NCERT Solutions for class 12 Maths chapter 5 Continuity & Differentiability
NCERT Solutions for class 12 Maths chapter 5 Continuity and Differentiability exercise 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8 and miscellaneous exercise in PDF form to free download. NCERT Solutions, Assignments, study material, chapter wise tests and mixed chapters tests with answers are given to download. Questions from CBSE board papers will also be uploaded related to continuity and differentiability in the form of practice tests. Click here to move next chapter: Application of Derivatives.
NCERT Solutions for class 12 Maths chapter 5
NCERT Solutions of Continuity and Differentiability
- Exercise 5.1
- Exercise 5.2
- Exercise 5.3
- Exercise 5.4
- Exercise 5.5
- Exercise 5.6
- Exercise 5.7
- Exercise 5.8
NCERT Chapter to study online and answers given in the end of NCERT books.
Revision books for practice, confined to NCERT Syllabus.
Assignments for practice
Mixed Chapter Tests
Chapter 5 & 6
Level 1 Test 1
Level 2 Test 1
- The main points of the chapter are continuous functions, algebra of continuous functions, differentiation and continuity, chain rule, rules for derivative of inverse functions, derivative of implicit function, parametric and logarithmic functions. Second order derivatives, mean value theorems (LMV) and Rolle’s Theorem.
- Calculus (differentiability) deals with related variables and constants. In differential calculus we investigate the way in which ‘one quantity varies when the other related quantity is made to vary’. We find the rate of change (in Application of derivatives) one variable quantity relative to another variable. The relation between a function and its derivative is same as between displacement of a particle and its velocity.
- The derivative of f(x) at x = a is represented by the slope (gradient) of the tangent to the curve y = f(x) at the point P[a, f(a)].
Let f be a real function defined in the closed interval [a, b] such that
- (i) f is continuous in the closed interval [ a, b ]
- (ii) f is differentiable in the open interval ( a, b )
- (iii) f (a) = f (b)
- Then f'(c) = 0, where c lies in (a, b).
Mean Value Theorem
Let f be a real valued function defined on the closed interval [a, b] such that
- (a) f is continuous on [a, b], and
- (b) f is differentiable in (a, b)
- Then there exists a point c in the open interval (a, b) such that f ‘(c) = [f (b) – f (a)]/[b – a]
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