Home > CBSE > CBSE Class 12 Maths Revision Notes Chapter 5 Continuity and Differentiability

CBSE Class 12 Maths Revision Notes Chapter 5 Continuity and Differentiability

Continuity checks whether a function has no break at a point, while differentiability checks whether a derivative exists at that point. For CBSE Class 12 Maths, this chapter covers continuity, differentiability, derivative rules, logarithmic differentiation, second-order derivatives and mean value theorems.

Continuity and Differentiability extend the differentiation concepts studied in Class 11. This chapter explains how functions behave near a point, when they are continuous, when they are differentiable and how derivatives can be found using different rules.

Use these CBSE Class 12 Maths Revision Notes Chapter 5 for the 2026–27 academic year to revise definitions, conditions, formulas, derivative rules and theorems. These Class 12 Mathematics Chapter 5 notes are useful for solving questions based on piecewise functions, inverse trigonometric functions, logarithmic differentiation and Rolle’s Theorem.

Key Takeaways

Continuity: A function is continuous at x = c when lim f(x) as x approaches c = f(c).
Differentiability: A function is differentiable at x = c when LHD = RHD.
Main relation: Differentiability implies continuity, but continuity does not always imply differentiability.
Theorems: Rolle’s Theorem and Lagrange’s Mean Value Theorem require continuity on [a, b] and differentiability on (a, b).

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Access Class 12 Maths Chapter 5 Continuity and Differentiability Notes in 30 Minutes

Chapter 5 is formula-heavy. Start with continuity and differentiability conditions, then revise derivative rules, special methods and theorems.

Revision Area	What to Revise
Continuity	LHL = RHL = f(c)
Discontinuity	Removable and non-removable cases
Differentiability	LHD = RHD
Relation	Differentiability implies continuity
Derivative Rules	Sum, product, quotient and chain rules
Special Derivatives	Trigonometric, inverse trigonometric, exponential and logarithmic
Logarithmic Differentiation	Used for variable powers and products
Parametric Differentiation	dy/dx = (dy/dt)/(dx/dt)
Second-Order Derivative	Derivative of first derivative
Theorems	Rolle’s Theorem and LMVT

Continuity and Differentiability Class 12 Notes: What This Chapter Covers

Continuity and differentiability are two connected ideas in calculus. Continuity studies whether a function is unbroken at a point. Differentiability studies whether the function has a derivative at that point.

Continuity

A function is continuous at a point if its graph does not break at that point.

Mathematically, a function f is continuous at x = c if:

lim f(x) as x approaches c = f(c)

This means:

LHL = RHL = f(c)

Differentiability

A function is differentiable at a point if its derivative exists at that point.

For differentiability at x = c:

LHD at c = RHD at c

Here, LHD means left-hand derivative and RHD means right-hand derivative.

Relation Between Continuity and Differentiability

Continuity and differentiability are related, but they are not the same.

Situation	Result
Function is differentiable at x = c	Function is continuous at x = c
Function is continuous at x = c	Function may or may not be differentiable
Function is not continuous at x = c	Function is not differentiable at x = c

Memory point:
Differentiability is a stronger condition than continuity.

Continuity in Class 12 Mathematics Chapter 5 Notes

Continuity is checked using the value of the function and its limit near a point.

Continuity at a Point

Let f be a real function and c be a point in its domain.

f is continuous at x = c if:

lim f(x) as x approaches c = f(c)

This is equivalent to:

LHL = RHL = f(c)

Where:

Term	Meaning
LHL	Left-hand limit
RHL	Right-hand limit
f(c)	Value of function at x = c

How to Check Continuity at x = c

Use these steps:

Find f(c).
Find LHL at x = c.
Find RHL at x = c.
Compare all three values.

Result	Conclusion
LHL = RHL = f(c)	Function is continuous at c
LHL = RHL but not equal to f(c)	Function is discontinuous at c
LHL ≠ RHL	Function is discontinuous at c
One of the limits does not exist	Function is discontinuous at c

Continuity in an Interval

A function f is continuous in an interval [a, b] if it is continuous at every point of the interval.

At the end points:

At x = a, use right-hand continuity.
At x = b, use left-hand continuity.

For f to be continuous on [a, b]:

Point	Condition
At a	lim f(x) as x approaches a+ = f(a)
Between a and b	lim f(x) as x approaches c = f(c)
At b	lim f(x) as x approaches b- = f(b)

Algebra of Continuous Functions

If f and g are continuous at x = c, then their sum, difference, product and quotient are also continuous under suitable conditions.

Function	Continuity Condition
f + g	Continuous at c
f - g	Continuous at c
f × g	Continuous at c
f/g	Continuous at c if g(c) ≠ 0
kf	Continuous at c, where k is a constant
f o g	Continuous if g is continuous at c and f is continuous at g(c)

Common Continuous Functions

Function	Interval of Continuity
Constant function	R
Polynomial function	R
Rational function p(x)/q(x)	Where q(x) ≠ 0
sin x	R
cos x	R
tan x	R except odd multiples of π/2
cot x	R except integral multiples of π
sec x	R except odd multiples of π/2
cosec x	R except integral multiples of π
ex	R
log x	(0, ∞)

Types of Discontinuity

A function is discontinuous at a point when continuity fails at that point. Discontinuity can be removable or non-removable.

Removable Discontinuity

A function has removable discontinuity at x = c if the limit exists but the function value is missing or different from the limit.

There are two common cases.

Type	Meaning
Missing point discontinuity	lim f(x) exists, but f(c) is not defined
Isolated point discontinuity	lim f(x) exists and f(c) exists, but both are unequal

In removable discontinuity, the function can be made continuous by redefining f(c).

Non-Removable Discontinuity

A function has non-removable discontinuity when the limit does not exist at the point.

Type	Meaning
Finite discontinuity	LHL and RHL are finite but unequal
Infinite discontinuity	LHL or RHL tends to infinity
Oscillatory discontinuity	Function values oscillate and limit does not exist

Quick Discontinuity Check Table

Observation at x = c	Type
LHL = RHL, but f(c) missing	Removable
LHL = RHL, but f(c) different	Removable
LHL and RHL finite but unequal	Non-removable
Limit tends to infinity	Non-removable
Limit oscillates	Non-removable

Differentiability and Derivatives

Differentiability checks whether the derivative of a function exists at a point.

Differentiability at a Point

A function f is differentiable at x = c if the following limit exists:

f′(c) = lim h→0 [f(c + h) - f(c)] / h

This limit gives the derivative of f at x = c.

Right-Hand Derivative

Right-hand derivative at x = c is:

RHD = lim h→0+ [f(c + h) - f(c)] / h

Left-Hand Derivative

Left-hand derivative at x = c is:

LHD = lim h→0- [f(c + h) - f(c)] / h

It may also be written as:

LHD = lim h→0+ [f(c) - f(c - h)] / h

Differentiability Condition

f is differentiable at x = c if:

LHD = RHD

If LHD and RHD are not equal, the function is not differentiable at that point.

Differentiability Implies Continuity

If a function is differentiable at x = c, then it is continuous at x = c.

But if a function is continuous at x = c, it may not be differentiable there.

Example: |x|

The function f(x) = |x| is continuous at x = 0.

But it is not differentiable at x = 0 because the graph has a sharp corner there.

Function Behaviour	Continuity	Differentiability
Smooth graph	Usually continuous	Usually differentiable
Break or jump	Not continuous	Not differentiable
Sharp corner	Continuous possible	Not differentiable
Vertical tangent	Continuous possible	Differentiability may fail

Rules of Differentiation

Derivative rules help find derivatives of complex functions quickly.

