The curve y - e xy - x -0 has a vertical tangent at the pointA- 1-1B- at no pointC- 0-1D- 1-0

The correct option is C (0, 1)x^cos y + y^cos x = 5 ⇒ e^cosy log_e x + e^cosy log_e y = 5 ∴ e^cos y log_e x{cosy/x log_e sin y dy/dx} + e^cosy log_e y{cos x ...

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Standard XII

Mathematics

Integration by Substitution

The curve y -...

Question

The curve $y - e^{x y} + x = 0$ has a vertical tangent at the point

(1, 1)

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at no point

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(0, 1)

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(1, 0)

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Solution

The correct option is C (0, 1)
$x^{c o s y} + y^{c o s x} = 5 \Rightarrow e^{c o s y l o g_{e} x} + e^{c o s y l o g_{e} y} = 5 ∴ e^{c o s y l o g_{e} x} {\frac{c o s y}{x} - l o g_{e} s i n y \frac{d y}{d x}} + e^{c o s y l o g_{e} y} {\frac{c o s x}{y} \frac{d y}{d x} - s i n x l o g_{e} y} = O (c o s 1 - 0) + (c o s 1 \frac{d y}{d x} - 0) = 0 \frac{d y}{d x} = - 1 y^{'} = - 1$

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The curve y−exy+x=0 has a vertical tangent at the point

The correct option is C (0, 1)xcosy+ycosx=5⇒ecosy logex+ecosy logey=5∴ecosy logex{cosyx−logesin ydydx}+ecosy logey{cosxydydx−sin x logey}=O(cos1−0)+(cos 1dydx−0)=0dydx=−1y′=−1

The curve $y - e^{x y} + x = 0$ has a vertical tangent at the point