If ab neq 0 and the sum of the coefficients o | vedclass

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

Question

(D) The general term in the expansion of $\left(\frac{x^2}{a}-\frac{b}{x}\right)^{11}$ is given by $T_{r+1} = {}^{11}C_r \left(\frac{x^2}{a}\right)^{1

Vedclass · Accepted Answer

$ab=1$

Vedclass · Answer

(D) The general term in the expansion of $\left(\frac{x^2}{a}-\frac{b}{x}ight)^{11}$ is given by $T_{r+1} = {}^{11}C_r \left(\frac{x^2}{a}ight)^{11-r} \left(-\frac{b}{x}ight)^r = {}^{11}C_r \left(\frac{1}{a}ight)^{11-r} (-b)^r x^{22-3r}$.For the coefficient of $x^7$,we set $22-3r = 7$,which gives $3r = 15$,so $r = 5$. The coefficient is $C_1 = {}^{11}C_5 \left(\frac{1}{a}ight)^6 (-b)^5$.For the coefficient of $x^4$,we set $22-3r = 4$,which gives $3r = 18$,so $r = 6$. The coefficient is $C_2 = {}^{11}C_6 \left(\frac{1}{a}ight)^5 (-b)^6$.Given $C_1 + C_2 = 0$,we have ${}^{11}C_5 \frac{(-b)^5}{a^6} + {}^{11}C_6 \frac{(-b)^6}{a^5} = 0$.Since ${}^{11}C_5 = {}^{11}C_6$,we can divide by ${}^{11}C_5 \frac{(-b)^5}{a^6}$ (assuming $ab 
eq 0$):$1 + \frac{(-b)}{a} \cdot a = 0$ is incorrect; let us simplify: $\frac{-b^5}{a^6} + \frac{b^6}{a^5} = 0$ $\Rightarrow \frac{b^6}{a^5} = \frac{b^5}{a^6}$ $\Rightarrow \frac{b}{a^5} = \frac{1}{a^6}$ $\Rightarrow ab = 1$.

If $ab \neq 0$ and the sum of the coefficients of $x^7$ and $x^4$ in the expansion of $\left(\frac{x^2}{a}-\frac{b}{x}\right)^{11}$ is $0$,then

Similar Questions

The terms containing $x^r y^s$ (for certain $r$ and $s$) are present in both the expansions of $(x+y^2)^{13}$ and $(x^2+y)^{14}$. If $\alpha$ is the number of such terms,then the sum $\alpha \sum_{r, s}(r+s) =$

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

If the sum of the coefficients of $x^r$ $(r=0, 1, 2, \ldots, 2n)$ in the expansion of $(1+3x-2x^2)^n$ is $128$,then $\sum_{r=1}^{2n} r \frac{^{2n}C_r}{^{2n}C_{r-1}} = $

Let $[t]$ denote the greatest integer $\leq t$. If the constant term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is $\alpha$,then $[\alpha]$ is equal to $............$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module