Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.2
Last Updated : 05 Sep, 2024
In Class 12 Mathematics, Chapter 1, "Relations and Functions" students delve into the foundational concepts of relations and functions. Exercise 1.2 focuses on the various problems to enhance understanding of these concepts. This exercise is crucial for grasping how different functions relate to each other and how they can be represented mathematically.
Relations and Functions
Relations: A relation between two sets is a collection of ordered pairs where each element from the first set is associated with the elements in the second set. The Relations can be represented using sets, tables, or graphs. For example: if we have two sets A and B a relation from the A to B can be described as a subset of the Cartesian product A × B.
Functions: A function is a special type of relation where each element in the domain is associated with exactly one element in the codomain. The Functions can be represented using the function notation such as the f(x) where x is an element of the domain and f(x) is the corresponding element in the codomain. The Functions are essential for defining mathematical relationships and are used extensively in the various branches of mathematics.
Question 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 a co-domain the same as R*?
Solution:
One-one:
f(x)=f(y)
⇒1/x =1/y
⇒x=y
Therefore, f is one-one.
Onto:
It is clear that for y∈ R* there exists x=(1/y)∈ R* (exists as y ≠ 0) such that f(x)=1/(1/y)=y
Therefore, f is onto.
Thus, consider function g: N⇢R* defined by g(x)=1/x
We have, f(x1)=g(x2)⇒1/x1=1/x2⇒x1=x2
Therefore, g is one-one.
Further, it is clear that g is not onto as for 1.2∈ R* there does not exist any x in N such that g(x)=1/(1.2)
Hence, function g is one-one but not onto.
Question 2. Check the injectivity and surjectivity of the following functions:
(i) f: N⇢N given by f(x)=x2
Solution:
It is seen that for x, y ∈ N, f(x)=f(y) ⇒x2=y2⇒x=y
Therefore, f is injective.
Now, 2 ∈ N but there does not exist any x in N such that f(x)=x2=2.
Therefore, f is not surjective.
(ii) f: Z⇢Z given by f(x)=x2
Solution:
It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.
-2 ∈ Z. But, there does not exist any x in Z such that f(x)= x2=-2.Therefore, f is not surjective
(iii) f: R⇢ R given by f(x)=x2
Solution:
It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.
-2 ∈ R. But, there does not exist any x in R such that f(x)= x2=-2.Therefore, f is not surjective.
(iv)f: N⇢N given by f(x)=x3
Solution:
It is seen that for x, y ∈ N, f(x)=f(y)⇒x3=y3⇒x=y. Therefore, f is injective.
2∈ N. But, there does not exist any element x in domain N such that f(x)=x3=2. Therefore, f is not surjective.
(v) f: Z⇢Z given by f(x)=x3
Solution:
It is seen that for x, y ∈Z, f(x)=f(y)⇒x3=y3⇒x=y. Therefore, f is injective.
2∈Z. But, there does not exist any element x in domain Z such that f(x)=x3=2. Therefore, f is not surjective.
Question 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.
Solution:
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. Therefore, f is not one-one.
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. Therefore, f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Question 4. Show that the Modulus Function f:R⇢R given by f(x)=|x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.
Solution:
It is seen that f(-1)=|-1|=1, f(1)=|1|=1.
f(-1)=f(1), but -1≠1. Therefore, f is not one-one.
Consider, -1∈R. It is known that f(x)=|x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x)=|x|=-1. Therefore, f is not onto.
Hence, the modulus function is neither one-one nor onto.
Question 5. Show that the signum function f: R⇢R given by, f(x)={ (1, if x>0), (0, if x=0), (-1, if x<0)} is neither one-one nor onto.
Solution:
It is seen that f(1)=f(2)=1, but 1≠2. Therefore, f is not one-one.
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. Therefore, f is not onto.
Hence, the signum function is neither one-one nor onto.
Question 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.
Solution:
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)}
Therefore, 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.
Question 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
Solution:
Let x1, x2 ∈R such that f(x1)=f(x2)
⇒3-4x1=3-4x2
⇒-4x1=-4x2
⇒x1=x2
Therefore, f is one-one.
For any real number (y) in R, there exists {(3-y)/4} in R such that f((3-y)/4)=3-4((3-y)/4)=y.
Therefore, f is onto
Hence, f is bijective.
(ii) f:R⇢R defined b f(x)=1+x2
Solution:
Let x1, x2 ∈ R such that f(x1)=f(x2)
⇒1+x12=1+x22
⇒x12=x22
⇒x1=±x2
Therefore, f(x1)=f(x2) does not imply that x1=x2
For instance, f(1)=f(-1)=2
Therefore, f is not one-one.
