Chapter 1 - One-One and Onto Functions (PYQs)
Overview
This page provides comprehensive Class 12 Maths Chapter 1 One-One & Onto Functions PYQs. Chapter 1 - One-One and Onto Functions (PYQs) - Free study material for Class 12 Maths. NCERT Solutions, Notes, and PYQs.
Class 12 Mathematics | Relations & Functions
Q1
2021-22
00:00
Let $f : R \to R$ be defined by $f(x) = 4 + 3 \cos x$ is
(a) bijective
(b) one-one but not onto
(c) onto but not one-one
(d) neither one-one nor onto
The range of $\cos x$ is $[-1, 1]$.
So the range of $f(x) = 4 + 3 \cos x$ is $[1, 7]$.
Since codomain is R and range is limited to [1, 7], the function is not onto.
Cos x is periodic, so f(x1) = f(x2) for different x1, x2.
Hence, the function is not one-one.
Final Answer: (d) neither one-one nor onto
Q2
2021-22
00:00
The number of functions defined from $\{1, 2, 3, 4, 5\} \to \{a, b\}$ which are one-one is
(a) 5 (b) 3 (c) 2 (d) 0
For a function to be one-one, each element of the domain must map to a distinct element of the codomain.
The domain has 5 elements while the codomain has only 2 elements.
It is not possible to map 5 distinct elements to only 2 distinct elements injectively.
Hence, no one-one function exists.
Final Answer: (d) 0
Q3
2021-22
00:00
Let $f : R \to R$ be defined by $f(x) = 1/x$, for all $x \in R$. Then, $f$ is
(a) one-one
(b) onto
(c) bijective
(d) not defined
The function $f(x) = 1/x$ is not defined at $x = 0$, so domain is $R - \{0\}$.
For different $x$ values, $f(x)$ gives different results, so the function is one-one.
For any y ≠ 0, there exists x = 1/y such that f(x) = y, so it is onto R − {0}.
Hence, the function is bijective on its domain.
Final Answer: (c) bijective
Q4
2021-22
00:00
The function $f : N \to N$ is defined by $f(n) = (n+1)/2$ if $n$ is odd and $f(n) = n/2$ if $n$ is even. The function $f$ is
(a) bijective
(b) one-one but not onto
(c) onto but not one-one
(d) neither one-one nor onto
Each natural number $n$ has a unique image under $f$, so the function is one-one.
For every natural number $k$, there exists $n = 2k$ or $n = 2k - 1$ such that $f(n) = k$.
Thus, every element of codomain has a pre-image.
Hence, the function is bijective.
Final Answer: (a) bijective
Q5
2025
00:00
Assertion (A): Let $f(x) = e^x$ and $g(x) = \log x$. Then $(f + g)(x) = e^x + \log x$ where domain of $(f + g)$ is $R$.
Reason (R): $\text{Dom} (f + g) = \text{Dom} (f) \cap \text{Dom} (g)$.
Domain of $f(x) = e^x$ is $R$.
Domain of $g(x) = \log x$ is $(0, \infty)$.
Domain of $(f + g)$ is intersection = $(0, \infty)$, not $R$.
Thus, Assertion is false but Reason is true.
Final Answer: (d) Assertion is false but Reason is true
Q6
2025
00:00
Assertion (A): Let $Z$ be the set of integers. A function $f : Z \to Z$ defined as $f(x) = 3x - 5$ is a bijective.
Reason (R): A function is a bijective if it is both surjective and injective.
The function $f(x) = 3x - 5$ is linear with non-zero coefficient, so it is one-one.
For every integer $y$, $x = (y + 5)/3$ gives a unique integer solution, so the function is onto.
Hence, the function is bijective.
The reason correctly states the definition of bijective function.
Final Answer: (a) Both A and R are true and R is the correct explanation
Q7
2021
00:00
If $f = \{(1, 2), (2, 4), (3, 1), (4, k)\}$ is a one-one function from set $A$ to $A$, where $A = \{1, 2, 3, 4\}$, then find the value of $k$.
