RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

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Question : 18 of 120

Marks: +1, -0

If

2x^2 cos 60º – 4 cot^2 45º – 2

tan

60º = 0,

then what is the value of x?

2
3
$√ 3 -1$
$√ 3 +1$

Solution:

(d)

Given, 2x^2 cos 60º – 4 cot^2 45º – 2 tan 60º = q

⇒2x^2 ×1/2– 4(1)^2 – 2 ×√ 3 = 0

⇒

x^2 – 4 – 2√ 3 = 0

⇒x^2 = 4 + 2 √3

⇒x^2 = 3 + 1 + 2√ 3

⇒ x^2 = (√3 )^2+ (1 )^ 2 +2√ 3 .1

⇒ x^2 = (√3_1 )^2

⇒x = √3 + 1

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