Question

For real value of x, x +1/x = 2 cos 0 , then find cos 0. 1​

Answer 1

Answer:

Correct option is

D

no value of θ is possible

Given equation is cosθ=x+

x

1

⇒x

2

−xcosθ+1=0

For real values of x,b

2

>4ac

∵a=1,b=−cosθ,c=1

∴cos

2

θ>4(1)(1)⇒∣cosθ∣>2

Which is not possible as ∣cosθ∣≤1

∴ No real value of θ is possible.

Answer 2

✦✧✧ Answer ✧✧✦

Given,

cos θ = x + 1/x

cos θ = (x2 + 1)/x

⇒ x2 + 1 = x cos θ

⇒ x2 – x cos θ + 1 = 0

We know that for any real root of the equation ax2 + bx + c = 0,

b2 – 4ac ≥ 0

⇒ (-cos θ)2 – 4 ≥ 0

⇒ cos2θ – 4 ≥ 0

⇒ cos2θ ≥ 4

⇒ cos θ ≥ ± 2

As we know -1 ≤ cos θ ≤ 1

Hence, no value of θ is possible