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Let f(x)=x^(2)-ax+b, 'a' is odd positive...

Let `f(x)=x^(2)-ax+b`, `'a'` is odd positive integar and the roots of the equation `f(x)=0` are two distinct prime numbers. If `a+b=35`, then the value of `f(10)=`

A

`-8`

B

`-10`

C

`-4`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` `a="sum of the roots"="odd"`
Therefore, both the roots cannot be odd.
Therefore, one root is `2`.
`:. 4-2a+b=0`………`(i)`
Given `a+b=35`…..`(ii)`
by solving `(1)` and `(2)` `a=13`, `b=22`
`:. F(x)=x^(2)-13x+22`
`:. f(10)=-8`
`:. "Required value"=880`
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