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  • Number of solutions of x  where its satisfy <img alt="(\sin ^{-1} x)^2-2 \sin^{-1}x +1 \leq 0" src="https://learn.careers360.com/latex-image/?%28%5Csin%20%5E%7B-1%7D%20x%29%5E2-2%20%

Number of solutions of x  where its satisfy (\sin ^{-1} x)^2-2 \sin^{-1}x +1 \leq 0

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

\infty


Answers (1)

best_answer

 

 

Domain and range of Inverse Trigonometric Function (Part 1) -

Domain and range of Inverse Trigonometric Function (Part 1)

y = sin-1(x)

The function y = sin(x) is many one so it is not invertible. Now consider the small portion of the function \mathrm{y=\sin x,\;x\in\left [ -\frac{\pi}{2},\frac{\pi}{2}\right ]\;\;and\;\;y=[-1,1]}

 

             


 

Which is strictly increasing, Hence, one-one and inverse is y = sin-1(x)

 

\mathrm{Domain\;is\;[-1,1]\;\;and\;\;Range\;\;is\;\;\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]}
 

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(\sin ^{-1} x)^2-2 \sin^{-1}x +1 \leq 0\\ (\sin ^{-1} x-1)^2 \leq 0\\ \because \text{ L.H.S. is } \geq 0\\ \therefore \text{only possible solution } is \\ \sin ^{-1} x=1\\ x= \sin 1\\

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Deependra Verma

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