IPMAT Indore 2025 (MCQ) - Let P(x) be a quadratic polynomial such that |arrayll P(0) & P(1) \\ P(0) & P(2) array|=0 Let P(0)=2 and P(1)+P(2)+P(3)=14. Then P(4) equals | PYQs + Solutions

IPMAT Indore 2025

Algebra

Quadratic Equations

Easy

Let $P(x)$ be a quadratic polynomial such that
$\left|\begin{array}{ll} P(0) & P(1) \\ P(0) & P(2) \end{array}\right|=0$
Let $P(0)=2$ and $P(1)+P(2)+P(3)=14$ . Then $P(4)$ equals

-14

-6

✅ Correct Option: 3

Let's evaluate the determinant in the first condition:

\left|\begin{array}{ll}P(0) & P(1) \\P(0) & P(2)\end{array}\right| = P(0)P(2) - P(0)P(1) = P(0)[P(2)-P(1)]

Since this equals zero, and

P(0)=2 \neq 0

, we must have

P(2) = P(1)

Since

P(x)

is quadratic, we can write it as:

P(x) = ax^2 + bx + c

With

P(0) = c = 2

, our polynomial is:

P(x) = ax^2 + bx + 2

Using

P(2) = P(1)

P(1) = a + b + 2

P(2) = 4a + 2b + 2

Setting them equal:

a + b + 2 = 4a + 2b + 2

0 = 3a + b

b = -3a

So our polynomial is now:

P(x) = ax^2 - 3ax + 2

From

P(1) + P(2) + P(3) = 14

P(1) = -2a + 2

P(2) = -2a + 2

P(3) = 9a - 9a + 2 = 2

This gives us:

(-2a + 2) + (-2a + 2) + 2 = 14

-4a + 6 = 14

-4a = 8

a = -2

With

a = -2

and

b = -3a = 6

, our polynomial is:

P(x) = -2x^2 + 6x + 2

Therefore:

P(4) = -2(16) + 6(4) + 2 = -32 + 24 + 2 = -6