Question

$f\left(x\right)=\frac{a{x}^2+b}{x^2+1},{lim}_{x\to 0}f\left(x\right)=1and{lim}_{x\to \infty }f\left(x\right)=1,thenprovethatf(-2)=f\left(2\right)=1.$

Answer 1

$f\left(x\right)=\frac{a{x}^2+b}{x^2+1}$

Also ${lim}_{x\to 0}f\left(x\right)=1...\left(i\right)\left[given\right]$

$\Rightarrow {lim}_{x\to 0}\frac{a{x}^2+b}{x^2+1}=1$

$\Rightarrow \frac{{lim}_{x\to 0}a{x}^2+b}{{lim}_{x\to 0}{x}^2+1}$

$\Rightarrow b=1\left[from\right(i\left)\right]$

Also,it is given that ${lim}_{x\to \infty }f\left(x\right)=1$

$\Rightarrow {lim}_{x\to 0}\frac{a{x}^2+b}{x^2+1}=1...\left(ii\right)$

$\Rightarrow {lim}_{x\to 0}\frac{a{x}^2+b}{x^2+1}=1\Rightarrow {lim}_{x\to 0}\frac{x^2(a+\frac{1}{x^2})}{x^2(1+\frac{1}{x^2})}=1$

$\Rightarrow {lim}_{x\to \infty }\frac{a+\frac{1}{x^2}}{1+\frac{1}{x^2}}=1\left[\frac{\infty }{\infty }form\right]$

$\Rightarrow a=1$

$Then,f\left(x\right)=\frac{a{x}^2+1}{x^2+1}=\frac{x^2+1}{x^2+1}=1\left[from\right(ii\left)\right]$

f(-2)=1

f(2)=1

f(-2)=1=f(2)

Hence,proved.