Verified answer

Answer:

Step-by-step explanation:

We have,

[tex]f(x)=\dfrac{{x}^{2}+1}{{x}^{2}+3}[/tex]

[tex]\rm{Let\,\,\,y=\dfrac{{x}^{2}+1}{{x}^{2}+3}}[/tex]

[tex]\rm{\implies\,y\left({x}^{2}+3\right)={x}^{2}+1}[/tex]

[tex]\rm{\implies\,y\,{x}^{2}+3\,y={x}^{2}+1}[/tex]

[tex]\rm{\implies\,y\,{x}^{2}-{x}^{2}=1-3\,y}[/tex]

[tex]\rm{\implies\,(y-1){x}^{2}=1-3\,y}[/tex]

[tex]\rm{\implies\,{x}^{2}=\dfrac{1-3y}{y-1}}[/tex]

[tex]\rm{\implies\,x=\sqrt{\dfrac{1-3y}{y-1}}}[/tex]

Since the square root is a positive quantity, so, the stuff inside it will be greater than or equal to 0,

[tex]\rm{\implies\dfrac{1-3y}{y-1}\ge0}[/tex]

[tex]\rm{\implies\dfrac{3y-1}{y-1}\le0}[/tex]

CASE I :

[tex]\tt{3y-1\le0\,\,\,\,\,and\,\,\,\,\,y-1>0}[/tex]

[tex]\tt{\implies\,y\le\dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\,y>1}[/tex]

So, no value of satisfies the given inequality,

CASE II :

[tex]\tt{3y-1\ge0\,\,\,\,\,and\,\,\,\,\,y-1<0}[/tex]

[tex]\tt{\implies\,y\ge\dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\,y<1}[/tex]

So, the required solution

[tex]\boxed{\green{\sf{range\Big(f(x)\Big)\in\left[\dfrac{1}{3},1\right)}}}[/tex]