JEE Main 2022 — Binomial Theorem Question with Solution

Given, Binomial expansion,

$${\left(a{x}^{\frac{1}{8}}+b{x}^{-\frac{1}{12}}\right)}^{10}$$

General term,

$$T_{r+1}={}^{10}{C}_r.{\left(a{x}^{\frac{1}{8}}\right)}^{10-r}.{\left(b{x}^{-\frac{1}{12}}\right)}^r$$

$$={}^{10}{C}_r.a^{10-r}.b^r.x^{\frac{10-r}{8}}.x^{-\frac{r}{12}}$$

$$={}^{10}{C}_r.a^{10-r}.b^r.x^{\left(\frac{10-r}{8}-\frac{r}{12}\right)}$$

$$={}^{10}{C}_r.a^{10-r}.b^r.x^{\frac{30-3r-2r}{24}}$$

$$={}^{10}{C}_r.a^{10-r}.b^r.x^{\frac{30-5r}{24}}$$

For constant term,

$$\frac{30-5r}{24}=0$$

$$\Rightarrow r=6$$

$$\therefore$$ Constant term,

$$T_{r+1}=T_{6+1}={}^{10}{C}_6.a^4.b^6$$

$$=\frac{10!}{6!4!}{a}^4.b^6$$

$$=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}.a^4.b^6$$

$$=210a^4{b}^6$$

We know, $$GM\ge HM$$

For terms a² and b³,

$$\sqrt{a^2{b}^3}\ge \frac{2}{\frac{1}{a^2}+\frac{1}{b^3}}$$

$$\Rightarrow \sqrt{a^2{b}^3}\ge \frac{2}{4}$$

$$\Rightarrow a^2{b}^3\ge \frac{1}{4}$$

$$\Rightarrow (a^2{b}^3{)}^2\ge \frac{1}{16}$$

$$\therefore$$ $$a^4{b}^6\ge \frac{1}{16}$$

$$\therefore$$ Minimum value of $$a^4{b}^6=\frac{1}{16}$$

$$\therefore$$ Minimum value of constant term

$$T_7=210\times a^4{b}^6$$

$$=210\times \frac{1}{16}$$

$$=\frac{105}{8}$$