If the coefficient of x^7 in [a x^2+ ( 1 b x ) ]^11 equals the coefficient of x^-7 in [a x^2- ( 1 b x ) ]^11, then a and b satisfy the relation

$$\begin{aligned} & T_{r+1} \text { in the expansion }\left[a x^2+\frac{1}{b x}\right]^{11}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{b x}\right)^r \\ & ={ }^{11} C_r(a)^{11-r}(b)^{-r}(x)^{22-2 r-r} \\ & \Rightarrow 22-3 r=7 \Rightarrow r=5 \\ & \therefore \text { coefficient of } x^7={ }^{11} C_5(a)^6(b)^{-5} ....(i) \end{aligned}$$​Tr+1​ in the expansion [ax2+bx1​]11=11Cr​(ax2)11−r(bx1​)r=11Cr​(a)11−r(b)−r(x)22−2r−r⇒22−3r=7⇒r=5∴ coefficient of x7=11C5​(a)6(b)−5....(i)​ Again $T_{r+1}$Tr+1​ in the expansion $\left[a x-\frac{1}{b x^2}\right]^{11}={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{b x^2}\right)^r$[ax−bx21​]11=11Cr​(ax)11−r(−bx21​)r $={ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{11-\mathrm{r}}(-1)^{\mathrm{r}} \times(\mathrm{b})^{-\mathrm{r}}(\mathrm{x})^{-2 \mathrm{r}}(\mathrm{x})^{11-\mathrm{r}}$=11Cr​a11−r(−1)r×(b)−r(x)−2r(x)11−r Now $11-3 r=-7 \Rightarrow 3 r=18 \Rightarrow r=6$11−3r=−7⇒3r=18⇒r=6 $$\begin{aligned} & \therefore \text { coefficient of } x^{-7}={ }^{11} C_6 a^5 \times 1 \times(b)-6 \\ & \Rightarrow{ }^{11} C_5(a)^6(b)^{-5}={ }^{11} C_6 a^5 \times(b)^{-6} \\ & \Rightarrow a b=1 \end{aligned}$$​∴ coefficient of x−7=11C6​a5×1×(b)−6⇒11C5​(a)6(b)−5=11C6​a5×(b)−6⇒ab=1​