Advertisements
Advertisements
प्रश्न
The domain of the function defined by f(x) = sin–1x + cosx is ______.
पर्याय
[–1, 1]
[–1, π + 1]
`(– oo, oo)`
φ
उत्तर
The domain of the function defined by f(x) = sin–1x + cosx is [–1, 1].
Explanation:
The domain of cos is R and the domain of sin–1 is [–1, 1].
Therefore, the domain of cosx + sin–1x is R ∩ [–1,1]
i.e., [–1, 1].
APPEARS IN
संबंधित प्रश्न
Find the value of the following:
`cos^(-1) (1/2) + 2 sin^(-1)(1/2)`
Find the principal value of `sin^-1(1/sqrt2)`
`sin^-1{cos(sin^-1 sqrt3/2)}`
Find the domain of the following function:
`f(x)sin^-1sqrt(x^2-1)`
Find the domain of the following function:
`f(x)=sin^-1x+sin^-1 2x`
Prove that:
cot−1 7 + cot−1 8 + cot−1 18 = cot−1 3 .
Prove the following:
`tan^-1(1/2) + tan^-1(1/3) = pi/(4)`
Find the principal solutions of the following equation:
cot 2θ = 0.
Find the principal value of the following:
tan-1 (-1)
The value of cot `(tan^-1 2x + cot^-1 2x)` is ______
lf `sqrt3costheta + sintheta = sqrt2`, then the general value of θ is ______
sin[3 sin-1 (0.4)] = ______.
Which of the following function has period 2?
The value of cot (- 1110°) is equal to ______.
Prove that `cot(pi/4 - 2cot^-1 3)` = 7
Show that `cos(2tan^-1 1/7) = sin(4tan^-1 1/3)`
When `"x" = "x"/2`, then tan x is ____________.
`"tan"(pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.
`"cos"^-1 ("cos" ((7pi)/6))` is equal to ____________.
If A = `[(cosx, sinx),(-sinx, cosx)]`, then A1 A–1 is
Which of the following functions is inverse of itself?
If `(-1)/sqrt(2) ≤ x ≤ 1/sqrt(2)` then `sin^-1 (2xsqrt(1 - x^2))` is equal to
If f(x) = x5 + 2x – 3, then (f–1)1 (–3) = ______.
`cot^-1(sqrt(cos α)) - tan^-1 (sqrt(cos α))` = x, then sin x = ______.
Derivative of `tan^-1(x/sqrt(1 - x^2))` with respect sin–1(3x – 4x3) is ______.
sin [cot–1 (cos (tan–1 x))] = ______.
If y = `tan^-1 (sqrt(1 + x^2) - sqrt(1 - x^2))/(sqrt(1 + x^2) + sqrt(1 - x^2))`, then `dy/dx` is equal to ______.
If sin–1x – cos–1x = `π/6`, then x = ______.
Find the value of `sin(2cos^-1 sqrt(5)/3)`.