Continuity and Differentiability Class 12th Notes - Free NCERT Class 12 Maths Chapter 5 Notes - Download PDF

Continuity and Differentiability Notes:

Continuity:

DEFINITION 1: Suppose $f$ is a real function, $c$ is a point in the domain of function $f$ in the range of real numbers. $f$ is said to be continuous if it satisfies

$\lim\limits_{x \rightarrow c} f(x)=f(c)$

Theoretical: If the left-hand side limit, right-hand side limits are equal, and x=c, then the 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 its domain.

Let us understand this with an example, suppose $f$ is a function defined on a closed interval $[a, b]$, then for $f$ to be continuous, it needs to be continuous at every point in $[a, b]$, including the end points $a$ and $b$. Continuity of $f$ at $a$ means

$\lim\limits_{x \rightarrow a^{+}} f(x)=f(a)$

and continuity of $f$ at $b$ means

$\lim\limits_{x \rightarrow b^{-}} f(x)=f(b)$

Observe that $\lim\limits_{x \rightarrow a^{-}} f(x)$ and $\lim\limits_{x \rightarrow b^{+}} f(x)$ do not make sense. As a consequence of this definition, if $f$ is defined only at one point, it is continuous there.

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) f/g is continuous at x=c and g(c) ≠ 0

Proof:

From equation (a) :

f + g is said to be continuous at x = c
$\begin{aligned} & \lim\limits_{x \rightarrow c}(f+g)(x)=\lim\limits_{x \rightarrow c} f(x)+\lim\limits_{x \rightarrow c} g(x)\end{aligned}$

$= f(c) + g(c)$

$= (f + g) (c)$

Hence proved that it is continuous.

Differentiability:

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

$\lim\limits_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$

Derivative of $f(x)$ is given by $f^{\prime}(x)$

$f^{\prime}(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

RIGHT HAND DERIVATIVE:

$R f^{\prime}(a)=\lim\limits_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

LEFT HAND DERIVATIVE:

$L f^{\prime}(a)=\lim\limits_{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$

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

$\frac{d y}{d x}=\frac{d}{d x} f(x) \pm \frac{d}{d x} g(x)$

Product rule: Let $\mathrm{y}=\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$

$\frac{d y}{d x}=\frac{d f(x)}{d x} g(x)+\frac{d g(x)}{d x} f(x)$

Quotient rule: Let $\mathrm{y}=\mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x}) ; \mathrm{g}(\mathrm{x}) \neq 0$

$\frac{d y}{d x}=\frac{g(x) \frac{d f(x)}{d x}-f(x) \frac{d g(x)}{d x}}{g(x)^2}$

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

$\begin{aligned} & \frac{d y}{d x}=\frac{d y}{d U} \frac{d U}{d x} \\ & \lim _{x \rightarrow 0} \sin x=0 \\ & \lim _{x \rightarrow 0} \cos x=1 \\ & \lim _{x \rightarrow 0} \frac{\sin x}{x}=1=\lim _{x \rightarrow 0} \frac{x}{\sin x} \\ & \lim _{x \rightarrow 0} \frac{\tan x}{x}=1=\lim _{x \rightarrow 0} \frac{x}{\tan x} \\ & \lim _{x \rightarrow 0} \frac{\log (x+1)}{x}=1 \\ & \lim _{x \rightarrow 0} e^x=1 \\ & \lim _{x \rightarrow 0} \frac{e^x-1}{x}=1 \\ & \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log _e a \\ & \lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e \\ & \lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e=\lim _{x \rightarrow-\infty}\left(1+\frac{1}{x}\right)^x\end{aligned}$

Differentiation of Functions in Parametric Form: The 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,

$\frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}$

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

$\frac{d^2 y}{d x^2}=\frac{d}{d x} \cdot\left(\frac{d y}{d x}\right)$

Rolle’s Theorem:

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

$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

This is called the mean value theorem.

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

NCERT Class 12 Notes Chapter Wise:

Subject-Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

• NCERT Exemplar Class 12 Solutions
• NCERT Exemplar Class 12 Maths
• NCERT Exemplar Class 12 Physics
• NCERT Exemplar Class 12 Chemistry
• NCERT Exemplar Class 12 Biology

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

• NCERT Solutions for Class 12 Mathematics
• NCERT Solutions for Class 12 Chemistry
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Biology

NCERT Books and Syllabus

Students should always analyze the latest CBSE syllabus before making a study routine. The following links will help them check the syllabus. Also, here is access to more reference books.

• NCERT Book for Class 12
• NCERT Syllabus for Class 12

Important points to note:

• NCERT problems are very important in order to perform well in the exams. Students must try to solve all the NCERT problems, including miscellaneous exercises, and if needed, refer to the NCERT solutions for class 12 maths chapter 5, Continuity and Differentiability.
• Students are advised to go through the NCERT Class 12 Maths Chapter 3 Notes before solving the questions.
• To boost your exam preparation as well as for a quick revision, these NCERT notes are very useful.

Happy learning !!!