Functions

Step 1: Put (x=5). Step 2: (f(5)=5^2-2\cdot5=25-10=15). Step 3: Calculate the square and multiplication carefully.

Step 1: On putting (x=7), the numerator becomes (7+1=8). Step 2: Hence (f(7)=\frac{8}{2}=4). Step 3: In fraction-based functions, simplify the numerator first.

Step 1: Put (x=-1). Step 2: (f(-1)=3(-1)^2+1=3\cdot1+1=4). Step 3: Remember that the square of a negative number is positive.

Step 1: (A) is the domain of the function. Step 2: Elements of the domain are treated as inputs. Step 3: The image of an input lies in (B).

Step 1: In an ordered pair, the first component is the input and the second component is the image. Step 2: Since ((4,7)) is given, the image of (4) is (7). Step 3: Do not interchange the first and second components while reading a pair.

Step 1: For being a function, every input must have exactly one image. Step 2: Here (1,2,3) each have one image. Step 3: Two different inputs may have the same image; that does not stop it from being a function.

Step 1: In a function, one input cannot have two different images. Step 2: In the first option, (1) has images (2) and (3). Step 3: Therefore the first option is not a function.

Step 1: Apply the function to every element of (A). Step 2: (f(1)=3), (f(2)=4), and (f(3)=5). Step 3: Therefore (f(A)={3,4,5}).

Step 1: For (f(a+b)), replace (x) by the whole expression (a+b). Step 2: This gives (f(a+b)=5(a+b)). Step 3: Put the entire input in brackets.

Step 1: For (f(a)), replace (x) by (a). Step 2: (x^2-1) becomes (a^2-1). Step 3: While changing the symbol, keep the power correctly.

The governing concept is substitution followed by algebraic simplification of a function expression. Since f(x) = x + 2, replace f(x) in the requested expression with x + 2. This gives f(x) − 2 = (x + 2) − 2. The constants +2 and −2 cancel, leaving x. Therefore option A is correct. Option B adds 2 instead of subtracting it, option C incorrectly doubles the variable, and option D subtracts 2 from x even though the function already contains +2 that must also be included. The result is an identity: for every value of x in the domain, evaluating f(x) and then subtracting 2 returns the original input x.

Step 1: Replace (x) by (x+1). Step 2: (f(x+1)=2(x+1)-1=2x+2-1). Step 3: Simplifying gives (2x+1).

Step 1: (f(x+1)=(x+1)^2) and (f(x)=x^2). Step 2: The difference is ((x+1)^2-x^2). Step 3: Simplifying gives (x^2+2x+1-x^2=2x+1).

The governing concept is addition of functions, defined pointwise by (f + g)(x) = f(x) + g(x). Evaluate each function at x = 2: f(2) = 2 + 3 = 5 and g(2) = 2(2) = 4. Add the results: (f + g)(2) = f(2) + g(2) = 5 + 4 = 9. Hence option A is correct. Option B is f(2) plus an incorrect value, option C may result from using only 2 + 2, and option D is only f(2), omitting g(2). It is important not to confuse function addition with composition; the notation f + g means adding corresponding outputs, not placing one function inside the other.

The governing concept is subtraction of functions, which is defined pointwise as (f − g)(x) = f(x) − g(x). Substitute the given expressions: (f − g)(x) = 3x − (x + 1). The parentheses are essential because the minus sign applies to both terms inside g(x). Distribute the subtraction: 3x − x − 1 = 2x − 1. Thus option A is correct. Option C results from failing to change the sign of the constant, option B adds the functions instead of subtracting them, and option D incorrectly multiplies expressions and introduces a square. The final expression is linear, as expected from subtracting two linear functions.

Here (fg)(3) denotes the product of the two function values, so the governing rule is (fg)(x) = f(x)g(x). Evaluate both functions at 3: f(3) = 3 + 2 = 5 and g(3) = 3 − 2 = 1. Multiply them: (fg)(3) = 5 × 1 = 5. Therefore option A is correct. Option B is only g(3), while option C and option D arise from other incorrect combinations of the input or outputs. This should not be confused with composition, written f(g(3)); composition would give f(1) = 3, which is not what the product notation (fg)(3) asks for in this context. The direct product calculation uniquely confirms 5.

The governing concept is composition of functions: (f ∘ g)(x) means that g is applied first and f is applied to the result. Therefore, (f ∘ g)(2) = f(g(2)). First calculate g(2) = 2 + 1 = 3. Next apply f to this output: f(3) = 3² = 9. Hence the required value is 9, so option A is correct. Option B, 5, could result from adding the two function values incorrectly; option C is only 2² and ignores g; option D does not follow the composition rule. The order of composition is essential because f ∘ g generally differs from g ∘ f.

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x-1) into (f). Step 3: (f(x-1)=2(x-1)+3=2x+1).

The governing concept is finding an inverse function by interchanging the input and output and solving for the original input. Let y = f(x) = x − 5. Add 5 to both sides: y + 5 = x, or x = y + 5. To express the inverse in the usual variable x, replace y by x. Thus f⁻¹(x) = x + 5, which is option A. Option B repeats the original function rather than reversing it. Option C changes the sign incorrectly and would represent 5 − x, while option D multiplies by 5 instead of undoing the subtraction. A quick check confirms the result: f(f⁻¹(x)) = (x + 5) − 5 = x.

Step 1: Let (y=\frac{x}{3}). Step 2: Multiplying both sides by (3), we get (x=3y). Step 3: Hence the inverse is (f^{-1}(x)=3x).

Step 1: (f^{-1}(4)) means the (x) for which (f(x)=4). Step 2: From (2x-6=4), we get (2x=10). Step 3: Hence (x=5), so (f^{-1}(4)=5).

Step 1: Different values of (x) give different values of (x-7). Step 2: For any real (y), choosing (x=y+7) gives (f(x)=y). Step 3: Hence it is both one-one and onto.

Step 1: Since (x^2\ge 0), (x^2+4\ge 4). Step 2: The codomain (R) contains (2), but no real (x) can map to it. Step 3: Therefore the range is not all of (R), so the function is not onto.

Step 1: For every real (x), (x^2\ge 0). Step 2: Therefore (x^2+4\ge 4). Step 3: The minimum value is (4), so the range is ([4,\infty)).

Step 1: For every (x), the value of the function remains (2). Step 2: Such a function is called a constant function. Step 3: In a constant function, all inputs give the same image.