Introduction to Relations

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.

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).

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.

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.

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.

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.

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).

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).

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.

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.

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.

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.

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.

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}).

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}).

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.

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}).

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.

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)).

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.

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).

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}).

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}).

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.