Sum and Difference Rule

If y = f(x) ± g(x), then:

dy/dx = f′(x) ± g′(x)

Product Rule

If y = u × v, then:

dy/dx = u(dv/dx) + v(du/dx)

d(uv)/dx = u v′ + v u′

Quotient Rule

If y = u/v, where v ≠ 0, then:

dy/dx = [v(du/dx) - u(dv/dx)] / v²

d(u/v)/dx = (v u′ - u v′) / v²

Chain Rule

If y = f(u) and u = g(x), then:

dy/dx = dy/du × du/dx

For three linked functions:

dy/dx = dy/du × du/dv × dv/dx

The chain rule is used for composite functions.

Derivatives of Trigonometric Functions

Function	Derivative
d/dx(sin x)	cos x
d/dx(cos x)	-sin x
d/dx(tan x)	sec²x
d/dx(cot x)	-cosec²x
d/dx(sec x)	sec x tan x
d/dx(cosec x)	-cosec x cot x

Derivatives of Inverse Trigonometric Functions

Function	Derivative
d/dx(sin⁻¹x)	1/√(1 - x²)
d/dx(cos⁻¹x)	-1/√(1 - x²)
d/dx(tan⁻¹x)	1/(1 + x²)
d/dx(cot⁻¹x)	-1/(1 + x²)
d/dx(sec⁻¹x)	1/(
d/dx(cosec⁻¹x)	-1/(

Use these formulas only where the functions are defined.

Derivatives of Exponential and Logarithmic Functions

Function	Derivative
d/dx(ex)	ex
d/dx(ax)	ax log a, where a > 0
d/dx(log x)	1/x, where x > 0
d/dx(xn)	nxn-1
d/dx(constant)	0

Here, log x means natural logarithm unless stated otherwise.

Logarithmic Differentiation

Logarithmic differentiation is useful when the function contains products, quotients, powers or variable exponents.

When to Use Logarithmic Differentiation

Use logarithmic differentiation when:

y = [f(x)]g(x)
y is a product of many factors
y is a quotient with powers
direct differentiation looks lengthy

Formula for y = [f(x)]g(x)

Let:

y = [f(x)]g(x)

Taking log on both sides:

log y = g(x) log f(x)

Differentiate both sides:

(1/y) dy/dx = g′(x) log f(x) + g(x) f′(x)/f(x)

So:

dy/dx = [f(x)]g(x) [g′(x) log f(x) + g(x) f′(x)/f(x)]

Parametric Differentiation

Parametric differentiation is used when x and y are both expressed in terms of a third variable.

Let:

x = f(t)
y = g(t)

Then:

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0

Second Derivative in Parametric Form

If dy/dx is known as a function of t, then:

d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

This is used in higher-order derivative questions.

Second-Order Derivatives in Class 12 Maths Chapter 5

The first derivative gives the rate of change of a function. The second derivative is the derivative of the first derivative.

If:

dy/dx = y′

Then:

d²y/dx² = d/dx(dy/dx)

y″ = d²y/dx²

Quick Example

If y = x³, then:

dy/dx = 3x²

d²y/dx² = 6x

So, the second-order derivative of x³ is 6x.

Derivative of Infinite Series

Some expressions define y using an infinite repeated pattern.

If:

y = f(x) + f(x) + f(x) + ...

Then the expression can often be rewritten in terms of y.

After rewriting, differentiate both sides using normal rules.

Use this only after reducing the infinite expression into a solvable equation.

Rolle’s Theorem

Rolle’s Theorem gives a condition under which the derivative becomes zero at some point inside an interval.

Statement of Rolle’s Theorem

Let f be a real-valued function on [a, b].

If:

f is continuous on [a, b]
f is differentiable on (a, b)
f(a) = f(b)

Then there exists at least one c in (a, b) such that:

f′(c) = 0

Meaning of Rolle’s Theorem

If a curve starts and ends at the same height and is smooth between the two points, then at least one tangent inside the interval is parallel to the x-axis.

Lagrange’s Mean Value Theorem

Lagrange’s Mean Value Theorem is also called LMVT. It is a wider result than Rolle’s Theorem.

Statement of Lagrange’s Mean Value Theorem

Let f be a real-valued function on [a, b].

If:

f is continuous on [a, b]
f is differentiable on (a, b)

Then there exists at least one c in (a, b) such that:

f′(c) = [f(b) - f(a)] / (b - a)

Meaning of LMVT

LMVT says that at some point inside the interval, the instantaneous rate of change equals the average rate of change over the interval.

Difference Between Rolle’s Theorem and LMVT

Point	Rolle’s Theorem	Lagrange’s Mean Value Theorem
Continuity condition	Continuous on [a, b]	Continuous on [a, b]
Differentiability condition	Differentiable on (a, b)	Differentiable on (a, b)
Extra condition	f(a) = f(b)	No need for f(a) = f(b)
Conclusion	f′(c) = 0	f′(c) = [f(b) - f(a)] / (b - a)
Relation	Special case of LMVT	General theorem

Useful Substitutions for Differentiation

Some expressions become easier after trigonometric substitution.

Expression	Useful Substitution
√(a² + x²)	x = a tan θ or x = a cot θ
√(a² - x²)	x = a sin θ or x = a cos θ
√(x² - a²)	x = a sec θ or x = a cosec θ
√((a - x)/(a + x))	x = a cos θ
√((a² - x²)/(a² + x²))	x² = a² cos θ

Quick Revision Tables for Continuity and Differentiability

Continuity Check Table

Step	What to Find
1	f(c)
2	LHL at x = c
3	RHL at x = c
4	Check LHL = RHL = f(c)
5	If yes, f is continuous at c

Differentiability Check Table

Step	What to Find
1	Check continuity first
2	Find LHD
3	Find RHD
4	Check LHD = RHD
5	If yes, f is differentiable at c

Derivative Rules Table

Rule	Formula
Sum Rule	d(f + g)/dx = f′ + g′
Difference Rule	d(f - g)/dx = f′ - g′
Product Rule	d(uv)/dx = u v′ + v u′
Quotient Rule	d(u/v)/dx = (v u′ - u v′)/v²
Chain Rule	dy/dx = dy/du × du/dx
Parametric Rule	dy/dx = (dy/dt)/(dx/dt)

Theorem Conditions Table

Theorem	Conditions	Result
Rolle’s Theorem	Continuous on [a, b], differentiable on (a, b), f(a) = f(b)	f′(c) = 0
LMVT	Continuous on [a, b], differentiable on (a, b)	f′(c) = [f(b) - f(a)]/(b - a)

Important Terms in Continuity and Differentiability

Term	Meaning
Continuity	No break in function value at a point
LHL	Limit from the left side
RHL	Limit from the right side
Differentiability	Existence of derivative at a point
LHD	Left-hand derivative
RHD	Right-hand derivative
Removable Discontinuity	Discontinuity that can be fixed by redefining f(c)
Non-Removable Discontinuity	Discontinuity where the limit does not exist
Chain Rule	Rule for differentiating composite functions
Second-Order Derivative	Derivative of the first derivative
Rolle’s Theorem	Theorem where f′(c) = 0 under specific conditions
LMVT	Theorem relating average and instantaneous rate of change

Common Mistakes in Class 12 Mathematics Chapter 5 Notes

Mistake	Correct Point
Assuming continuity means differentiability	Continuity does not always imply differentiability
Checking only f(c)	Also check LHL and RHL
Ignoring LHD and RHD	Differentiability needs both to be equal
Using quotient rule with wrong signs	Formula is (v u′ - u v′)/v²
Forgetting conditions in theorems	Continuity and differentiability conditions are compulsory
Applying log differentiation without positive base	log f(x) requires f(x) > 0
Ignoring dx/dt in parametric form	dy/dx = (dy/dt)/(dx/dt)

Useful Links for Class 12 Maths

Section	Useful Links
Syllabus	CBSE Class 12 Maths Syllabus
Revision Notes	CBSE Class 12 Maths Revision Notes
Maths Notes	CBSE Class 12 Maths Revision Notes Chapter 1
NCERT Solutions	NCERT Solutions Class 12 Maths
Sample Papers	CBSE Sample Papers for Class 12 Maths
Important Questions	Important Questions Class 12 Maths
NCERT Books	NCERT Books for Class 12 Maths
Previous Year Papers	CBSE Maths Question Paper Class 12

