Question

If $x + \dfrac{1}{x} = 2$, then find the principal value of ${\sin ^{ - 1}}x$.
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{2}$
C. $\pi $
D. $\dfrac{{3\pi }}{4}$

Answer 1

Hint: We will first form a quadratic equation using the equation given us and then find the roots of it. Now, put those roots in the function whose principal value we require.

Complete step-by-step answer:
We are given that $x + \dfrac{1}{x} = 2$.
Multiplying by x the whole equation, we will get a quadratic equation as follows:-
$ \Rightarrow {x^2} + \dfrac{x}{x} = 2x$
The equation given above can be re – written as following expression:-
$ \Rightarrow {x^2} + 1 - 2x = 0$
Now, we can modify the terms and re – write the same equation like the following equation:-
$ \Rightarrow {\left( x \right)^2} + {\left( 1 \right)^2} - 2 \times 1 \times x = 0$ …………..(1)
Now, we will use the formula given by the following expression:-
$ \Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab$
Replacing a by x and b by 1 in the above formula, we will get:-
$ \Rightarrow {(x - 1)^2} = {x^2} + {1^2} - 2x$
Putting the above derived expression in equation 1, we will then assume the following:-
$ \Rightarrow {\left( {x - 1} \right)^2} = 0$
Therefore, we have the roots as x = 1, 1.
Now, let us put x = 1 in ${\sin ^{ - 1}}x$, we have to find the principal value of ${\sin ^{ - 1}}1$ which is definitely equal to $\dfrac{\pi }{2}$.

Hence, the correct option is (B).

Note:
The students must note that in the last second step, where we found the value of ${\sin ^{ - 1}}1$.
We can do the same by assuming z = ${\sin ^{ - 1}}1$.
Now, taking sin from right hand side to left hand side, we will then obtain:-
 $ \Rightarrow \sin z = 1$
And, this is true when z is equal to $\dfrac{\pi }{2}$. Hence, we have our answer.
The students must also note that in starting few step only, we multiplied the equation by x, we could do that because we know that x is not equal to 0 since we are given a function with x in denominator which is $x + \dfrac{1}{x} = 2$. Here, if x would have been zero, this function would not have been defined in the initial only. So, you must always find a way to discard the possibility of x being zero, while multiplying or dividing by it.