If 2x² 60^{} - 4 ² 45^{} - 2 tan60^{} = 0, then what is the value of x?

Correct Answer is : 3+1{3} + 13+1

Correct Answer is : 3+1{3} + 13+1

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RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

Question : 18 of 120

Marks: +1, -0

2x^2 \cos 60^{\circ} - 4 \cot^2 45^{\circ} - 2

60^{\circ} = 0,

then what is the value of x?

Solution:

Concept:

Use standard trigonometric values and simplify the equation.

Explanation:

Given:

2x^2 \cos 60^\circ - 4 \cot^2 45^\circ - 2\tan 60^\circ = 0

Substitute

\cos 60^\circ = \frac{1}{2}

\cot 45^\circ = 1

, and

\tan 60^\circ = \sqrt{3}

2x^2 \times \frac{1}{2} - 4(1)^2 - 2\sqrt{3} = 0

x^2 - 4 - 2\sqrt{3} = 0

x^2 = 4 + 2\sqrt{3}

Now,

4 + 2\sqrt{3} = (\sqrt{3})^2 + (1)^2 + 2\sqrt{3}(1)

4 + 2\sqrt{3} = (\sqrt{3} + 1)^2

So,

x = \pm(\sqrt{3} + 1)

From the given options, the positive value is

\sqrt{3} + 1

Answer:

x = \sqrt{3} + 1

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