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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}
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B
{1}
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C
{2}
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D
{2}
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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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