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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2011 · Mathematics · Application of Derivatives

The curve y-e^ x y +x=0 has a vertical tangent at

The curve y−exy+x=0y-e^{x y}+x=0 has a vertical tangent at

Options

  1. A(1,1)(1,1)
  2. B(0,1)(0,1)
  3. C(1,0)(1,0)
  4. Dno point

Correct answer

C. (1,0)(1,0)

Step-by-step solution

y−exy+x=0∴dydx−exy(y+xdydx)+1=0i.e., dydx−y(x+y)−x(x+y)dydx+1=0i.e., [1−x(x+y)]dydx=y(x+y)−1\begin{aligned}& y-e^{x y}+x=0 \\ & \therefore \frac{d y}{d x}-e^{x y}\left(y+x \frac{d y}{d x}\right)+1=0 \\ & \text{i.e., } \frac{d y}{d x}-y(x+y)-x(x+y) \frac{d y}{d x}+1=0 \\ & \text{i.e., }[1-x(x+y)] \frac{d y}{d x}=y(x+y)-1 \end{aligned})forforthe vertical tangents 1−x(x+y)=01-x(x+y)=0 i.e., (\quady=y=\frac{1-x²}{x} (\therefore \mathrm{x}=1)and(and (\mathrm{y}=0)$
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