Math, asked by rahasohail578, 19 days ago

number of positive integers in the range of f(x)=x²+1/x²+3 will be​

Answers

Answered by senboni123456
0

Answer:

Step-by-step explanation:

We have,

f(x)=\dfrac{{x}^{2}+1}{{x}^{2}+3}

\rm{Let\,\,\,y=\dfrac{{x}^{2}+1}{{x}^{2}+3}}

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

\rm{\implies\,y\,{x}^{2}+3\,y={x}^{2}+1}

\rm{\implies\,y\,{x}^{2}-{x}^{2}=1-3\,y}

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

\rm{\implies\,{x}^{2}=\dfrac{1-3y}{y-1}}

\rm{\implies\,x=\sqrt{\dfrac{1-3y}{y-1}}}

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

\rm{\implies\dfrac{1-3y}{y-1}\ge0}

\rm{\implies\dfrac{3y-1}{y-1}\le0}

CASE I :

\tt{3y-1\le0\,\,\,\,\,and\,\,\,\,\,y-1>0}

\tt{\implies\,y\le\dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\,y>1}

So, no value of satisfies the given inequality,

CASE II :

\tt{3y-1\ge0\,\,\,\,\,and\,\,\,\,\,y-1<0}

\tt{\implies\,y\ge\dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\,y<1}

So, the required solution

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

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