Question

The curve $y=(\lambda +1)x^2+2$ intersects the curve $y=\lambda x+3$ in exactly one point, if $\lambda$ is

• A. $\{-2,2\}$
• B. $\left\{1\right\}$
• C. $\{-2\}$
• D. $\left\{2\right\}$

Answer 1

The correct option is C $\{-2\}$

Given that $y=(\lambda +1)x^2+2$ intersects $y=\lambda x+3$ in exactly one point

$\therefore \lambda x+3=(\lambda +1)x^2+2$ has exactly one solution

$\therefore (\lambda +1)x^2-\lambda x-1=0$

$△={(-\lambda )}^2+4(\lambda +1)=0$

$△={(\lambda +2)}^2=0$

$\therefore \lambda =-2$