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))
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y=xtan−1(log(xe))
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y=xtan−1(log(ex))
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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 ⇒dvcos2v=−dxx⇒sec2udu=−1xdx On integration, we get tanu=−logx+logC ⇒tan(yx)=−logx+logC This passes through (1,π4), therefore 1=logC. so, tan(yx)=−logx+1 ⇒tan(yx)=−logx+1⇒tan(yx)=−logx+loge⇒y=xtan−1(log(ex))