Functions

Step 1: ((f\circ g)(x)) means (f(g(x))). Step 2: Substitute (g(x)=3x-2) into (f), giving ((3x-2)^2+1=9x^2-12x+5). Step 3: In composition, the order matters.

Step 1: Total functions from (A) to (B) are (2^4=16). Step 2: Not onto functions use only (a) or only (b), giving (2) such functions. Step 3: Hence onto functions are (16-2=14).

Step 1: In an onto function, every element of the codomain must be an image of some domain element. Step 2: The (5) elements of (A) can cover at most (5) distinct elements. Step 3: Hence the maximum possible size of (B) is (5).

Step 1: (f(1)=1) and (f(-1)=1), so distinct inputs have the same image; it is not one-one. Step 2: No negative real number is an output, so it is not onto (\mathbb{R}). Step 3: For absolute value functions, check both domain and codomain carefully.

Step 1: First find (f(-3)), which is (-1). Step 2: Then compute (g(-1)=(-1)^2=1). Step 3: In composition, apply the right-side function first.

Step 1: In a one-one function, the three elements of (A) must get distinct images. Step 2: The choices are (4), then (3), then (2). Step 3: Total number is (4\times3\times2=24).

Step 1: Write (\frac{x-1}{x+1}=2). Step 2: Then (x-1=2x+2), so (x=-3). Step 3: For rational functions, also check that the denominator is not zero.

Step 1: (f) is a bijective linear function, so its inverse exists. Step 2: The inverse reverses the effect of (f), hence (f^{-1}(f(10))=10). Step 3: Recognize the property first; full calculation is not always needed.

Step 1: In a one-one function, every distinct element of (A) needs a distinct image. Step 2: For (6) distinct images, (B) must have at least (6) elements. Step 3: For injective functions, the codomain cannot be smaller than the domain.

Step 1: Complete the square: (x^2-4x+7=(x-2)^2+3). Step 2: The least value of ((x-2)^2) is (0). Step 3: Therefore, the minimum value of the function is (3).

Step 1: In a one-one function, distinct inputs must have distinct images. Step 2: Here (1\neq -1), but (f(1)=1) and (f(-1)=1). Step 3: Once two different inputs give the same output, the function is not one-one.

Step 1: In a constant function, all elements of (A) have the same image. Step 2: This common image can be (1), (2), or (3). Step 3: The number of constant functions equals the number of elements in the codomain.

Step 1: The denominator is (x-1). Step 2: The denominator cannot be zero, so (x-1\neq 0), meaning (x\neq 1). Step 3: The domain includes only those real numbers for which the function is defined.

Step 1: For every (x), the function value is always (5). Step 2: So it is a constant function and not one-one, because many inputs have the same image. Step 3: To identify a constant function, check whether the value depends on (x).

Step 1: (f^{-1}(7)) means the value of (x) for which (f(x)=7). Step 2: (x^3-1=7), so (x^3=8) and (x=2). Step 3: For inverse value questions, set the original function equal to the given output.

Step 1: The expression inside the square root must be non-negative. Step 2: (x-3\geq 0), so (x\geq 3). Step 3: For square-root functions, always set the inside expression (\geq 0).

Step 1: Each element of (A) can be mapped to any one of the (n) elements of (B). Step 2: There are (m) independent choices. Step 3: By the multiplication rule, the total number of functions is (n^m).

Step 1: If (a=0), then (f(x)=b) becomes a constant function. Step 2: A constant function cannot be one-one. Step 3: Therefore, a linear function (ax+b) is one-one only when (a\neq 0).

Step 1: ((f\circ g)(x)=f(x+1)=\frac{1}{x+1}). Step 2: The denominator (x+1) must not be zero. Step 3: Hence (x\neq -1), so the domain is (\mathbb{R}-{-1}).

The direct answer is A: \(h^2\). A function gives an output for each input. Put \(1+h\) into \(f(x)=x^2-2x\): \(f(1+h)=(1+h)^2-2(1+h)=1+2h+h^2-2-2h=h^2-1\). Also, \(f(1)=1^2-2(1)=-1\). Therefore, \(f(1+h)-f(1)=(h^2-1)-(-1)=h^2\). Option A is correct because it is exactly the simplified difference. Option B, \(h^2-1\), is only \(f(1+h)\), not the difference after subtracting \(f(1)\). Option C, \(2h+h^2\), does not result after the cancellation of the \(2h\) terms. Option D, \(h\), is missing the square term and is not obtained by the calculation. Remember: substitute both inputs, calculate both function values, and then subtract.

Step 1: (x^5+x^3+x) keeps increasing with (x), so it is one-one. Step 2: For very negative (x), values are very negative, and for very positive (x), values are very positive, so all real values occur. Step 3: Increasing odd-degree polynomial functions often have range (\mathbb{R}).

Step 1: Write (y=\frac{x+2}{x-2}). Step 2: (y(x-2)=x+2), so (x(y-1)=2y+2) and (x=\frac{2y+2}{y-1}). Step 3: Replace (y) by (x) at the end to get the inverse.

Step 1: For every (y\geq 0), there is an (x), such as (x=\sqrt{y}), so the function is onto. Step 2: (f(2)=f(-2)=4), so it is not one-one. Step 3: Changing the codomain can change whether a function is onto.

Step 1: On ([0,\infty)), (x^2) is increasing, so distinct inputs give distinct outputs. Step 2: For every (y\geq 0), (x=\sqrt{y}) lies in the domain. Step 3: Restricting the domain can make the same formula one-one.

Step 1: First apply (f(x)=x+1). Step 2: Then (g(f(x))=2(x+1)=2x+2). Step 3: Finally (h(2x+2)=(2x+2)^2=4x^2+8x+4).