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Let f(x)=x^(2)+bx+c and g(x)=x^(2)+b(1)x...

Let `f(x)=x^(2)+bx+c` and `g(x)=x^(2)+b_(1)x+c_(1)` Let the real roots of `f(x)=0` be `alpha, beta` and real roots of `g(x)=0` be `alpha +k, beta+k` fro same constant `k`. The least value fo `f(x)` is `-1/4` and least value of `g(x)` occurs at `x=7/2` The roots of `g(x)=0` are

A

3,4

B

`-3,4`

C

`-3,-4`

D

`3,-4`

Text Solution

Verified by Experts

The correct Answer is:
A
We have `(alpha-beta)=(alpha+k)-(beta+k)`
`implies(sqrt(b^(2)-4c))/1=(sqrt(b_(1)^(2)-4c_(1)))/1`
`impliesb^(2)-4c=b_(1)^(3)-4c_(1)`…i
Given least value of `f(x)=-1/4-((b^(2)-4c))/(4xx1)=-1/4`
`impliesb^(2)-4c=1`
`:.b^(2)-4c=1=b_(1)^(2)-4c_(1)` [from Eq. (i) ]..ii
Also given least value of `g(x)` occurs at `x=7/2`
`:.-(b_(1))/(2xx1)=7/2`
`:.b_(1)=-7`
`:'g(x)=0`
`:.x^(2)+b_(1)x+c_(1)=0`
`impliesx=(-b_(1)+-sqrt(b_(1)^(2)-4c_(1)))/2`
`=(7+-1)/2=3,4`
`:.` Roots of `g(x)=0` are 3,4.
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