Solution: All the choices are correct. A = {1, 2, 3, 4} On verifying with the given options, (1) {1, 2, 4} 1, 2, 4 ∈ A 1 * 2 = 2 1 * 4 = 4... View Article
Solution: (3) y = 3 sin (√{[π2 / 16] - x2}) The domain is [0, π / 4] Put x = 0, y = 3 sin (√{[π2 / 16] - 0}) = 3 sin (π / 4) = 3 * (1 /... View Article
Solution: (3) F (x) = √cosx √cosx = positive or 0 cosx = (-1, 1) cosx lies between 0 and 1. x lies in the first or fourth quadrant. x ∈... View Article
Solution: (3) bx - For very large negative values of x, the function is very close to 0. The domain of the function is R, the set of real... View Article
Solution: (2) sin2θ = [1 - cos 2θ] / 2 = (1 / 2) - (1 / 2) cos 2θ If f (x) is a periodic fucntion with period T, then the period of a + f... View Article
Question Solution: (4) f (x) = x, when x is rational and 0 when x is irrational g (x) = 0, when x is rational and x when x is irrational... View Article
Solution: (3) f : X → Y X : domain and Y : codomain One-one function definition: x1 ≠ x2 f (x1) ≠f (x2) Onto function definition:... View Article
Solution: (2) f : R → R Domain = R Codomain = R f (x) = (x - 1) (x - 2) (x - 3) The coeffiecient of x3 is 1. f (x1) = f (x2) = f (x3) =... View Article
Solution: (4) Given that A = [-1, 1] and f : A → A is defined as f (x) = x|x|,∀ x ∈ A f (x) = x2 for x > 0 and f (x) = −x2 for x < 0 f... View Article
Solution: (4) f (x) = e2ix = cos 2x + i sin 2x The domain is R and the codomain is C. The periodic functions are not one-one, hence, the... View Article
Solution: (3) f (n) = 1 + n2 If n = 1, f (n) = 2 If n = 2, f (n) = 5 If n = 3, f (n) = 10 f (n) is one-one. As range ≠codomain, f (n)... View Article
Question Solution: (4) f (n) = n, n2 and 2n+1 is odd ∀ n ∈ N f (n) is not even. ⇒ It is not onto or not surjective. Since, f (3) =... View Article
