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Mathematics Test - 34
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Mathematics Test - 34
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  • Question 1/10
    5 / -1

    If f : R → R is given by f(x) = (5 + x4)1/4, then fοf(x) is _______

    Solutions

    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

  • Question 2/10
    5 / -1

    Let ‘*’ be a binary operation on N defined by a * b =a - b + ab2, then find 4 * 5.

    Solutions

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

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

  • Question 3/10
    5 / -1

    [-1, 1] is the domain for which of the following inverse trigonometric functions?

    Solutions

    [-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,∞).

  • Question 4/10
    5 / -1

    Solutions

    Given that,

  • Question 5/10
    5 / -1

    A function is invertible if it is ____________

    Solutions

    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.

  • Question 6/10
    5 / -1

    Let M={7,8,9}. Determine which of the following functions is invertible for f:M→M.

    Solutions

    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.

  • Question 7/10
    5 / -1

    Let R be a relation in the set N given by R={(a,b): a+b=5, b>1}. Which of the following will satisfy the given relation?

    Solutions

    (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

  • Question 8/10
    5 / -1

    If f: N→N, g: N→N and h: N→R is defined f(x) = 3x - 5, g(y) = 6y2 and h(z) = tan⁡z, find ho(gof).

    Solutions

    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)

  • Question 9/10
    5 / -1

    Let ‘*’ be defined on the set N. Which of the following are both commutative and associative?

    Solutions

    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).

  • Question 10/10
    5 / -1

    Let ‘*’ be a binary operation defined by a * b = 4ab. Find (a * b) * a.

    Solutions

    Given that, a * b = 4ab.

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

    = 4(4ab)(a) = 16a2 b.

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