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A polynomial $p(x)$ is such that $p(0) = 5, p(1) = 4, p(2) = 9$ and $p(3) = 20$. The minimum degree it should have is

1. $1$
2. $2$
3. $3$
4. $4$

### First read  Sachin Mittal  sir's comment that is written below the best answer, too good approach. Although best answer that is written by Arjun sir is also good but it is like formal approach.

Lets take $p(x) = ax +b$
Now, $p(0) = 5 \implies b = 5.$

$p(1) = 4 \implies a + b = 4, a = -1$
$p(2) = 9 \implies 2a + b = 9 \implies -2 + 5 = 9$, which is false. So, degree $1$ is not possible.

Let $p(x) =$ $ax^{2}$ $+$ $bx$ $+$ $c$

$p(0) = 5 \implies c = 5$
$p(1) = 4 \implies a + b + c = 4 \implies a + b = -1 \quad \to(1)$
$p(2) = 9 \implies 4a + 2b + c = 9 \implies 2a + b = 2 \quad \to (2)$

$(2) - (1) \implies a = 3, b = -1 - 3 = -4$

$p(3) = 20 \implies 9a + 3b + c = 20, 27 -12 + 5 = 20$, equation holds.
So, degree $2$ also will suffice.

Correct Answer: $B$
by

nice one Thanks.. !
is this still in gate syllabus for 2021 ?
Since value of P(x) seems to be decreasing b/w 0 and 1 and then 1 to 2 it is increasing. So, there must be atleast two roots (may be both real and both imaginary). Because polynomials are smooth and continous on real numbers. So, there has to be one U-turn.
(b) is the answer. we can take a generalized quadratic equation ax2 +bx+c and verify.