QJEEM18Ma6e02: JEE Main 2018 Mathematics - Class 11 JEEM18Ma6e02 Solution: Let S = { R : 0 and }. Then S

Question

Verified Solution • 2025–26 Session moderate

Let S = {

x

\in

R :

x

\ge

0 and

2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0

}. Then S

Accepted Answer

Given,

2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0

Case 1 :

When

\sqrt x - 3 \ge 0,

then equation becomes

2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0

\Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0

\Rightarrow \,\,\,\,x\, - 4\sqrt x = 0

\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0

\therefore\,\,\,

\sqrt x = 0,4

but as

\sqrt x - 3 \ge 0

or

\sqrt x \ge 3

then

\sqrt x \ne 0

\therefore\,\,\,

\sqrt x = 4

value is possible.

Case 2 :

When

\sqrt x - 3 &lt; 0

or

\sqrt x &lt; 3.

There equation becomes

- 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0

\Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0

\Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0

\Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0

\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0

\Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0

\therefore\,\,\,

\sqrt x = 2,6

as

\sqrt x &lt; 3

so

\sqrt x \ne 6

\therefore\,\,\,

\sqrt x = 2

is possible.

So, total possible value of

\sqrt x = 2,4

or for x possible values are 4, 16.

\therefore\,\,\,

Set S contains exactly two elements 4 and 16.

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