Continuity and Differentiability

What is Continuity and Differentiability?

Continuity and differentiability class 12 are fundamental concepts in calculus. They help analyze changes, optimize processes, and predict trends in fields like physics, engineering, and economics. Understanding these concepts is essential in class 12 maths continuity and differentiability for solving critical problems and class 12 continuity and differentiability solutions.

Continuity

A function $f$ is continuous at a point $x=c$ if

$ \lim_{x \to c} f(x) = f(c) $

In other words, if the left-hand limit (LHL), right-hand limit (RHL), and the function value at $x=c$ exist and are equal, then $f$ is continuous at $x=c$.

• If $f$ is not continuous, $c$ is called a point of discontinuity.

• Continuous functions have graphs without breaks, while discontinuous functions have jumps or holes.

Continuity can be categorized as:

1. Continuity at a point

2. Continuity over an interval

Continuity at a point

Let us see different types of conditions to see continuity at point $x = a$

We see that the graph of $f(x)$ has a hole at $x=a$, which means that $f(a)$ is undefined. At the very least, for $f(x)$ to be continuous at $x=a$, we need the following conditions:

(i) $f(a)$ is defined

Next, for the graph given below, although $f(a)$ is defined, the function has a gap at $x=a$. In this graph, the gap exists because lim $\lim\limits _{x → a }f(x)$ does not exist. We must add another condition for continuity at $x=a$, which is

(ii) $\lim\limits _{x \rightarrow a} f(x)$ exists

The above two conditions by themselves do not guarantee continuity at a point. The function in the figure given below satisfies both of our first two conditions but is still not continuous at $a$. We must add a third condition to our list:

(iii) $\lim\limits _{x \rightarrow a} f(x)=f(a)$

So, a function $f(x)$ is continuous at a point $x = a$ if and only if the following three conditions are satisfied:

i) $f(a)$ is defined

ii) $\lim\limits_{x \rightarrow a} f(x)$ exists

iii) $\lim\limits_{x \rightarrow a} f(x)=f(a)$ or

$\begin{aligned} & \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a) \\ & \text { i.e. } \text { L.H.L. }=\text { R.H.L. }=\text { value of the function at } x=a\end{aligned}$

A function is discontinuous at a point $a$ if it fails to be continuous at $a$.

Continuity over an Interval

In class 12 maths continuity and differentiability, a function can be continuous over an interval. The rules differ for open intervals $(a, b)$ and closed intervals $[a, b]$.

Continuity over an Open Interval $(a, b)$

A function $f(x)$ is continuous over an open interval $(a, b)$ if it is continuous at every point in the interval.

For any $c \in (a, b)$: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

Continuity over a Closed Interval $[a, b]$

A function $f(x)$ is continuous over a closed interval $[a, b]$ if:

• It is continuous at every point in $(a, b)$

• It is right-continuous at $x = a$, i.e., $f(a) = \lim_{x \to a^+} f(x) = \lim_{h \to 0^+} f(a+h)$

• It is left-continuous at $x = b$, i.e., $f(b) = \lim_{x \to b^-} f(x) = \lim_{h \to 0^+} f(b-h)$

Note: For $x = a$, LHL is not evaluated; for $x = b$, RHL is not evaluated.

Example: $f(x) = [x]$ on $[2,3]$

Step 1: Continuity in $(2,3)$
For any $c \in (2,3)$:

$f(c) = [c] = 2$

Left-hand limit (LHL):

$\lim_{x \to c^-} [x] = 2$

Right-hand limit (RHL):

$\lim_{x \to c^+} [x] = 2$

Hence, $f(x)$ is continuous for all $c \in (2,3)$.

Step 2: Right-continuity at $x=2$

$f(2) = 2, \quad \lim_{x \to 2^+} [x] = \lim_{h \to 0^+} [2+h] = 2$

So $f(x)$ is right-continuous at $x=2$.

Step 3: Left-continuity at $x=3$

$f(3) = 3, \quad \lim_{x \to 3^-} [x] = \lim_{h \to 0^+} [3-h] = 2$

Since $f(3) \ne \lim_{x \to 3^-} [x]$, $f(x)$ is not left-continuous at $x=3$.

