Question

The point of intersection of the tangents drawn to the curve $x^{2}y=1-y$ at the points where it is meet by the curve xy = 1 – y, is given by

• A. (0, –1)
• B. (1, 1)
• C. (0, 1)
• D. none of these

Answer 1

The correct option is C (0, 1)

Given, f’’’(x) = 0
On integrating
$f"\left(x\right)=c\Rightarrow f"\left(1\right)=c\Rightarrow -1=c$ (given)
$\therefore f"\left(x\right)=-1f^{\prime }\left(x\right)=-x+d\because f^{\prime }\left(x\right)=-1+d=2\therefore d=3f^{\prime }\left(x\right)=3-xorf\left(x\right)=-\frac{x^2}{2}+3x+e\because f\left(0\right)=0+0+e=1\therefore e=1Hence,f\left(x\right)=-\frac{x^2}{2}+3x+1$