Class 12 NCERT Mathematics Solutions : Chapter 2 Inverse Trigonometric Functions - Exercise 2.1
Last Updated : 03 Sep, 2024
In this article, we will be going to solve the entire exercise 2.1 of our NCERT textbook which is Inverse Trigonometric Functions. Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's fundamental in many areas of mathematics and applied sciences, including physics, engineering, and astronomy.
Inverse Trigonometric Functions
Inverse Trigonometric Functions are functions that reverse the action of the basic trigonometric functions (sine, cosine, tangent, etc.). They allow us to find the angle when the value of the trigonometric function is known. Understanding these functions is crucial for solving equations and modelling situations where the angle needs to be found from a given trigonometric value.
Knowing How to Interpret Inverse Trigonometric Functions
The idea of inverse trigonometric functions—the opposites of common trigonometric functions like sine, cosine, tangent, etc.—is first presented to students in Exercise 2.1. The definition of these functions, their domains and ranges, and the process of determining principal values for particular inputs are the main points of emphasis in this section.
Qualities of Trigonometric Functions in Reverse
Understanding the fundamental characteristics of inverse trigonometric functions—including periodicity, symmetry, and behavior under composition with trigonometric functions—is essential for this exercise. Comprehending these attributes is crucial for resolving issues related to modifying and simplifying expressions that incorporate inverse trigonometric functions.
Methodical Solutions for Exercise 2.1 Issues
This section ensures that students understand the process needed to solve problems involving inverse trigonometric functions by offering comprehensive, step-by-step solutions to every problem in Exercise 2.1. The solutions deal with identifying principal values, reducing the complexity of expressions, and using identities to resolve equations.
Class 12 NCERT Mathematics Solutions: Inverse Trigonometric Functions - Exercise 2.1
Find the principal values of the following:
Question 1. sin-1(-1/2)
Solution:
Let sin-1(-1/2) = y then, sin y = -1/2
Range of principal value for sin-1 is [-π/2,π/2] and sin(-π/6)=-1/2.
Therefore, principal value of sin-1(-1/2)=-π/6.
Question 2. cos-1(√3/2)
Solution:
Let cos-1(√3/2) = y then, cos y = √3/2
Range of principal value for cos-1 is [0, π] and cos(π/6) = √3/2
Therefore, principal value of cos-1(√3/2) = π/6.
Question 3. cosec-1(2)
Solution:
Let cosec-1(2) = y then, cosec y = 2
Range of principal value for cosec-1 is [-π/2, π/2] -{0} and cosec(π/6) = 2
Therefore, principal value of cosec-1(2) = π/6.
Question 4: tan-1(-√3)
Solution:
Let tan-1(-√3) = y then, tan y = -√3
Range of principal value for tan-1 is (-π/2, π/2) and tan(-π/3) = -√3
Therefore, principal value of tan-1(-√3) = -π/3.
Question 5. cos-1(-1/2)
Solution:
Let cos-1(-1/2) = y then, cos y = -1/2
Range of principal value for cos-1 is [0, π] and cos(2π/3) = -1/2
Therefore, principal value of cos-1(-1/2) = 2π/3.
Question 6. tan-1(-1)
Solution:
Let tan-1(-1) = y then, tan y = -1
Range of principal value for tan-1 is (-π/2, π/2) and tan(-π/4) = -1
Therefore, principal value of tan-1(-1) = -π/4.
Question 7. sec-1(2/√3)
Solution:
Let sec-1(2/√3) = y then, sec y = 2/√3
Range of principal value for sec-1 is [0, π] - {π/2} and sec(π/6) = 2/√3
Therefore, principal value of sec-1(2/√3) = π/6.
Question 8. cot-1(√3)
Solution:
Let cot-1(√3) = y then, cot y = √3
Range of principal value for cot-1 is (0, π) and cot(π/6) = √3
Therefore, principal value of cot-1(√3) = π/6.
