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In the given figue vertices of DeltaABC ...

In the given figue vertices of `DeltaABC` lie on `y=f(x)=ax^(2)+bx+c`. The `DeltaAB` is right angled isosceles triangle whose hypotenuse `AC=4sqrt(2)` units.

Number of integral value of `lamda` for which `(lamda)/2` lies between the roots of `f(x)=0`, is

A

9

B

10

C

11

D

12

Text Solution

Verified by Experts

The correct Answer is:
C
Given that `AC=4sqrt(2)` units
`:.AB=BC=(AC)/(sqrt(2))=4` units andd `OB=sqrt((BC)^(2)-(OC)^(2))`
`=sqrt((4)^(2)-(2sqrt(2))^(2))[ :' OC=(AC)/2]`
`=2sqrt(2)` units
`:.` Vertices are `A=(-2sqrt(2),0)`,
`B=-(0,-2sqrt(2))`
and `C=(2sqrt(2),0)`
`f(x)=0`
`implies(x^(2))/(2sqrt(2))-2sqrt(2)=0impliesx=+-2sqrt(2)`
Given `-2sqrt(2) lt (lamda)/2lg2sqrt(2)`
or `-4sqrt(2)lt lamda lt 4sqrt(2)`
`:.` Initial values of `lamda` are ,brgt `-5,-4,-3,-2,-1,0,1,2,3,4,5`
`:.` Number of integral values is 11.
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