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Question

Question: If for real values of \(x,\cos\theta = x + \frac{1}{x},\) then...

If for real values of x,cosθ=x+1x,x,\cos\theta = x + \frac{1}{x}, then

A

θ\theta is an acute angle

B

θ\theta is a right angle

C

θ\thetais an obtuse angle

D

No value of θ\thetais possible

Answer

No value of θ\thetais possible

Explanation

Solution

The quadratic equation is x2xcosθ+1=0x^{2} - x\cos\theta + 1 = 0

But x is real, therefore B24AC0B^{2} - 4AC \geq 0

cos2θ4(1)(1)cos2θ4\Rightarrow \cos^{2}\theta \geq 4(1)(1) \Rightarrow \cos^{2}\theta \geq 4, which is impossible.