If the fourth term in the expansion of (x + x^\log _2x)^7 is 4480, then the value of x where x \in N is equal to :

Binomial Theorem PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q31 with Solution

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Correct answer: D

For

(a+b)^7,

the general term is

T_{r+1} = {7 \choose r} a^{7-r} b^r.

Here,

a=x, \qquad b=x^{\log_2 x}.

So,

T_{r+1}={7 \choose r}x^{7-r}\left(x^{\log_2 x}\right)^r.

The fourth term corresponds to $r=3$ .

Thus,

T_4={7 \choose 3}x^{7-3}\left(x^{\log_2 x}\right)^3.

Now,

{7 \choose 3}=35,

T_4=35x^4\cdot x^{3\log_2 x} = 35x^{4+3\log_2 x}.

Given that this term is $4480$ ,

35x^{4+3\log_2 x}=4480.

Divide by $35$ :

x^{4+3\log_2 x}=128.

Since

128=2^7,

we check the options.

Option A: $x=3$
$3^{4+3\log_2 3} \neq 128.$
Not correct.
Option B: $x=1$
$1^{4+3\log_2 1}=1 \neq 128.$
Not correct.
Option C: $x=4$ Since $\log_2 4=2$ ,
$4^{4+3(2)}=4^{10} \neq 128.$
Not correct.
Option D: $x=2$ Since $\log_2 2=1$ ,
$2^{4+3(1)}=2^7=128.$
Hence,
$35\cdot 128=4480.$
Correct.

Therefore, the required value is

x=2.