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NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

Edited By Ramraj Saini | Updated on Sep 12, 2022 06:52 PM IST

Introduction: Class 12 Maths chapter 5 notes are concerning continuity and differentiability. In continuity and differentiability, Class 12 notes include finding continuity and differentiability nothing but finding the limits and differentiating. This Class 12 Maths chapter 5 notes contains the following topics: continuity at a point, continuity in the interval, few standard limits notations, differentiability, right-hand derivative, left-hand derivative, rules of differentiation like sum, difference, product, quotient, chain rules, second-order derivatives, standard formulas, Rolle’s theorem, Mean value theorem and many more.

Notes for Class 12 Maths chapter 5 also contain standard formulas that are to be memorised for the implementation in problems. NCERT Class 12 Math chapter 5 contains a detailed explanation of topics, theorems, examples, exercises. By going through the document students can cover all the topics that are in NCERT Notes for Class 12 Math chapter 5 textbook. It also contains examples, exercises, a few interesting points and most importantly contains FAQs that are frequently asked questions by students which can clarify many other students with the same doubt.

Every concept that is in CBSE Class 12 Maths chapter 5 notes is explained here in a simple and understanding way that can reach students easily. All these concepts can be downloaded from Class 12 Maths chapter 5 notes pdf download, Class 12 notes continuity and differentiability, continuity and differentiability Class 12 notes pdf download.

Students can also refer,

NCERT Class 12 Chapter 5 Notes

Continuity:

DEFINITION: suppose f be a real function, c be a point in the domain of function f in the range of real numbers. f is said to be continuous if it satisfies

x→c fx=f(c)

Theoretical: If the left-hand side limit, right-hand side limits are equal, and x=c then function f is said to be continuous else not continuous.

DEFINITION 2: A function f is said to be continuous if it is continuous at every point in domain f.

Continuity of a: \lim_{x \to a} f(x) =f(a)

Algebra of continuous functions:

Theorem:

f and g be two real numbers that are continuous and real at point c, then

a) f + g is continuous at x = c

b) f-g is continuous at x = c

c) f. g is continuous at x = c

d) fg is continuous at x=c and g(c) ≠ 0

Proof:

From equation a :

f + g is said to be continuous at x = c

\lim_{x \to c} (f+g)(x)=\lim_{x \to c} f(x)+\lim_{x \to c} g(x)

=\lim_{x \to c} f(x)+\lim_{x \to c} g(x)

= f(c) + g(c)

= (f+ g) (c)

Hence proved that it is continuous.

Differentiability:

suppose f be a real function, c be a point in the domain of function f in the range of real numbers. f is said to be differentiable if it satisfies:

\lim_{h \to 0}\frac{f(c+h)-f(c)}{h}

Derivative of f(x) is given by f’(x)

f'(x)= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}

RIGHT HAND DERIVATIVE: Rf'(a)= \lim_{h \to 0} \frac{f(a+h)-f(a)}h

LEFT HAND DERIVATIVE: Lf'(a)= \lim_{h \to 0} \frac{f(a-h)-f(a)}{-h}

Sum and difference rule: let y = f(x) ± g(x)

\frac{dy}{dx} = \frac{d}{dx}{f(x)} \pm \frac{d}{dx}g(x)

Product rule: Let y = f(x) g(x)

\frac{dy}{dx} = \frac{d \ f(x)}{dx}g(x)+ \frac{d \ g(x)}{dx}f(x)

Quotient rule: Let y = f(x)/g(x); g(x) ≠ 0

\frac{dy}{dx} = \frac{g(x)\frac{d \ f(x)}{dx}- f(x) \frac{d \ g(x)}{dx}}{g(x)^{2}}

Chain rule: Let y = f(u) and u = f(x)

\frac{dy}{dx}=\frac{dy}{dU}\frac{dU}{dx}

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1644486472423

Differentiation of Functions in Parametric Form: Relationship between two values x and y that are expressed in the forms x = f(t), y = g(t) is called to be in parametric form when,

\frac{dy}{dx}= \frac{dy}{dt}.\frac{dt}{dx}

Second-order Derivative: Deriving First order derivative will result in a second-order derivative.

\frac{d^2y}{dx^2}= \frac{d}{dx}.(\frac{dy}{dx})

Rolle’s Theorem:

Let f : [a, b] → R be a real number and is continuous on [a, b] satisfies following:

a)differentiable in the interval (a, b) so that f(a) = f(b); a, b are real numbers. Then,

b) there will be at least one number c in (a, b) that satisfies the condition f'(c) = 0.

This is called Rolle’s Theorem.

Mean Value Theorem:

Let f : [a, b] → R be a real number that is continuous in the interval [a, b] satisfies the following:

a) differentiable in the interval (a, b). Then,

b) there will be at least one number c in the interval (a, b) that satisfies the condition

f'(c) = \frac{f(b)-f(a)}{b-a}

This is called the mean value theorem.

With this topic we conclude NCERT class 12 chapter 5 notes.

The link for the NCERT textbook pdf is given below:

URL: ncert.nic.in/ncerts/l/lemh105.pdf

Significance of NCERT Class 12 Maths Chapter 5 Notes:

NCERT Class 12 Maths chapter 5 notes will be very much helpful for students to obtain maximum marks in their 12 board exams. In Class 12 continuity and differentiability notes, we have discussed many topics like continuity, differentiability, derivatives, standard limits formulas, sum, difference, product, quotient, chain rules with their mathematical representations, Rolle’s and Mean value theorem along with their conditions and rules of solving the problems. NCERT Class 12 Mathematics chapter 5 is also very helpful to cover major topics of the Class 12 CBSE Mathematics Syllabus.

The CBSE Class 12 Maths chapter 5 will enable us to understand the theorems, statements, rules with their conditions in detail. This pdf also contains previous year’s questions and NCERT Textbook pdf. The next part contains FAQs' most frequently asked questions along with a topic-wise explanation. By referring to the document you can get a complete idea of all the topics of Class12 chapter 5 continuity and differentiability pdf download.

NCERT Class 12 Notes Chapter Wise.

Go through the above link to learn more examples topic-wise.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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