Introduction to Relations

Correct answer: A. (g) is onto

Step 1: If (g\circ f) is onto, every element of the final codomain is an image. Step 2: That final image is produced through (g). Step 3: Therefore (g) reaches every element of its codomain, so (g) is onto.

Correct answer: A. (x)

Step 1: Inverse functions undo each other's action. Step 2: Applying (g) first and then (f) brings the original value back. Step 3: Therefore ((f\circ g)(x)=x).

Correct answer: A. 6

Since \(f(a)=38\), substitute \(a\) for \(x\) in the function: \(6a+2=38\). Thus, \(6a=36\) and \(a=6\), so option A is correct. Option B is incorrect because \(f(5)=6\times5+2=32\), not 38. Exam tip: When the function value is given, substitute the unknown input into the function, solve the resulting equation, and verify the answer by substitution.

Correct answer: A. 4

The governing concept is using the definition of a function to form and solve an equation. Substituting a into f(x) = x² − 3 gives f(a) = a² − 3. Because f(a) = 13, we set a² − 3 = 13. Adding 3 to both sides gives a² = 16. Taking square roots yields a = 4 or a = −4. The additional condition a > 0 is decisive, so the acceptable value is a = 4. Hence option A is correct. Option B is the squared value a², not the required input. Option C is the other algebraic root but is excluded because it is negative. Option D fails the original condition, since 5² − 3 = 22 rather than 13. Both substitution and the sign condition must be used.

Correct answer: A. 5

Substitute the input into the linear function: \(f(5)=5k+4\). Given \(5k+4=29\), so \(5k=25\) and hence \(k=5\). Note on distractors: option C (6) would arise from an arithmetic mistake like treating \(5k=30\); option B (4) comes from a wrong subtraction such as assuming \(5k+4=24\). Exam tip: substitute carefully, perform addition/subtraction first to isolate the coefficient, then divide to find the parameter.

Correct answer: A. 5

Core idea: Substitute the given x-value into the quadratic to get a linear equation in the parameter k. Solution: \(f(2)=2^2+2k+1=4+2k+1=5+2k\). Given \(5+2k=15\), so \(2k=10\) and \(k=5\). Why the closest distractor is wrong: for option 4, \(f(2)=5+2\times4=13\), not 15. Exam tip: always substitute the given x directly and simplify step-by-step to avoid arithmetic mistakes.

Correct answer: A. (k\ne0)

Step 1: If (k=0), then (f(x)=8) becomes a constant function. Step 2: A constant function is not one-one. Step 3: Therefore the linear function is one-one when (k\ne0).

Correct answer: A. (k\ne0)

Step 1: If (k=0), the function always has value (6). Step 2: Then the range is only ({6}), not all of (R). Step 3: Therefore onto requires (k\ne0).

Correct answer: A. (1)

Step 1: Write (y=\frac{x-3}{x+1}). Step 2: From (y(x+1)=x-3), we get (x(y-1)=-3-y). Step 3: Putting (y=1) gives an impossible statement, so (1) is not in the range.

Correct answer: A. (R-{1})

Step 1: Let (y=\frac{x-3}{x+1}). Step 2: Solving gives (x=\frac{-3-y}{y-1}), if (y\ne1). Step 3: Hence all real values except (1) are in the range.

Correct answer: A. (3)

Step 1: The squares of the given elements are (16,1,0,1,16). Step 2: Repeated values are written only once in the range. Step 3: The range is ({0,1,16}), so it has (3) elements.

Correct answer: A. Onto but not one-one

Step 1: All three elements (a,b,c) appear as images, so the function is onto. Step 2: Both (2) and (3) map to (b), so it is not one-one. Step 3: While deciding the type, check both the range and repeated images.

Correct answer: A. 25

Composition is applied from the inside out. Step 1: \(f(1)=1+2=3\). Step 2: apply \(g\) to that result: \(g(3)=2\cdot3-1=5\). Step 3: apply \(h\): \(h(5)=5^2=25\). So the value is 25. The closest wrong choice 16 often arises from an arithmetic/order mistake (for example getting 4 instead of 5 and then squaring); avoid that by writing intermediate results. Exam tip: evaluate one function at a time and record each intermediate value to prevent order errors.

Correct answer: A. (\sqrt{x^2})

Step 1: ((g\circ f)(x)=g(f(x))). Step 2: Put (x^2-9) in place of (x) in (g). Step 3: (g(f(x))=\sqrt{(x^2-9)+9}=\sqrt{x^2}).

Correct answer: A. (\sqrt[3]{x+7})

Step 1: Write (y=x^3-7). Step 2: Then (x^3=y+7), so (x=\sqrt[3]{y+7}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x+7}).

Correct answer: A. (3)

Step 1: To find (f^{-1}(14)), solve (x^2+5=14). Step 2: (x^2=9), so (x=3) or (x=-3). Step 3: Since the domain is ([0,\infty)), only (3) is valid.

Correct answer: A. ({-4,4})

Step 1: For the preimage, solve (x^2-4=12). Step 2: This gives (x^2=16). Step 3: Hence (x=4) or (x=-4), so the preimage is ({-4,4}).

Correct answer: A. (f(3)) and (f(4))

Step 1: (f(x)=x^2-7x+12=(x-3)(x-4)). Step 2: At (x=3) and (x=4), the product becomes zero. Step 3: At (x=5), (f(5)=2), so it is not zero.

Correct answer: A. ([2,\infty))

Step 1: (|x+3|\ge0) for every real (x). Step 2: Therefore (|x+3|+2\ge2). Step 3: At (x=-3), the minimum value (2) is obtained, so the range is ([2,\infty)).

Correct answer: A. Because (1) cannot be its image

Step 1: The value of (|x+3|+2) is always at least (2). Step 2: The codomain (R) contains (1), but it cannot be the image of any (x). Step 3: Therefore the function is not onto.

Correct answer: A. (9)

Step 1: In a bijective function, every element is matched exactly one-to-one. Step 2: Hence the domain and codomain have the same number of elements. Step 3: Since (A) has (9) elements, (B) also has (9).

Correct answer: A. (g=f^{-1})

Step 1: Both compositions give identity functions. Step 2: This means (f) and (g) completely undo each other's action. Step 3: Therefore (g) is the inverse of (f), so (g=f^{-1}).

Correct answer: A. ([-1,\infty)-{3})

Step 1: First, for (f(x)=\sqrt{x+1}), we need (x+1\ge0), so (x\ge-1). Step 2: ((g\circ f)(x)=\frac{1}{\sqrt{x+1}-2}), so the denominator must not be zero. Step 3: (\sqrt{x+1}=2) gives (x=3), hence the domain is ([-1,\infty)-{3}).

Correct answer: B. [3, ∞)

Solve the defining inequality step by step: 2x − 1 ≥ 5. Adding 1 to both sides gives 2x ≥ 6, and dividing by the positive number 2 gives x ≥ 3. Thus A contains 3 and every real number greater than 3. In interval notation, inclusion of 3 is shown by a square bracket, while infinity always uses a parenthesis. Therefore A = [3,∞), so B is correct.