Functions

Correct answer: A. It is neither one-one nor onto

Step 1: Both (1) and (2) map to (a), so it is not one-one. Step 2: (b) is in the codomain but is not an image, so it is not onto. Step 3: Still, every input has one image, so it is a function.

Correct answer: A. Because (f(a)=f(b)) gives (a=b)

Step 1: Assume (f(a)=f(b)). Step 2: Then (5a-2=5b-2), so (5a=5b). Step 3: This gives (a=b), so the function is one-one.

Correct answer: A. Because (f(3)=f(-3))

Step 1: In a one-one function, different inputs have different images. Step 2: (3\ne -3), but (f(3)=0) and (f(-3)=0). Step 3: Therefore this function is not one-one.

Correct answer: A. Yes, because for every real (y), (x=\sqrt[3]{y+2}) exists

Step 1: For onto, every real (y) must have a preimage. Step 2: From (x^3-2=y), we get (x=\sqrt[3]{y+2}). Step 3: This is real for every real (y), so the function is onto.

Correct answer: A. Because (f(-3)=f(-1))

Step 1: (f(-3)=|-3+2|=1). Step 2: (f(-1)=|-1+2|=1). Step 3: Two different inputs have the same image, so the function is not one-one.

Correct answer: A. ([0,\infty))

Step 1: An absolute value is never negative. Step 2: At (x=-2), (|x+2|=0). Step 3: Therefore the range is ([0,\infty)).

Correct answer: A. (f(a_1)\ne f(a_2))

Step 1: In a one-one function, different inputs have different images. Step 2: Here (a_1\ne a_2) is given. Step 3: Therefore (f(a_1)\ne f(a_2)).

Correct answer: A. It has at least one preimage in (A)

Step 1: In an onto function, no element of the codomain is left out. Step 2: Therefore each element of (B) is reached by at least one element of (A). Step 3: This is the main idea of onto.

Correct answer: A. (x)

Step 1: (f(x)=x) is the identity function. Step 2: ((f\circ f)(x)=f(f(x))=f(x)). Step 3: Therefore ((f\circ f)(x)=x).

Correct answer: A. (x)

The direct answer is A, \\(x\\). The notation \\(g\\circ f\\) means \\(g(f(x))\\), so apply \\(f\\) first. We have \\(f(x)=x+4\\). Put this complete expression into \\(g(t)=t-4\\): \\(g(x+4)=(x+4)-4=x\\). The positive shift of 4 and the negative shift of 4 cancel. Option A is correct because the final value is exactly \\(x\\). Option B, \\(x+8\\), wrongly adds the shifts, instead of noticing that the second shift is negative. Option C, \\(x-8\\), wrongly subtracts 4 twice. Option D, \\(2x\\), introduces multiplication that is not present in either definition. The order is important even though these particular opposite shifts cancel; in other examples changing the order may change the answer. Remember: for \\(g\\circ f\\), start with \\(f\\), then substitute into \\(g\\).

Correct answer: A. They are inverses of each other

Step 1: (f(g(x))=f(\frac{x}{3})=x). Step 2: (g(f(x))=g(3x)=x). Step 3: Both composites give the identity function, so they are inverses.

Correct answer: A. (f^{-1}:B\to A) is defined

Step 1: A bijective function is both one-one and onto. Step 2: Therefore every image has exactly one preimage. Step 3: Hence the inverse function is defined from (B) to (A).