Reflexive relation

Correct answer: A. Because negative real numbers are not images

Step 1: The range of (e^x) is \((0,\infty)\). Step 2: The codomain is \(\mathbb{R}\), but values like (-1) are never attained. Step 3: If even one codomain element is missed, the function is not onto.

Correct answer: A. (f) is onto but not one-one

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so the function is onto. Step 3: Since (\sin x) is periodic, different inputs can give the same output.

Correct answer: A. Both one-one and onto

Step 1: On the given domain, (\sin x) is strictly increasing. Step 2: Its range is ([-1,1]), equal to the codomain. Step 3: On restricted trigonometric intervals, check both monotonicity and range.

Correct answer: A. It is both one-one and onto

Step 1: On ([0,\pi]), (\cos x) is strictly decreasing. Step 2: (\cos 0=1) and (\cos \pi=-1), so all values in ([-1,1]) occur. Step 3: A strictly monotonic function is one-one and becomes onto when its range equals the codomain.

Correct answer: A. Because its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)

Step 1: Values of \(\tan^{-1}x\) lie only in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Step 2: The codomain is \(\mathbb{R}\), so many real values are missed. Step 3: Remember the principal range of inverse trigonometric functions.

Correct answer: A. (f) is one-one and onto

Step 1: \(\tan^{-1}x\) is strictly increasing, so it is one-one. Step 2: Its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), equal to the codomain. Step 3: Choosing the exact range as codomain often makes a function onto.

Correct answer: A. No because only odd integers are obtained

Step 1: (2n+1) is always an odd integer. Step 2: The codomain (\mathbb{Z}) also contains even integers, which are not images. Step 3: For integer functions, parity checks are very useful for onto questions.

Correct answer: A. Onto

Step 1: (f(n)=2n+1) gives all odd integers. Step 2: For any odd (y), (n=\frac{y-1}{2}) is an integer. Step 3: Onto is proved when every target element has a preimage.

Correct answer: A. Because (1,2,3) are not images

Step 1: For (n\in\mathbb{N}), (f(n)\ge4). Step 2: The codomain (\mathbb{N}) contains (1,2,3), but they are not attained. Step 3: For natural number functions, check the smallest possible image.

Correct answer: A. It is onto

Step 1: The first value of (f(n)=n+3) is (4). Step 2: For every (y\ge4), (n=y-3) is a natural number. Step 3: In onto proofs, take an arbitrary (y) and find its preimage.

Correct answer: A. No because the domain has fewer elements

Step 1: In an onto function, every element of the codomain must be hit at least once. Step 2: Here the domain has (3) elements and the codomain has (4), so covering all four elements is impossible. Step 3: For finite sets, first compare the number of elements.

Correct answer: A. (24)

Step 1: For finite sets with equal size, an onto function is automatically one-one. Step 2: Hence such functions are bijections, counted by (4!). Step 3: (4!=24); for equal finite sizes, onto counting becomes permutation counting.

Correct answer: A. (6)

Step 1: Total functions are (2^3=8). Step 2: The non-onto functions are the (2) constant functions. Step 3: Onto functions are (8-2=6); for small sets, subtract the missed cases.

Correct answer: A. (g\circ f) is onto

Step 1: Take any (z\in C). Step 2: Since (g) is onto, (g(y)=z) for some (y\in B), and since (f) is onto, (f(x)=y) for some (x\in A). Step 3: Thus ((g\circ f)(x)=z), so the composition is onto.

Correct answer: A. (g) is onto

Step 1: Since (g\circ f) is onto, every element of (C) is ((g\circ f)(x)) for some (x). Step 2: But ((g\circ f)(x)=g(f(x))), so that element is attained by (g). Step 3: If the composition is onto, the outer function must be onto.

Correct answer: A. (f) is onto

Step 1: If (g\circ f) is onto, (g) must be onto. Step 2: But (f) need not cover all of (B), because (g) may cover (C) using only part of (B). Step 3: In composition questions, keep the roles of inner and outer functions separate.

Correct answer: A. (a\ne0)

Step 1: If (a\ne0), then for any (y\in\mathbb{R}), (x=\frac{y-b}{a}) is real. Step 2: Hence every (y) has a preimage. Step 3: A linear function on (\mathbb{R}) is onto when its slope is non-zero.

Correct answer: A. For no real (k)

Step 1: If (k>0), values are at least (1). Step 2: If (k<0), values are at most (1), and if (k=0), the function is constant. Step 3: In no case does the range become all of (\mathbb{R}), so it cannot be onto.

Correct answer: A. Yes

Step 1: Since (x^2\ge0), (x^2+2\ge2). Step 2: For every (y\ge2), taking (x=\sqrt{y-2}) gives (f(x)=y). Step 3: For square functions, identify the minimum value to get the range.

Correct answer: A. Because negative real values are not attained

Step 1: (\sqrt{x}) is always non-negative. Step 2: The codomain (\mathbb{R}) includes negative values, which are not attained. Step 3: Remember that the range of the square root function is ([0,\infty)).

Correct answer: A. One-one and onto

Step 1: (\sqrt{x}) is increasing on ([0,\infty)), so it is one-one. Step 2: For every (y\ge0), choose (x=y^2), then (f(x)=y). Step 3: Thinking through the inverse helps prove onto quickly.

Correct answer: A. (f) is one-one and onto

Step 1: (\frac{1}{x}) is never (0), and (0) is excluded from the domain. Step 2: For any (y\ne0), choosing (x=\frac{1}{y}) gives (f(x)=y). Step 3: For reciprocal functions, solve for (x) in terms of (y).

Correct answer: A. Yes

Step 1: Let (y=\frac{x}{x+1}). Step 2: From (y(x+1)=x), we get (x=\frac{y}{1-y}), defined for (y\ne1). Step 3: Since (1) is excluded from the codomain, every codomain value has a preimage.

Correct answer: A. Because (1) is never attained

Step 1: Setting (\frac{x}{x+1}=1) gives (x=x+1), which is impossible. Step 2: So (1) is not in the range, while the codomain is (\mathbb{R}). Step 3: In rational functions, identify the impossible output value.

Correct answer: A. Yes it is onto

Step 1: The function value is always between (0) and (1).
Step 2: For any (0<y<1), solving gives \(x=\ln\frac{y}{1-y}\).
Step 3: When the codomain is an open interval, endpoints need not be attained.