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Question: If \(\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0\), where ‘x’ is a n...

If [sin⁡−1x]+[cos⁡−1x]=0\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)

Explanation

Solution

According to given in the question that x’ is a non-negative real number hence, we can say that x⩾0x \geqslant 0. Now, we will find the domain of [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right] and [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right]
Hence, for [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right] x will lie between 0, 1 because as we know that sin⁡00=0\sin {0^0} = 0and sin⁡900=1\sin {90^0} = 1 therefore, for [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right], x∈[0,1]x \in [0,1]
And, for [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right] x will lie between 1, 0 because as we know that cos⁡0=1{\cos ^0} = 1and cos⁡900=0\cos {90^0} = 0
Therefore, for [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right], x∈[1,0]x \in [1,0]
So, we can say that,
sin⁡−1x∈[0,π2]{\sin ^{ - 1}}x \in \left[ {0,\dfrac{\pi }{2}} \right] and cos⁡−1x∈[π2,0]{\cos ^{ - 1}}x \in \left[ {\dfrac{\pi }{2},0} \right]
Now, According to given in the question that [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right]= 0 and [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right]= 0 so, from here we can find the value of x.

Complete step by step answer:
Given,
[sin⁡−1x]+[cos⁡−1x]=0\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0
Where, ‘x’ is a non-negative real number
Means x⩾0x \geqslant 0
And [.]\left[ . \right] denotes the greatest integer function.
Step 1: First of all we will find the domain of [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right] as we know that x will lie between 0, 1 because as we know that sin⁡00=0\sin {0^0} = 0and sin⁡π2=1\sin \dfrac{\pi }{2} = 1 therefore, [sin⁡−1x]\left[ {{{\sin }^{ - 1}}x} \right], x∈[0,1]x \in [0,1]
Step 2: Now, we will find the domain of [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right] as we know that x will lie between 0, 1 because as we know that cos⁡00=1\cos {0^0} = 1 and sin⁡π2=0\sin \dfrac{\pi }{2} = 0 therefore, [cos⁡−1x]\left[ {{{\cos }^{ - 1}}x} \right], x∈[1,0]x \in [1,0]
Step 3: Hence, from the step 1 and step 2 we can obtain that,
sin⁡−1x∈[0,π2]{\sin ^{ - 1}}x \in \left[ {0,\dfrac{\pi }{2}} \right] and, cos⁡−1x∈[π2,0]{\cos ^{ - 1}}x \in \left[ {\dfrac{\pi }{2},0} \right]
Step 4: Now, as given in the question that [sin⁡−1x]+[cos⁡−1x]=0\left[ {{{\sin }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0 hence, we can say that
⇒[sin⁡−1x]=0\Rightarrow \left[ {{{\sin }^{ - 1}}x} \right] = 0When, x∈[0,sin⁡1]x \in [0,\sin 1]………………………………….(1)and,
⇒[cos⁡−1x]=0\Rightarrow \left[ {{{\cos }^{ - 1}}x} \right] = 0When, x∈[cos⁡1,1]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,
⇒x∈(cos⁡1,sin⁡1)\Rightarrow x \in (\cos 1,\sin 1)

If [sin⁡−1x]+[cos⁡−1x]=0\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⁡\cos graphs to find the liability of x for sin and cos.
For this case, the maximum value of sin is 1 at sin⁡π2\sin \dfrac{\pi }{2}and the minimum value of sin is 0 at sin⁡00\sin {0^0}same as maximum value of cos⁡\cos is 1 at cos⁡00\cos {0^0}and the minimum value of cos⁡\cos is 0 at cos⁡π2\cos \dfrac{\pi }{2}
‘x’ is a non-negative real number hence, we can say that x⩾0x \geqslant 0