If the third term in the binomial expansion of $\left(1+x^{\log _{2} x}\right)^{5}$ equals 2560 , then a possible value of $x$ is:

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Binomial Theorem

If the third term in the binomial expansion of

(1 + x^{l o g_{2} x})^{5}

equals 2560 , then a possible value of

x

is:

Answer: (A)

The third term of

(1 + x^{l o g_{2} x})^{5}

(2 5) (x^{l o g_{2} x})^{2} = 10 x^{2 l o g_{2} x}

. Setting it to

2560

gives

x^{2 l o g_{2} x} = 256

, and taking

lo g_{2}

of both sides,

2 (lo g_{2} x)^{2} = 8

, so

(lo g_{2} x)^{2} = 4

and

lo g_{2} x = \pm 2

. That yields

x = 4

x = \frac{1}{4}

, and the value among the choices is

\frac{1}{4}

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option