Question 9. cos-1(-1/√2)
Solution:
Let cos-1(-1/√2) = y then, cos y = -1/√2
Range of principal value for cos-1 is [0, π] and cos(2π/3) = -1/2
Therefore, principal value of cos-1(-1/2) = 3π/4.
Question 10. cosec-1(-√2)
Solution:
Let cosec-1(-√2) = y then, cosec y = -√2
Range of principal value for cosec-1 is [-π/2, π/2] -{0} and cosec(-π/4) = -√2
Therefore, principal value of cosec-1(-√2) = -π/4.
Find the values of the following:
Question 11. tan-1(1) + cos-1(-1/2) + sin-1(-1/2)
Solution:
For solving this question we will use principal values of sin-1, cos-1 & tan-1
Let sin-1(-1/2) = y then, sin y = -1/2
Range of principal value for sin-1 is [-π/2, π/2] and sin(-π/6) = -1/2.
Therefore, principal value of sin-1(-1/2) = -π/6.
Let cos-1(-1/2) = x then, cos x = -1/2
Range of principal value for cos-1 is [0, π] and cos(2π/3) = -1/2
Therefore, principal value of cos-1(-1/2) = 2π/3.
Let tan-1(1) = z then, tan z = -1
Range of principal value for tan-1 is (-π/2, π/2) and tan(π/4) = 1
Therefore, principal value of tan-1(1) = π/4.
Now, tan-1(1) + cos-1(-1/2) + sin-1(-1/2) = π/4 + 2π/3 - π/6
Adding them we will get,
= (3π + 8π - 2π)/12
= 9π/12
= 3π/4
Question 12. cos-1(1/2) + 2 sin-1(1/2)
Solution:
For solving this question we will use principal values of sin-1 & cos-1
Let sin-1(1/2) = y then, sin y = -1/2
Range of principal value for sin-1 is [-π/2, π/2] and sin(π/6) = 1/2.
Therefore, principal value of sin-1(1/2) = π/6.
Let cos-1(1/2) = x then, cos x = 1/2
Range of principal value for cos-1 is [0, π] and cos(π/3) = 1/2
Therefore, principal value of cos-1(1/2) = π/3.
Now, cos-1(1/2) + 2 sin-1(1/2) = π/3 + 2π/6
Adding them we will get,
= (2π + 2π)/6
= 4π/6
= 2π/3
Question 13. If sin–1 x = y, then
(A) 0 ≤ y ≤ π (B) -π / 2 ≤y ≤ π / 2 (C) 0 < y < π (D) -π / 2 <y < π / 2
Solution:
We know that the principal range for sin-1 is [-π / 2, π / 2]
Hence, if sin-1 x = y, y € [-π / 2, π / 2]
Therefore, -π / 2 ≤y ≤ π / 2.
Hence, option (B) is correct.
Question 14. tan–1(√3) - sec-1(-2) is equal to
(A) π (B) -π/3 (C) π/3 (D) 2π/3
Solution:
For solving this question we will use principal values of sec-1 & tan-1
Let tan-1(√3) = y then, tan y = √3
Range of principal value for tan-1 is (-π/2, π/2) and tan(π/3) = √3
Therefore, principal value of tan-1(√3) = π/3.
Let sec-1(-2) = y then, sec y = -2
Range of principal value for sec-1 is [0, π] - {π/2} and sec(2π/3) = - 2
Therefore, principal value of sec-1(-2) = 2π/3.
Now, tan–1 (√3) - sec -1(-2)
= π/3 - 2π/3
= -π/3
Hence, option (B) is correct.
Related Articles,
Conclusion
The purpose of Exercise 2.1 in the Class 12 NCERT Mathematics textbook is to introduce the inverse trigonometric functions and highlight their principal values and domains and their properties. Students will obtain a basic understanding of manipulating and applying inverse trigonometric functions in a variety of mathematical contexts by completing this exercise.
