Functions

Step 1: Put (y=\frac{3x+1}{x-2}) and isolate (x). Step 2: From (y(x-2)=3x+1), we get (x(y-3)=2y+1). Step 3: Hence (x=\frac{2y+1}{y-3}), and replacing (y) by (x) gives the inverse.

Step 1: (f(f(x))=a(ax+b)+b=a^2x+b(a+1)). Step 2: Equating it with (x) gives (a^2=1) and (b(a+1)=0). Step 3: Hence (a=1,b=0), or (a=-1) with any real (b).

Step 1: Write (x^2-6x+11=(x-3)^2+2). Step 2: Since ((x-3)^2\geq0), the minimum value is (2). Step 3: Completing the square is the safest method for range questions.

Step 1: The expression is the sum of distances of (x) from (1) and (-1). Step 2: For (-1\leq x\leq1), the sum is (2), and outside this interval it increases. Step 3: A distance interpretation helps solve absolute value range questions quickly.

Step 1: Since (x^2\geq0), the function value cannot be negative. Step 2: At (x=0), the value is (0), and for large (|x|), it approaches (1) but never equals (1). Step 3: Distinguish between approaching a value and attaining it.

Step 1: Choose the (2) elements of the range in (\binom{3}{2}=3) ways. Step 2: Onto functions onto these two chosen elements are (2^5-2=30). Step 3: Total number is (3\cdot30=90).

Step 1: For a piecewise function, values from different pieces must also be compared. Step 2: (f(-1)=(-1)^2+2=3) and (f(1)=1+2=3). Step 3: Equal outputs for different inputs break one-one behaviour.

Step 1: (x^2+2x=x(x+2)), and for (x\geq0) it starts at (0) and increases. Step 2: The minimum value is (0) at (x=0), so the range is ([0,\infty)). Step 3: A quadratic can become one-one after restricting its domain.

Step 1: ((g\circ f)(-x)=g(f(-x))). Step 2: Since (f) is odd, (f(-x)=-f(x)); since (g) is even, (g(-f(x))=g(f(x))). Step 3: Therefore (g\circ f) is even.

Step 1: (\frac{1}{x}) is defined only when (x\neq0). Step 2: At (x=0), the denominator becomes zero, so it must be removed. Step 3: For fractional functions, always exclude values that make the denominator zero.

Step 1: For positive (x), (x+\frac{1}{x}\geq2). Step 2: At (x=1), the value (2) is attained. Step 3: The AM-GM idea is very useful for such range questions.

Step 1: The (x)-coordinate of the vertex is (-\frac{p}{2}). Step 2: From (-\frac{p}{2}=2), we get (p=-4). Step 3: Using (f(2)=5), (4-8+q=5), so (q=9).

Step 1: If (m\neq0), then (f(1)=f(-1)). Step 2: If (m=0), the function becomes the constant (1). Step 3: Therefore no real (m) makes it one-one on all of (\mathbb{R}).

Step 1: The denominator is (x+3). Step 2: Solving (x+3=0) gives (x=-3). Step 3: A value making the denominator zero must be removed from the domain.

Step 1: From (y=\frac{2x-1}{x+3}), we get (x=\frac{1+3y}{2-y}). Step 2: For every (y\neq2), there is a unique (x), and (x\neq-3). Step 3: A function with exactly one preimage for every codomain value is bijective.

Step 1: ((f\circ g)(x)=f(x+1)=(x+1)^2). Step 2: ((g\circ f)(x)=g(x^2)=x^2+1). Step 3: The difference is ((x+1)^2-(x^2+1)=2x).

Step 1: Put (x+(-x)=0). Step 2: Then (f(0)=f(x)f(-x)=1). Step 3: Since (f(x)>0), (f(-x)=\frac{1}{f(x)}).

Step 1: By the additive property, (f(5)=f(1+1+1+1+1)). Step 2: This becomes (5f(1)=5\cdot3=15). Step 3: In such questions, use repeated addition of the given value.

Step 1: To minimize the maximum, make (x) and (1-x) equal. Step 2: Solving (x=1-x) gives (x=\frac{1}{2}). Step 3: At this point both values are (\frac{1}{2}), so the minimum is (\frac{1}{2}).

Step 1: (x^2) starts from (0) and increases. Step 2: When (x^2\geq4), the function takes the value (4). Step 3: Hence all values from (0) to (4) are attained, including both endpoints.