Question

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Find the domain of the following function

$f(x) = {sin}^{ - 1}( \frac{1 + {x}^{2} }{2x})$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {sin}^{ - 1}\bigg[\dfrac{1 + {x}^{2} }{2x} \bigg]$

We know,

$\boxed{\tt{ Domain \: of \: {sin}^{ - 1}x \: is \: x \: \in \: [ - 1,1] \: }}$

So, using this

$\rm :\longmapsto\: - 1 \leqslant \dfrac{1 + {x}^{2} }{2x} \leqslant 1$

We know,

$\boxed{\tt{ - y \leqslant x \leqslant y\rm \implies\: |x| \leqslant y \: }}$

So, using this

$\rm \implies\:\bigg |\dfrac{1 + {x}^{2} }{2x} \bigg| \leqslant 1$

$\rm \implies\:\dfrac{1 + {x}^{2} }{2 |x| } \leqslant 1$

$\rm \implies\:\dfrac{2 |x| }{1 + {x}^{2} } \geqslant 1$

$\rm \implies\:1 + {x}^{2} \leqslant 2 |x|$

$\rm \implies\:1 + {x}^{2} - 2 |x| \leqslant 0$

$\rm \implies\: { |x| }^{2} - 2 |x| + 1\leqslant 0$

$\rm \implies\: {( |x| - 1) }^{2} \leqslant 0$

As square of number can never be negative.

$\rm \implies\: {( |x| - 1) }^{2} = 0$

$\rm \implies\: {|x| - 1 } = 0$

$\rm \implies\: {|x|} = 1$

$\bf\implies \:x = 1 \: \: or \: \: - 1$

Hence,

$\boxed{\tt{ Domain \: of \: f(x) = {sin}^{ - 1}\bigg[\dfrac{1 + {x}^{2} }{2x} \bigg] \: is \: \{ - 1, \: 1 \} \: }}$

Answer 2

Answer:

{-1, 1}

Step-by-step explanation:

      arcsin of g(x) is definable only for the real values of x which lie between -1 and 1 (including -1 and 1).          

 if we have  sin⁻¹g(x),  then

                - 1 ≤ g(x) ≤ 1

In the given question,  g(x) = (1 + x²)/2x

        1 ≥ (1 + x²)/2x ≥ -1

Multiply and divide by (2x)² :

      (2x)² ≥ 2x(1 + x²) ≥ -(2x)²

      4x² ≥ 2x + 2x³   or 2x + 2x³ ≥ -4x²

      0 ≥ x - 2x² + x³   or   x + 2x² + x³ ≥ 0

Case 1:     0 ≥ x - 2x² + x³  

   0 ≥ x(1 - 2x + x²)

   0 ≥ (x - 0)(x - 1)²

∴ x ∈ (-∞, 0] U {1}    

As for x = 0,  f(x) is undefined,  x ∈ (-∞, 0) U {1}        [from this case]

Case 2:     x + 2x² + x³ ≥ 0

      x(1 - 2x + x²) ≥ 0

      (x - 0)(x + 1)² ≥ 0  

x ∈ [0, ∞)  U {-1}       But f(x) is undefined for x = 0,  

     therefore,  x ∈ (0, ∞) U {-1}

Combining both the cases, we get

x ∈ {1, -1}

      Therefore, domain of this function is {1, -1}