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If f(x) is a polynomial function of the second degree such that, `f(-3)=6,f(0)=6" and "f(2)=11`, then the graph of the function, f(x) cuts the ordinate x = 1 at the point

A

`(1,8)`

B

`(1,4)`

C

`(1,-2)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Let `f(x)=ax^(2)+bx+c`
`because f(0)=6impliesc=6`
`:.f(x)=ax^(2)+bx+6`
`:.f(-3)=6`
`implies" "9a-3b+6=6`
`implies" "3a=b" "...(i)`
`f(2)=11`
`implies" "4a+2b+6=11`
`implies" "4a+6a=5`
`implies" "a=1//2`
`implies" "b=3//2" "["from Eq. (i)"]`
`:." "f(x)=(x^(2))/(2)+(3)/(2)x+6`
Now, `" "f(1)=(1)/(2)+(3)/(2)+6=8`
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