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Class 12 Science (PCM)

Mathematics Part 1

Continuity and Differentiability

Continuity and Differentiability

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Second Order Derivative Identities
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Algebra of Derivatives
Accuracy: 14%
Total Attempted7
Total Correct1
Total Incorrect6
Difficulty Level:
High
Derivatives of Composite Functions
Accuracy: 17%
Total Attempted6
Total Correct1
Total Incorrect5
Difficulty Level:
High
Rolle’s Theorem
Accuracy: 18%
Total Attempted11
Total Correct2
Total Incorrect9
Difficulty Level:
High
Parametric Form and First Derivative
Accuracy: 25%
Total Attempted4
Total Correct1
Total Incorrect3
Difficulty Level:
High
Implicit Differentiation
Accuracy: 27%
Total Attempted11
Total Correct3
Total Incorrect8
Difficulty Level:
High
Algebra of Continuous Functions
Accuracy: 33%
Total Attempted18
Total Correct6
Total Incorrect12
Difficulty Level:
High
Continuity at a Point and Points of Discontinuity
Accuracy: 38%
Total Attempted8
Total Correct3
Total Incorrect5
Difficulty Level:
High
Lagrange’s Mean Value Theorem
Accuracy: 40%
Total Attempted5
Total Correct2
Total Incorrect3
Difficulty Level:
High
Variable-Base and Variable-Exponent Differentiation
Accuracy: 40%
Total Attempted5
Total Correct2
Total Incorrect3
Difficulty Level:
High
Derivatives of Exponential and Logarithmic Composites
Accuracy: 43%
Total Attempted7
Total Correct3
Total Incorrect4
Difficulty Level:
High
Continuity of Modulus and Greatest Integer Functions
Accuracy: 44%
Total Attempted9
Total Correct4
Total Incorrect5
Difficulty Level:
High
Logarithmic Differentiation
Accuracy: 47%
Total Attempted15
Total Correct7
Total Incorrect8
Difficulty Level:
High
Second Order Derivatives
Accuracy: 48%
Total Attempted25
Total Correct12
Total Incorrect13
Difficulty Level:
High
Continuity of Composite Functions
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Continuity of Functions
Accuracy: 50%
Total Attempted20
Total Correct10
Total Incorrect10
Difficulty Level:
Medium
Parameter Values for Continuity
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Verification of Rolle’s and Mean Value Theorems
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Differentiability and Derivatives
Accuracy: 57%
Total Attempted23
Total Correct13
Total Incorrect10
Difficulty Level:
Medium
Logarithmic Function and Log Rules
Accuracy: 60%
Total Attempted15
Total Correct9
Total Incorrect6
Difficulty Level:
Medium
Parametric Differentiation
Accuracy: 67%
Total Attempted6
Total Correct4
Total Incorrect2
Difficulty Level:
Medium
Inverse Trigonometric Functions
Accuracy: 71%
Total Attempted21
Total Correct15
Total Incorrect6
Difficulty Level:
Medium
Chain Rule for Differentiation
Accuracy: 72%
Total Attempted25
Total Correct18
Total Incorrect7
Difficulty Level:
Medium
Continuity of Standard Functions
Accuracy: 79%
Total Attempted19
Total Correct15
Total Incorrect4
Difficulty Level:
Low
Derivatives of Inverse Trigonometric Functions Table
Accuracy: 84%
Total Attempted19
Total Correct16
Total Incorrect3
Difficulty Level:
Low
Exponential and Logarithmic Functions
Accuracy: 88%
Total Attempted32
Total Correct28
Total Incorrect4
Difficulty Level:
Low
Continuity on Intervals and Domains
Accuracy: 100%
Total Attempted11
Total Correct11
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the concepts of continuity and differentiability.
  • Differentiate polynomial, trigonometric, inverse trigonometric, exponential, and logarithmic functions.
  • Apply the chain rule for differentiation.
  • Identify and prove the continuity of various functions at specific points.
  • Utilize logarithmic differentiation for complex functions.
  • Explore the relationships between continuity and differentiability.
  • Recognize the importance of fundamental theorems in differential calculus.

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Revision Notes & Summary

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Chapter 5: Continuity and Differentiability

5.1 Introduction

  • Continuation of differentiation study from Class XI.
  • Concepts introduced:
    • Continuity
    • Differentiability
    • Inverse trigonometric functions
    • Exponential and logarithmic functions
  • Fundamental theorems in differential calculus.

5.2 Continuity

  • Definition: A function is continuous at a point if the limit at that point equals the function's value.
  • Example: Function defined as:
    • f(x) = 1, if x ≤ 0
    • f(x) = 2, if x > 0
  • Graph Analysis: At x = 0, the function is not continuous.

Exercises

  1. Prove that f(x) = 5x - 3 is continuous at x = 0, x = -3, and x = 5.
  2. Examine continuity of f(x) = 2x² - 1 at x = 3.
  3. Examine the following functions for continuity:
    • (a) f(x) = 5
    • (b) f(x) = x
    • (c) f(x) = |x - 5|

5.3 Differentiability

  • Definition: The derivative of a real function at a point c is defined as:
    • lim (f(c+h) - f(c)) / h as h approaches 0.
  • Key Point: Every differentiable function is continuous, but not vice versa.

