Category: Inter 1st Year

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 1st Year Physics Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Jr Inter 1st Year Physics Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 1st Year Physics Study Material and AP Inter 1st Year Physics Important Questions for exam preparation.

AP Intermediate 1st Year Physics Notes

• Chapter 1 Physical World Notes
• Chapter 2 Units and Measurements Notes
• Chapter 3 Motion in a Straight Line Notes
• Chapter 4 Motion in a Plane Notes
• Chapter 5 Laws of Motion Notes
• Chapter 6 Work, Energy and Power Notes
• Chapter 7 Systems of Particles and Rotational Motion Notes
• Chapter 8 Oscillations Notes
• Chapter 9 Gravitation Notes
• Chapter 10 Mechanical Properties of Solids Notes
• Chapter 11 Mechanical Properties of Fluids Notes
• Chapter 12 Thermal Properties of Matter Notes
• Chapter 13 Thermodynamics Notes
• Chapter 14 Kinetic Theory Notes

These TS AP Intermediate 1st Year Physics Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 1st Year Physics Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

TS AP Intermediate 1st Year Physics Important Questions Chapter Wise 2022: Here we have created a list of Telangana & Andhra Pradesh BIEAP TS AP Intermediate Jr Inter 1st Year Physics Important Questions Chapter Wise with Answers 2022-2023 Pdf Download just for you. Those who are preparing for Inter exams should practice the Intermediate 1st Year Physics Important Questions Pdf and do so will clear their doubts instantly. These Jr Inter Physics Important Questions with Answers enhance your conceptual knowledge and prepares you to solve different types of questions in the exam.

Students must practice these AP Inter 1st Year Physics Important Questions Pdf to boost their preparation for the exam paper. These AP Intermediate 1st Year Physics Important Questions with Answers are prepared as per the latest exam pattern. Each of these chapters contains a set of solved questions and additional questions for practice.

Students can also read AP Inter 1st Year Physics Study Material for exam preparation. Students can also go through AP Inter 1st Year Physics Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Physics Important Questions with Answers Chapter Wise 2022

Intermediate 1st Year Physics Important Questions Chapter Wise 2022

• Chapter 1 Physical World Important Questions
• Chapter 2 Units and Measurements Important Questions
• Chapter 3 Motion in a Straight Line Important Questions
• Chapter 4 Motion in a Plane Important Questions
• Chapter 5 Laws of Motion Important Questions
• Chapter 6 Work, Energy and Power Important Questions
• Chapter 7 Systems of Particles and Rotational Motion Important Questions
• Chapter 8 Oscillations Important Questions
• Chapter 9 Gravitation Important Questions
• Chapter 10 Mechanical Properties of Solids Important Questions
• Chapter 11 Mechanical Properties of Fluids Important Questions
• Chapter 12 Thermal Properties of Matter Important Questions
• Chapter 13 Thermodynamics Important Questions
• Chapter 14 Kinetic Theory Important Questions

We are providing well-organized Physics Important Questions Inter 1st Year that will help students in their exam preparation. These Physics Inter 1st Year Important Questions are designed by subject experts. So Students should utilise the Intermediate 1st Year Physics Important Questions with Answers 2022 to pass their board exams with flying colours.

The questions given in the Inter First Year Physics Important Questions are designed and laid out chronologically and as per the syllabus. Students can expect that these TS AP Intermediate Inter 1st Year Physics Important Questions Chapter Wise Pdf 2022-2023 might be covered in the final exam paper.

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Physics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 1st Year Physics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 1st Year Physics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 1st Year Study Material Pdf.

