If maximum value of term independent of t in expansion of left( t ^{2} x ^{frac{1}{5}}+frac{(1- x )^

English

Gujarati

Hindi

7.Binomial Theorem

hard

If the maximum value of the term independent of $t$ in the expansion of $\left( t ^{2} x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}, x \geq 0$, is $K$, then $8\,K$ is equal to $....$

A

$6006$

B

$6005$

C

$6007$

D

$6008$

(JEE MAIN-2022)

Solution

$\left( t ^{2} x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}$

$T_{r+1}={ }^{15} C_{r}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^{r}}$

For independent of $t$,

$30-2 r-r=0$

$r =10$

So, Maximum value of ${ }^{15} C _{10} x (1- x )$ will be at

$x=\frac{1}{2}$

i.e. $6006$

Standard 11

Mathematics

Share

Similar Questions

If the second, third and fourth term in the expansion of ${(x + a)^n}$ are $240, 720$ and $1080$ respectively, then the value of $n$ is

hard

The middle term in the expansion of ${\left( {x + \frac{1}{{2x}}} \right)^{2n}}$, is

easy

If the $6^{th}$ term in the expansion of the binomial ${\left[ {\sqrt {{2^{\log (10 – {3^x})}}} + \sqrt[5]{{{2^{(x – 2)\log 3}}}}} \right]^m}$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first, third and fifth terms of an $A.P$. (the symbol log stands for logarithm to the base $10$), then $x = $

hard

Let $\alpha > 0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in N$. Then $\alpha$ is equal to $………….$.

hard

(JEE MAIN-2023)

Two middle terms in the expansion of ${\left( {x – \frac{1}{x}} \right)^{11}}$ are

easy