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Question: If \(x + \dfrac{1}{x} = 2\), then find the principal value of \({\sin ^{ - 1}}x\). A. \(\dfrac{\pi...

If x+1x=2x + \dfrac{1}{x} = 2, then find the principal value of sin⁡−1x{\sin ^{ - 1}}x.
A. π4\dfrac{\pi }{4}
B. π2\dfrac{\pi }{2}
C. π\pi
D. 3π4\dfrac{{3\pi }}{4}

Explanation

Solution

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+1x=2x + \dfrac{1}{x} = 2.
Multiplying by x the whole equation, we will get a quadratic equation as follows:-
⇒x2+xx=2x\Rightarrow {x^2} + \dfrac{x}{x} = 2x
The equation given above can be re – written as following expression:-
⇒x2+1−2x=0\Rightarrow {x^2} + 1 - 2x = 0
Now, we can modify the terms and re – write the same equation like the following equation:-
⇒(x)2+(1)2−2×1×x=0\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:-
⇒(a−b)2=a2+b2−2ab\Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab
Replacing a by x and b by 1 in the above formula, we will get:-
⇒(x−1)2=x2+12−2x\Rightarrow {(x - 1)^2} = {x^2} + {1^2} - 2x
Putting the above derived expression in equation 1, we will then assume the following:-
⇒(x−1)2=0\Rightarrow {\left( {x - 1} \right)^2} = 0
Therefore, we have the roots as x = 1, 1.
Now, let us put x = 1 in sin⁡−1x{\sin ^{ - 1}}x, we have to find the principal value of sin⁡−11{\sin ^{ - 1}}1 which is definitely equal to π2\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⁡−11{\sin ^{ - 1}}1.
We can do the same by assuming z = sin⁡−11{\sin ^{ - 1}}1.
Now, taking sin from right hand side to left hand side, we will then obtain:-
⇒sin⁡z=1\Rightarrow \sin z = 1
And, this is true when z is equal to π2\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+1x=2x + \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.