Students can also read AP Inter 1st Year Physics Syllabus & AP Inter 1st Year Physics Important Questions for exam preparation. Students can also go through AP Inter 1st Year Physics Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Physics Study Material Pdf Download | Jr Inter 1st Year Physics Textbook Solutions

• Chapter 1 Physical World
• Chapter 2 Units and Measurements
• Chapter 3 Motion in a Straight Line
• Chapter 4 Motion in a Plane
• Chapter 5 Laws of Motion
• Chapter 6 Work, Energy and Power
• Chapter 7 Systems of Particles and Rotational Motion
• Chapter 8 Oscillations
• Chapter 9 Gravitation
• Chapter 10 Mechanical Properties of Solids
• Chapter 11 Mechanical Properties of Fluids
• Chapter 12 Thermal Properties of Matter
• Chapter 13 Thermodynamics
• Chapter 14 Kinetic Theory

TS AP Inter 1st Year Physics Weightage Blue Print 2022-2023

TS AP Inter 1st Year Physics Weightage 2022-2023 | TS AP Inter 1st Year Physics Blue Print 2022

Intermediate 1st Year Physics Syllabus

TS AP Inter 1st Year Physics Syllabus

Chapter 1 Physical World
What is Physics?, Scope and excitement of Physics, Physics, technology, and society, Fundamental forces in nature, Nature of Physical laws.

Chapter 2 Units and Measurements
Introduction, The international system of units, Measurement of length, Measurement of mass, Measurement of time, Accuracy, the precision of instruments, and errors in measurement, Significant figures, Dimensions of physical quantities, Dimensional formulae and dimensional equations, Dimensional analysis and its applications.

Chapter 3 Motion in a Straight Line
Introduction, Position, path length, and displacement, Average velocity and average speed, Instantaneous velocity and speed, Acceleration, Kinematic equations for uniformly accelerated motion, Relative velocity.

Chapter 4 Motion in a Plane
Introduction, Scalars and vectors, Multiplication of vectors by real numbers, Addition and subtraction of vectors, graphical method, Resolution of vectors, Vector addition, analytical method, Motion in a plane, Motion in a plane with constant acceleration, Relative velocity in two dimensions, Projectile motion, Uniform circular motion.

Chapter 5 Laws of Motion
Introduction, Aristotle’s fallacy, The law of inertia, Newton’s first law of motion, Newton’s second law of motion, Newton’s third law of motion, Conservation of momentum, Equilibrium of a particle, Common forces in mechanics, friction, Circular motion, Solving problems in mechanics.

Chapter 6 Work, Energy and Power
Introduction, Notions of work and kinetic energy: The work-energy theorem, Work, Kinetic energy, Work done by a variable force, The work-energy theorem for a variable force, The concept of potential energy, The conservation of mechanical energy, The potential energy of a spring, Various forms of energy: the law of conservation of energy, Power, Collisions.

Chapter 7 System of Particles and Rotational Motion
Introduction, Centre of mass, Centre of Gravity, The motion of centre of mass, Linear momentum of a system of particles, Vector product of two vectors, Angular velocity and its relation with linear velocity, Kinematics of rotational motion about a fixed axis, Torque and angular momentum, Equilibrium of a rigid body, Moment of inertia, Theorems of perpendicular and parallel axes, Dynamics of rotational motion about a fixed axis, Angular momentum in case of rotations about a fixed axis, Rolling motion.

Chapter 8 Oscillations
Introduction, Periodic and oscillatory motions, Simple harmonic motion, Simple harmonic motion and uniform circular motion, Velocity and acceleration in simple harmonic motion, Force law for Simple Harmonic Motion, Energy in simple harmonic motion, Some systems executing Simple Harmonic Motion, Damped simple harmonic motion, Forced oscillations and resonance.

Chapter 9 Gravitation
Introduction, Kepler’s laws, The universal law of gravitation, The gravitational constant, Acceleration due to the gravity of the earth, Acceleration due to gravity below and above the surface of the earth, Gravitational potential energy, Escape speed, Earth satellite, The energy of an orbiting satellite, Geostationary and polar satellites, Weightlessness.

Chapter 10 Mechanical Properties of Solids
Introduction, Elastic behaviour of solids, Stress and strain, Hooke’s law, Stress-strain curve, Elastic moduli, Applications of elastic behaviour of materials.

Chapter 11 Mechanical Properties of Fluids
Introduction, Pressure, Streamline flow, Bernoulli’s principle, Viscosity, Reynolds number, Surface tension.

