JEE Main 2022 — Functions Question with Solution

Given,

$$f(x)={\left(2\left(1-\frac{x^{25}}{2}\right)\left(2+x^{25}\right)\right)}^{\frac{1}{50}}$$

and $$g(x)=f\left(f\left(f\left(x\right)\right)\right)+f\left(f\left(x\right)\right)$$

$$\therefore$$ $$g(1)=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$$

Now, $$f(1)={\left(2\left(1-\frac{1^{25}}{2}\right)\left(2+1^{25}\right)\right)}^{\frac{1}{50}}$$

$$={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}$$

$$={\left(3\right)}^{\frac{1}{50}}$$

$$\therefore$$ $$f\left(f\left(1\right)\right)=f\left(3^{\frac{1}{50}}\right)$$

$$={\left(2\left(1-\frac{{\left(3^{\frac{1}{50}}\right)}^{25}}{2}\right)\left(2+{\left(3^{\frac{1}{50}}\right)}^{25}\right)\right)}^{\frac{1}{50}}$$

$$={\left(2\left(1-\frac{3^{\frac{1}{2}}}{2}\right)\left(2+3^{\frac{1}{2}}\right)\right)}^{\frac{1}{50}}$$

$$={\left(2\times \left(\frac{2-\sqrt{3}}{2}\right)\left(2+\sqrt{3}\right)\right)}^{\frac{1}{50}}$$

$$={\left[\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\right]}^{\frac{1}{50}}$$

$$={\left(4-3\right)}^{\frac{1}{50}}$$

$$=1^{\frac{1}{50}}=1$$

Now, $$f\left(f\left(f\left(1\right)\right)\right)=f(1)=3^{\frac{1}{50}}$$

$$\therefore$$ $$g(1)=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$$

$$=3^{\frac{1}{50}}+1$$

Now, greatest integer less than or equal to $$g(1)$$

$$=\left[g(1)\right]$$

$$=\left[3^{\frac{1}{50}}+1\right]$$

$$=\left[3^{\frac{1}{50}}\right]+\left[1\right]$$

$$=[1.02]+1$$

$$=1+1=2$$