Exercise 1.4 delves into the advanced concepts of function composition and invertible functions, building upon the foundational knowledge of relations and functions established earlier in the chapter. This exercise is crucial for developing a deeper understanding of how functions interact and transform, paving the way for more complex mathematical analysis. Students will explore the process of combining functions through composition, understand the conditions under which functions can be inverted, and learn to find and work with inverse functions. These skills are fundamental in various branches of mathematics, including calculus, abstract algebra, and complex analysis, and have wide-ranging applications in fields such as physics, engineering, and computer science. Mastering the concepts in this exercise will equip students with powerful tools for solving sophisticated problems and understanding the intricate relationships between mathematical entities.
Question 1: Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(i) On Z+, define ∗ by a ∗ b = a – b
Solution:
If a, b belongs to Z+
a * b = a – b which may not belong to Z+
For eg: 1 – 3 = -2 which doesn’t belongs to Z+
Therefore, * is not a Binary Operation on Z+
(ii) On Z+, define * by a * b = ab
Solution:
If a, b belongs to Z+
a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
(iii) On R, define * by a * b = ab²
Solution:
If a, b belongs to R
a * b = ab2 which belongs to R
Therefore, * is Binary Operation on R
(iv) On Z+, define * by a * b = |a – b|
Solution:
If a, b belongs to Z+
a * b = |a – b| which belongs to Z+
Therefore, * is Binary Operation on Z+
(v) On Z+, define * by a * b = a
Solution:
If a, b belongs to Z+
a * b = a which belongs to Z+
Therefore, * is Binary Operation on Z+
Question 2: For each binary operation * defined below, determine whether * is binary, commutative or associative.
(i) On Z, define a * b = a – b
Solution:
a) Binary:
If a, b belongs to Z
a * b = a – b which belongs to Z
Therefore, * is Binary Operation on Z
b) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – b + c
RHS = (a – b) * c = a – b- c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) On Q, define a * b = ab + 1
Solution:
a) Binary:
If a, b belongs to Q, a * b = ab + 1 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab + 1
RHS = b * a = ba + 1 = ab + 1
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc + 1) = abc + a + 1
RHS = (a * b) * c = abc + c + 1
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) On Q, define a ∗ b = ab/2
Solution :
a) Binary:
If a, b belongs to Q, a * b = ab/2 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab/2
RHS = b * a = ba/2
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc/2) = (abc)/2
RHS = (a * b) * c = (ab/2) * c = (abc)/2
Since, LHS is equal to RHS
Therefore, * is Associative
(iv) On Z+, define a * b = 2ab
Solution:
a) Binary:
If a, b belongs to Z+, a * b = 2ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = 2ab
RHS = b * a = 2ba = 2ab
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)
RHS = (a * b) * c = 2ab * c = 22abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) On Z+, define a * b = ab
Solution:
a) Binary:
If a, b belongs to Z+, a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = ab
RHS = b * a = ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc = ab^c
RHS = (a * b) * c = ab * c = abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(vi) On R – {– 1}, define a ∗ b = a / (b + 1)
Solution:
a) Binary:
If a, b belongs to R, a * b = a / (b+1) which belongs to R
Therefore, * is Binary Operation on R
b) Commutative:
If a, b belongs to R, a * b = b * a
LHS = a * b = a / (b + 1)
RHS = b * a = b / (a + 1)
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to A, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1
RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)
Since, LHS is not equal to RHS
Therefore, * is not Associative
Question 3. Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}. Write the operation table of the operation ∧.
Solution:
| ^ | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 3 | 3 | 3 |
| 4 | 1 | 2 | 3 | 4 | 4 |
| 5 | 1 | 2 | 3 | 4 | 5 |
Question 4: Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(Hint: use the following table)
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
Solution:
Here, (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) Is ∗ commutative?
Solution:
The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
Solution:
(2 * 3) * (4 * 5) = 1 * 1 = 1
Question 5: Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.
