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

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(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

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$ab=1$

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(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

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