Q.1

$Differentiate : y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}}$

Ans

$\begin{array}{l} y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}} \\ Squaring both sides \\ y^{2} = \log + y \\ Differentiating w . r . t . x \\ \frac{d}{dx} (y^{2}) = \frac{d}{dx} (logx + y) \\ \Rightarrow 2 y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \\ \Rightarrow 2 y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1}{x (2 y - 1)} \end{array}$

Q.2

$If x = a (θ - sinθ), y = a (1 + cosθ), find \frac{d^{2} y}{{dx}^{2}} at θ = \frac{π}{2} .$

Ans

$\begin{array}{l} x = a (θ - sinθ) \Rightarrow \frac{dx}{dθ} = a (1 - cosθ) \\ y = a (1 + cosθ) \Rightarrow \frac{dx}{dθ} = - asinθ \\ \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}} \frac{- asinθ}{a (1 - cosθ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = - \cot \frac{θ}{2} \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} . \frac{dθ}{dx} \\ = {cosec}^{2} \frac{θ}{2} . \frac{dθ}{dx} = \frac{1}{2} {cosec}^{2} \frac{θ}{2} . \frac{1}{a (1 - cosθ)} \\ \Rightarrow {\frac{d^{2} y}{{dx}^{2}}|}_{θ = \frac{π}{2}} = \frac{1}{2} {cosec}^{2} \frac{π}{4} . \frac{1}{a (1 - \cos \frac{π}{2})} \\ = \frac{1}{2} {(\sqrt{2})}^{2} \frac{1}{a (1 - 0)} = \frac{2}{2 a} = \frac{1}{a} \end{array}$

Q.3

$Differentiate \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}] w . r . t \tan^{- 1} x .$

Ans

$\begin{array}{l} Let y = \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}], u = \tan^{- 1} x \\ y = \tan^{- 1} [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}} \times [\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}]] \\ = \tan^{- 1} [\frac{{(\sqrt{1 + sinx} + \sqrt{1 - sinx})}^{2}}{\sqrt{1 + sinx} - \sqrt{1 + sinx}}] \\ = \tan^{- 1} [\frac{1 + sinx + 1 - sinx + 2 \sqrt{1 - \sin^{2} x}}{2 sinx}] \\ = \tan^{- 1} [\frac{2 + 2 cosx}{2 sinx}] \\ = \tan^{- 1} [\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] \\ = \tan^{- 1} [\cot \frac{x}{2}] \\ = \tan^{- 1} [\tan (\frac{π}{2} - \frac{x}{2})] \\ = \frac{π}{2} - \frac{x}{2} \\ \frac{dy}{dx} = - \frac{1}{2} . \dots.. (1) \\ Now, u = \tan^{- 1} x \\ \Rightarrow \frac{du}{dx} = \frac{1}{1 + x^{2}} . \dots.. (2) \\ By (1) and (2) \\ \frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}} = \frac{- \frac{1}{2}}{\frac{1}{1 + x^{2}}} = - \frac{1}{2} (1 + x^{2}) \end{array}$

Q.4

${If y = e}^{{acos}^{- 1} x}, - 1 \leq x \leq 1, show that (1 - x^{2}) \frac{d^{2 y}}{{dx}^{2}} - x \frac{dy}{dx} - a^{2} y = 0$

Ans

$\begin{array}{l} Here, y = e^{a \cos^{- 1}} \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} (e^{{acos}^{- 1} x}) \\ = {e^{{acos}^{- 1} x}}^{} \frac{d}{dx} ({acos}^{- 1} x) \\ = \frac{- ay}{\sqrt{1 - x^{2}}} \\ \sqrt{1 - x^{2}} \frac{dy}{dx} = - ay \\ again differentiating w . r . tx \\ \Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} \frac{d}{dx} (\sqrt{1 - x^{2}}) = - a \frac{dy}{dx} \\ \Rightarrow \sqrt{1 - x^{2}} \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} . [\frac{1}{2 \sqrt{1 - x^{2}}} . (- 2 x)] = - a \frac{dy}{dx} \\ \Rightarrow (\sqrt{1 - x^{2}}) \frac{d^{2} y}{{dx}^{2}} - . (\frac{x}{\sqrt{1 - x^{2}}}) \frac{dy}{dx} = - a (\frac{- ay}{\sqrt{1 - x^{2}}}) \\ \Rightarrow (\sqrt{1 - x^{2}}) \frac{d^{2} y}{{dx}^{2}} - x \frac{dy}{dx} - a^{2} y = 0 \end{array}$

Q.5

$If \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y) then prove that \frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}} .}$

Ans

$\begin{array}{l} Here, \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y) \\ Let x = sinθ \Rightarrow θ = \sin^{- 1} x \\ y = sinÏ• \Rightarrow Ï• = \sin^{- 1} y \\ \Rightarrow \sqrt{1 - \sin^{2} θ} + \sqrt{1 - \sin^{2} Ï•} = a (sinθ - sinÏ•) \\ \Rightarrow cosθ + cosÏ• = a (sinθ - sinÏ•) \\ \Rightarrow \frac{cosθ + cosÏ•}{sinθ - sinÏ•} = a \\ \Rightarrow \frac{2 \cos (\frac{θ + Ï•}{2}) \cos (\frac{θ - Ï•}{2})}{2 \cos (\frac{θ + Ï•}{2}) \sin (\frac{θ - Ï•}{2})} = a \\ \Rightarrow \cot (\frac{θ - Ï•}{2}) = a \\ \Rightarrow θ - Ï• = 2 \cot^{- 1} a \\ \Rightarrow \sin^{- 1} x - \sin^{- 1} y = 2 \cot^{- 1} a \\ differentiating w . r . t . x \\ \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} \end{array}$

Q.6

$\begin{array}{l} Using mathematical induction prove that \\ \frac{d}{dx} (x^{n}) = {nx}^{n - 1}, \forall n \in I^{+} \end{array}$

Ans

$\begin{array}{l} Let p (n) : \frac{d}{dx} (x^{n}) = {nx}^{n - 1} \\ At n = 1 \\ P (1) : \frac{d}{dx} (x^{1}) = 1 = 1 x^{1 - 1} = x^{0} \\ ∴ P (1) is true \\ Suppose p (k) is true \\ P (k) : \frac{d}{dx} (x^{k}) = {kx}^{k - 1} \\ We have to prove P (k + 1) is true \\ i . e . P (k) : \frac{d}{dx} (x^{k + 1}) = (k + 1) x^{k} \\ P (k + 1) : \frac{d}{dx} (x^{k + 1}) = \frac{d}{dx} (x^{k} . x) \\ = x^{k} \frac{d}{dx} (x) + x \frac{d}{dx} (x^{k}) \\ = x^{k} . 1 + {kxx}^{k - 1} \\ = x^{k} + {kxx}^{k - 1} \\ = x^{k} + {kx}^{k} \\ = (k + 1) x^{k} \\ ∴ P (k + 1) is true \\ Hence by mathematical induction \\ P (n) : \frac{d}{dx} (x^{n}) = {nx}^{n - 1} is true \end{array}$

Q.7

$\begin{array}{l} If {(x - a)}^{2} + {(y - b)}^{2} = c^{2} for some c > 0, \\ Prove that {\frac{\{1 + {(\frac{dy}{dx})}^{2}\}}{\frac{d^{2} y}{{dx}^{2}}}}^{\frac{3}{2}} is a constant, independent of a and b . \end{array}$