$f(x) = [x]$ is not continuous on the closed interval $[2,3]$.

So the third condition is not satisfied and hence $f(x)$ is not continuous in $[2,3]$

Properties of Continuous Functions

In class 12 maths continuity and differentiability, continuous functions have several important properties that are useful in solving problems and understanding the behavior of functions.

1. Algebraic Operations on Continuous Functions

Suppose $f$ and $g$ are two real functions continuous at a point $x = a$ of their common domain $D$. Then:

1. The sum of two continuous functions is continuous at $x = a$: $(f+g) \text{ is continuous at } x = a$

2. The difference of two continuous functions is continuous at $x = a$: $(f-g) \text{ is continuous at } x = a$

3. The product of two continuous functions is continuous at $x = a$: $(f \cdot g) \text{ is continuous at } x = a$

4. The quotient of two continuous functions is continuous at $x = a$, provided $g(a) \ne 0$: $\left(\frac{f}{g}\right) \text{ is continuous at } x = a, \quad \text{if } g(a) \ne 0$

2. Continuity and Sign of a Function

If $f$ is continuous at $x = a$ and $f(a) \ne 0$, then there exists an open interval $(a-\delta, a+\delta)$ such that: $f(x) \text{ has the same sign as } f(a) \quad \forall x \in (a-\delta, a+\delta)$

This property ensures that continuous functions do not change sign abruptly near points where they are nonzero.

3. Boundedness of Continuous Functions on a Closed Interval

If a function $f$ is continuous on a closed interval $[a, b]$, then:

$f(x)$ is bounded on $[a, b]$, i.e., there exist real numbers $k$ and $K$ such that:

$k \le f(x) \le K, \quad \forall x \in [a, b]$

This property is fundamental in class 12 continuity and differentiability solutions and is often used in proving maximum and minimum values of continuous functions.

Differentiability

The instantaneous rate of change of a function with respect to its independent variable is called the derivative. Let $f(x)$ be a function of one variable and $\Delta x$ denote a small change in $x$.

The corresponding change in $f$ is: $\Delta f = f(x+\Delta x) - f(x)$

If the ratio $\frac{\Delta f}{\Delta x}$ approaches a finite limit as $\Delta x \to 0$, this limit is called the derivative of $f$ at $x$. The derivative can be denoted by:

$f'(x), \quad \frac{df}{dx}, \quad \frac{d}{dx}f(x), \quad \frac{df(x)}{dx}$

Definition of Derivative at a Point

The derivative of $f(x)$ at $x = a$ is defined as: $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

If this limit does not exist, the function $f(x)$ is not differentiable at $x = a$.

A function $f(x)$ is differentiable at $x = a$ if the right-hand derivative (RHD) and left-hand derivative (LHD) exist and are equal: $R f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}, \quad L f'(a) = \lim_{h \to 0} \frac{f(a-h)-f(a)}{-h}$

The condition for differentiability at $x = a$ is: $R f'(a) = L f'(a)$

Or equivalently: $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = \lim_{h \to 0} \frac{f(a-h)-f(a)}{-h}$

How to examine Differentiability?

1. Using Differentiation (for Continuous Functions)

For piecewise functions: $f(x) = \begin{cases} g_1(x), & x < a \\ g_2(x), & x \ge a \end{cases}$

• Step 1: Check if $f(x)$ is continuous at $x = a$. If not continuous, it is not differentiable.

• Step 2: If continuous, differentiate each branch: $f'(x) = \begin{cases} (g_1(x))', & x < a \\ (g_2(x))', & x > a \end{cases}$

• Step 3: Compare left-hand and right-hand derivatives at $x = a$: $\lim_{x \to a^-} (g_1(x))' \quad \text{and} \quad \lim_{x \to a^+} (g_2(x))'$

2. Differentiability Using Graphs

A function $f(x)$ is not differentiable at $x = a$ if:

1. $f(x)$ is discontinuous at $x = a$.

2. The graph of $f(x)$ has a sharp turn at $x = a$.

3. The graph of $f(x)$ has a vertical tangent at $x = a$.

Illustration 1

Check the differentiability of the following function.
1. $f(x)=\sin |x|$

Method 1

Using graphical transformation, we can draw its graph

Using the graph we can tell that at $x=0$, the graph has a sharp turn, so it is not differentiable at $x=0$.

