JEE Main 2022 — Functions Question with Solution

Given,

$$f(x)=\frac{2e^{2x}}{e^{2x}+e}$$

$$\therefore$$ $$f(1-x)=\frac{2e^{2(1-x)}}{e^{2(1-x)}+e}$$

$$=\frac{2.\frac{e^2}{e^{2x}}}{\frac{e^2}{e^{2x}}+e}$$

$$=\frac{2e^2}{e^2+e^{2x}.e}$$

$$=\frac{2e^2}{e(e+e^{2x})}$$

$$=\frac{2e}{e+e^{2x}}$$

$$\therefore$$ $$f(x)+f(1-x)=\frac{2e^{2x}}{e^{2x}+e}+\frac{2e}{e^{2x}+e}$$

$$=\frac{2(e^{2x}+e)}{e^{2x}+e}$$

$$=2$$ ...... (1)

Now,

$$f\left(\frac{1}{100}\right)+f\left(\frac{99}{100}\right)$$

$$=f\left(\frac{1}{100}\right)+f\left(1-\frac{1}{100}\right)$$

$$=2$$ [as $$f(x)+f(1-x)=2$$]

Similarly,

$$f\left(\frac{2}{100}\right)+f\left(1-\frac{2}{100}\right)=2$$

$$⋮$$

$$f\left(\frac{49}{100}\right)+f\left(1-\frac{49}{100}\right)=2$$

$$\therefore$$ Total sum $$=49\times 2$$

Remaining term $$=f\left(\frac{50}{100}\right)=f\left(\frac{1}{2}\right)$$

Put $$x=\frac{1}{2}$$ in equation (1), we get

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

$$\Rightarrow 2f\left(\frac{1}{2}\right)=2$$

$$\Rightarrow f\left(\frac{1}{2}\right)=1$$

$$\therefore$$ Sum $$=49\times 2+1=99$$