Ans

$\begin{array}{l} Here, {(x - a)}^{2} + {(y - b)}^{2} = c^{2} \\ differentiating w . r . t . x \\ \frac{d}{dx} [{(x - a)}^{2} + {(y - b)}^{2}] = 0 \\ \Rightarrow 2 (x - a) + 2 (y - b) \frac{dy}{dx} = 0 \\ \Rightarrow 2 (y - b) \frac{dy}{dx} = - 2 (x - a) \\ \Rightarrow \frac{dy}{dx} = - \frac{(x - a)}{(y - b)} . \dots\dots. (1) \\ again differentiating w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = - \frac{(y - b) . 1 - (x - a) . \frac{dy}{dx}}{{(y - b)}^{2}} \\ = - \frac{(y - b) + (x - a) . \frac{(x - a)}{(y - b)}}{{(y - b)}^{2}} \\ = - \frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}} \\ = - \frac{c^{2}}{{(y - b)}^{3}} . \dots.. (2) \\ Now, \frac{{\{1 + {(\frac{dy}{dx})}^{2}\}}^{\frac{3}{2}}}{\frac{d^{2} y}{{dx}^{2}}} = \frac{{\{1 + {(\frac{x - a}{y - b})}^{2}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} \\ \frac{{\{1 + {(\frac{x - a}{y - b})}^{2}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} = \frac{{\{\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{2}}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} \\ = \frac{{\{\frac{c^{2}}{{(y - b)}^{2}}\}}^{\frac{3}{2}}}{\frac{c^{2}}{{(y - b)}^{3}}} = \frac{\frac{c^{3}}{{(y - b)}^{3}}}{\frac{c^{2}}{{(y - b)}^{3}}} = - \frac{c^{3}}{c^{2}} = - c \end{array}$

Q.8

$If x \sqrt{1 + y} + y \sqrt{1 + x} = 0 for - 1 < x < 1, prove that \frac{dy}{dx} = - \frac{1}{{(1 + x)}^{2}}$

Ans

$\begin{array}{l} Here x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \\ \Rightarrow x \sqrt{1 + y} = - y \sqrt{1 + x} \\ squaring both sides we get \\ x^{2} (1 + y) = y^{2} (1 + x) \\ \Rightarrow x^{2} + x^{2} y = y^{2} + y^{2} x \\ \Rightarrow x^{2} - y^{2} = y^{2} x - x^{2} y \\ \Rightarrow (x - y) (x + y) = - xy (x - y) \\ \Rightarrow x + y = - xy \\ \Rightarrow x =- (xy + y) \\ \Rightarrow \frac{x}{x + 1} = - y \\ differentiating w . r . t x \\ - \frac{dy}{dx} = \frac{(x + 1) . 1 - x . 1}{{(x + 1)}^{2}} = \frac{1}{{(x + 1)}^{2}} \\ - \frac{dy}{dx} = = \frac{1}{{(x + 1)}^{2}} \end{array}$

Q.9 Verify the Mean value theorem for the function: f(x) = log_ex on [1, 2].

Ans

(a) f(x) is continuous on [1, 2].

(b) f‘(x) = 1/x therefore function is differentiable on [1, 2].

Both the two conditions of Mean Value Theorem are true

∴∃ at least one point c ∈ (1, 2)

$\begin{array}{l} s . t . f ‘ (c) = \frac{f (2) - f (1)}{2 - 1} \\ \Rightarrow \frac{1}{c} = \frac{\log_{e} 2 - \log_{e} 1}{2 - 1} = \log_{e} \frac{2}{1} \\ \Rightarrow c = \frac{1}{\log_{e} 2} = \log_{2} e . \end{array}$

Q.10 Verify Rolle’s Theorem for the following functions: f(x) = (x – 1)(x – 2)² on [1, 2].

Ans

(a) f(x) is a polynomial function, therefore the function is continuous on [1, 2].

(b) f‘(x) = 1.(x – 2)² + (x – 1).2(x – 2)
= (x – 2)[x – 2 + 2x – 2]
= (x – 2)(3x – 4)

Thus, function is differentiable on (1, 2)

∴ ∃ at least one point c (1, 2) s.t. f‘(∈ c) = 0

⇒ (c – 2)(3c – 4) = 0

⇒ c = 4/3 or c = 2
but 2 ∉(1, 2), hence this choice is rejected and the value of c is 4/3.

Q.11

$Differentiate \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} w . r . t . x .$

Ans

$\begin{array}{l} Let y = \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} \\ Taking \log both sides we have \\ \log y = \log \sqrt{\frac{(x - 3 (x^{2} + 4))}{3 x^{2} + 4 x + 5}} \\ = \log {(\frac{(x - 3) (x^{2} + 4)}{3 x^{2} + 4 x + 5})}^{\frac{1}{2}} \\ = \frac{1}{2} [\log (x - 3) + \log (x^{2} + 4) - \log (3 x^{2} + 4 x + 5)] \\ differentiating w . r . t . x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} [\frac{1}{x - 3} + \frac{1}{x^{2} + 4} \frac{d}{dx} (x^{2 + 4}) - \frac{1}{3 x^{2} + 4 x + 5} \frac{d}{dx} (3 x^{2} + 4 x + 5)] \\ \frac{dy}{dx} = \frac{y}{2} [\frac{1}{x - 3} + \frac{2 x}{x^{2} + 4} - \frac{6 x + 4}{3 x^{2} + 4 x + 5}] \\ \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x - 3) (x^{2 + 4})}{3 x^{2} + 4 x + 5} -} [\frac{1}{x - 3} + \frac{2 x}{x^{2} + 4} - \frac{6 x + 4}{3 x^{2} + 4 x + 5}] \end{array}$

Q.12

$If y = {(\tan^{- 1} x)}^{2}, show that {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2$

Ans

$\begin{array}{l} Here, y = {(\tan^{- 1} x)}^{2} \\ differentiating y w . r . t . x \\ y_{1} = \frac{dy}{dx} = \frac{d}{dx} {(\tan^{- 1} x)}^{2} \\ = 2 (\tan^{- 1} x) \frac{d}{dx} (\tan^{- 1} x) \\ = \frac{2 \tan^{- 1} x}{x^{2} + 1} \\ (x^{2} + 1) y_{1} = 2 \tan^{- 1} x \\ again differentiating w . r . t . x \\ \frac{d}{dx} [(x^{2} + 1) y_{1}] = \frac{d}{dx} (2 \tan^{- 1} x) \\ \Rightarrow (x^{2} + 1) \frac{d}{dx} y_{1} + y_{1} \frac{d}{dx} (x^{2} + 1) = \frac{d}{dx} (2 \tan^{- 1} x) \\ \Rightarrow (x^{2} + 1) y_{2} + y_{1} (2 x) = \frac{2}{x^{2} + 1} [as y_{2} = \frac{d^{2 y}}{{dx}^{2}}] \\ \Rightarrow {(x^{2} + 1)}^{2} y_{2} + y_{1} (x^{2} + 1) (2 x) = 2 \\ \Rightarrow {(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2 . \end{array}$

Q.13

$If x = \sqrt{a^{\sin^{- 1}},} y = \sqrt{a^{\cos^{- 1}}} show that \frac{dy}{dx} = - \frac{y}{x} .$

Ans

$\begin{array}{l} Here, x = \sqrt{a^{\sin^{- 1}},} \Rightarrow x^{2} = a^{\sin - 1} t \\ d i f f e r e n t i a t i n g x w.r.t.t \\ 2x \frac{d x}{d t} = a^{\sin^{- 1}} \log a . \frac{d}{d t} [\sin^{- 1} t] \\ 2x \frac{d x}{d t} = x^{2} . \log a . \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{d x}{d t} = \frac{1}{2} x . \log a . \frac{1}{\sqrt{1 - t^{2}}} ….. (1) \\ Now y = \sqrt{a^{\cos^{- 1} t}} \Rightarrow y^{2} = a^{\cos^{- 1} t} \\ d i f f e r e n t i a t i n g y w . r . t . t \\ 2 y \frac{d y}{d t} = a^{\cos^{- 1} t} \log a \frac{a}{d t} [\cos^{- 1} t] \\ 2 y \frac{d y}{d t} = y^{2} \log a . \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{d y}{d t} = – \frac{1}{2} y \log a . \frac{1}{\sqrt{1 - t^{2}}} ……. (2) \\ As \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- \frac{1}{2} y \log a . \frac{1}{\sqrt{1 - t^{2}}}}{\frac{1}{2} x . \log a . \frac{1}{\sqrt{1 - t^{2}}}} b y (1) a n d (2) \\ = - \frac{y}{x} \end{array}$

