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AP TET Paper 2 Math and Science Model Paper 4

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Question : 109 of 150

Marks: +1, -0

If

\sqrt{1+\frac{\sqrt{3}}{2}}- \sqrt{1-\frac{\sqrt{3}}{2}}= c

, then the value of c is:

1
4
3
2

Solution:

Concept:

Squaring the difference of two conjugate square roots eliminates the outer radicals.

Use the identity

(x-y)^2 = x^2 + y^2 - 2xy

.

Explanation:

Let

A = 1 + \frac{\sqrt{3}}{2}

and

B = 1 - \frac{\sqrt{3}}{2}

.

Then

c = \sqrt{A} - \sqrt{B}

.

Square both sides:

c^2 = A + B - 2\sqrt{AB}

.

Now

A + B = \left(1 + \frac{\sqrt{3}}{2}\right) + \left(1 - \frac{\sqrt{3}}{2}\right) = 2

.

Also

AB = \left(1 + \frac{\sqrt{3}}{2}\right)\left(1 - \frac{\sqrt{3}}{2}\right) = 1 - \frac{3}{4} = \frac{1}{4}

.

So

c^2 = 2 - 2\sqrt{\frac{1}{4}} = 2 - 2\left(\frac{1}{2}\right) = 2 - 1 = 1

.

Since

\sqrt{1+\frac{\sqrt{3}}{2}} > \sqrt{1-\frac{\sqrt{3}}{2}}

, the value of

c

is positive.

Therefore

c = \sqrt{1} = 1

.

Answer:

c = 1

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