JEE Mathematics 2023 Question with Solution

Given:

$$(a+bx+cx^2)^{10}=\sum_{i=0}^{20} p_i x^i,$$

with $p_1=20$ and $p_2=210$.

Find: $2(a+b+c)$.

For the coefficient of $x^1$, we choose $bx$ once and $a$ from the remaining $9$ factors.

$$p_1=\binom{10}{1}a^9 b=10a^9 b$$

Given $p_1=20$,

$$10a^9 b=20 \Rightarrow a^9 b=2$$

Since $a,b\in\mathbb{N}$, we get

$$a=1,\quad b=2$$

For the coefficient of $x^2$, there are two possibilities:

1. choose $cx^2$ once and $a$ from the remaining $9$ factors,
2. choose $bx$ twice and $a$ from the remaining $8$ factors.

So,

$$p_2=\binom{10}{1}a^9 c+\binom{10}{2}a^8 b^2$$

Substituting $a=1$ and $b=2$,

$$p_2=10c+45(2^2)$$ $$=10c+180$$

Given $p_2=210$,

$$10c+180=210 \Rightarrow c=3$$

Now,

$$a+b+c=1+2+3=6$$

Therefore,

$$2(a+b+c)=2\times 6=12$$

So, the correct option is C.

Given: $p_1=20$ and $p_2=210$.

Find: $2(a+b+c)$.

Use coefficient counting directly.

From the coefficient of $x$,

$$10a^9b=20 \Rightarrow a^9b=2$$

Because $a$ and $b$ are natural numbers, the only possibility is

$$a=1,\quad b=2$$

Now for $x^2$,

$$p_2=10a^9c+45a^8b^2$$

Substitute immediately:

$$210=10c+45\cdot 4=10c+180$$

Hence,

$$c=3$$

Thus,

$$2(a+b+c)=2(1+2+3)=12$$

This works because coefficients in multinomial expansions come from all distinct ways of selecting terms whose powers add to the required exponent.