Functions

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.

Step 1: A bijection from (A) to (A) is a permutation of the elements. Step 2: The number of permutations of (n) elements is (n!). Step 3: For bijections on the same finite set, use (n!).

Step 1: Total functions are (3^3=27). Step 2: One-one functions are (3!=6). Step 3: Not one-one functions are (27-6=21).

Step 1: To be not onto, one element of (B) must be completely missed. Step 2: There are two possibilities: all elements map to (a), or all map to (b). Step 3: Therefore the number of not onto functions is (2).

Step 1: Put \(y=\frac{x}{1+x^2}\). Step 2: Treat \(yx^2-x+y=0\) as a quadratic in (x) and require its discriminant to be non-negative. Step 3: From \(1-4y^2\geq0\), we get \(-\frac{1}{2}\leq y\leq\frac{1}{2}\).

For \(f(x)=e^x\), the output is positive for every real \(x\), and the function is strictly increasing. Hence, distinct inputs give distinct outputs, so it is one-one. However, it never produces 0 or a negative value; its range is \((0,\infty)\). Therefore, it is not onto \(\mathbb{R}\). The function \(x^3\) is both one-one and onto, whereas \(x^2\) is not one-one. Exam tip: To test whether a function is onto, check whether every element of the codomain occurs in its range.

Step 1: (x^3-6x^2+12x+1=(x-2)^3+9). Step 2: The cubic ((x-2)^3) is strictly increasing on (\mathbb{R}) and takes all real values. Step 3: Shifting does not change bijectivity.

Step 1: For (x<1), (2x+1<3), giving the range part ((-\infty,3)). Step 2: For (x\geq1), (x^2+2\geq3), and (3) occurs at (x=1). Step 3: Combining both parts gives all of (\mathbb{R}).

Step 1: (f^{-1}(5)) means the input (x) for which (f(x)=5). Step 2: From (\frac{2x+3}{x-1}=5), we get (2x+3=5x-5), so (3x=8). Step 3: Hence (x=\frac{8}{3}); for inverse values, set the original function equal to the target value and solve.