Functions

Step 1: Write (x^2-4x+5=(x-2)^2+1). Step 2: Since ((x-2)^2\ge0), (f(x)\ge1). Step 3: The minimum value is (1), attained at (x=2).

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

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

Step 1: Write (y=\frac{x}{x+1}). Step 2: From (y(x+1)=x), we get (x=\frac{y}{1-y}), if (y\ne1). Step 3: Hence every real (y) is possible except (1).

Step 1: Since (g) is onto, every element of (C) is the image of some element of (B). Step 2: Since (f) is onto, it can reach every needed element of (B). Step 3: Therefore (g\circ f) is onto from (A) to (C).

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

Step 1: The composition of inverse functions gives the identity function. Step 2: This means applying (g) first and then (f) returns the original value. Step 3: Therefore ((f\circ g)(x)=x).

The governing concept is evaluating a function at a specified input and then solving the resulting equation. Since f(x) = x² − 1, replacing x by a gives f(a) = a² − 1. The condition f(a) = 8 therefore produces a² − 1 = 8. Adding 1 to both sides gives a² = 9, so a = 3 or a = −3. The question specifically states a > 0, which eliminates −3 and leaves a = 3. Thus option A is correct. Option B is the value of a², not a itself. Option C satisfies the squared equation but violates the positive condition, while option D does not satisfy the equation because 4² − 1 = 15, not 8.

Step 1: If (k=0), then (f(x)=1) becomes a constant function. Step 2: A constant function is not one-one. Step 3: Therefore, for the linear function to be one-one, (k\ne0).

Step 1: If (k=0), the function always has value (3). Step 2: Then the range is only ({3}), not all of (R). Step 3: Therefore, onto requires (k\ne0).

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

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

Step 1: The squares of the given elements are (9,4,1,0,1,4,9). Step 2: Repeated values are written only once in the range. Step 3: The range is ({0,1,4,9}), so it has (4) elements.

Step 1: Both (a) and (b) 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: Onto and one-one should be checked separately.

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

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

Step 1: To find (f^{-1}(9)), solve (x^2=9). Step 2: The real solutions are (x=3) and (x=-3). Step 3: Since the domain is ([0,\infty)), only (3) is valid.

Step 1: For the preimage, solve (x^2=4). Step 2: This gives (x=2) or (x=-2). Step 3: Therefore (f^{-1}({4})={-2,2}).

Step 1: (f(x)=x^2-3x+2=(x-1)(x-2)). Step 2: At (x=1) and (x=2), the product becomes zero. Step 3: At (x=0), (f(0)=2), so it is not zero.

Step 1: (|x|\ge0) for every real (x). Step 2: Therefore (|x|+2\ge2). Step 3: At (x=0), the minimum value (2) is obtained, so the range is ([2,\infty)).