JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $(a + bx + cx^2)^{10} = \sum_{i=0}^{20} p_i x^i$ , $a, b, c \in \mathbb{N}$ . If $p_1 = 20$ and $p_2 = 210$ , then $2(a + b + c)$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: General term nikalna

Multinomial expansion ka upyog karte hue, general term hoga:

T = \frac{10!}{n_1! n_2! n_3!} (a)^{n_1} (bx)^{n_2} (cx^2)^{n_3}

Jahan $n_1 + n_2 + n_3 = 10$ .

Step 2: $p_1$ ki value (Coefficient of $x^1$ )

$x^1$ ke liye, $n_2 + 2n_3 = 1$ hona chahiye. Iske liye ek hi case sambhav hai:

$n_3 = 0, n_2 = 1 \implies n_1 = 9$

p_1 = \frac{10!}{9! 1! 0!} a^9 b^1 = 10 a^9 b = 20

\implies a^9 b = 2

Kyunki $a, b \in \mathbb{N}$ (natural numbers), isliye $a = 1$ aur $b = 2$ .

Step 3: $p_2$ ki value (Coefficient of $x^2$ )

$x^2$ ke liye, $n_2 + 2n_3 = 2$ hona chahiye. Iske do cases hain:

$n_3 = 0, n_2 = 2 \implies n_1 = 8$
$n_3 = 1, n_2 = 0 \implies n_1 = 9$

p_2 = \frac{10!}{8! 2! 0!} a^8 b^2 + \frac{10!}{9! 0! 1!} a^9 c = 210

45 a^8 b^2 + 10 a^9 c = 210

Ab $a = 1$ aur $b = 2$ ki values rakhte hain:

45(1)^8 (2)^2 + 10(1)^9 c = 210

45(4) + 10c = 210

180 + 10c = 210

10c = 30 \implies \mathbf{c = 3}

Step 4: Final calculation

Hume $2(a + b + c)$ nikalna hai:

2(1 + 2 + 3) = 2(6) = 12

Correct Option: (2) $12$