Similar Reads
NCERT Solutions for Class 12 Maths Maths is among the highest-scoring subjects in the Class 12th board exam 2024-25. The syllabus of the CBSE Maths exam is based on the latest NCERT Maths syllabus. So, GeeksforGeeks has curated the NCERT Class 12 Maths Solution for you to prepare. Students can also download the NCERT Solution PDF of
11 min read
Chapter 1 - Relations and Functions
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.1Relation and Function are two ways of establishing links between two sets in mathematics. Relation and Function in maths are analogous to the relation that we see in our daily lives i.e., two persons are related by the relation of father-son, mother-daughter, brother-sister, and many more. On a simi
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.1 | Set 2Content of this article has been merged with Chapter 1 Relations And Functions - Exercise 1.1 as per the revised syllabus of NCERT. Question 11. Show that the relation R in the set A of points in a plane given by R ={ (P,Q): distance of the point P from the origin is the same as the distance of the
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.2In 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 eac
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions - Exercise 1.3Please note that Exercise 1.3 from Chapter 1, “Relations and Functions†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. Solve The Following Questions NCERT Solutions for Class 12 Math's Chapter 1 Exerci
10 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 1 Relations and Functions - Exercise 1.4Exercise 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 transf
15+ min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 1 Relations and Functions - Exercise 1.4 | Set 2Exercise 1.4 | Set 2 builds upon the concepts introduced in the first set, focusing on more advanced aspects of function composition and inverse functions. This set challenges students to apply their understanding to more complex scenarios, including composite functions with trigonometric, exponenti
10 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 1Question 1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = 1R. Solution: As, it is mentioned here f : R → R be defined as f(x) = 10x + 7 To, prove the function one-one Let's take f(x) = f(y) 10x + 7 = 10y + 7 x = y Hence f is one-one. To, prove the
9 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 2Chapter 1 of NCERT Class 12 Mathematics Part I focuses on Relations and Functions, building upon the foundational concepts introduced in earlier classes. This chapter delves deeper into various types of relations and functions, exploring their properties, compositions, and applications. Students wil
8 min read
Chapter 2 - Inverse Trigonometric Functions
Class 12 NCERT Mathematics Solutions : Chapter 2 Inverse Trigonometric Functions - Exercise 2.1In this article, we will be going to solve the entire exercise 2.1 of our NCERT textbook which is Inverse Trigonometric Functions. Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's fundamental in many areas of mathematics and appli
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 1Exercise 2.2 | Set 1 focuses on the properties and applications of inverse trigonometric functions. This set of problems is designed to deepen students' understanding of these functions, their domains, ranges, and various identities. Inverse trigonometric functions, also known as arc functions, are
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 2Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 as per the revised syllabus of NCERT. Find the values of each of the following: Question 11. tan−1[2cos(2sin−11/2​)] Solution: Let us assume that sin−11/2 = x So, sinx = 1/2 Therefore, x = π​/6 = si
3 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 1Question 1. Find the value of {\cos }^{-1}(\cos \frac {13\pi} {6}) Solution: We know that \cos^{-1} (\cos x)=x Here, \frac {13\pi} {6} \notin [0,\pi]. Now, {\cos }^{-1}(\cos \frac {13\pi} {6}) can be written as : {\cos }^{-1}(\cos \frac {13\pi} {6})={\cos }^{-1}[\cos( 2\pi+\frac {\pi} {6})] , where
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 2Chapter 2 of Class 12 NCERT Mathematics Part I delves deeper into Inverse Trigonometric Functions, building upon the foundational concepts introduced earlier. This set of miscellaneous exercises challenges students to apply their understanding of inverse trigonometric functions in more complex scena
4 min read
Chapter 3 - Matrices
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1In this article, we will be going to solve the entire Miscellaneous Exercise 3.1 of Chapter 3 of the NCERT textbook. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The numbers, symbols, or expressions in the matrix are called elements or entries. M
11 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2Matrices are a fundamental concept in linear algebra, used extensively in mathematics, physics, engineering, computer science, and various other fields. A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The numbers or elements inside a matrix are enclo
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2 | Set 2The content of this article has been merged with Chapter 3 Matrices- Exercise 3.2 as per the revised syllabus of NCERT.Chapter 3 of the Class 12 NCERT Mathematics Part I textbook, titled "Matrices," introduces the fundamental concepts and operations related to matrices. This chapter covers various m
15 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 3 Matrices - Exercise 3.3Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," delves into the fundamental concepts of matrices, including their types, operations, and applications. Exercise 3.3 focuses on practical problems involving matrix operations, such as addition, subtraction, and multiplication of
