The number of integral value(s) of x in the domain of the function f(x)
sin-[sin-1x], is
(a) Zero
(b) One
(c) Two
(d) Three
Answers
Given : f(x) = sin⁻¹( sin ⁻¹(x) )
To Find : The number of integral value(s) of x in the domain of the function f(x)
(a) Zero
(b) One
(c) Two
(d) Three
Solution:
f(x) = sin⁻¹( sin ⁻¹(x) )
sin ⁻¹(x) Hence x ∈ [-1 , 1 ]
and sin ⁻¹(x) belongs to [-π/2 , π/2]
as π = 3.14
=> sin ⁻¹(x) belongs to [-1.57 , 1.57]
but again sin⁻¹ is there hence inside sin ⁻¹(x) range must lie with in [-1 , 1 ]
Hence x ∉ - 1 , 1
so we left with only x = 0
f(x) = sin⁻¹( sin ⁻¹(x) )
=> f(x) = sin⁻¹( sin ⁻¹(0) )
=> f(x) = sin⁻¹( 0 )
=> f(x) = 0
if x = - 1
then f(x) = sin⁻¹( sin ⁻¹(-1) )
=> f(x) = sin⁻¹( -π/2 )
=> f(x) = sin⁻¹( -1.57 ) while - 1.57 < - 1
=> f(x) not defined
if x = 1
then f(x) = sin⁻¹( sin ⁻¹( 1) )
=> f(x) = sin⁻¹( π/2 )
=> f(x) = sin⁻¹( 1.57 ) while 1.57 > 1
=> f(x) not defined
Hence The number of integral value(s) of x in the domain of the function f(x) sin⁻¹( sin ⁻¹(x) ) is 1
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