If for real values of x,cos theta = x + frac{ | vedclass

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(D) Given the equation $\cos 	heta = x + \frac{1}{x}$.Multiplying by $x$,we get $x^2 - x \cos 	heta + 1 = 0$.Since $x$ is a real value,the discrimin

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No value of $	heta$ is possible

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(D) Given the equation $\cos 	heta = x + \frac{1}{x}$.Multiplying by $x$,we get $x^2 - x \cos 	heta + 1 = 0$.Since $x$ is a real value,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.$D = b^2 - 4ac = (-\cos 	heta)^2 - 4(1)(1) \ge 0$.$\cos^2 	heta - 4 \ge 0 \Rightarrow \cos^2 	heta \ge 4$.However,the range of $\cos 	heta$ is $[-1, 1]$,so $\cos^2 	heta$ must be in the range $[0, 1]$.Since $\cos^2 	heta \ge 4$ is impossible for any real $	heta$,there is no real value of $x$ that satisfies the given equation for any $	heta$.Therefore,no value of $	heta$ is possible.

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

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