Q.14

$Find \frac{dy}{dx} if x = a [cost + logtan \frac{t}{2}, and y = a sint]$

Ans

$\begin{array}{l} Here x = a [cost + logtan \frac{t}{2}] \\ differentiating x w . r . t . t \\ \frac{dx}{dt} = a \frac{d}{dt} [cost + logtan \frac{t}{2}] \\ = a [- sint + \frac{1}{\tan \frac{t}{2}} . \frac{d}{dt} (tam \frac{t}{2})] \\ = a [- sint + \frac{1}{\tan \frac{t}{2}} . \frac{1}{2} \sec^{2} \frac{t}{2}] \\ = a [- sint + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}] \\ = a [sint + \frac{1}{sint}] [\sin 2 x = 2 sinxcosx] \\ = a [\frac{- \sin^{2} t + 1}{sint}] = [\frac{\cos^{2} t}{sint}] . \dots. (1) \\ Now y = a sint \\ differentiating y w . r . t . t \\ \frac{dy}{dx} = a cost . \dots (2) \\ As \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{acost}{\cos^{2} t} (by (1) and (2)) \\ = tant \end{array}$

Q.15

$Find \frac{dy}{dx}, if x^{y} + y^{x} =1.$

Ans

$\begin{array}{l} Here x^{y} + y^{x} =1 \\ \Rightarrow e^{{logx}^{y}} + e^{{logy}^{x}} = 1 \\ diff w . r . t x \\ \Rightarrow \frac{d}{dx} (e^{{logx}^{y}} + e^{{logy}^{x}}) = 0 \\ \Rightarrow e^{{logx}^{y}} \frac{d}{dx} ({logx}^{y}) + e^{{logy}^{x}} \frac{d}{dx} ({logy}^{x}) = 0 \\ \Rightarrow e^{{logx}^{y}} \frac{d}{dx} (y . logx) + e^{{logy}^{x}} \frac{d}{dx} (x . logy) = 0 \\ \Rightarrow x^{y} [y \frac{d}{dx} (logx) + logx \frac{dy}{dx}] + x^{y} [x \frac{d}{dx} (logy) + logy \frac{dy}{dx} (x)] = 0 \\ \Rightarrow x^{y} [\frac{y}{x} + logx \frac{dy}{dx}] + y^{x} [\frac{x}{y} . \frac{dy}{dx} + (logy)] = 0 \\ \Rightarrow x^{y - 1} y + x^{y} logx \frac{dy}{dx} + {xy}^{x - 1}, \frac{dy}{dx} + y^{x} logy = 0 \\ \Rightarrow [x^{y} logx + {xy}^{x - 1}] \frac{dy}{dx} = - [y^{x} logy + x^{y - 1} y] \\ \Rightarrow \frac{dy}{dx} = - \frac{y^{x} logy + x^{y - 1} y}{x^{y} logx + {xy}^{x - 1}} \end{array}$

Q.16

${Differentiate y=x}^{sinx} + {(sinx)}^{cosx} w . r . t . x$

Ans

$\begin{array}{l} y = e^{{logx}^{sinx}} + e^{\log {(sinx)}^{cosx}} [e^{loga} = a] \\ diff y w . r . t . x \\ \frac{dy}{dx} = \frac{d}{dx} [e^{{logx}^{sinx}} + e^{\log {(sinx)}^{cosx}}] \\ = e^{{logx}^{sinx}} \frac{d}{dx} ({logx}^{sinx}) + e^{\log {(sinx)}^{cosx}} \frac{d}{dx} (\log {(sinx)}^{cosx}) \\ = x^{sinx} \frac{d}{dx} (sinx . logx) + {(sinx)}^{cosx} \frac{d}{dx} (cosx . logsinx) \\ = x^{sinx} [sinx \frac{d}{dx} (logx) + logx \frac{d}{dx} (sinx)] + {(sinx)}^{cosx} [cosx \frac{d}{dx} (logsinx) + logsinx \frac{d}{dx} (cosx)] \\ = x^{sinx} [\frac{sinx}{x} cosx . logx] + {(sinx)}^{cosx} [cosx . cotx - sinx . logsinx] \end{array}$

Q.17

$Determine, if f defined by f (x) = \{\begin{cases} x^{2} \sin \frac{1}{x} if x \neq 0 \\ 0 if x = 0 \end{cases} is a continuous function ?$

Ans

$\begin{array}{l} Here f (0) = 0, \\ L . H . L . = \lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) . Putting x = 0 - h \\ = \lim_{h \to 0^{-}} {(0 - h)}^{2} \sin \frac{1}{0 - h} = - \lim_{h \to 0^{-}} h^{2} \sin \frac{1}{h} \\ \sin \frac{1}{h} lies between -1 and 1 \\ Thus L . H . L .= - \lim_{h \to 0^{-}} h^{2} \sin \frac{1}{h} = 0 \\ R . H . L . = \lim_{h \to 0^{+}} (x^{2} \sin \frac{1}{x}), Putting x = 0 + h \\ = \lim_{h \to 0^{}} (0 + h^{2}) \sin \frac{1}{0 + h} = \lim_{h \to 0^{}} h^{2} \sin \frac{1}{h} \\ ∴ R . H . L . = \lim_{h \to 0^{+}} h^{2} \sin \frac{1}{h} = 0 \\ R . H . L . = L . H . L = f (o) \\ Hence, f is continuous for all x \in R \end{array}$

Q.18

$\begin{array}{l} Find the values ofa and b so that the function defined by \\ f (x) = \{\begin{cases} 5 if x \leq 2 \\ ax + b if 2< x <10 \\ 21 if x \geq 10 \end{cases} is continous function . \end{array}$

Ans

$\begin{array}{l} Since f is continuous at x = 10, we obtain \\ \lim_{x \to 10^{-}} f (x) = \lim_{x \to 10^{+}} f (x) = f (10) \\ \Rightarrow \lim_{x \to 10^{-}} (ax + b) = \lim_{x \to 10^{+}} f (21) = (21) \\ \Rightarrow 10 a + b = 21 = 21 \\ \Rightarrow 10 a + b = 21 . \dots. (2) \end{array}$

Q.19

$\begin{array}{l} Find the relationship betweena andb so that the function defined by \\ f (x) = \{\begin{cases} ax +1 if x \leq 3 \\ bx +3 if x > 3 \end{cases} is continuous at x = 3. \end{array}$

Ans

$\begin{array}{l} Here f (3) = a (3) + 1 = 3 a + 1 \dots\dots (1) \\ R . H . L . = \lim_{x \to 3^{+}} (bx + 3) . \\ Putting x = 3 + h \\ R . H . L . = \lim_{h \to 0} b (3 + h) + 3 = 3 b + 3 . \dots (2) \\ As f is continuous at x = 3 \\ By (1) and (2) \\ R . H . L = f (3) \\ 3 b + 3 = 3 a + 1 \\ a - b = \frac{2}{3} or a = b + \frac{2}{3} \end{array}$

