Question

( f(x)=0 quad ) at ( x=1 quad ) and ( x=-2 )
( f(x)=(a x+b)(x-1)(x+2) )
[ cutric polynonimal ( ] )
( f(-1)=4 quad Rightarrow quad(-a+b)(-2)(1)=4 )
( -2 quad a-b=2 rightarrow(2) )
( f(2)=28 Rightarrow(2 a+b)(1)(4)=28 )
( Rightarrow quad 2 a+b=7 quad rightarrow(1) )
( a-b=2 rightarrow(2) )
[
begin{array}{r}3 a=9 a=3end{array}
]
( therefore f(1)=(3(1)+1)(1-1)(1+2)=0 )

# SECTION-III(i) Numerical Grid Type (Single digit Ranging from 0 to 9) (4 Marks each, -1 for wrong answer) A polynomial in x of degree three vanishes when x = 1 and x = -2 and has the values 4 and 28 when x = -1 and x = 2 respectively, then f(1) is 1 2 . 12. 10 1. 2. A :

Solution