Let p(x) be a quadratic polynomial such that p(0) = 1 . If p(x) leaves remainder 4 when divided by x - 1 and it leaves remainder 6 when divided by x + 1 then:

Correct answer: A. p(- 2) = 19 Assume , p(x) = a x ^ 2 + bx + c Given , p(0) = 1 ⇒ c = 1 Now, array ll p(1) = 4 p(- 1) = 6 array ] array ll a + b + c = 4 a - b + c = 6 array ] ⇒ array ll a + b + 1 = 4 a - b + 1 = 6 array ] array ll a + b = 3 a - b = 5 array ] ⇒ array ll a = 4 b = - 1 array ] p(x) = 4 x ^ 2 - x + 1 p(- 2) = 16 + 2 + 1 = 19

JEE Main 20178 AprilOnline

Exam: JEE Main
Year: 2017
Date: 8 April
Mode: Online
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2017 · 8 April · Online · Mathematics · Quadratic Equation

Let p(x) be a quadratic polynomial such that p(0) = 1 . If p(x)...

Let

p\left(x\right)

be a quadratic polynomial such that

p\left(0\right) = 1 .

p\left(x\right)

leaves remainder

4

when divided by

x - 1

and it leaves remainder

6

when divided by

x + 1

then:

Options

A $p\left(- 2\right) = 19$
B $p\left(2\right) = 19$
C $p\left(- 2\right) = 11$
D $p\left(2\right) = 11$

Correct answer

A. $p\left(- 2\right) = 19$

Step-by-step solution

Assume ,

p\left(x\right) = a{x}^{2} + bx + c

Given ,

p\left(0\right) = 1 ⇒ c = 1

Now,\left

\begin{array}{ll}p\left(1\right) = 4 \\ p\left(- 1\right) = 6\end{array}

\right]

<br /><br />\left

\begin{array}{ll}a + b + c = 4 \\ a - b + c = 6\end{array}

\right] ⇒ \left

\begin{array}{ll}a + b + 1 = 4 \\ a - b + 1 = 6\end{array}

\right]\left\left

\begin{array}{ll}a + b = 3 \\ a - b = 5\end{array}

\right] ⇒

\begin{array}{ll}a = 4 \\ b = - 1\end{array}

\right]

p\left(x\right) = 4{x}^{2} - x + 1

p\left(- 2\right) = 16 + 2 + 1 = 19

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