Q.20

$If y = \cos^{- 1} x, find \frac{d^{2} y}{{dx}^{2}} in terms of y alone .$

Ans

$\begin{array}{l} Here y = \cos^{- 1} x \Rightarrow x = \cos y \\ differentiating w . r . t x \\ \frac{d}{dx} (x) = \frac{d}{dx} (\cos y) \\ 1 = - siny . \frac{dy}{dx} \\ ∴ \frac{dy}{dx} = - cosec y . \dots.. (1) \\ differentiating \frac{dy}{dx} w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (- cosec y) \\ = cosecy . coty . \frac{dy}{dx} \\ = - {cosec}^{2} y . coty . .. by (1) \end{array}$

Q.21

$Find \frac{d^{2} y}{{dx}^{2}} if y = \sin^{- 1} x .$

Ans

$\begin{array}{l} Here y = \sin^{- 1} x \\ differentiating y w . r . t . x \\ \frac{dy}{dx} = \frac{d}{dx} (\sin^{- 1} x) \\ = \frac{1}{\sqrt{1 - x^{2}}} \\ diff \frac{dy}{dx} w . r . t . x \\ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (\frac{1}{\sqrt{1 - x^{2}}}) = \frac{d}{dx} {(1 - x^{2})}^{- \frac{1}{2}} \\ = - \frac{1}{2} {(1 - x^{2})}^{- \frac{1}{2} - 1} \frac{d}{dx} (1 - x^{2}) \\ = - \frac{1}{2} {(1 - x^{2})}^{- \frac{1}{2} - 1} . (- 2 x) \\ = x {(1 - x^{2})}^{- \frac{3}{2}} = \frac{x}{{(1 - x^{2})}^{\frac{3}{2}}} \end{array}$

Q.22

$Find \frac{dy}{dx} if y = \sin (\tan^{- 1} e^{- x}) .$

Ans

$\begin{array}{l} Here y = \sin (\tan^{- 1} e^{- 1}) \\ differentiating y w . r . t x \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (\tan^{- 1} e^{- 1})] \\ = \cos (\tan^{- 1} e^{- 1}) \frac{d}{dx} (\tan^{- 1} e^{- x}) \\ = \cos (\tan^{- 1} e^{- 1}) . \frac{1}{1 + {(e^{- x})}^{2}} \frac{d}{dx} (e^{- x}) \\ = \cos (\tan^{- 1} e^{- x}) . \frac{1}{1 + {(e^{- x})}^{2}} . e^{- x} \frac{d}{dx} (- x) \\ = \cos (\tan^{- 1} e^{- x}) . \frac{1}{1 + {(e^{- x})}^{2}} . e^{- x} . (- 1) \\ = - \frac{e^{- x} \cos (\tan^{- 1} e^{- x})}{1 + e^{- 2 x}} \end{array}$

Q.23

$Find \frac{dy}{dx}, if y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) .$

Ans

$\begin{array}{l} Let x = tanθ \Rightarrow θ = \tan^{- 1} x \\ y = \tan^{- 1} (\frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) \\ = \tan^{- 1} (\tan 3 θ) \\ = 3 θ = 3 \tan^{- 1} x \\ Now differentiating y w . r . t . x \\ \frac{dy}{dx} = \frac{3}{1 + x^{2}} . \end{array}$

Q.24

$\begin{array}{l} If f (x) = \{\begin{cases} \frac{kcosx}{π - 2 x} if x \neq \frac{π}{2}, find the value of k, \\ 3 if x = \frac{π}{2}, \end{cases} \\ if f is continuous at x = \frac{π}{2} . \end{array}$

Ans

$\begin{array}{l} Here, f \frac{π}{2} = 3 \\ Left hand limit of f at x = \frac{π}{2} is \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} \\ Putting x - \frac{π}{2} - h \\ \lim_{x \to \frac{π}{2}} f (x) = \lim_{x \to \frac{π}{2}} \frac{kcosx}{π - 2 x} = \lim_{x \to 0} \frac{kcos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} = \lim_{x \to 0} \frac{ksin h}{2 h} = \frac{k}{2} . \\ As f is continuous \\ ∴ \frac{k}{2} = 3 or k = 6 . \end{array}$

Q.25

$Examine the continuity of f (x) = x -5$

Ans

$\begin{array}{l} f (x) = |x - 5| or f (x) = \{\begin{cases} (x - 5) if x \geq 5 \\ (x - 5) if x < 5 \end{cases} \\ This function f is defined at all points of the real line . \\ Let c be a point on a real line . \\ Then, c < 5 or c =0 or c > 5 \\ Case I : c < 5 \\ Then, f (c) = 5- c \\ \lim_{x \to c} f (x) = \lim_{x \to c} f (5 - x) = 5- c = f (c) \\ Therefore, f is continuous at all real numbers less than 5. \\ Case II : c = 5 \\ Then, f (c) = f (5) = (5 - 5) = 0 \\ \lim_{x \to 5} f (x) =0 = \lim_{x \to 5} f (5 - x) = (5 - 5) = 0 \\ \lim_{x \to 5} f (x) =0 = \lim_{x \to 5} (x - 5) = (5 - 5) = 0 \\ ∴ f is continuous at x = 5 \\ Case III : c > 5 \\ Then, f (c) = c - 5 \\ \lim_{x \to c} f (x) = \lim_{x \to 5} (x - 5) = c - 5 = f (c) \\ Therefore, f is continuous at all real numbers greater than 5. \\ Hence, f is continuous at every real number and therefore, \\ it is a continuous function . \end{array}$

Q.26

$Differentiate : y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}}$

Ans

$\begin{array}{l} y = \sqrt{logx + \sqrt{logx + \sqrt{logx + . \dots}}} \\ y = \sqrt{logx + y} \\ Squaring both sides \\ y^{2} = logx + y \\ Differentiating w . r . t . x \\ \frac{d}{dx} (y^{2}) = \frac{d}{dx} (logx + y) \\ \Rightarrow 2 y \frac{dy}{dx} = \frac{1}{x} + \frac{dy}{dx} \\ \Rightarrow 2 y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow (2 y - 1) \frac{dy}{dx} = \frac{1}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{1}{x (2 y - 1)} \end{array}$

Q.27

$Differentiate y = \frac{x}{x +1} with respect to x .$

Ans

$\begin{array}{l} y = \frac{x}{x +1} \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{x}{x + 1}) \\ \frac{dy}{dx} = \frac{(x + 1) \frac{d}{dx} (x) - x \frac{d}{dx} (x + 1)}{{(x + 1)}^{2}} \\ = \frac{x + 1 - x}{{(x + 1)}^{2}} = \frac{\frac{1}{{(x + 1)}^{2}}}{} \end{array}$

Q.28

$Find the points where the constant function f (x) = k is continuous .$

Ans

$\begin{array}{l} The function is defined at all real numbers and by definition, its \\ value at any real number equalsk . Letcbe any real number . Then \\ \lim_{x \to c} f (x) = \lim_{x \to c} k = k \\ Since, f (c) = k = \lim_{x \to c} f (x) for any real number c, \\ the function f is continuous at every real number . \end{array}$

Q.29

$\begin{array}{l} Examine the continuity of functionfgiven by \\ f (x) = 2 x + 5 at x = 1. \end{array}$

Ans

$\begin{array}{l} Function is defined on x = 1 at its value is 7 \\ \lim_{x \to 1} f (x) = \lim_{x \to 1} (2 x + 5) = 2 (1) + 5 = 7 \\ \lim_{x \to 1} f (x) = f (1) = 7 \\ Hence, f is continuous at x = 1 \end{array}$

Q.30

$\begin{array}{l} Prove that the identity function on real numbers \\ given byf (x) = x is continuous at every real number . \end{array}$

Ans

$\begin{array}{l} Here, f (c) = cfor every real numberc . Also, \\ \lim_{x \to c} f (x) = c = f (c) and hence the function is \\ continuous at every real number . \end{array}$

Q.31 If f and g be two real functions continuous at a real number c, then f + g is continuous at x = c.

