
If \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\] the value of \[{{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}}\] is ?
A. 0
B. 1
C. 2
D. 13
Answer
162k+ views
Hint:
We have to know the range and values of inverse sine functions to solve the given problem. We know the domain of function varies from ( -1, 1) and ranges from \[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]. To solve the question, we divide the RHS and on the LHS we have a sum of three inverse functions that take the maximum value of \[ \dfrac{\pi }{2}\]. By putting the value equal to \[\dfrac{\pi }{2}\], we get the value of x, y and z. Then we solve the equations to get our answer.
Complete Step- by- step Solution
Given \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\]
This will happen only when \[{{\sin }^{-1}}x={{\sin }^{-1}}y={{\sin }^{-1}}z=\dfrac{\pi }{2}\]
We have to find the value of \[{{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}}\]
We know \[{{\sin }^{-1}}x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]
Sum of three inverse of sine is \[3\times \dfrac{\pi }{2}\]
That is every sine inverse function is equal to \[\dfrac{\pi }{2}\]
That is $x = \dfrac{3\pi }{2} \sin \dfrac{\pi }{2}, y = \sin \dfrac{\pi }{2}, z = \sin \dfrac{\pi }{2}$
This means $\sin \dfrac{\pi }{2}=1$
That is $x = y = z = 1$
Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 1 + 1 + 1-\dfrac{9}{1+1+1}$
By solving the above equation, we get $3-\dfrac{9}{3}= 0$
Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 0$
Thus, Option (A) is correct.
Note The domain of function of sin inverse is ( -1,1) and range is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. That means the maximum value that the inverse sine function can take is $\dfrac{\pi }{2}$. If we observe that the given problem on the RHS value is $\dfrac{3\pi }{2}$ and on the LHS we have a sum of three inverse functions. the sum can be achieved a value of $\dfrac{3\pi }{2}$, if and only if each inverse sine function takes their maximum value. That is the main thing we take care while solving these type of questions.
We have to know the range and values of inverse sine functions to solve the given problem. We know the domain of function varies from ( -1, 1) and ranges from \[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]. To solve the question, we divide the RHS and on the LHS we have a sum of three inverse functions that take the maximum value of \[ \dfrac{\pi }{2}\]. By putting the value equal to \[\dfrac{\pi }{2}\], we get the value of x, y and z. Then we solve the equations to get our answer.
Complete Step- by- step Solution
Given \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\]
This will happen only when \[{{\sin }^{-1}}x={{\sin }^{-1}}y={{\sin }^{-1}}z=\dfrac{\pi }{2}\]
We have to find the value of \[{{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}}\]
We know \[{{\sin }^{-1}}x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]
Sum of three inverse of sine is \[3\times \dfrac{\pi }{2}\]
That is every sine inverse function is equal to \[\dfrac{\pi }{2}\]
That is $x = \dfrac{3\pi }{2} \sin \dfrac{\pi }{2}, y = \sin \dfrac{\pi }{2}, z = \sin \dfrac{\pi }{2}$
This means $\sin \dfrac{\pi }{2}=1$
That is $x = y = z = 1$
Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 1 + 1 + 1-\dfrac{9}{1+1+1}$
By solving the above equation, we get $3-\dfrac{9}{3}= 0$
Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 0$
Thus, Option (A) is correct.
Note The domain of function of sin inverse is ( -1,1) and range is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. That means the maximum value that the inverse sine function can take is $\dfrac{\pi }{2}$. If we observe that the given problem on the RHS value is $\dfrac{3\pi }{2}$ and on the LHS we have a sum of three inverse functions. the sum can be achieved a value of $\dfrac{3\pi }{2}$, if and only if each inverse sine function takes their maximum value. That is the main thing we take care while solving these type of questions.
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