Solution:
Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.
| *’ | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
We see that the operation *’ is the same as the operation * in Exercise 4 above.
Question 6: Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
(i) 5 ∗ 7, 20 ∗ 16
Solution:
If a, b belongs to N
a * b = LCM of a and b
5 * 7 = 35
20 * 16 = 80
(ii) Is ∗ commutative?
Solution:
If a, b belongs to N
LCM of a * b = ab
LCM of b * a = ba = ab
a*b = b*a
Thus, * binary operation is commutative.
(iii) Is ∗ associative?
Solution:
a * (b * c) = LCM of a, b, c
(a * b) * c = LCM of a, b, c
Since, a * (b * c) = (a * b) * c
Thus, * binary operation is associative.
(iv) Find the identity of ∗ in N
Solution:
Let ‘e’ is an identity
a * e = e * a, for a belonging to N
LCM of a * e = a, for a belonging to N
LCM of e * a = a, for a belonging to N
e divides a
e divides 1
Thus, e = 1
Hence, 1 is an identity element
(v) Which elements of N are invertible for the operation ∗?
Solution:
a * b = b * a = identity element
LCM of a and b = 1
a = b = 1
only ‘1’ is invertible element in N.
Question 7: Is ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b a binary operation? Justify your answer.
Solution:
The operation * on the set {1, 2, 3, 4, 5} is defined as
a * b = L.C.M. of a and b
Let a=3, b=5
3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set
Thus, * is not a Binary Operation.
Question 8: Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N?
Solution:
If a, b belongs to N
LHS = a * b = HCF of a and b
RHS = b * a = HCF of b and a
Since LHS = RHS
Therefore, * is Commutative
Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = HCF of a, b and c
RHS = (a – b) * c = HCF of a, b and c
Since, LHS = RHS
Therefore, * is Associative
Now, 1 * a = a * 1 ≠ a
Thus, there doesn’t exist any identity element.
Question 9: Let ∗ be a binary operation on the set Q of rational numbers as follows:
(i) a ∗ b = a – b
(ii) a ∗ b = a2 + b2
(iii) a ∗ b = a + ab
(iv) a ∗ b = (a – b)2
(v) a ∗ b = ab / 4
(vi) a ∗ b = ab2
Find which of the binary operations are commutative and which are associative.
Solution:
(i) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – (b – c) = a – b + c
RHS = (a – b) * c = a – b – c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a2 + b2
RHS = b * a = b2 + a2
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2
RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a + ab
RHS = b * a = b + ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)
RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iv) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = (a – b)2
RHS = b * a = (b – a)2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b – c)2 = [a – (b – c)2]2
RHS = (a * b) * c = (a – b)2 * c = [(a – b)2 – c]2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab / 4
RHS = b * a = ba / 4
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc/4 = abc/16
RHS = (a * b) * c = ab/4 * c = abc/16
Since, LHS is equal to RHS
Therefore, * is Associative
(vi) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab2
RHS = b * a = ba2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc)2 = a(bc2)2
RHS = (a * b) * c = (ab2) * c = ab2c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
Question 10: Find which of the operations given above has identity
Solution:
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, for a ∈ Q
for (v) a * b = ab/4
Let e be an identity element
a * e = a = e * a
LHS : ae/4 = a
=> e = 4
RHS : ea/4 = a
=> e = 4
LHS = RHS
Thus, Identity element exists
Other operations doesn’t satisfy the required conditions.
Hence, other operations doesn’t have identity.
Question 11: Let A = N × N and ∗ be the binary operation on A defined by :
(a, b) ∗ (c, d) = (a + c, b + d)
Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.
Solution:
Given (a, b) * (c, d) = (a+c, b+d) on A
Let (a, b), (c, d), (e,f) be 3 pairs ∈ A
Commutative :
LHS = (a, b) * (c, d) = (a+c, b+d)
RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)
RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)
Since, LHS is equal to RHS
Therefore, * is Associative
Existence of Identity element:
For a, e ∈ A, a * e = a
(a, b) * (e, e) = (a, b)
(a+e, b+e) = (a, b)
a + e = a
=> e = 0
b + e = b
=> e = 0
As 0 is not a part of set of natural numbers. So, identity function does not exist.