Method 2

As $L H L=R H L=f(0)=0$, so the function is continuous at $x=0$
So we can use differentiation to check differentiability

$
\begin{aligned}
& \quad \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}
-\sin x, & x<0 \\
\sin x, & x \geq 0
\end{array}\right. \\
& \therefore \quad \quad \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}
-\cos x, & x<0 \\
\cos x, & x>0
\end{array}\right. \\
& \therefore \quad \text { LHD }=\mathrm{f}^{\prime}\left(0^{-}\right)=-1 \text { and } \mathrm{RHD}=\mathrm{f}^{\prime}\left(0^{+}\right)=1
\end{aligned}
$
As these are not equal, so, $f(x)=\sin |x|$ is not differentiable at $x=0$

Illustration 2

$
f(x)=\|\log \mid x\|, x \text { not equal to } 0
$
Plot the graph of | log $|\mathrm{x}|$ | using graphical transformation

We can see that graph has a sharp turn at +1 and -1 so the function is not differentiable at these points.

Properties of Differentiability

1. A function $f(x)$ is differentiable in an open in interval ( $a, b$ ) if it is differentiable at every point on the open interval $(a, b)$.

2. If $f(x)$ is differentiable at every point on the open interval (a,b). And, It is differentiable from the right at " $a$ " and the left at " $b$ ".(In other words, $\lim\limits_{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a}$ and $\lim\limits_{x \rightarrow b^{-}} \frac{f(x)-f(b)}{x-b}$ both exists), then $f(x)$ is said to be differentiable in $[a, b]$

3. If a function $f(x)$ is differentiable at every point in an interval, then it must be continuous in that interval. But the converse may or may not be true.

Continuity and Differentiability Formulas

Continuity and differentiability class 12 formulas include algebraic properties of continuity and discontinuity, algebraic properties of differentiability, differentiation of implicit functions, and differentiation of functions in parametric form.

Algebra of Continuity and Discontinuity

The algebraic properties of continuity and discontinuity are,

1. If $f(x)$ and $g(x)$ are continuous functions in the given interval, then the following functions are continuous at $\mathrm{x}=\mathrm{a}$.
   (i) $f(x) \pm g(x)$
   (ii) $\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})$
   (iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a}) \neq 0$

2. If $f(x)$ is continuous and $g(x)$ is discontinuous, then $f(x) \pm g(x)$ is a discontinuous function.

   Let $f(x)=x$, which is continuous at $x=0$ and $g(x)=[x]$ (greatest integer function) which is discontinuous at $x=0$, are to function.

   Now, $f(x)-g(x)=x-[x]=\{x\}$ (fractional part of $x$ )
   discontinuous at $\mathrm{x}=0$

3. If $f(x)$ is continuous and $g(x)$ is discontinuous at $x=a$ then the product of the functions, $h(x)=f(x) g(x)$ is may or may not be continuous at $x=a$.

   For example,
   Consider the functions, $f(x)=x^3$. And $g(x)=\operatorname{sgn}(x)$.
   $f(x)$ is continuous at $x=0$ and $g(x)$ is discontinuous at $x=0$
   Now,

   $
   h(x)=f(x) \cdot g(x)=\left\{\begin{array}{cl}
   x^3, & x>0 \\
   0, & x=0 \\
   -x^3, & x<0
   \end{array}\right.
   $

   $h(x)$ is continuous at $x=0$
   Take another example, consider $f(x)=x$ and $g(x)=1 /|x|$
   $f(x)$ is continuous at $x=0$ and $g(x)$ is discontinuous at $x=0$
   Now,

   $
   h(x)=f(x) \cdot g(x)=x \cdot \frac{1}{|x|}=\operatorname{sgn}(\mathrm{x})
   $
   And we know that signum function is discontinuous at $x=0$.

4. If $f(x)$ and $g(x)$, both are discontinuous at $x=a$ then the the function obtained by algebraic operation of $f(x)$ and $g(x)$ may or may not be continuous at $x=a$.

Rules of Differentiation

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following rules of differentiation holds.

Sum Rule

The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$.

