The term independent of x in the expansion of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Simplify the expression inside the bracket:First term: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(C) Simplify the expression inside the bracket:First term: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$Second term: $\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$Subtracting the terms: $(x^{1/3}+1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$The expression becomes $(x^{1/3} - x^{-1/2})^{10}$.The general term $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.For the term to be independent of $x$,the exponent must be zero:$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$.The term is ${}^{10}C_4 (-1)^4 = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.

The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}, x \neq 1,$ is equal to ....... .

Similar Questions

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

The coefficient of $x^5$ in the expansion of $(x + 3)^6$ is

If the coefficients of $x^3$ and $x^4$ in the expansion of $(1 + ax + bx^2)(1 - 2x)^{18}$ in powers of $x$ are both zero,then $(a, b)$ is equal to

If the coefficients of $x^7$ in $\left(ax^2+\frac{1}{2bx}\right)^{11}$ and $x^{-7}$ in $\left(ax-\frac{1}{3bx^2}\right)^{11}$ are equal,then

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2x^3 + \frac{3}{x^k}\right)^{12}, x \neq 0$ is $2^8 \cdot \ell$,where $\ell$ is an odd integer,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module