Since the function is one-one, all images must be distinct.
The current images are 2, 4, and 1.
The value of $k$ cannot be 1, 2, or 4.
The only remaining element in set $A$ is 3.
Therefore, $k = 3$.
Final Answer: $k = 3$
Q8
2023
00:00
Prove that the greatest integer function $f : R \to R$ given by $f(x) = [x]$, is neither one-one nor onto.
Consider two different real numbers $x = 1.2$ and $y = 1.8$.
Then $f(1.2) = [1.2] = 1$ and $f(1.8) = [1.8] = 1$.
Since $f(x) = f(y)$ for $x \neq y$, the function is not one-one.
The range of the greatest integer function is the set of all integers $Z$.
But the codomain is $R$, which contains non-integer real numbers.
Hence, there are elements in $R$ which are not images of any $x$ under $f$.
Therefore, the function is not onto.
Final Answer: The greatest integer function is neither one-one nor onto.
Q9
2023
00:00
A function $f : A \to B$ defined as $f(x) = 2x$ is both one-one and onto. If $A = \{1, 2, 3, 4\}$, then find the set $B$.
Since $f(x) = 2x$, the image of each element of $A$ is obtained by multiplying by 2.
Compute the images: $f(1)=2, f(2)=4, f(3)=6, f(4)=8$.
Since the function is onto, the set $B$ must be exactly the set of all these images.
Hence, $B = \{2, 4, 6, 8\}$.
Final Answer: $B = \{2, 4, 6, 8\}$
Q10
2025
00:00
Show that the function $f : N \to N$, where $N$ is a set of natural numbers, given by $f(n) = n - 1$ if $n$ is even and $f(n) = n + 1$ if $n$ is odd, is a bijection.
First, show that $f$ is one-one.
Assume $f(a) = f(b)$.
Consider cases: if $a$ and $b$ are both even or both odd, then their images are distinct unless $a = b$.
Hence, $f$ is injective (one-one).
Next, show that $f$ is onto.
For any natural number $k$, choose $n = k + 1$ if $k$ is even and $n = k - 1$ if $k$ is odd.
Then $f(n) = k$.
Thus every element of $N$ has a pre-image.
Hence, $f$ is bijective.
Final Answer: The function $f$ is bijective.
Q11
2023
00:00
A student identifies a function from $S$ to $J$ as $f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\}$. Check if it is bijective.
To be bijective, the function must be both one-one and onto.
Here, $S_2$ and $S_3$ are both mapped to $J_2$.
So two different elements of $S$ have the same image.
Hence, the function is not one-one.
Therefore, the function cannot be bijective.
Final Answer: The function is not bijective.
Q12
2023
00:00
How many one-one functions can be there from set $S$ to set $J$?
Let $|S| = m$ and $|J| = n$, with $m \le n$.
The number of one-one (injective) functions from $S$ to $J$ is ${}^nP_m = \frac{n!}{(n - m)!}$.
This counts the number of ways to assign distinct images in $J$ to each element of $S$.
Final Answer: Number of one-one functions = ${}^nP_m = \frac{n!}{(n - m)!}$
Q13
2023
00:00
Let $f : R - \{4/3\} \to R$ be a function defined as $f(x) = \frac{4x}{3x + 4}$. Show that $f$ is a one-one function. Also, check whether $f$ is an onto function or not.
To prove one-one, assume $f(x_1) = f(x_2)$.
So, $\frac{4x_1}{3x_1+4} = \frac{4x_2}{3x_2+4}$.
Cross multiplying: $4x_1(3x_2+4) = 4x_2(3x_1+4)$.
Simplifying: $12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2$.
So, $16x_1 = 16x_2 \Rightarrow x_1 = x_2$.
Hence, $f$ is one-one.
To check onto, let $y \in R$. Then $y = \frac{4x}{3x+4}$.
Solving for $x$: $y(3x+4)=4x \Rightarrow 3xy+4y=4x \Rightarrow x(3y-4)=-4y \Rightarrow x = \frac{-4y}{3y-4}$.
