JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Mock Test - Complex Numbers and Quadratic Equations

The curve y=(\[\lambda \]+1)\[{{x}^{2}}\]+2 intersects the curve y=\[\lambda x+3\]in exactly one point, if \[\lambda \]equals

A) \[\left\{ -2,2 \right\}\]

B) \[\left\{ 1 \right\}\]

C) \[\left\{ -2 \right\}\]

D) \[\left\{ 2 \right\}\]

Correct Answer: C

Solution :

[c] As \[(\lambda +1){{x}^{2}}+2=\lambda x+3\]has only one solution, so D=0

\[\Rightarrow {{\lambda }^{2}}-4(\lambda +1)(-1)=0\] Or \[{{\lambda }^{2}}+4\lambda +4=0\] Or \[{{(\lambda +2)}^{2}}=0\]

\[\therefore \lambda =-2\]