QJEEM17Mac149: JEE Main 2017 Mathematics - Class 11 JEEM17Mac149 Solution: Let p(x) be a quadratic polynomial such that p(0)=1. If p(x)…

Let, P(x) = ax² + bx + c

As, P(0) = 1,

$\therefore\,\,\,$ a(0)² + b(0) + c = 1

$\Rightarrow$ $\,\,\,$ c = 1

$\therefore\,\,\,$ P(x) = ax² + bx + 1

If   P(x) is divided by x $-$ 1, remainder = 4

$\Rightarrow$ $\,\,\,$ P $\left( 1 \right) = 4$

$\therefore\,\,\,$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$\Rightarrow$ $\,\,\,$ P( $-$ 1) = 6

$\therefore\,\,\,$ a $-$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $-$ 1

$\therefore\,\,\,$ P(x) = 4x² $-$ x + 1

P(2) = 4(2)² $-$ 2 + 1 = 15

P( $-$ 2) = 4 ( $-$ 2)² $-$ ( $-$ 2) + 1 = 19