$
\frac{d}{d x}(f(x)+g(x))=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))
$
In general,

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

Difference Rule

The derivative of the difference of a function $f$ and $a$ function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$.

$
\begin{aligned}
& \frac{d}{d x}(f(x)-g(x))=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x)) \\
& \frac{d}{d x}(f(x)-g(x)-h(x)-\ldots \ldots)=\frac{d}{d x}(f(x))-\frac{d}{d x}(g(x))-\frac{d}{d x}(h(x))-\ldots \ldots
\end{aligned}
$

Constant Multiple Rule

The derivative of a constant $k$ multiplied by a function $f$ is the same as the constant multiplied by the derivative of $f$

$
\frac{d}{d x}(k f(x))=k \frac{d}{d x}(f(x))
$

Product rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then,

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

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Extending the Product Rule

If three functions are involved, i.e let $k(x)=f(x) \cdot g(x) \cdot h(x)$
Let us have a function $\mathrm{k}(\mathrm{x})$ as the product of the function $\mathrm{f}(\mathrm{x}), \mathrm{g}(\mathrm{x})$ and $\mathrm{h}(\mathrm{x})$. That is, $k(x)=(f(x) \cdot g(x)) \cdot h(x)$. Thus,

$
k^{\prime}(x)=\frac{d}{d x}(f(x) g(x)) \cdot h(x)+\frac{d}{d x}(h(x)) \cdot(f(x) g(x))
$

[By applying the product rule to the product of $f(x) g(x)$ and $h(x)$.]

$
\begin{aligned}
& =\left(f^{\prime}(x) g(x)+g^{\prime}(x) f(x)\right) h(x)+h^{\prime}(x) f(x) g(x) \\
& =f^{\prime}(x) g(x) h(x)+f(x) g^{\prime}(x) h(x)+f(x) g(x) h^{\prime}(x)
\end{aligned}
$

Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$
\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot \frac{d}{d x}(f(x))-f(x) \cdot \frac{d}{d x}(g(x))}{(g(x))^2}
$
OR
if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{(g(x))^2}$
As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives.

Chain Rule

If $u(x)$ and $v(x)$ are differentiable funcitons, then $u o v(x)$ or $u[v(x)]_{\text {is also differentiable. }}$
If $y=u o v(x)=u[v(x)]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} u\{v(x)\}}{\mathrm{d}\{v(x)\}} \times \frac{\mathrm{d}}{\mathrm{d} x} v(x)
$

is known as the chain rule. Or,

$
\text { If } y=f(u) \text { and } u=g(x) \text {, then } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \cdot \frac{\mathrm{d} u}{\mathrm{~d} x}
$
The chain rule can be extended as follows
If $y=[\operatorname{uovow}(x)]=u[v\{w(x)\}]$, then

$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}[u[v\{w(x)\}]}{\mathrm{d} v\{w(x)\}} \times \frac{\mathrm{d}[v\{w(x)\}]}{\mathrm{d} w(x)} \times \frac{\mathrm{d}[w(x)]}{\mathrm{d} x}
$

Algebra of Differentiability

The algebraic properties of differentiablity is

1. If $f(x)$ and $g(x)$ are both differentiable functions at $x=a$, then the following functions are also differentiable at $\mathrm{x}=\mathrm{a}$.
(i) $\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})$
(ii) $\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})$
(iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a}) \neq 0$

2. If $f(x)$ is differentiable at $x=a$ and $g(x)$ is not differentiable at $x=a$, then $f(x) \pm g(x)$ will not be differentiable at $x=a$.

For example, $\cos (x)+|x|$ is not differentiable at $x=0$, as $\cos (x)$ is differentiable at $x=0$, but $|x|$ is not differentiable at $x=0$.

In other cases, $f(x) \cdot g(x)$ and $f(x) / g(x)$ may or may not be differentiable at $x$ = a, and hence should be checked using LHD-RHD, continuity or graph.

For example, if $f(x)=0$ (differentiable at $x=0$ ) and $g(x)=|x|$ (nondifferentiable at $x=0$ ). Their product is $f(x) \cdot g(x)=0$ which is differentiable. But if $f(x)=2, g(x)=|x|$, then $f(x) \cdot g(x)=2|x|$ is non-differentiable at $x=0$.