Chapter 12 Thermal Properties of Matter
Introduction, Temperature and heat, Measurement of temperature, Ideal-gas equation and absolute temperature, Thermal expansion, Specific heat capacity, Calorimetry, Change of state, Heat transfer, Newton’s law of cooling.

Chapter 13 Thermodynamics
Introduction, Thermal equilibrium, Zeroth law of thermodynamics, Heat, internal energy, and work, The first law of thermodynamics, Specific heat capacity, Thermodynamic state variables and equation of State, Thermodynamic processes, Heat engines, Refrigerators and heat pumps, The second law of thermodynamics, Reversible and irreversible processes, Carnot engine, Carnot’s theorem.

Chapter 14 Kinetic Theory
Introduction, Molecular nature of matter, Behaviour of gases, Kinetic theory of an ideal gas, Law of equipartition of energy, Specific heat capacity, Mean free path.

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Physics Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Jr Inter 1st Year Physics Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 1st Year Chemistry Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Jr Inter 1st Year Chemistry Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 1st Year Chemistry Study Material and AP Inter 1st Year Chemistry Important Questions for exam preparation.

AP Intermediate 1st Year Chemistry Notes

• Chapter 1 Atomic Structure Notes
• Chapter 2 Classification of Elements and Periodicity in Properties Notes
• Chapter 3 Chemical Bonding and Molecular Structure Notes
• Chapter 4 States of Matter: Gases and Liquids Notes
• Chapter 5 Stoichiometry Notes
• Chapter 6 Thermodynamics Notes
• Chapter 7 Chemical Equilibrium and Acids-Bases Notes
• Chapter 8 Hydrogen and its Compounds Notes
• Chapter 9 The s-Block Elements Notes
• Chapter 10 The p-Block Elements – Group 13 Notes
• Chapter 11 The p-Block Elements – Group 14 Notes
• Chapter 12 Environmental Chemistry Notes
• Chapter 13 Organic Chemistry-Some Basic Principles and Techniques Notes

These TS AP Intermediate 1st Year Chemistry Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 1st Year Chemistry Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

TS AP Intermediate 1st Year Chemistry Important Questions Chapter Wise 2022: Here we have created a list of Telangana & Andhra Pradesh BIEAP TS AP Intermediate Jr Inter 1st Year Chemistry Important Questions Chapter Wise with Answers 2022-2023 Pdf Download just for you. Those who are preparing for Inter exams should practice the Intermediate 1st Year Chemistry Important Questions Pdf and do so will clear their doubts instantly. These Jr Inter Chemistry Important Questions with Answers enhance your conceptual knowledge and prepares you to solve different types of questions in the exam.

Students must practice these AP Inter 1st Year Chemistry Important Questions Pdf to boost their preparation for the exam paper. These AP Intermediate 1st Year Chemistry Important Questions with Answers are prepared as per the latest exam pattern. Each of these chapters contains a set of solved questions and additional questions for practice.

Students can also read AP Inter 1st Year Chemistry Study Material for exam preparation. Students can also go through AP Inter 1st Year Chemistry Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Chemistry Important Questions with Answers Chapter Wise 2022

Intermediate 1st Year Chemistry Important Questions Chapter Wise 2022

• Chapter 1 Atomic Structure Important Questions
• Chapter 2 Classification of Elements and Periodicity in Properties Important Questions
• Chapter 3 Chemical Bonding and Molecular Structure Important Questions
• Chapter 4 States of Matter: Gases and Liquids Important Questions
• Chapter 5 Stoichiometry Important Questions
• Chapter 6 Thermodynamics Important Questions
• Chapter 7 Chemical Equilibrium and Acids-Bases Important Questions
• Chapter 8 Hydrogen and its Compounds Important Questions
• Chapter 9 The s-Block Elements Important Questions
• Chapter 10 The p-Block Elements – Group 13 Important Questions
• Chapter 11 The p-Block Elements – Group 14 Important Questions
• Chapter 12 Environmental Chemistry Important Questions
• Chapter 13 Organic Chemistry-Some Basic Principles and Techniques Important Questions

We are providing well-organized Chemistry Important Questions Inter 1st Year that will help students in their exam preparation. These Chemistry Inter 1st Year Important Questions are designed by subject experts. So Students should utilise the Intermediate 1st Year Chemistry Important Questions with Answers 2022 to pass their board exams with flying colours.