Ans

$\begin{array}{l} \lim_{x \to c} (f + g) (x) = \lim_{x \to c} [f (x) + g (x)] \\ = \lim_{x \to c} f (x) + \lim_{x \to c} g (x) \\ = f (c) + g (c) \\ = (f + g) (c) \end{array}$

Q.32

$\begin{array}{l} Find the point of discontinuity of f where f is defined by \\ f (x) = \{\begin{cases} 2 x + 3, if x \leq 2 \\ 2 x - 3 if x > 2 \end{cases} \end{array}$

Ans

$\begin{array}{l} f (2) = 2 \times 2 +3 = 7 \\ R . H . L = f (2 + h) = \lim_{h \to 0} (2 + h) = (2 \times 2 - 3) = 1 \\ f (2) is not equal to its R . H . L \\ hence, function is discontinus at x = 2 \end{array}$

Q.33 Prove that every rational function is continuous.

Ans

Every rational function is defined by

f(x) = p(x)/q(x), q(x) ≠ 0

Where p and q are the polynomial functions. The domain of f is all real numbers except points at which q is zero. Since polynomial functions are continuous, f is continuous.

Q.34

$Differentiate \cos (sinx) with respect to x .$

Ans

$\begin{array}{l} Let y = \cos (sinx) \\ \Rightarrow \frac{dy}{dx} = \frac{d}{dx} \cos (sinx) \\ = - \sin (sinx) \frac{d}{dx} (sinx) \\ = - \sin (sinx) cosx \end{array}$

Q.35

$Differentiate y = e^{- 2 x} with respect to x .$

Ans

$\begin{array}{l} y = e^{- 2 x} \\ \frac{dy}{dx} = \frac{d}{dx} (e^{- 2 x}) \\ \frac{dy}{dx} = e^{- 2 x} \frac{d}{dx} = (- 2 x) \\ = e^{- 2 x} \times - 2 \\ = - 2 e^{- 2 x} \end{array}$

Q.36 Differentiate y = sin (xy) with respect to x.

Ans

$\begin{array}{l} y = \sin (xy) \\ \frac{dy}{dx} = \frac{d}{dx} [\sin (xy)] \\ \frac{dy}{dx} = \cos (xy) \frac{d}{dx} (xy) \\ = \cos (xy) (x \frac{dy}{dx} + y) \\ = \cos (xy) \frac{dy}{dx} + ycos (xy) \\ \frac{dy}{dx} = \frac{y \cos (xy)}{1 - xcos (xy)} \end{array}$

Q.37

$If y = 5cos x-3sinx, prove that \frac{d^{2} y}{{dx}^{2}} + y = 0 .$

Ans

$\begin{array}{l} y = 5 \cos x - 3 sinx \\ \Rightarrow \frac{dy}{dx} = 5 \frac{d}{dx} cosx - 3 \frac{d}{dx} sinx \\ = - 5 sinx - 3 \cos x \\ \Rightarrow \frac{d^{2} y}{d^{2} x} = - 5 \frac{d}{dx} sinx - 3 \frac{d}{dx} cosx \\ = - 5 cosx + 3 sinx = - (5 cosx - 3 sinx) = - y \\ \Rightarrow \frac{d^{2} y}{d^{2} x} + y = 0 \end{array}$

Q.38

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = \frac{1}{x}, x \neq 0 . \end{array}$

Ans

$\begin{array}{l} Let c be any non - zero real number . \\ We have, \\ \lim_{x \to c} f (x) = \lim_{x \to c} (\frac{1}{x}) = (\frac{1}{c}) \\ Also, for c \neq 0, f (c) = \frac{1}{c} \\ \lim_{x \to c} f (x) = f (c) \\ \Rightarrow f is continuous . \end{array}$

Q.39

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = \frac{1}{x}, at x = 0. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{0}} f (x) = \lim_{x \to \bar{0}} (\frac{1}{x}) = (\frac{1}{- 0}) = - \infty \\ RHL = \lim_{x \to \bar{\overset{+}{0}}} f (x) = \lim_{x \to \bar{\bar{\overset{+}{0}}}} (\frac{1}{x}) = (\frac{1}{+ 0}) = + \infty \\ LHL \neq RHL = f (0) \\ \Rightarrow f is discontinuous at x = 0. \end{array}$

Q.40

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = |x|, at x = 0. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{0}} f (x) = \lim_{x \to \bar{0}} |x| = |- 0| = 0 \\ RHL = \lim_{x \to \bar{\overset{+}{0}}} f (x) = \lim_{x \to \overset{+}{0}} |x| = |+ 0| = 0 \\ LHL = RHL = f (0) \\ \Rightarrow f is continuous at x = 0. \end{array}$

Q.41

$\begin{array}{l} Discuss the differentiability of the function f defined \\ by f (x) = |x| at x =0. \end{array}$

Ans

$\begin{array}{l} LHD = \lim_{h \to 0} \frac{f (a - h) - f (a)}{- h} = \lim_{h \to \bar{0}} \frac{f (0 - h) - f (0)}{- h} \\ = \lim_{h \to \bar{0}} \frac{|- h| - |0|}{- h} = \lim_{h \to \bar{0}} \frac{|h|}{- h} = - 1 \\ RHD = \lim_{h \to 0} \frac{f (a + h) - f (a)}{- h} = \lim_{h \to \bar{0}} \frac{f (0 + h) - f (0)}{- h} \\ = \lim_{h \to \bar{0}} \frac{|h| - |0|}{- h} = \lim_{h \to \bar{0}} \frac{|h|}{- h} = 1 \\ LHD \neq RHD \\ \Rightarrow Function f (x) = |x| is not differentiable at x = 0. \end{array}$

Q.42

$\begin{array}{l} Discuss the continuity of the function f defined \\ by f (x) = 5 x +3 at x =1. \end{array}$

Ans

$\begin{array}{l} We have, \\ LHL = \lim_{x \to \bar{1}} f (x) \\ = \lim_{x \to \bar{1}} f (5 x + 3) \\ = (5 \times 1 + 3) = 8 \\ RHL = \lim_{x \overset{Ë™}{\to} \overset{+}{1}} f (x) \\ = \lim_{x \overset{Ë™}{\to} \overset{+}{1}} f (5 x + 3) \\ = (5 \times 1 + 3) = 8 \\ f (1) = (5 \times 1 + 3) = 8 \\ LHL = RHL = f (1) \\ \Rightarrow f is continuous at x = 1 \end{array}$

Q.43

$\begin{array}{l} Find the set of all points where the function \\ f (x) = x |x| is differentiable . \end{array}$

Ans

$\begin{array}{l} Since domain of the function, f (x) = x |x| is the set of all \\ real numbers, therefore, it is differentiable at the set of all \\ real numbers, that is, (- \infty, \infty) . \end{array}$

Q.44

$\begin{array}{l} Let f be a continuous function on satisfying \\ f (x + y) = f (x) + f (y) \forall x, y \in R and f (1) = 5, \\ then evaluate \lim_{x \to 4} f (x) . \end{array}$

Ans

$\begin{array}{l} \lim_{x \to 4} f (x) = f (4) \\ = f (2 + 2) \\ = f (2) + f (2) [f (x + y) = f (x) + f (y)] \\ = f (1 + 1) + f (1 + 1) \\ = f (1) + f (1) + f (1) + f (1) \\ = 5 + 5 + 5 + 5 [f (1) = 5] \\ = 20 \end{array}$

Q.45

$\begin{array}{l} Let f be a continuous function on satisfying \\ f (x) f (y) = f (x) + f (y) + f (xy) - 2 \forall x, y \in R and f (2) = 5, \\ then evaluate \lim_{x \to 4} f (x) . \end{array}$

Ans

$\begin{array}{l} \lim_{x \to 4} f (x) = f (4) \\ = f (2 \times 2) [f (xy) = f (x) f (y) + 2 - f (x) - f (y)] \\ = f (2) f (2) + 2 - f (2) - f (2) \\ = 5 \times 5 + 2 - 5 - 5 \\ = 25 + 2 - 10 [f (2) - 5] \\ = 17 \end{array}$

Q.46

$If f (x) = \log_{7} x, then f ‘ (x) .$

Ans

$\begin{array}{l} We have, \\ f (x) = \log_{7} x \\ = \frac{logx}{\log 7} [by change of base formula] \\ Differentiating w . r . t . x, we get \\ f ‘ (x) = \frac{d}{dx} (\frac{logx}{\log 7}) \\ = \frac{1}{\log 7}, \frac{1}{x} \\ = \frac{1}{xlog 7} \end{array}$

Q.47 Prove that sine function is continuous.

