BITSAT2011MathematicsApplication of DerivativesActual
The curve (y-e^ x y +x=0 ) has a vertical tangent at
Options
- A((1,1) )
- B((0,1) )
- C((1,0) )
- Dno point
Correct answer
C. ((1,0) )
Step-by-step solution
( aligned & y-e^ x y +x=0 & d y d x -e^ x y (y+x d y d x )+1=0 & i.e., d y d x -y(x+y)-x(x+y) d y d x +1=0 & i.e., [1-x(x+y)] d y d x =y(x+y)-1 aligned ) for the vertical tangents (1-x(x+y)=0 ) i.e., ( y= 1-x^2 x ) ( x =1 ) and ( y =0 )