The questions given in the Inter First Year Chemistry Important Questions are designed and laid out chronologically and as per the syllabus. Students can expect that these TS AP Intermediate Inter 1st Year Chemistry Important Questions Chapter Wise Pdf 2022-2023 might be covered in the final exam paper.

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Chemistry Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 1st Year Chemistry Blue Print Weightage 2022-2023, Telugu Academy Intermediate 1st Year Chemistry Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 1st Year Study Material Pdf.

Students can also read AP Inter 1st Year Chemistry Syllabus & AP Inter 1st Year Chemistry Important Questions for exam preparation. Students can also go through AP Inter 1st Year Chemistry Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Chemistry Study Material Pdf Download | Jr Inter 1st Year Chemistry Textbook Solutions

• Chapter 1 Atomic Structure
• Chapter 2 Classification of Elements and Periodicity in Properties
• Chapter 3 Chemical Bonding and Molecular Structure
• Chapter 4 States of Matter: Gases and Liquids
• Chapter 5 Stoichiometry
• Chapter 6 Thermodynamics
• Chapter 7 Chemical Equilibrium and Acids-Bases
• Chapter 8 Hydrogen and its Compounds
• Chapter 9 The s-Block Elements
• Chapter 10 The p-Block Elements – Group 13
• Chapter 11 The p-Block Elements – Group 14
• Chapter 12 Environmental Chemistry
• Chapter 13 Organic Chemistry-Some Basic Principles and Techniques

TS AP Inter 1st Year Chemistry Weightage Blue Print 2022-2023

TS AP Inter 1st Year Chemistry Weightage 2022-2023 | TS AP Inter 1st Year Chemistry Blue Print 2022

Intermediate 1st Year Chemistry Syllabus

TS AP Inter 1st Year Chemistry Syllabus

Chapter 1 Atomic Structure (20 Periods)

• 1.1 Sub-atomic particles
• 1.2 Atomic models-Rutherford’s Nuclear model of the atom
• 1.3 Developments to Bohr’s model of the atom.
  • 1.3.1 Nature of electromagnetic radiation
  • 1.3.2 Particle nature of electromagnetic radiation-Planck’s quantum theory
• 1.4 Bohr’s model for the hydrogen atom
  • 1.4.1 Explanation of line spectrum of hydrogen
  • 1.4.2 Limitations of Bohr’s model
• 1.5 Quantum mechanical considerations of sub-atomic particles
  • 1.5.1 Dual behaviour of matter
  • 1.5.2 Heisenberg’s uncertainty principle
• 1.6 Quantum mechanical model of an atom. Important features of the Quantum mechanical model of the atom.
  • 1.6.1 Orbitals and quantum numbers
  • 1.6.2 Shapes of atomic orbitais
  • 1.6.3 Energies of orbitals
  • 1.6.4 Filling of orbitals in atoms, Aufbau principle, Pauli’s exclusion principle, and Hund’s rule of maximum multiplicity
  • 1.6.5 Electronic configurations of atoms
  • 1.6.6 Stability of half-filled and completely filled orbitals.

Chapter 2 Classification of Elements and Periodicity in Properties (20 Periods)

• 2.1 Need to classify elements
• 2.2 Genesis of periodic classification
• 2.3 Modem periodic law and present form of the periodic table
• 2.4 Nomenclature of elements with atomic number greater than 100
• 2.5 Electronic configuration of elements and the periodic table
• 2.6 Electronic configuration and types of elements s.p.d. and f blocks
• 2.7 Trends in physical properties:
  • 2.7.1 (a) Atomic radius, (b) Ionic radius, (c) Variation of size in inner transition elements, (d) Ionization enthalpy, (e) Electron gain enthalpy, (f) Electro negativity
  • 2.7.2 Periodic trends in chemical properties: (a) Valence or Oxidation states, (b) Anomalous properties of second-period elements – diagonal relationship.
  • 2.7.3 Periodic trends and chemical reactivity.