Consider, an element -2 in co-domain R.
It is seen that f(x)=1+x2 is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x)=-1.
Therefore, f is not onto.
Hence, f is neither one-one nor onto.
Question 8. Let A and B be sets. Show that f: A x B ⇢B x A such that (a, b)=(b, a) is bijective function.
Solution:
Let (a1, b1), (a2, b2) ∈ A x b such that f(a1, b1)=f(a2, b2)
⇒(b1, a1)=(b2, a2)
⇒b1=b2 and a1=a2
⇒(a1, b1)=(a2, b2)
Therefore, f is one-one.
Let (b,a) ∈ B x A such that f(a, b)=(b,a).
Therefore, f is onto.
Hence, f is bijective.
Question 9. Let f: N⇢ N defined by f(n)={((n+1)/2, if n is odd), (n/2, if n is even) for all n ∈ N. State whether the function f is bijective. Justify your answer.
Solution:
It can be observed that:
f(1)=(1+1)/2=1 and f(2)=2/2=1
So, f(1)=f(2), where, 1≠2
Therefore, f is not one-one.
Therefore, it is not bijective. (Since, it needs to be both one-one and onto to be bijective).
Question 10. Let A=R-{1}. Consider the function f: A⇢B defined by f(x)=(x-2)/(x-3). Is f one-one and onto? Justify your answer.
Solution:
Let x, y ∈ A such that f(x)=f(y)
⇒ (x-2)/(x-3)=(y-2)/(y-3)
⇒(x-2)(y-3)=(y-2)(x-3)
⇒ xy-3x-2y+6=xy-3y-2x+6
⇒ -3x-2y=-3y-2x
⇒ 3x-2x=3y-2y
⇒ x=y
Therefore, 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,
f(x)=y
⇒ (x-2)/(x-3)=y
⇒ x-2=xy-3y
⇒ x(1-y)=-3y+2
⇒ x=(2-3y)/(1-y) ∈ A
Thus, for any y ∈ B, there exists (2-3y)/(1-y) ∈ A such that f((2-3y)/(1-y))={((2-3y)/(1-y))-2}/{((2-3y)/(1-y))-3}=(2-3y-2+2y)/(2-3y-3+3y)=(-y)/(-1)=y
Therefore, f is onto.
Hence, function f is one-one and onto.
Question 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
Solution:
Let x, y ∈ R such that f(x)=f(y)
⇒ x4=y4
⇒ x=±y
Therefore, f(x1)=f(x2) does not imply that x1=x2
For instance, f(1)=f(-1)=1
Therefore, f(1)=f(-1)=1
Therefore, 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
Therefore, f is not onto.
The correct answer is D.
Question 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
Solution:
Let x, y ∈ R such that f(x)=f(y)
⇒ 3x = 3y
⇒ x=y
Therefore, f is one-one.
Also, for any real number (y) in co-domain R, there exists y/3 in R such that f(y/3) = 3(y/3) = y
Therefore, f is onto.
Hence, the correct answer is A.
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Conclusion
Understanding relations and functions is fundamental to advancing in the mathematics as these concepts form the basis for the more complex topics. Exercise 1.2 in Chapter 1 of the Class 12 NCERT Mathematics book provides the essential practice to the solidify these foundational ideas. Mastery of these concepts will enhance problem-solving skills and prepare students for the higher-level mathematical challenges.
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 1Question 1. Find the maximum and minimum values, if any, of the following function given by(i) f(x) = (2x - 1)2 + 3 Solution: Given that, f(x) = (2x - 1)2 + 3 From the given function we observe that (2x - 1)2 ≥ 0 ∀ x∈ R, So, (2x - 1)2 + 3 ≥ 3 ∀ x∈ R, Now we find the minimum value of function f when
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 2In Class 12 Mathematics, the chapter on Applications of Derivatives is one of the most important topics. This chapter focuses on how derivatives are applied to solve real-world problems such as finding the rate of change determining the slope of a curve, or calculating the maximum and minimum values
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 1Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," focuses on how derivatives are used in various real-life situations and mathematical problems. The chapter covers concepts such as finding the rate of change of quantities, determining maxima and minima, and a
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 2Content of this article has been merged with Chapter 6 Application of Derivatives - Miscellaneous Exercise as per the revised syllabus of NCERT.Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," is essential for understanding how derivatives are applied in rea
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