JEE Main 2022 — Binomial Theorem Question with Solution

General term of $${\left(t^2{x}^{\frac{1}{5}}+\frac{{(1-x)}^{\frac{1}{10}}}{t}\right)}^{15}$$ is

$$T_{r+1}={}^{15}{C}_r.{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}.{\left(\frac{{(1-x)}^{\frac{1}{10}}}{t}\right)}^r$$

$$={}^{15}{C}_r.t^{30-2r}.x^{\frac{15-r}{5}}.{\left(1-x\right)}^{\frac{r}{10}}.t^{-r}$$

$$={}^{15}{C}_r.t^{30-3r}.x^{\frac{15-r}{5}}.{\left(1-x\right)}^{\frac{r}{10}}$$

Term will be independent of $$\mathrm{t}$$ when $$30-3r=0\Rightarrow r=10$$

$$\therefore$$ $$T_{10+1}=T_{11}$$ will be independent of $$\mathrm{t}$$

$$\therefore$$ $$T_{11}={}^{15}{C}_{10}.x^{\frac{15-10}{5}}.{\left(1-x\right)}^{\frac{10}{10}}$$

$$={}^{15}{C}_{10}.x^1.{\left(1-x\right)}^1$$

$${\mathrm{T}}_{11}$$ will be maximum when $$x(1-x)$$ is maximum.

Let $$f(x)=x(1-x)=x-x^2$$

$$f(x)$$ is maximum or minimum when $$f^{\prime }(x)=0$$

$$\therefore$$ $$f^{\prime }(x)=1-2x$$

For maximum/minimum $$f^{\prime }(x)=0$$

$$\therefore$$ $$1-2x=0$$

$$\Rightarrow x=\frac{1}{2}$$

Now, $$f^{\prime \prime }(x)=-2<0$$

$$\therefore$$ At $$x=\frac{1}{2}$$, $$f(x)$$ maximum

$$\therefore$$ Maximum value of $${\mathrm{T}}_{11}$$ is

$$={}^{15}{C}_{10}.\frac{1}{2}\left(1-\frac{1}{2}\right)$$

$$={}^{15}{C}_{10}.\frac{1}{4}$$

Given $$K={}^{15}{C}_{10}.\frac{1}{4}$$

Now, $$8K=2\left({}^{15}{C}_{10}\right)$$

$$=6006$$