Introduction to Relations

Step 1: Write the function as ((x-4)^2+4). Step 2: A square is always (0) or more. Step 3: Therefore the least value is (4), and the range is ([4,\infty)).

Step 1: Assume (\frac{3x}{x+2}=3). Step 2: This gives (3x=3x+6), which is impossible. Step 3: Therefore (3) is not in the range of this function.

Step 1: Write (y=\frac{3x}{x+2}). Step 2: From (y(x+2)=3x), we get (x=\frac{-2y}{y-3}), if (y\ne3). Step 3: Hence every real (y) is possible except (3).

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.

Step 1: (f(a)=a^2-3). Step 2: From (a^2-3=13), we get (a^2=16). Step 3: Since (a>0), (a=4).

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}).