Question

The Solution set of $x - \sqrt {1 - \left| x \right|} < 0$ is
1. $\left[ { - 1,\dfrac{{\left( { - 1 + \sqrt 5 } \right)}}{2}} \right]$
2. $\left[ { - 1,1} \right]$
3. $\left[ { - 1,\dfrac{{\left( {1 + \sqrt 5 } \right)}}{2}} \right]$
4. $\left[ {1, - 1} \right]$

Answer 1

Hint:Start by solving the square root condition of the given inequality. Now Solving inequality till the required equation will be a quadratic equation. Find the roots and compare both situations and write the values.

Formula Used:
Quadratic inequality –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step Solution:
Given that,
$x - \sqrt {1 - \left| x \right|} < 0 - - - - - (1)$
From above inequality,
$1 + \left| x \right| \geqslant 0$
$1 \geqslant \left| x \right|$
$1 \geqslant \pm x$
$ \Rightarrow - 1 \leqslant x \leqslant 1 - - - - - (2)$
Now from equation (1),
$x < \sqrt {1 - \left| x \right|} $
${x^2} > 1 - \left| x \right|$
${x^2} > 1 - x$
${x^2} + x - 1 > 0$
Compare the required quadratic inequality with general quadratic equation $a{x^2} + bx + c$
$ \Rightarrow a = 1,b = 1,c = - 1$
Using quadratic formula,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
$x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$
$x = \dfrac{{ - 1 + \sqrt 5 }}{2}\left( {x \ne \dfrac{{ - 1 - \sqrt 5 }}{2} < - 1} \right) - - - - - (3)$
From equation (2) and (3)
$x \in \left[ { - 1,\dfrac{{\left( { - 1 + \sqrt 5 } \right)}}{2}} \right]$
$ \Rightarrow $Option (1) is the correct .


Therefore, the correct option is 1.

Note: In such type of question, the value inside the square root should be greater than or equal to zero. Solve the inequality the same as equality but multiplying and dividing any number in inequality change the sign of inequality. If the required values are equal also use the square(closed) bracket and if the values are only less or greater then apply a circular(open) bracket.