Ans

$\begin{array}{l} Since, \lim_{x \to 0} sinx = 0 \\ Let f (x) = \sin x is defined for every real number . \\ Let c be a real number . Put c = c + h . \\ If x \to c, we know that h \to 0 . Therefore, \\ \lim_{x \to c} f (x) = \lim_{x \to c} sinx \\ = \lim_{h \to 0} \sin (c + h) \\ = \lim_{h \to 0} (sinc \cosh - cosc \sinh) \\ = \lim_{h \to 0} (sinc \cosh) + \lim_{h \to 0} (cosc \sinh) \\ = sinc + 0 \\ = sinc = f (c) \\ Therefore, \\ \lim_{x \to c^{-}} f (x) = f (c) \\ and hence f is a continuous function . \end{array}$

Q.48 Prove that cosine function is continuous.

Ans

$\begin{array}{l} Let f (x) = cosx is defined for every real number . \\ Let c be a real number . Put c = c + h . \\ If x \to c, we know that h \to 0 . Therefore, \\ \lim_{x \to c^{+}} f (x) = \lim_{x \to c^{+}} cosx \\ = \lim_{h \to 0} \cos (c + h) \\ = \lim_{h \to 0} (cosc \cosh - sinc \sinh) \\ = \lim_{h \to 0} (cosc \cosh) + \lim_{h \to 0} (sinc \sinh) \\ = cosc + 0 [\begin{array}{l} \lim_{h \to 0} \cosh = 1 and \\ \underset{h \to 0}{\lim \sinh} = 0 \end{array}] \\ = cosc \\ Similarly, \\ \lim_{x \to c^{-}} f (x) = f (c) \\ ∴ \lim_{x \to c^{+}} f (x) = \lim_{x \to c^{-}} f (x) = f (c) \\ Hence f (x) is a continuous at x = c . \\ Since, c is arbitrary real number, so f (x) is \\ everywhere continuous . \end{array}$

Q.49 Prove that the function is defined by g(x) = x – [x] is discontinuous at all integral points.

Ans

$\begin{array}{l} Let a be an integer, then [a - h] = a - 1, [a + h] = a \\ and [a] = a \\ At x = a, \\ LHL = \lim_{x \to a^{-}} g (x) \\ = \lim_{x \to a^{-}} (x - [x]) \\ Putting x = a - h, as x \to a^{+}, h \to 0 \\ ∴ \lim_{x \to a^{-}} (x - [x]) = \lim_{h \to 0} (a - h - (a - 1)) [[a - h] = (a - 1)] \\ = \lim_{h \to 0} (1 - h) \\ = 1 \\ RHL = \lim_{x \to a} (x - [x]) \\ [Putting x = a + h as x \to a^{+} when h \to 0] \\ = \lim_{h \to 0} (a + h - (a + h)) \\ = \lim_{h \to 0} (a + h - a) [[a + h] = a] \\ = \lim_{h \to 0} (h) \\ = 0 \\ ∴ RHL \neq LHL \\ Thus, g (x) is discontinuous at all integral points . \end{array}$

Q.50

$\begin{array}{l} f (x) = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} becomes \\ continuous for all x . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} \\ = \frac{\sqrt{a^{2} - ax + x^{2}} - \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} \times \frac{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} + ax + x^{2}}}{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} + ax + x^{2}}} \\ \times \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} \\ = \frac{(a^{2} - ax + x^{2}) - (a^{2} + ax + x^{2})}{\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}}} \times \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} \\ = \frac{- 2 ax (\sqrt{a + x} + \sqrt{a - x})}{2 x (\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}})} \\ \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{- a (\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a^{2} - ax + x^{2}} + \sqrt{a^{2} - ax + x^{2}})} \\ = \frac{- a (\sqrt{a + 0} + \sqrt{a - 0})}{(\sqrt{a^{2} - a) (0) + 0^{2}} + \sqrt{a^{2} - a {+ 0}^{2}})} \\ = \frac{- a (2 \sqrt{a})}{(2 a)} = - \sqrt{a} \end{array}$

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FAQs (Frequently Asked Questions)

Find the left-hand limit, right-hand limit and function value at that point. If all three are equal, the function is continuous there. If any one of them differs or does not exist, the function is discontinuous.

A derivative can exist only when the function behaves smoothly near the point. That requires continuity first. So, every differentiable function is continuous at that point, but every continuous function need not be differentiable.

First check continuity at the joining point. Then find the left-hand derivative and right-hand derivative. If LHD = RHD, the function is differentiable at that point. If they differ, it is not differentiable.

Use logarithmic differentiation when the function has variable powers, repeated products, quotients or complex exponents. It converts multiplication and powers into simpler derivative steps using logarithms.

Rolle’s Theorem needs f(a) = f(b) and gives f′(c) = 0. LMVT does not need f(a) = f(b). It gives f′(c) = [f(b) – f(a)]/(b – a).

CBSE Class 12 Maths Revision Notes Chapter 5 Continuity and Differentiability

Key Takeaways

Access Class 12 Maths Chapter 5 Continuity and Differentiability Notes in 30 Minutes

Continuity and Differentiability Class 12 Notes: What This Chapter Covers

Continuity

Differentiability

Relation Between Continuity and Differentiability

Continuity in Class 12 Mathematics Chapter 5 Notes

Continuity at a Point

How to Check Continuity at x = c

Continuity in an Interval

Algebra of Continuous Functions

Common Continuous Functions

Types of Discontinuity

Removable Discontinuity

Non-Removable Discontinuity

Quick Discontinuity Check Table

Differentiability and Derivatives

Differentiability at a Point

Right-Hand Derivative

Left-Hand Derivative

Differentiability Condition

Differentiability Implies Continuity

Example: |x|

Rules of Differentiation

Sum and Difference Rule

Product Rule

Quotient Rule

Chain Rule

Derivatives of Trigonometric Functions

Derivatives of Inverse Trigonometric Functions

Derivatives of Exponential and Logarithmic Functions

Logarithmic Differentiation

When to Use Logarithmic Differentiation

Formula for y = [f(x)]g(x)

Parametric Differentiation

Second Derivative in Parametric Form

Second-Order Derivatives in Class 12 Maths Chapter 5

Quick Example

Derivative of Infinite Series

Rolle’s Theorem

Statement of Rolle’s Theorem

Meaning of Rolle’s Theorem

Lagrange’s Mean Value Theorem

Statement of Lagrange’s Mean Value Theorem

Meaning of LMVT

Difference Between Rolle’s Theorem and LMVT

Useful Substitutions for Differentiation

Quick Revision Tables for Continuity and Differentiability

Continuity Check Table

Differentiability Check Table

Derivative Rules Table

Theorem Conditions Table

Important Terms in Continuity and Differentiability

Common Mistakes in Class 12 Mathematics Chapter 5 Notes

Useful Links for Class 12 Maths

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FAQs (Frequently Asked Questions)

1. What is the fastest way to check continuity at a point?

2. Why does differentiability imply continuity?

3. How do I check differentiability in a piecewise function?

4. When should logarithmic differentiation be used?

5. What is the main difference between Rolle’s Theorem and LMVT?

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