NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

Suppose you are riding a bike on a smooth and long road without any interruptions. That would be a pretty nice and pleasant journey, right? Now, imagine riding a bike on a rough road with sudden breaks, sharp turns, and multiple obstacles. How would that feel? In mathematics, functions are quite similar to these types of roads, where some are smooth and uninterrupted, while some have breaks or many obstacles. This is where the concept of Continuity comes in. From NCERT Class 12 Maths, the chapter Continuity and Differentiability contains the concepts of continuity, differentiability, derivatives of functions, logarithmic differentiation, etc. These concepts will help the students grasp more advanced calculus topics easily and will also enhance their problem-solving ability in real-world applications.

This article on NCERT Class 12 Maths Notes Chapter 5 Continuity and Differentiability offers well-structured NCERT notes to help the students grasp the concepts of Continuity and Differentiability easily. Students who want to revise the key topics of Continuity and Differentiability quickly will find this article very useful. It will also boost the exam preparation of the students by many folds. These NCERT Notes are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 12 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

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NCERT Notes for Class 12 Chapter 5 Continuity and Differentiability:

NCERT Class 12 Chapter 5 Continuity and Differentiability builds on the concepts from the previous class and introduces key ideas like continuity, differentiability, and relations between them. It also includes differentiation of inverse trigonometric, exponential, and logarithmic functions, along with important theorems and their applications.

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:

If f and g are 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, and 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.

Continuity and Differentiability: Previous Year Question and Answer

Given below are some previous year question answers of various examinations from the NCERT class 12 chapter 5, Continuity and Differentiability:

Question 1: If $x=a\left(\cos \theta+\log \tan \frac{\theta}{2}\right)$ and $y=\sin \theta$, then find $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{4}$.

Solution:
Given:
$
x = a\left(\cos \theta + \log \tan \frac{\theta}{2} \right), \quad y = \sin \theta
$

Differentiate both with respect to $ \theta $:

$
\frac{dx}{d\theta} = a\left(-\sin \theta + \frac{1}{\sin \theta} \right) = a \cdot \frac{-\sin^2 \theta + 1}{\sin \theta} = a \cdot \frac{\cos^2 \theta}{\sin \theta}
$

$
\frac{dy}{d\theta} = \cos \theta
$

$
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos \theta}{a \cdot \frac{\cos^2 \theta}{\sin \theta}} = \frac{\sin \theta}{a \cos \theta}
$

Differentiate again to find $ \frac{d^2 y}{dx^2} $:

$
\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{\sin \theta}{a \cos \theta} \right ) = \frac{d}{d\theta} \left( \frac{\sin \theta}{a \cos \theta} \right ) \cdot \frac{1}{\frac{dx}{d\theta}}
$

$
\frac{d}{d\theta} \left( \frac{\sin \theta}{a \cos \theta} \right ) = \frac{a \cos^2 \theta + a \sin^2 \theta}{a^2 \cos^2 \theta} = \frac{1}{a \cos^2 \theta}
$

So,
$
\frac{d^2 y}{dx^2} = \frac{1}{a \cos^2 \theta} \cdot \frac{1}{\frac{dx}{d\theta}} = \frac{1}{a \cos^2 \theta} \cdot \frac{1}{a \cdot \frac{\cos^2 \theta}{\sin \theta}} = \frac{\sin \theta}{a^2 \cos^4 \theta}
$

At $ \theta = \frac{\pi}{4} $, we have $ \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} $, so:

$
\frac{d^2 y}{dx^2} = \frac{1/\sqrt{2}}{a^2 (1/\sqrt{2})^4} = \frac{1/\sqrt{2}}{a^2 \cdot \frac{1}{4}} = \frac{4}{a^2 \sqrt{2}} = \frac{2\sqrt{2}}{a^2}
$

Hence, the correct answer is $\frac{2\sqrt{2}}{a^2}$.

Question 2: If $\tan ^{-1}\left(x^2+y^2\right)=a^2$, then find $\frac{d y}{d x}$.

Solution:
Given $\tan ^{-1}\left(x^2+y^2\right)=a^2$

where, $a$ is a constant

Taking $\tan$ of both sides:

$\tan(\tan ^{-1}\left(x^2+y^2\right))=\tan a^2$

$x^2+y^2 = \tan(a^2) = C$ (constant)-------- Using identity $\tan( \tan^{-1} x)=x$
Differentiate with respect to $x$
$\Rightarrow2x + 2y \frac{dy}{dx} = 0$
$\Rightarrow2y \frac{dy}{dx} = -2x$
$\Rightarrow\frac{dy}{dx} = -\frac{x}{y}$
Hence, the correct answer is $-\frac{x}{y}$.

Question 3: Differentiate $2^{\cos ^2 x}$ w.r.t $\cos ^2 x$.

Solution:
Given:
The function $2^{\cos ^2 x}$ to be differentiated with respect to $\cos^2x$
Let $y = 2^{\cos^2 x}$ and $u = \cos^2 x$.
$\frac{dy}{du} = \frac{d}{du}(2^u) $
$= 2^u \ln 2$--------Using the identity $\frac{da^x}{dx}=a^xlna$
Substitute back $u = \cos^2 x$:
$\frac{d}{d(\cos^2 x)}(2^{\cos^2 x}) = 2^{\cos^2 x} \ln 2$
Hence, the correct answer is $2^{\cos^2x} \ln 2$.

NCERT Class 12 Notes Chapter Wise:

All the links of chapter-wise notes for NCERT class 12 maths are given below:

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