Chapter 3 Chemical Bonding and Molecular Structure (20 Periods)

• 3.1 Kossel – Lewis approach to chemical bonding
• 3.2 Ionic or electrovalent bond – Factors favourable for the formation of the ionic compounds-Crystal structure of Sodium chloride – General Properties of ionic compounds.
• 3.3 Bond Parameters – bond length, bond angle, bond enthalpy, bond order, and resonance – Polarity of bonds dipole moment.
• 3.4 Valence Shell Electron Pair Repulsion (VSEPR) theories. Predicting the geometry of simple molecules.
• 3.5 Valence bond theory – Orbital overlap concept-Directional properties of bonds-overlapping of atomic orbitals strength of sigma and pi bonds-Factors favouring the formation of covalent bonds.
• 3.6 Hybridisation-different types of hybridization involving s, p, and d orbitals -shapes of simple covalent molecules.
• 3.7 Coordinate bond – definition with examples.
• 3.8 Molecular orbital theory – Formation of molecular orbitals. Linear combination of atomic orbitals (LCAO)- conditions for a combination of atomic orbitals – Energy level diagrams for molecular orbitals – Bonding in some homo nuclear diatomic molecules – H₂, He₂, Li₂, B₂, C₂, N₂ and O₂
• 3.9 Hydrogen bonding-cause of formation of hydrogen bond-Types of hydrogen bonds-inter and intra molecular – General properties of hydrogen bonds.

Chapter 4 States of Matter: Gases and Liquids (15 Periods)

• 4.1 Intermolecular forces
• 4.2 Thermal Energy
• 4.3 Intermolecular forces Vs Thermal interactions
• 4.4 The Gaseous State
• 4.5 The Gas Laws
• 4.6 Ideal gas equation
• 4.7 Graham’s law of diffusion – Dalton’s Law of partial pressures
• 4.8 Kinetic molecular theory of gases
• 4.9 Kinetic gas equation of an ideal gas(No derivation)- Deduction of gas laws from Kinetic gas equation.
• 4. 10 Distribution of molecular speeds – rms, average and most probable speeds -Kinetic energy of gas molecules.
• 4.11 Behaviour of real gases – Deviation from Ideal gas behaviour – Compressibility factor Vs Pressure diagrams of real gases.
• 4.12 Liquefaction of gases
• 4.13 Liquid State – Properties of Liquids in terms of intermolecular interactions – Vapour pressure, Viscosity, and Surface tension (Qualitative idea only. No mathematical derivation).

Chapter 5 Stoichiometry (15 Periods)

• 5.1 Some Basic Concepts – Properties of matter – uncertainty in Measurement – significant figures, dimensional analysis
• 5.2 Laws of Chemical Combinations – Law of Conservation of mass, the law of definite proportions, the law of multiple proportions, Gay Lussac’s law of gaseous volumes, Dalton’s atomic theory, Avogadro law, principles, and examples.
• 5.3 Atomic and molecular masses- mole concept and molar mass concept of equivalent weight.
• 5.4 Percentage composition of compounds and calculations of empirical and molecular formulae of compounds.
• 5.5 Stoichiometry and stoichiometric calculations.
• 5.6 Methods of expressing concentrations of solutions-mass percent, mole fraction, molarity, and normality.
• 5.7 Redox reactions-classical idea of redox reactions, oxidation, and reduction reactions-redox reactions in terms of electron transfer.
• 5.8 Oxidation number concept.
• 5.9 Types of Redox reactions-combination, decomposition, displacement, and disproportionation reactions.
• 5.10 Balancing of redox reactions- oxidation number method, half-reaction(ion-electron) method.
• 5.11 Redox reactions in titrimetry

Chapter 6 Thermodynamics (10 Periods)