For every real $y \neq 4/3$, $x$ exists in $R - \{4/3\}$.
Hence, $f$ is onto.
Final Answer: $f$ is both one-one and onto, hence bijective.
Q14
2020
00:00
Show that the function $f : (-\infty, 0) \to (-1, 0)$ defined by $f(x) = \frac{x}{1 + |x|}$, $x \in (-\infty, 0)$ is one-one and onto.
First show one-one: Assume $f(x_1)=f(x_2)$.
So, $\frac{x_1}{1+|x_1|} = \frac{x_2}{1+|x_2|}$.
Cross multiplying gives $x_1(1+|x_2|) = x_2(1+|x_1|)$.
Since both $x_1$ and $x_2$ are negative, $|x_1| = -x_1$ and $|x_2| = -x_2$.
Substituting and simplifying leads to $x_1 = x_2$.
Hence, $f$ is one-one.
To prove onto, let $y \in (-1,0)$.
Solve $y = \frac{x}{1+|x|}$ for $x$ and get $x = \frac{y}{1-|y|}$.
Since $y \in (-1,0)$, $x \in (-\infty,0)$.
Hence, every $y$ has a preimage and $f$ is onto.
Final Answer: $f$ is one-one and onto, hence bijective.
Q15
2024
00:00
Show that a function $f : R \to R$ defined by $f(x) = \frac{2x}{1 + x^2}$ is neither one-one nor onto. Further, find set $A$ so that the given function $f : R \to A$ becomes an onto function.
To check one-one, take $x$ and $-x$. Then $f(x)=\frac{2x}{1+x^2}$ and $f(-x)=\frac{-2x}{1+x^2}$.
So different $x$ give different values in sign but $f(1)=f(-1)$ in magnitude, hence not one-one.
To find range, use calculus or inequality: $|\frac{2x}{1+x^2}| \le 1$.
So range is $[-1,1]$.
Since codomain is $R$, not all real numbers are obtained, so $f$ is not onto.
If codomain $A$ is chosen as $[-1,1]$, then $f$ becomes onto.
Final Answer: $f$ is neither one-one nor onto; choosing $A = [-1, 1]$ makes $f$ onto.
Q16
2024
00:00
Show that a function $f : R \to R$ defined as $f(x) = x^2 + x + 1$ is neither one-one nor onto. Also find all the values of $x$ for which $f(x) = 3$.
Since $f(x)$ is a quadratic function, it is not one-one over $R$.
Its minimum value occurs at $x = -1/2$ and minimum $f(x) = 3/4$.
So range is $[3/4, \infty)$. Hence, $f$ is not onto $R$.
To find $f(x)=3$: $x^2 + x + 1 = 3 \Rightarrow x^2 + x - 2 = 0$.
Solving: $(x+2)(x-1)=0 \Rightarrow x = -2 \text{ or } x = 1$.
Final Answer: $f$ is neither one-one nor onto; solutions of $f(x)=3$ are $x = -2$ and $x = 1$.
Q17
2024
00:00
Let $A = R - \{5\}$ and $B = R - \{1\}$. Consider the function $f : A \to B$, defined by $f(x) = \frac{x - 3}{x - 5}$. Show that $f$ is one-one and onto.
To show one-one, assume $f(x_1)=f(x_2)$.
So $\frac{x_1-3}{x_1-5} = \frac{x_2-3}{x_2-5}$.
Cross multiplying and simplifying gives $x_1 = x_2$.
Hence, $f$ is one-one.
To show onto, take any $y \in B, y \neq 1$.
Solve $y = \frac{x-3}{x-5} \Rightarrow y(x-5)=x-3 \Rightarrow x(y-1)=5y-3 \Rightarrow x = \frac{5y-3}{y-1}$.
Since $y \neq 1$, $x$ exists in $A$.
Hence, $f$ is onto.
Final Answer: $f$ is both one-one and onto, hence bijective.