Question

Give Explanation. Don't post irrelevant answers ❌. Class 11th ​

Answer 1

$\sf\bold{\underline{\pink{Answer:-}}}$

Option (3)

→ Contains exactly Two elements.

⇒Possible Values of x are 4 and 16.

Answer 2

ANSWER :

❖ (3) Contains exactly two elemets.

• ✎ Let S = {x ∈ R: x ≥ 0 and 2 |√x - 3| + √x (√x - 6) + 6 = 0}. Then S contains exactly two elements.

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SOLUTION :

❒ Given :-

• The set is S = {x ∈ R : x ≥ 0 and 2 |√x - 3| + √x (√x - 6) + 6 = 0}

Here, we have,

• x ∈ R : x ≥ 0

• 2 |√x - 3| + √x (√x - 6) + 6 = 0

Suppose,

• √x - 3 = a ----------> (1)

Then

• √x = a + 3 ----------> (2)

Now,

• ★ 2 |√x - 3| + √x (√x - 6) + 6 = 0

⇒ 2 |a| + (a + 3) {(a + 3) - 6} + 6 = 0 [From (1) and (2)]

⇒ 2 |a| + (a + 3) (a + 3 - 6) + 6 = 0

⇒ 2 |a| + (a + 3) (a - 3) + 6 = 0

⇒ 2 |a| + (a² - 3²) + 6 = 0

⇒ 2 |a| + a² - 9 + 6 = 0

⇒ 2 |a| + a² - 3 = 0

⇒ a² + 2 |a| - 3 = 0

This equation can be written as :

• ★ |a|² + 2 |a| - 3 = 0

⇒ |a|² + 3 |a| - |a| - 3 = 0

⇒ |a| (|a| + 3) - 1 (|a| + 3) = 0

⇒ (|a| + 3) (|a| - 1) = 0

So,

• |a| + 3 = 0

➝ |a| = - 3; but |a| ≠ - 3

Or,

• |a| - 1 = 0

➝ |a| = 1

➝ a = ± 1

Again,

• √x - 3 = a

➜ √x - 3 = ± 1

Therefore,

• ✠ √x - 3 = 1

➨ √x = 3 + 1

➨ √x = 4

➨ x = (4)²

∴ x = 16

And,

• ✠ √x - 3 = - 1

➨ √x = 3 - 1

➨ √x = 2

➨ x = (2)²

∴ x = 4

Thus,

• x = 16, 4

It implies that, if set S = {x ∈ R : x ≥ 0 and 2 |√x - 3| + √x (√x - 6) + 6 = 0}, then S = {16, 4}.

Hence, this set S contains exactly two elements.