5.4 Exponential and Logarithmic Functions

  • Introduction to exponential functions and logarithmic functions.
  • Natural Exponential Function: y = e^x.
  • Logarithm Definition: For b > 1, log_b(a) = x if b^x = a.

Important Observations

  • Logarithm function is defined from positive real numbers to all real numbers.
  • Common logarithm (base 10) and natural logarithm (base e).

5.5 Logarithmic Differentiation

  • Technique for differentiating functions of the form f(x) = [u(x)]^v(x).
  • Both f(x) and u(x) must be positive for this technique.

5.6 Derivatives of Functions in Parametric Forms

  • Relationships between variables expressed via a third variable.
  • Example: x = a cos³(θ), y = a sin³(θ).

5.7 Miscellaneous Exercises

  • Differentiate various functions and find second-order derivatives.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

It is used to differentiate functions of the form [u(x)]v(x)[u(x)]^{v(x)} by taking the natural logarithm first.

B.

It can be applied to any function regardless of whether the function is positive or negative.

C.

It is primarily used for differentiating polynomial functions.

D.

It is a technique used to integrate exponential functions.
Correct Answer: A

Solution:

Logarithmic differentiation is used to differentiate functions such as [u(x)]v(x)[u(x)]^{v(x)}, products, quotients, and variable-exponent expressions by taking the natural logarithm first. The method requires the relevant function/base to be positive.

Chapter Concept:

Logarithmic Differentiation

A.

f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}

B.

g(x)=sin⁡(x)g(x) = \sin(x)

C.

h(x)=∣x−1∣h(x) = |x - 1|

D.

k(x)=1x−1k(x) = \frac{1}{x - 1}
Correct Answer: D

Solution:

The function k(x)=1x−1k(x) = \frac{1}{x - 1} is not continuous at x=1x = 1 because it is undefined at that point.

Chapter Concept:

Chain Rule for Differentiation

A.

a=2,b=−5a = 2, b = -5

B.

a=3,b=−1a = 3, b = -1

C.

a=4,b=−7a = 4, b = -7

D.

a=1,b=0a = 1, b = 0
Correct Answer: A

Solution:

To ensure continuity at x=1x = 1, the left-hand limit, right-hand limit, and the function value at x=1x = 1 must be equal. For the left-hand side, f(1)=a(1)+b=a+bf(1) = a(1) + b = a + b. For the right-hand side, f(1)=2(1)2−3=−1f(1) = 2(1)^2 - 3 = -1. Setting these equal gives a+b=−1a + b = -1. Solving this system with given options, a=2a = 2 and b=−5b = -5 satisfies the equation.

Chapter Concept:

Parameter Values for Continuity

A.

f(x)f(x) is differentiable everywhere.

B.

f(x)f(x) is differentiable at x=0x = 0.

C.

f(x)f(x) is not differentiable at x=0x = 0.

D.

f(x)f(x) is differentiable only at x=0x = 0.
Correct Answer: C

Solution:

The function f(x)=∣x∣f(x) = |x| is not differentiable at x=0x = 0 because the left-hand and right-hand derivatives at this point do not match.

Chapter Concept:

Differentiability and Derivatives

A.

There exists c∈(a,b)c \in (a, b) such that f′(c)=0f'(c) = 0.

B.

The function f(x)f(x) must be increasing on [a,b][a, b].

C.

The function f(x)f(x) must be decreasing on [a,b][a, b].

D.

The function f(x)f(x) is not differentiable at any point in (a,b)(a, b).
Correct Answer: A

Solution:

According to Lagrange's Mean Value Theorem, if f(a)=f(b)f(a) = f(b), then there exists a point c∈(a,b)c \in (a, b) such that f′(c)=f(b)−f(a)b−a=0f'(c) = \frac{f(b) - f(a)}{b - a} = 0. This implies that the derivative at some point cc is zero.

Chapter Concept:

Lagrange’s Mean Value Theorem

A.

1

B.

2

C.

Does not exist

D.

0
Correct Answer: A

Solution:

The left-hand limit as xx approaches 0 is 1 because for x≤0x \leq 0, f(x)=1f(x) = 1.

Chapter Concept:

Continuity at a Point and Points of Discontinuity

A.

f(x)=∣x∣f(x) = |x|

B.

f(x)=x3f(x) = x^3

C.

f(x)=sin⁡(x)f(x) = \sin(x)

D.

f(x)=x2f(x) = x^2
Correct Answer: A

Solution:

The function f(x)=∣x∣f(x) = |x| is continuous everywhere but not differentiable at x=0x = 0 because the left-hand derivative and right-hand derivative at x=0x = 0 are not equal.

Chapter Concept:

Derivatives of Composite Functions

A.

Expand the product to obtain a single polynomial and then differentiate.

B.

Use the product rule directly.

C.

Apply logarithmic differentiation.