3. If $f(x)$ and $g(x)$ both are nondifferentiable functions at $x=a$, then the function obtained by the algebraic operation of $f(x)$ and $g(x)$ may or may not be differentiable at $x=a$. Hence they should be checked.

For example, Let $f(x)=|x|$, not differentiable at $x=0$, and $g(x)=-|x|$ which is also not differentiable at $x=0$. Their sum $=0$ is differentiable and the difference $=2|x|$ is not differentiable. So there is no definite rule.

4. Differentiation of a continuous function may or may not be continuous.

Differentiation of Implicit Functions

An implicit function is a function that includes both dependent and independent variables such as $F(x, y)=0$. For example, the equation of a circle $x^2+y^2=r^2$ is an implicit function because $y$ is not explicitly expressed as a function of $x$. Implicit differentiation means differentiation on both sides of the equation concerning the independent variable with the help of the Chain rule.

To find $\frac{d y}{d x}$ in such a case, we differentiate both sides of the given relation concerning $x$ keeping in mind that the derivative of $\Phi(\mathrm{y})$ concerning $x$ is $\frac{d \phi}{d y} \times \frac{d y}{d x}$

For example

$\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}, \frac{d}{d x}\left(y^2\right)=2 y \frac{d y}{d x}$

It should be noted that $\frac{d}{d y}(\sin y)=\cos y$ but $\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}$

Differentiation of Functions in Parametric Form

Parametric differentiation is the process of finding the derivative of the equation in which the dependent variable $y$ and independent variable $x$ are equated to another variable $t$. To find the derivative of $y$ for $x$ we use the chain rule.

Sometimes, $x$ and $y$ are given as functions of a single variable, i.e., $x=g(t)$ and $y=f(t)$ are two functions and $t$ is a variable. In such cases, $x$ and $y$ are called parametric functions or parametric equations and $t$ is called the parameter.

To find $\frac{d y}{d x}$ in such cases, first find the relationship between $x$ and $y$ by eliminating the parameter $t$ and then differentiate concerning $t$.

We have $x=f(t)$ and $y=f(t)$, in this case, we will differentiate both functions separately. We will first differentiate $x$ and $y$ concerning '$t$ separately. On differentiating $x$ for ' $t$ ' we get $\frac{d x}{d t}$ and on differentiating $y$ by ' $t$ ' we get $\frac{d y}{d t}$.

But sometimes it is not possible to eliminate $t$, then in that case use

$
\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{f^{\prime}(t)}{g^{\prime}(t)}
$

For example

If $x=a(1-\cos \theta)$ and $y=a(\theta-\sin \theta)$, then $d y / d x$ is
Solution.

$
\begin{aligned}
& \text { Given } \mathrm{x}=\mathrm{a}(1-\cos \theta) \text { and } \mathrm{y}=\mathrm{a}(\theta+\sin \theta) \\
& \Rightarrow \frac{d x}{d \theta}=a(\sin \theta) \text { and } \frac{d y}{d \theta}=a(1+\cos \theta) \\
& \Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{1+\cos \theta}{\sin \theta}
\end{aligned}
$
Using the circle example, we differentiate $(-\cot (t))$ concerning

$
\begin{aligned}
& (t):\left[\frac{d}{d t}(-\cot (t))=\csc ^2(t)\right] \text { Since }\left(\frac{d x}{d t}=-\sin (t)\right), \text { wefind }\left(\frac{d^2 y}{d x^2}\right):\left[\frac{d^2 y}{d x^2}\right. \\
& \left.=\frac{\csc ^2(t)}{-\sin (t)}=-\csc (t)\right]
\end{aligned}
$

Difference between Continuity and Differentiability

List of Topics related to continuity and differentiability according to NCERT/JEE MAIN

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Important Books and Resources for Class 12 Continuity and Differentiability

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NCERT Resources

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• NCERT Maths Notes for Class 12th Chapter 5 - Continuity and Differentiability

• NCERT Maths Solutions for Class 12th Chapter 5 - Continuity and Differentiability

• NCERT Maths Exemplar Solutions for Class 12th Chapter 5 - Continuity and Differentiability

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Practice Questions based on Continuity and Differentiability

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