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The curve y=(lambda=1)x^2+2 intersects t...

The curve `y=(lambda=1)x^2+2` intersects the curve `y=lambdax+3` in exactly one point, if `lambda` equals

A

`{-2, 2}`

B

`{1}`

C

`{-2}`

D

`{2}`

Text Solution

Verified by Experts

The correct Answer is:
3
As `(lambda + 1)x^(2) + 2 = lambdax + 3` has only one solution, so
D = 0
`rArr lambda^(2) - 4(lambda + 1)(-1) = 0`
or `lambda^(2) + 4lambda + 4 = 0`
or `(lambda + 2)^(2) = 0`
`therefore lambda = -2`
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