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
{1}
C
{2}
D
{2}

Solution

## The correct option is C $$\{-2\}$$Given that $$y=(\lambda+1){ x }^{ 2 }+2$$ intersects $$y=\lambda x+3$$ in exactly one point$$\therefore \lambda x+3=\left( \lambda +1 \right) { x }^{ 2 }+2$$ has exactly one solution$$\therefore \left( \lambda +1 \right) { x }^{ 2 }-\lambda x-1=0$$$$\triangle ={ \left( -\lambda \right) }^{ 2 }+4\left( \lambda +1 \right) =0$$$$\triangle={ \left( \lambda +2 \right) }^{ 2 }=0$$$$\therefore \lambda =-2$$Mathematics

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