdvdx=vcos2v or dvcos2v=dxx
On integrating, we get tan v=log x+log c
tan(yx)=log x+log C
This passes through (1,π4), therefore 1 =log c
tan(yx)=log x+log ey=xtan1[log(ex)].

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