TS ICET 23 May 2019 Shift 2 Paper

Question : 124 of 200

Marks: +1, -0

If p (x) is a quadratic polynomial such that
p (1) = 7 , p (- 1) = 3, p (0) = 4, then p (10) =

Solution:

Let P(x)

=ax^2+bx+c

P (1)= a + b + c =1-

→ (1)

P (-1)= a - b + c =3

→ (2)

P(0)=c=4

→(3)
from (1), (2) and (3)

a = 1, b = 2, c = 4

∴ P(x)=x^{2}+2 x+4

P(10)=100+20+4=124

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