Question

If for the real value of $x$,$\cos \theta = x + \dfrac{1}{x}$, then
A. $\theta $ is an acute angle
B. $\theta $is a right angle
C. $\theta $ is an obtuse angle
D. No value of $\theta $ is possible

Answer 1

Hint: We are asked to find the value of $\theta $for the real value of $x$So, we will consider a quadratic equation in form of $x$. Now we know that the value of $x$belongs to the real number set. So, we find the discrimination of the quadratic equation to find the value of $\theta $

Formula Used:
We have used the following formula:
$D = {b^2} - 4ac$

Complete step by step solution:
We are given that $\cos \theta = x + \dfrac{1}{x}$
Now we consider the given equation in quadratic form as:
$
 \cos \theta = x + \dfrac{1}{x} \\
 \cos \theta = \dfrac{{{x^2} + 1}}{x} \\
 x\,\cos \theta = {x^2} + 1 \\
 $
$
 {x^2} + 1 - x\,\cos \,\theta = 0 \\
 {x^2} - x\,\cos \theta \, + 1\, = \,0 \\
 $
Now we know that the values of $x$ belong to the real number set. So, the discriminant of the quadratic equation must be greater than or equal to zero for $x$ to assume all the real values.
Thus, we first find the discriminant of the quadratic equation ${x^2} - x\,\cos \theta \, + 1\, = \,0$with the standard form of the quadratic equation $a{x^2} + bx + c = 0$, we get
$
 a = 1 \\
 b = - \cos \theta \\
 c = 1 \\
 $
We know that $D = {b^2} - 4ac$
$
 D = {\left( { - \cos \theta } \right)^2} - 4 \times 1 \times 1 \\
 = {\left( { - \cos \theta } \right)^2} - 4 \\
 $
Now express our discriminant greater than or equal to zero:
${\left( { - \cos \theta } \right)^2} - 4 \geqslant 0$
$
 {\left( { - \cos \theta } \right)^2} - 4 \geqslant 0 \\
 {\cos ^2}\theta - 4 \geqslant 0 \\
 {\cos ^2}\,\theta \geqslant 4 \\
 {\cos ^2}\theta \geqslant {\left( 2 \right)^2} \\
 $
Further solving we get,
$\cos \theta \geqslant \pm 2$
Now as we know that $ - 1 \leqslant \cos \theta \leqslant 1$
Therefore, no value of $\theta $is possible.

Option ‘D’ is correct

Note: A function's range is the set of values that the function assumes for different values of variables in the domain. Students should remember the standard formula of a quadratic equation and the discriminant formula to find the roots of the equation i.e, $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $b^2 - 4ac$ is the discriminant, also remember the range of cosine.