JIPMAT 2025 (QA) - 2+sqrt(3) x 2+2+sqrt(3) x 2+2+2+sqrt(3) x 2-2+2+sqrt(3) is equal to | PYQs + Solutions

JIPMAT 2025

Algebra

Indices

Hard

$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

1

2

4

\sqrt{6}

✅ Correct Option: 1

\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}

Let's define:

a = \sqrt{3}

b = \sqrt{2+a} = \sqrt{2+\sqrt{3}}

c = \sqrt{2+b} = \sqrt{2+\sqrt{2+\sqrt{3}}}

d = \sqrt{2+c} = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}

Our expression is then:

b \times c \times d \times \sqrt{2-c}

The key insight is that these nested square roots form a pattern where:

\newline

b^2 = 2+a

\newline

c^2 = 2+b

\newline

d^2 = 2+c

From this, we get:

\newline

c^2-2 = b

\newline

d^2-2 = c

For the term

\sqrt{2-c}

, note that:

\sqrt{2-c} \times d = \sqrt{2-c} \times \sqrt{2+c} = \sqrt{(2-c)(2+c)} = \sqrt{4-c^2} = \sqrt{4-(2+b)} = \sqrt{2-b}

Similarly:

\newline

\sqrt{2-b} \times c = \sqrt{2-b} \times \sqrt{2+b} = \sqrt{4-b^2} = \sqrt{4-(2+a)} = \sqrt{2-a}

And:

\newline

\sqrt{2-a} \times b = \sqrt{2-a} \times \sqrt{2+a} = \sqrt{4-a^2} = \sqrt{4-3} = \sqrt{1} = 1

Therefore:

\newline

b \times c \times d \times \sqrt{2-c}

\newline

\newline

b \times c \times \sqrt{2-b} = b \times 1 = 1

b \times \sqrt{2-a}=1

So the value of the given expression is

1