The slope of the tangent at (x, y) to a curve passing through (1,π4) is given by yx−cos2(yx), then the equation of the curve is
A
y=tan−1[log(ex)]
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B
y=xtan−1[log(xe)]
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C
y=xtan−1[log(ex)]
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D
Noneofthese
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Solution
The correct option is Cy=xtan−1[log(ex)]
We have dydx=yx−cos2(yx)
Putting y =vx so that dydx=v+xdvdx, we get v+xdvdx=v−cos2v or dvcos2v=−dxx
On integrating, we get tanv=−logx+logc ⇒tan(yx)=−logx+logC
This passes through (1,π4), therefore 1 =log c ⇒tan(yx)=−logx+loge⇒y=xtan−1[log(ex)].