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The solution set of \frac{a^{2} -3a+4}{a+1} > 1, a \epsilon R is 

  • Option 1)

    \left ( 3, \infty \right )

  • Option 2)

    \left ( -1, 1 \right ) U \left ( 3, \infty \right )

  • Option 3)

    \left [ -1, 1 \right ] U \left [ 3, \infty \right )

  • Option 4)

    None. 

 

Answers (1)

best_answer

As learnt in concept

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}

 

and wavy curves, 

\frac{a^{2}-3a+4}{a+1}-1 \geqslant 0

\frac{a^{2}-3a+4-a-1}{a+1} \geqslant 0

\frac{a^{2}-4a+3}{a+1} \geqslant 0

(a-1)(a-3)(a+1) \geqslant 0

a\:\in [-1,1]\:\cup [3, \infty]

 


Option 1)

\left ( 3, \infty \right )

This option is incorrect.

Option 2)

\left ( -1, 1 \right ) U \left ( 3, \infty \right )

This option is correct.

Option 3)

\left [ -1, 1 \right ] U \left [ 3, \infty \right )

This option is incorrect.

Option 4)

None. 

This option is incorrect.

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prateek

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