JEE Main 2022 — Functions Question with Solution

f(x) is polynomial

Put y = 1/x in given functional equation we get

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

$$\Rightarrow (c+1){\left(x+\frac{1}{x}\right)}^2+(1-c^2)\left(x+\frac{1}{x}\right)+2K$$

$$=(c+1)x^2+(1-c^2)x+2K+(c+1)\frac{1}{x^2}+(1-c^2)\frac{1}{x}+2K-1$$

$$\Rightarrow 2(c+1)=2K-1$$ ..... (1)

and put $$x=y=0$$ we get

$$f(0)=2+f(0)-0\Rightarrow f(0)=0\Rightarrow k=0$$

$$\therefore$$ $$k=0$$ and $$2c=-3\Rightarrow c=-3/2$$

$$f(x)=-\frac{x^2}{2}-\frac{5x}{4}=\frac{1}{4}(5x+2x^2)$$

$$\left|2\sum _{i=1}^{20}f(i)\right|=\left|\frac{-2}{4}\left(\frac{\mathrm{5.20.21}}{2}+\frac{\mathrm{2.20.21.41}}{6}\right)\right|$$

$$=\left|\frac{-1}{2}(2730+5740)\right|$$

$$=\left|-\frac{6790}{2}\right|=3395$$.