Looking for NCERT Solutions for Class 12 Maths Exercise 1.2? If yes then you are in the right place. Here we have provided you with the Question and Answers of Exercise 1.2. These solutions are written by expert teachers and are so accurate to rely on.
| Chapter | 1. Relations and Functions |
| Exercise | 1.2 |
| Class | Twelve |
| Subject | Maths |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.
NCERT Solutions for Class 12 Maths Exercise 1.2
Relations and Functions Solutions
Q 1) Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto, where : R* is the set of all non-zero real numbers. Is the result true, if the domain : R* is replaced by N with co-domain being same as : R*?
Answer)
Hence, function g is one-one but not onto.
Q 2) Check the injectivity and surjectivity of the following functions:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f: R → R given by f(x) = x2
(iv) f: N → N given by f(x) = x3
(v) f: Z → Z given by f(x) = x3
Answer)
(i) f: N → N is given by,
f(x) = x2
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x2 = y2 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by,
f(x) = x3
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x3 = y3 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by,
f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y)
⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
Q 3) Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answer)
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Q 4) Show that the Modulus Function f : R → R, given by f(x) = |x| is neither one-one nor onto, where is |x| if x is positive or 0 and |-x| is -x if x is negative.
Answer)
Q 5) Show that the Signum Function f: R → R , given by
is neither one-one nor onto.
Answer)
f: R → R
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) ) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R , there does not exist any x in domain R such that
f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
Q 6) Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
Show that f is one-one.
Answer)
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
Q 7) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x2
Answer)
(i) f: R → R is defined as f(x) = 3 − 4x.
∴ f is one-one.
For any real number (y) in R, there exists 
in R such that
∴f is onto.
Hence, f is bijective.
(ii) f: R → R is defined as
For instance,
∴ f is not one-one.
Consider an element −2 in co-domain R.
It is seen that 
is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f (x) = −2.
∴ f is not onto.
Hence, f is neither one-one nor onto.
Q 8) Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
Answer)
f: A × B → B × A is defined as f(a, b) = (b, a).
∴ f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]
∴ f is onto.
Hence, f is bijective.
Q 9) Let f: N → N be defined by 
State whether the function f is bijective. Justify your answer.
Answer) f: N → N is defined as
It can be observed that:
∴ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
∴n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈N such that
Case II: n is even
∴n = 2r for some r ∈ N. Then,there exists 4r ∈N such that 
∴ f is onto.
Hence, f is not a bijective function.
Q 10) Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by
Is f one-one and onto? Justify your answer.
Answer)
A = R − {3}, B = R − {1}
f: A → B is defined as
∴ f is one-one.
Let y ∈B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈A such that f(x) = y.
Now,
Thus, for any y ∈ B, there exists 
such that
Hence, function f is one-one and onto.
Q 11) Let f: R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
Answer) f: R → R is defined as f(x) = x4
Let x, y ∈ R such that f(x) = f(y).
∴ f (x1) = f (x2) does not imply that x1 = x2
For instance,
∴ f is not one-one.
Consider an element −-2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = −-2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.
Q 12) Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
Answer)
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
∴f is one-one.
Also, for any real number (y) in co-domain R, there exists 
in R such that
∴f is onto.
Hence, function f is one-one and onto.
The correct answer is A.
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