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If x is real, then \frac{x^{2}-2x+4}{x^{2}+2x+4}  takes value in the interval 

  • Option 1)

    \left [ \frac{1}{3}, 3 \right ]

  • Option 2)

    \left ( \frac{1}{3}, 3\right )

  • Option 3)

    \left ( 3,3 \right )

  • Option 4)

    \left ( \frac{-1}{3},3 \right )

 

Answers (1)

best_answer

As we learnt in 

Range -

The range of the relation R is the set of all second elements of the ordered pairs in a relation R.

- wherein

eg. R={(a,b),(c,d)}. Then Range is {b,d}

 

 y=\frac{x^{2}-2x+4}{x^{2}+2x+4}\\*\\*x^{2}y+2xy+4y=x^{2}-2x+4\\*\\*x^{2}\left ( y-1 \right )+x\left ( 2y+2 \right )+4\left ( y-1 \right )=0\\*\\*D=4\left ( y+1 \right )^{2}-16\left ( y-1 \right )^{2}\geq 0\\*\\*\left ( y+1 \right )^{2}-\left [ 2\left ( y-1 \right ) \right ]^{2}\geq 0\\*\\*\left ( y+1+2y-2 \right )\left ( y+1-2y+2 \right )\geq 0\\*\\*\left ( 3y-1 \right )\left ( y-3 \right )\leq 0\\*\\*y\in \left [ \frac{1}{3} ,3\right ]


Option 1)

\left [ \frac{1}{3}, 3 \right ]

Correct

Option 2)

\left ( \frac{1}{3}, 3\right )

Incorrect

Option 3)

\left ( 3,3 \right )

Incorrect

Option 4)

\left ( \frac{-1}{3},3 \right )

Incorrect

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