Question 12: State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation ∗ on a set N, a ∗ a = a ∀ a ∈ N.
(ii) If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a
Solution:
(i) Let * be an operation on N, defined as:
a * b = a + b ∀ a, b ∈ N
Let us consider b = a = 6, we have:
6 * 6 = 6 + 6 = 12 ≠ 6
Therefore, this statement is false.
(ii) Since, * is commutative
LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS
Therefore, this statement is true.
Question 13: Consider a binary operation ∗ on N defined as a ∗ b = a3+ b3. Choose the correct answer.
(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?
Solution:
On N, * is defined as a * b = a3 + b3
Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a3 + b3
RHS = b * a = b3 + a3
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3
Since, LHS is not equal to RHS
Therefore, * is not Associative
Thus, Option (B) is correct
Summary
Exercise 1.4 serves as a critical juncture in the study of functions, bridging basic concepts with more advanced applications. Through exploring function composition, students learn to create complex functions from simpler ones, a skill essential in modeling real-world phenomena. The focus on invertible functions introduces the idea of reversibility in mathematical operations, a concept with far-reaching implications in various fields of study. By working through problems involving finding inverses and composing functions, students develop analytical skills and a deeper intuition about functional relationships. This exercise not only reinforces previous learning but also prepares students for more advanced topics in calculus and beyond, such as differential equations and functional analysis. Mastery of these concepts provides a solid foundation for understanding transformation of functions, a key element in many areas of higher mathematics and its applications in science and engineering.
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Exercise 4.2 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook delves deeper into the properties and applications of determinants. This set builds upon the foundational concepts introduced earlier, challenging students with more complex problems that require a nuanced understanding of d
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3
Question 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8) Solution: (i) (1, 0), (6, 0), (4, 3) [Tex]Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6&0
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.4
Exercise 4.4 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook focuses on advanced applications of determinants and their properties. This exercise builds upon the foundational knowledge established in previous sections, challenging students to apply determinant techniques to more compl
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.5
NCERT Solutions for Class 9 Maths, Chapter 1, Number Systems, Exercise 1.3, meticulously crafted by our subject experts, facilitating effortless learning for students. These solutions serve as a valuable reference for students tackling exercise problems. Exercise 1.3 delves into the realm of real nu
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 1
Examine the consistency of the system of equations in Exercises 1 to 6.Question 1. x + 2y = 2 2x + 3y = 3 Solution: Matrix form of the given equations is AX = B where, A =[Tex]\begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix} [/Tex] , B = [Tex]\begin{bmatrix}2 \\3 \\\end{bmatrix} [/Tex] and, X =
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 2
Chapter 4 Determinants - Exercise 4.6 | Set 1Question 11. 2x + y + z = 1 x - 2y - z = 3/2 3y - 5z = 9 Solution: Matrix form of the given equation is AX = B i.e.[Tex]\begin{bmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatri
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Miscellaneous Exercises on Chapter 4
Question 1. Prove that the determinant [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix} [/Tex]is independent of θ. Solution: A = [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\
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Chapter 5 - Continuity and Differentiability
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1
Question 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. Solution: To prove the continuity of the function f(x) = 5x – 3, first we have to calculate limits and function value at that point. Continuity at x = 0 Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1 | Set 2
Chapter 5 on Continuity and Differentiability is a crucial part of calculus in Class 12 mathematics. It builds upon the concept of limits and introduces students to the fundamental ideas of continuous functions and differentiation. This chapter lays the groundwork for understanding rates of change,
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.2
Chapter 5 of the Class 12 NCERT Mathematics textbook focuses on the concepts of continuity and differentiability which are fundamental in understanding calculus. This chapter helps students grasp how functions behave concerning their limits and derivatives. Exercise 5.2 is designed to test and reinf
