Mathematics Test - 34

Given that f(x) = (5 + x⁴)^1/4

∴ fοf(x) = f(f(x)) = (5 + {(5 + x⁴)^1/4}⁴)^1/4

= (5 + (5 + x⁴))^1/4 = (10+x⁴)^1/4

The binary operation is defined by a * b = a - b + ab².

∴ 4 * 5 = 4 - 5 + 4(5²) = -1 + 100 = 99.

[-1, 1] is the domain for sin^-1⁡x.

The domain for cot^-1⁡x is (-∞,∞).

The domain for tan^-1⁡⁡x is (-∞,∞).

The domain for sec^-1⁡⁡x is (-∞,-1] ∪ [1,∞).

Given that,

A function is invertible if and only if it is bijective i.e. the function is both injective and surjective. If a function f: A → B is bijective, then there exists a function g: B → A such that f(x) = y ⇔ g(y) = x, then g is called the inverse of the function.

The function f = {(7,7),(8,8),(9,9)} is invertible as it is both one – one and onto. The function is one – one as every element in the domain has a distinct image in the co – domain. The function is onto because every element in the codomain M = {7,8,9} has a pre – image in the domain.

(2,3) ∈ R as 2+3 = 5, 3>1, thus satisfying the given condition.

(4,2) doesn’t belong to R as 4+2 ≠ 5.

(2,1) doesn’t belong to R as 2+1 ≠ 5.

(5,0) doesn’tbelong to R as 0⊁1

Given that, f(x) = 3x - 5, g(y) = 6y² and h(z) = tan⁡z,

Then, ho(gof) = hο(g(f(x)) = h(6(3x-5)²) = tan⁡(6(3x - 5)²)

∴ ho(gof) = tan⁡(6(3x - 5)²)

The binary operation ‘*’ is both commutative and associative for a * b = a + b.

The operation is commutative on a * b = a + b because a + b = b + a.

The operation is associative on a * b = a + b because (a + b) + c = a + (b + c).

Given that, a * b = 4ab.

Then, (a * b) * a = (4ab) * a

= 4(4ab)(a) = 16a² b.