13 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 1Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.Question 1.\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right] Solution: \begin{aligned} &\text { Since we know that, } A= IA \Rightarrow\left[\begin{array}{cc} 1 & -1
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 2Matrices are a fundamental concept in mathematics providing a compact way to represent and manipulate data and equations. They are used in various fields including algebra, calculus, and applied mathematics. This chapter focuses on the operations and properties of matrices essential for solving the
12 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Miscellaneous Exercise on Chapter 3Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," covers fundamental concepts related to matrices, including operations such as addition, multiplication, and finding determinants and inverses. The Miscellaneous Exercise in this chapter provides a range of problems that integra
12 min read
Chapter 4 - Determinants
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.1Chapter 4 of Class 12 NCERT Mathematics Part I introduces the concept of determinants, a crucial topic in linear algebra. Exercise 4.1 focuses on the fundamentals of determinants, including their definition, calculation for 2x2 and 3x3 matrices, and their basic properties. This exercise lays the gro
4 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.2 | Set 1Exercise 4.2 specifically focuses on the properties and applications of determinants, building upon the basic definitions introduced earlier in the chapter. This exercise aims to deepen students' understanding of how determinants can be manipulated and applied to solve various mathematical problems.
10 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants- Exercise 4.2 | Set 2Exercise 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
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3Question 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) Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6&0&
2 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.4Exercise 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
7 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.5NCERT 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
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 1Examine 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 =\begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix} , B = \begin{bmatrix}2 \\3 \\\end{bmatrix} and, X = \begin{bmatrix}x\\y\\\en
5 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 2Chapter 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.\begin{bmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Miscellaneous Exercises on Chapter 4Question 1. Prove that the determinant \begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix} is independent of θ. Solution: A = \begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1
15+ min read
Chapter 5 - Continuity and Differentiability
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1Question 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 = \lim_{x \to 0^-} f(x) = \lim_{x \to
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1 | Set 2Chapter 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,
15+ min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.2Chapter 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
4 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.3Chapter 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
5 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.4Differentiate the following w.r.t xQuestion 1. y = \frac{e^x}{\sin x} Solution: \frac{dy}{dx}=\frac{d}{dx}(\frac{e^x}{\sin x}) \frac{dy}{dx}=\frac{\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin x}{sin^2x} (\frac{d}{dx}(\frac{u}{v})=v\frac{\frac{du}{dx}-u\frac{dv}{dx}}{v^2}) \frac{dy}{dx}=\frac{\sin x e^x
1 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5Differentiate 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,
10 min read
Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2Exercise 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
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.6If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find \frac{dy}{dx}Question 1. x = 2at2, y = at4 Solution: Here, x = 2at2, y = at4 \frac{dx}{dt} = \frac{d(2at^2)}{dt} = 2a \frac{d(t^2)}{dt} = 2a (2t) = 4at And, now \frac{dy}{dt}
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.7Exercise 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
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.8Note: 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
6 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5In 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
13 min read
Chapter 6 - Applications of Derivatives
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1The 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
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2The 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
13 min read
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2| Set 2The 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
8 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2In 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
12 min read
Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.4Question 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)
10 min read
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
15+ min read
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
15 min read
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
10 min read
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
11 min read