Functions

Step 1: In an onto function, every element of the codomain must be an image. Step 2: Here the domain has (3) elements, but the codomain has (4). Step 3: Three inputs cannot cover four different codomain elements, so no onto function exists.

Step 1: To find (f^{-1}(10)), solve (x^3+2=10). Step 2: (x^3=8), so (x=2). Step 3: For a cubic function, the real cube root can be taken directly.

Step 1: Write (y=\frac{3x-2}{4}). Step 2: Then (4y=3x-2), so (x=\frac{4y+2}{3}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{4x+2}{3}).

Step 1: For an inverse function to exist, the original function must be one-one. Step 2: (f(0)=0) and (f(-2)=0), while (0\ne-2). Step 3: Since one image has two preimages, the inverse is not a well-defined function.

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

Step 1: Write (x^2+6x+10=(x+3)^2+1). Step 2: Since ((x+3)^2\ge0), (f(x)\ge1). Step 3: The minimum value is (1), attained at (x=-3).

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

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

Step 1: Write (y=\frac{2x}{x-1}). Step 2: From (y(x-1)=2x), we get (x=\frac{y}{y-2}), if (y\ne2). Step 3: Hence every real (y) is possible except (2).

Step 1: Bijective means both one-one and onto. Step 2: The composite of two one-one functions is one-one, and the composite of two onto functions is onto. Step 3: Therefore (g\circ f) is bijective.

Step 1: Suppose (f(a_1)=f(a_2)). Step 2: Then (g(f(a_1))=g(f(a_2))), so ((g\circ f)(a_1)=(g\circ f)(a_2)). Step 3: Since (g\circ f) is one-one, (a_1=a_2), so (f) is one-one.

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

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}).