    Question

# 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=tan1(log(ex))
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B
y=xtan1(log(xe))
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C
y=xtan1(log(ex))
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D
None of these
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Solution

## The correct option is C y=xtan−1(log(ex))we have, dydx=yx−cos2(yx) Putting y=vx, so that dydx=v+xdvdx, we get v+xdvdx=v−cos2v ⇒dvcos2v=−dxx⇒sec2udu=−1xdx On integration, we get tanu=−logx+logC ⇒tan(yx)=−logx+logC This passes through (1,π4), therefore 1=log⁡C. so, tan(yx)=−log x+1 ⇒tan(yx)=−logx+1⇒tan(yx)=−logx+loge⇒y=xtan−1(log(ex))  Suggest Corrections  0      Similar questions
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