Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁻¹(3x/5) + sin⁻¹(4x/5) = sin⁻¹x is equal to :
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁻¹(3x/5) + sin⁻¹(4x/5) = sin⁻¹x is equal to :
Option 1 -
2
Option 2 -
0
Option 3 -
1
Option 4 -
3
-
1 Answer
-
Correct Option - 3
Detailed Solution:Solve sin? ¹ (3x/5) + sin? ¹ (4x/5) = sin? ¹x.
Using the formula sin? ¹a + sin? ¹b = sin? ¹ (a√ (1-b²) + b√ (1-a²):
sin? ¹ ( (3x/5)√ (1 - (4x/5)²) + (4x/5)√ (1 - (3x/5)²) ) = sin? ¹x
(3x/5) * √ (1 - 16x²/25) + (4x/5) * √ (1 - 9x²/25) = x
x * [ (3/5) * √ (25-16x²)/5 + (4/5) * √ (25-9x²)/5 - 1 ] = 0
So, x = 0 is one solution.
For the other part:
3√ (25-16x²) + 4√ (25-9x²) = 25
Let's check integer solutions. If x = 1:
3√ (9) + 4√ (16) = 33 + 44 = 9 + 16 = 25. So x = 1 is a solution.
If x...more
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