Question

If $\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0$, where ‘x’ is a non-negative real number and $\left[ . \right]$ denotes the greatest integer function, then complete set of values of x is
A) (cos1, 1)
B) (-1, cos1)
C) (sin1, 1)
D) (cos1, sin1)

Answer 1

Hint: According to given in the question that x’ is a non-negative real number hence, we can say that $x \geqslant 0$. Now, we will find the domain of $\left[ {{{\sin }^{ - 1}}x} \right]$ and $\left[ {{{\cos }^{ - 1}}x} \right]$
Hence, for $\left[ {{{\sin }^{ - 1}}x} \right]$ x will lie between 0, 1 because as we know that $\sin {0^0} = 0$and $\sin {90^0} = 1$ therefore, for $\left[ {{{\sin }^{ - 1}}x} \right]$, $x \in [0,1]$
And, for $\left[ {{{\cos }^{ - 1}}x} \right]$ x will lie between 1, 0 because as we know that ${\cos ^0} = 1$and $\cos {90^0} = 0$
Therefore, for $\left[ {{{\cos }^{ - 1}}x} \right]$, $x \in [1,0]$
So, we can say that,
${\sin ^{ - 1}}x \in \left[ {0,\dfrac{\pi }{2}} \right]$ and ${\cos ^{ - 1}}x \in \left[ {\dfrac{\pi }{2},0} \right]$
Now, According to given in the question that $\left[ {{{\sin }^{ - 1}}x} \right]$= 0 and $\left[ {{{\cos }^{ - 1}}x} \right]$= 0 so, from here we can find the value of x.

Complete step by step answer:
Given,
$\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0$
Where, ‘x’ is a non-negative real number
Means $x \geqslant 0$
And $\left[ . \right]$ denotes the greatest integer function.
Step 1: First of all we will find the domain of $\left[ {{{\sin }^{ - 1}}x} \right]$ as we know that x will lie between 0, 1 because as we know that $\sin {0^0} = 0$and $\sin \dfrac{\pi }{2} = 1$ therefore, $\left[ {{{\sin }^{ - 1}}x} \right]$, $x \in [0,1]$
Step 2: Now, we will find the domain of $\left[ {{{\cos }^{ - 1}}x} \right]$ as we know that x will lie between 0, 1 because as we know that $\cos {0^0} = 1$ and $\sin \dfrac{\pi }{2} = 0$ therefore, $\left[ {{{\cos }^{ - 1}}x} \right]$, $x \in [1,0]$
Step 3: Hence, from the step 1 and step 2 we can obtain that,
${\sin ^{ - 1}}x \in \left[ {0,\dfrac{\pi }{2}} \right]$ and, ${\cos ^{ - 1}}x \in \left[ {\dfrac{\pi }{2},0} \right]$
Step 4: Now, as given in the question that $\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0$ hence, we can say that
$\Rightarrow \left[ {{{\sin }^{ - 1}}x} \right] = 0$When, $x \in [0,\sin 1]$………………………………….(1)and,
$\Rightarrow \left[ {{{\cos }^{ - 1}}x} \right] = 0$When, $x \in [\cos 1,1]$…………………………………..(2)
Step 5: Now, we will take the intersection of both (1) and (2) to find that, from where the value of x will lie.
Hence,
$\Rightarrow x \in (\cos 1,\sin 1)$

If $\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0$, where ‘x’ is a non-negative real number and $\left[ . \right]$denotes the greatest integer function, then complete set of values of x is: (cos1,sin1). Therefore option (D) is correct.

Note:
We can also take the help of the sin and $\cos $graphs to find the liability of x for sin and cos.
For this case, the maximum value of sin is 1 at $\sin \dfrac{\pi }{2}$and the minimum value of sin is 0 at $\sin {0^0}$same as maximum value of $\cos $ is 1 at $\cos {0^0}$and the minimum value of $\cos $ is 0 at $\cos \dfrac{\pi }{2}$
‘x’ is a non-negative real number hence, we can say that $x \geqslant 0$