Question

If for the real value of x, $\cos \left(\theta \right)=\mathrm{x}+\frac{1}{\mathrm{x}}$, then

• A. $\mathrm{\theta }$ is an acute angle
• B. $\mathrm{\theta }$ is a right angle
• C. $\mathrm{\theta }$ is an obtuse angle
• D. No value of $\mathrm{\theta }$ is possible

Answer 1

The correct option is D

No value of $\mathrm{\theta }$ is possible

Explanation for the correct option

Step 1: Formation of the equation

Since $\cos \left(\theta \right)=\mathrm{x}+\frac{1}{\mathrm{x}}$,

$\begin{array}{rcl}\therefore \cos \left(\mathrm{\theta }\right)& =& \frac{{\mathrm{x}}^2+1}{\mathrm{x}}\\ & \Rightarrow & x\cos \left(\mathrm{\theta }\right)={\mathrm{x}}^2+1\\ & \Rightarrow & {\mathrm{x}}^2-x\cos \left(\mathrm{\theta }\right)+1=0\end{array}$

Step 2: Determination of the real value for x

For the real value, we know that the discriminant$\left(D\right)$ of a quadratic equation $a{x}^2+bx+c=0$ is greater than or equal to zero.

Now, the discriminant is calculated as $b^2-4ac$.

From the quadratic equation $\begin{array}{rcl}{\mathrm{x}}^2-x\cos \left(\mathrm{\theta }\right)+1& =& 0\end{array}$ we have $\mathrm{a}=1,\mathrm{b}=-\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left(\mathrm{\theta }\right)\end{array},\text{and}\mathrm{c}=1$.

$\begin{array}{rcl}\therefore {\left(-\cos \left(\mathrm{\theta }\right)\right)}^2-4\left(1\right)\left(1\right)& \ge & 0\\ & \Rightarrow & {\cos }^2\left(\mathrm{\theta }\right)\ge 4\\ & \Rightarrow & \cos \left(\mathrm{\theta }\right)\ge \pm 2\end{array}$

Since $\cos \left(\theta \right)$ cannot be lesser than $-1$ or greater than $1$. Therefore, no real value of $\mathrm{\theta }$ is possible.

Hence, the correct option is (D).