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  • Can someone explain - Sets, Relations and Functions - JEE Main-2

Find the range of the function f(x)=\frac{2x-2}{x^{2}-2x+3},

  • Option 1)

    \left [ -1/2,1/2 \right ]

     

     

     

  • Option 2)

    \left [-1,1 \right ]

  • Option 3)

    \left [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ]

  • Option 4)

    \left [-2,2 \right ]

 

Answers (1)

best_answer

As we learned 

 

Rational Function -

f(x )= \frac{\rho \left ( x \right )}{q\left ( x \right )} Where \rho \left ( x \right )\: \: and\: \: q\left ( x \right ) polynomials in x.

 

 

- wherein

Domin of this function is R-\left \{ x:q\left ( x \right )= 0 \right \} Range depends on function.

 

 Here f(x)= \frac{2x-2}{x^{2}-2x+3}

Let y=f(x)= \; i.e\; y=\frac{2x-2}{x^{2}-2x+3}

or yx^{2}-2\left ( y+1 \right )x+3y+2=0

which is a quadratic in x. For the above quadratic equation to have real roots, 

\Delta \geq 0\Rightarrow \left ( y +1\right )^{2}-y\left ( 3y+2 \right )\geq 0\Rightarrow y^{2}\leq \frac{1}{2}\Rightarrow -\frac{1}{\sqrt{2}}\leq y\leq \frac{1}{\sqrt{2}}.

Hence the range of f(x) is \left [ -\frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right ].

 


Option 1)

\left [ -1/2,1/2 \right ]

 

 

 

Option 2)

\left [-1,1 \right ]

Option 3)

\left [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ]

Option 4)

\left [-2,2 \right ]

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gaurav

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