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.3
Chapter 5 of the Class 12 NCERT Mathematics textbook, "Continuity and Differentiability," introduces the concepts of continuity and differentiability of functions, which are fundamental in calculus. Exercise 5.3 focuses on applying these concepts to solve problems related to the continuity and diffe
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.4
Differentiate the following w.r.t xQuestion 1. y = [Tex]\frac{e^x}{\sin x}[/Tex] Solution: [Tex]\frac{dy}{dx}=\frac{d}{dx}(\frac{e^x}{\sin x})[/Tex] [Tex]\frac{dy}{dx}=\frac{\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin x}{sin^2x} [/Tex] ([Tex]\frac{d}{dx}(\frac{u}{v})=v\frac{\frac{du}{dx}-u\frac{dv}{dx
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5
Differentiate the functions given in question 1 to 10 with respect to x.Question 1. cos x.cos2x.cos3x Solution: Let us considered y = cos x.cos2x.cos3x Now taking log on both sides, we get log y = log(cos x.cos2x.cos3x) log y = log(cos x) + log(cos 2x) + log (cos 3x) Now, on differentiating w.r.t x,
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2
Exercise 5.5 focuses on the Mean Value Theorem and Rolle's Theorem, which are fundamental concepts in calculus. These theorems provide powerful tools for analyzing functions and their behavior over intervals. This exercise helps students understand how to apply these theorems to solve various mathem
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.6
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find [Tex]\frac{dy}{dx}[/Tex]Question 1. x = 2at2, y = at4 Solution: Here, x = 2at2, y = at4 [Tex]\frac{dx}{dt} = \frac{d(2at^2)}{dt}[/Tex] = 2a [Tex]\frac{d(t^2)}{dt}[/Tex] = 2a (
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.7
Exercise 5.7 focuses on the application of derivatives in approximation and errors. This exercise explores how derivatives can be used to estimate function values and calculate errors in measurements or approximations. It introduces students to concepts like absolute and relative errors, percentage
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.8
Note: Please note that Exercise 5.8 from Chapter 5, "Continuity and Differentiability" in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum. Exercise 5.8 focuses on the application of derivatives to find tang
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5
In Chapter 5 of the Class 12 NCERT Mathematics textbook, titled Continuity and Differentiability, students explore fundamental concepts related to the behavior of functions. The chapter emphasizes understanding continuity, differentiability, and their implications in calculus. It includes various ex
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Chapter 6 - Applications of Derivatives
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1
The study of derivatives is a cornerstone in calculus providing the essential tools for understanding and analyzing functions. The Application of Derivatives a key topic in Class 12 Mathematics involves using the derivatives to solve real-world problems. This topic helps in understanding how rates o
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Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2
The Application of derivatives is a crucial topic in calculus that involves using derivatives to solve practical problems and understand the various aspects of functions beyond their basic behavior. It helps in analyzing the rates of change optimizing the functions and understanding the geometric pr
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Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2| Set 2
The chapter "Application of Derivatives" in Class 12 Mathematics is a critical part of the NCERT curriculum. It focuses on using derivatives to the solve real-world problems including rate of change, maxima and minima, tangents and normals. Exercise 6.2 specifically deals with the problems related t
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2
In this article, we will see some problems of derivatives a fundamental concept in calculus, and mathematical analysis that measures how a function changes as input changes. Exercise 6.3 focuses on the application of derivatives to find the rate of change of quantities. This exercise builds upon the
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.4
Question 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i)√25.3 (ii)√49.5 (iii) √0.6 (iv) (0.009)1/3 (v) (0.999)1/10 (vi) (15)1/4 (vii) (26)1/3 (viii) (255)1/4 (ix) (82)1/4 (x) (401)1/2 (xi) (0.0037)1/2 (xii) (26.57)1/3 (xiii) (81.5)1/4 (xiv)
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 1
Question 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 2
In 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 1
Chapter 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 2
Content 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 re
13 min read