D.

Differentiate each term separately and add the results.
Correct Answer: C

Solution:

Logarithmic differentiation is efficient for products of functions, especially when the product involves complex expressions.

Chapter Concept:

Algebra of Derivatives

A.

u′v−uv′v2=2xtan⁡(x)−x2sec⁡2(x)tan⁡2(x)\frac{u'v - uv'}{v^2} = \frac{2x \tan(x) - x^2 \sec^2(x)}{\tan^2(x)}

B.

u′v+uv′v2=2xtan⁡(x)+x2sec⁡2(x)tan⁡2(x)\frac{u'v + uv'}{v^2} = \frac{2x \tan(x) + x^2 \sec^2(x)}{\tan^2(x)}

C.

u′vv2=2xtan⁡(x)tan⁡2(x)\frac{u'v}{v^2} = \frac{2x \tan(x)}{\tan^2(x)}

D.

uv′v2=x2sec⁡2(x)tan⁡2(x)\frac{uv'}{v^2} = \frac{x^2 \sec^2(x)}{\tan^2(x)}
Correct Answer: A

Solution:

Using the quotient rule for differentiation, the derivative of uv\frac{u}{v} is given by u′v−uv′v2\frac{u'v - uv'}{v^2}. For u(x)=x2u(x) = x^2 and v(x)=tan⁡(x)v(x) = \tan(x), we have u′(x)=2xu'(x) = 2x and v′(x)=sec⁡2(x)v'(x) = \sec^2(x). Substituting these into the quotient rule formula gives 2xtan⁡(x)−x2sec⁡2(x)tan⁡2(x)\frac{2x \tan(x) - x^2 \sec^2(x)}{\tan^2(x)}.

Chapter Concept:

Algebra of Derivatives

A.

f∘gf \circ g is continuous at cc.

B.

f∘gf \circ g is discontinuous at cc.

C.

f∘gf \circ g is continuous only if ff is differentiable.

D.

f∘gf \circ g is continuous only if gg is differentiable.
Correct Answer: A

Solution:

According to the theorem on the continuity of composite functions, if gg is continuous at cc and ff is continuous at g(c)g(c), then f∘gf \circ g is continuous at cc.

Chapter Concept:

Continuity of Composite Functions

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True or False

Correct Answer: True

Solution:

According to the theorem on continuity of composite functions, if gg is continuous at cc and ff is continuous at g(c)g(c), then f∘gf \circ g is continuous at cc. This is used to prove the continuity of functions like sin⁡(x2)\sin(x^2) and cos⁡(x2)\cos(x^2).

Chapter Concept :

Continuity of Composite Functions

Correct Answer: True

Solution:

The formula for the derivative dydx\frac{dy}{dx} in parametric differentiation is dydt÷dxdt=g′(t)f′(t)\frac{dy}{dt} \div \frac{dx}{dt} = \frac{g'(t)}{f'(t)}, which holds true as long as dxdt≠0\frac{dx}{dt} \neq 0.

Chapter Concept :

Parametric Differentiation

Correct Answer: True

Solution:

For a function to be continuous at a point x=cx = c, the left-hand limit, right-hand limit, and the function value f(c)f(c) must all be equal. This ensures that the graph of the function can be drawn without lifting the pen.

Chapter Concept :

Continuity at a Point and Points of Discontinuity

Correct Answer: True

Solution:

Logarithmic differentiation involves taking the natural logarithm of the function, which is only defined for positive values.

Chapter Concept :

Logarithmic Differentiation

Correct Answer: True

Solution:

Theorem 3 states that if a function ff is differentiable at a point cc, then it is also continuous at that point.

Chapter Concept :

Differentiability and Derivatives

Correct Answer: True

Solution:

The chain rule is a method for finding the derivative of composite functions by differentiating the outer function and multiplying by the derivative of the inner function.

Chapter Concept :

Chain Rule for Differentiation

Correct Answer: True

Solution:

For a function to be continuous on a closed interval [a,b][a, b], it must be continuous at every point in the interval, including the endpoints. This means checking right-hand continuity at aa and left-hand continuity at bb.

Chapter Concept :

Continuity on Intervals and Domains

Correct Answer: True

Solution:

The derivative of the exponential function exe^x with respect to xx is indeed exe^x, as stated in Theorem 5.

Chapter Concept :

Exponential and Logarithmic Functions

Correct Answer: True

Solution:

The second derivative, denoted as d2ydx2\frac{d^2y}{dx^2}, is calculated by differentiating the first derivative dydx\frac{dy}{dx} with respect to xx. This is a fundamental concept in calculus for analyzing the curvature of functions.

Chapter Concept :

Second Order Derivatives

Correct Answer: True

Solution:

The modulus function ∣x∣|x| is continuous for all real numbers, as it does not have any jumps or breaks in its graph. On the other hand, the greatest integer function [x][x], which returns the greatest integer less than or equal to xx, is discontinuous at every integer because there is a jump of 1 unit at each integer point.

Chapter Concept :

Continuity of Modulus and Greatest Integer Functions

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