Functions

The governing concept is substitution into a function followed by solving a quadratic equation. From f(x) = x² + 2, we obtain f(a) = a² + 2. Using the given value f(a) = 18 gives a² + 2 = 18. Subtracting 2 from both sides results in a² = 16. The square-root equation has two algebraic possibilities, a = 4 and a = −4, but the condition a > 0 selects only a = 4. Therefore option A is correct. Option B is a² rather than a. Option C is the negative root and conflicts with the stated positivity condition. Option D is obtained by an incorrect operation and gives 8² + 2, which is not 18. The sign restriction is necessary for an unambiguous answer.

Step 1: If (k=0), then (f(x)=-5) 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 (-1). Step 2: Then the range is only ({-1}), not all of (R). Step 3: Therefore onto requires (k\ne0).

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

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

Step 1: The absolute values of the given elements are (4,2,0,2,4). Step 2: Repeated values are written only once in the range. Step 3: The range is ({0,2,4}), so it has (3) elements.

Step 1: All three elements (a,b,c) appear as images, so the function is onto. Step 2: Both (1) and (2) map to (a), so it is not one-one. Step 3: In such questions, check both codomain coverage and distinct images.

Step 1: ((g\circ f)(x)=g(f(x))). Step 2: Put (x^2-4) in place of (x) in (g). Step 3: (g(f(x))=\sqrt{(x^2-4)+4}=\sqrt{x^2}).

Step 1: Write (y=x^3+5). Step 2: Then (x^3=y-5), so (x=\sqrt[3]{y-5}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x-5}).

Step 1: To find (f^{-1}(10)), solve (x^2+1=10). 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+1=5). Step 2: This gives (x^2=4). Step 3: Hence (x=2) or (x=-2), so the preimage is ({-2,2}).

Step 1: (f(x)=x^2-5x+6=(x-2)(x-3)). Step 2: At (x=2) and (x=3), the product becomes zero. Step 3: At (x=4), (f(4)=2), so it is not zero.

Step 1: (|x-4|\ge0) for every real (x). Step 2: Therefore (|x-4|+1\ge1). Step 3: At (x=4), the minimum value (1) is obtained, so the range is ([1,\infty)).

Step 1: The value of (|x-4|+1) is always at least (1). Step 2: The codomain (R) contains (0), 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 (B) has (8) elements, (A) also has (8).

Step 1: Both compositions give identity functions. Step 2: This means (f) and (g) completely undo each other's action. Step 3: Therefore (f) is bijective and (g) is the inverse of (f).