Question

Let $S = \left\{ {x \in R:x \geqslant 0 and2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0} \right\}$. Then $S:$
1. contains exactly two elements
2. contains exactly four elements
3. is an empty set
4. contains exactly one element

Answer 1

Hint: In this question, we are given the set $S = \left\{ {x \in R:x \geqslant 0 and 2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0} \right\}$ and we have to find the number of elements in given set. First step is to let $\sqrt x - 3 = y$ and put in the set $2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$. Solve until you will get the quadratic equation and calculate the roots of required equation. In last put the value of roots in the equation $\sqrt x - 3 = y$.

Complete step by step Solution:
Given that,
$S = \left\{ {x \in R:x \geqslant 0and2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0} \right\}$
Here, $2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 - - - - - \left( 1 \right)$
Let, $\sqrt x - 3 = y \Rightarrow \sqrt x = y + 3$
Put the above value in equation (1)
$2\left| y \right| + \left( {y + 3} \right)\left( {y + 3 - 6} \right) + 6 = 0$
$2\left| y \right| + \left( {y + 3} \right)\left( {y - 3} \right) + 6 = 0$
$2\left| y \right| + {y^2} - {3^2} + 6 = 0$
$2\left| y \right| + {y^2} - 3 = 0$
${\left| y \right|^2} + 2\left| y \right| - 3 = 0$
Write the term ${y^2}$ as ${\left| y \right|^2}$,
${\left| y \right|^2} + 3\left| y \right| - \left| y \right| - 3 = 0$
Solve the above quadratic equation,
$\left| y \right|\left( {\left| y \right| + 3} \right) - 1\left( {\left| y \right| + 3} \right) = 0$
$\left( {\left| y \right| - 1} \right)\left( {\left| y \right| + 3} \right) = 0$
$\left| y \right| = 1,\left| y \right| \ne - 3$
Here the value of $\left| y \right|$cannot be negative
Therefore, $\left| y \right| = 1 \Rightarrow y = \pm 1$
At $y = 1$,
$\sqrt x - 3 = 1$
$x = 16$
And at$y = - 1$,
$\sqrt x - 3 = - 1$
$x = 4$

Hence, the correct option is 1.

Note: Students must know that this question can also be solved by like we can take two cases means first we can modify the modulus of given conditions equation $2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$by taking first case as $\sqrt x < 3$and the second will be $\sqrt x > 3$. By solving the both the cases we’ll get two quadratic equations and when we will solve them same two elements will be there. Also, while solving the quadratic equation either we can find the roots by multiplying first$\left( a \right)$ and third $\left( c \right)$constant to break them as second constant $\left( b \right)$or we can apply the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where the general quadratic equation is $a{x^2} + bx + c = 0$.