Q

# Try this! - Algebra - BITSAT-7

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.

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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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