Functions

Step 1: For (x\geq0), (f(x)=x^2), and for (x<0), (f(x)=-x^2). Step 2: Negative inputs give negative outputs and positive inputs give positive outputs, with ordered growth. Step 3: Every real (y) has exactly one preimage, so the function is bijective.

Step 1: Since (4) is positive, the input (x) must be positive. Step 2: For positive (x), (f(x)=x^2), so (x^2=4). Step 3: The positive solution is (x=2).

Step 1: If (f(a_1)=f(a_2)), then (g(f(a_1))=g(f(a_2))). Step 2: Since (g\circ f) is one-one, (a_1=a_2). Step 3: Hence (f) must be one-one, while (g) need not be one-one on all of (B).

Step 1: Since (g\circ f) is onto, every element of (C) is of the form (g(f(a))). Step 2: That same element is also of the form (g(b)), where (b=f(a)). Step 3: Hence (g) must be onto.

Step 1: (x^2-1=(x-1)(x+1)). Step 2: For (x\neq1), (\frac{x^2-1}{x-1}=x+1). Step 3: Even after cancellation, the value making the original denominator zero must be excluded.

Step 1: For (x\neq1), the function equals (x+1). Step 2: Since (x=1) is removed, the value (x+1=2) cannot occur. Step 3: Even after simplification, removed domain points can affect the range.

Step 1: (f(x)=x^2+1) is not one-one on all of (\mathbb{R}). Step 2: If its domain is restricted to (x\geq0), then (y=x^2+1) gives (x=\sqrt{y-1}). Step 3: To define an inverse, the domain often needs a suitable restriction.

Step 1: Every real (x) is the sum of its greatest integer part and fractional part. Step 2: Therefore (\lfloor x\rfloor+{x}=x). Step 3: For greatest integer questions, remember this basic identity.

Step 1: For every integer (n), choosing (x=n) gives (\lfloor x\rfloor=n), so it is onto. Step 2: (\lfloor 2.1\rfloor=\lfloor 2.9\rfloor=2), so it is not one-one. Step 3: Taking codomain (\mathbb{Z}) changes onto behaviour.

Step 1: (2n+1) gives different values for different (n). Step 2: Its values are always odd integers, so even integers are not obtained. Step 3: One-one and onto must be checked separately.

Step 1: On integers, (n^3) is increasing, so different (n) give different values. Step 2: (2) is not the cube of any integer, so not every codomain value is reached. Step 3: For power functions on integers, check the range carefully.

Step 1: On natural numbers, (n^2) increases, so it is one-one. Step 2: (2) is not the square of any natural number, so it is not onto. Step 3: In natural numbers, not every value is a perfect square.

Step 1: (1+x^2\geq1) for every real (x). Step 2: Hence (\ln(1+x^2)\geq\ln1=0), and (0) occurs at (x=0). Step 3: For logarithmic functions, first check the minimum of the inside expression.

Step 1: (\ln x) is strictly increasing on ((0,\infty)). Step 2: For every real (y), choosing (x=e^y>0) gives (\ln x=y). Step 3: Hence it is bijective from ((0,\infty)) to (\mathbb{R}).

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

Step 1: Write \(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\). Step 2: The sine function lies in ([-1,1]). Step 3: Hence the range is \([-\sqrt{2},\sqrt{2}]\).

Step 1: On the given interval, (\sin x) is strictly increasing. Step 2: It attains every value from (-1) to (1). Step 3: With a suitable domain and codomain, a trigonometric function can be bijective.

Step 1: On the domain ([0,\pi]), (\cos x) is one-one. Step 2: The solution of (\cos x=0) is (x=\frac{\pi}{2}). Step 3: On a restricted domain, the inverse value is unique.

Step 1: For the inverse relation to be a function, each output must come from only one input. Step 2: If (f) is not one-one, one output has two inputs. Step 3: Therefore, if (f^{-1}) is a function, (f) must be one-one.

Step 1: A bijective function is both one-one and onto. Step 2: For finite sets, a bijection exists only when domain and codomain have equal sizes. Step 3: Hence (B) also has (6) elements.