Advertisements
Advertisements
प्रश्न
Find the domain of the following
`f(x) = sin^-1 ((x^2 + 1)/(2x))`
उत्तर
The range of sin–1x is – 1 to 1
`- 1 ≤ (x^2 + 1)/(2x) ≤ 1`
⇒ `(x^2 + 1)/(2x) ≥ - 1` or `(x^2 + 1)/(2x) ≤ 1`
⇒ x2 + 1 ≥ – 2x or x2 + 1 ≤ 2x
⇒ x2 + 1 + 2x ≥ 0 or x2 + 1 – 2x ≤ 0
⇒ (x + 1)2 ≥ 0 or (x – 1)2 ≤ 0 which is not possible
⇒ – 1 ≤ x ≤ 1 or
APPEARS IN
संबंधित प्रश्न
Find all the values of x such that – 10π ≤ x ≤ 10π and sin x = 0
Find all the values of x such that Find all the values of x such that −3π ≤ x ≤ 3π and sin x = −1
Find the period and amplitude of y = sin 7x
Find the period and amplitude of y = `- sin(1/3 x)`
Find the period and amplitude of y = 4 sin(– 2x)
Sketch the graph of y = `sin(1/3 x)` for 0 ≤ x ≤ 6π
Find the domain of the following
`g(x) = 2sin^-1(2x - 1) - pi/4`
Choose the correct alternative:
`sin^-1(cos x) = pi/2 - x` is valid for
Choose the correct alternative:
If `cot^-1x = (2pi)/5` for some x ∈ R, the value of tan-1 x is
Choose the correct alternative:
The domain of the function defined by f(x) = `sin^-1 sqrt(x - 1)` is