• 6.1 Thermodynamic terms
  • 6.1.1 The system and the surroundings
  • 6.1.2 Types of systems and surroundings
  • 6.1.3 The state of the system
  • 6.1.4 The Internal Energy as a state function, (a) Work (b) Heat (c) The general case, the first law of Thermodynamics
• 6.2 Applications
  • 6.2.1 Work
  • 6.2.2 Enthalpy. H- a useful new state function
  • 6.2.3 Extensive and intensive properties.
  • 6.2.4 Heat capacity
  • 6.2.5 The relationship between C_p and C_v
• 6.3 Measurement of “U and H”: calorimetry.
• 6.4 Enthalpy change, ‘rH’of reactions-reaction Enthalpy, (a) Standard enthalpy of reactions, (b) Enthalpy changes during transformations, (c) Standard enthalpy of formation, (d) Thermo chemical equations, (e) Hess’s law of constant heat summation.
• 6.5 Enthalpies for different types of reactions, (a) Standard enthalpy of combustion (“cH°) (b) Enthalpy of atomization (“a Ho). Phase transition, sublimation, and ionization, (c) Bond Enthalpy (“bond H6) (d) Enthalpy of solution (“sol Ho), and dilution.
• 6.6 Spontaneity, (a) Is a decrease in enthalpy a criterion for spontaneity? (b) Entropy and spontaneity are the second law of thermodynamics, (c) Gibbs Energy and spontaneity.
• 6.7 Gibbs Energy change and equilibrium.
• 6.8 Absolute entropy and the third law of thermodynamics.

Chapter 7 Chemical Equilibrium and Acids-Bases (15 Periods)

• 7.1 Equilibrium in a physical process
• 7.2 Equilibrium in chemical process – dynamic equilibrium.
• 7.3 Law of chemical equilibrium- Law of mass action and equilibrium constant.
• 7.4 Homogeneous equilibria, the equilibrium constant in the gaseous system, Relationship between K and Kc.
• 7.5 Heterogeneous equilibria.
• 7.6 Applications of equilibrium constant.
• 7.7 Relationship between equilibrium constant K reaction quotient Q arid Gibbs energy G.
• 7.8 Factors affecting equilibrium Le-chatelier’s principle applies to the industrial synthesis of ammonia and sulphur trioxide.
• 7.9 Ionic Equilibrium in solutions.
• 7.10 Acids, bases, and salts – Arrhenius, Bronsted- Lowry, and Lewis concepts of acids and bases.
• 7.11, Ionisation of acids and bases – Ionisation constant of water and ifs ionic product- pH scale -ionisation constants of weak acids-ionization of weak bases-relation between K_a and K_b -di and polybasic acids and di poly acidic bases-factors affecting acid strength-common ion effect in the ionization of acids and bases-hydrolysis of salts and pH of their solutions.
• 7.12 Buffer solutions-designing of buffer solution-preparation of the acidic buffer.
• 7.13 Solubility equilibria of sparingly soluble salts, solubility product constant common ion effect solubility of Ionic salts.

Chapter 8 Hydrogen and its Compounds (8 Periods)

• 8.1 Position of hydrogen in the periodic table.
• 8.2 Dihydrogen-occurrence and isotopes.
• 8.3 Preparation of dihydrogen
• 8.4 Properties of dihydrogen
• 8.5 Hydrides:ionic. covalent, and non-stiochiometrichydrides
• 8.6 Water-physical properties; structure of water, ice. Chemical properties of water; hard and soft water temporary and permanent hardness of water
• 8.7 Hydrogen peroxide: Preparation; physical properties; structure and chemical properties; storage and uses
• 8.8 Heavy water
• 8.9 Hydrogen as a fuel

Chapter 9 s-Block Elements (Alkali and Alkaline Earth Metals) (8 Periods)
Group 1 Elements:

• 9.1 Alkali metals; electronic configurations; atomic and ionic radii; ionization enthalpy hydration enthalpy; physical properties; chemical properties; uses.
• 9.2 General characteristics of the compounds of the alkali metals; oxides; halides; salts of oxy acids.
• 9.3 Anomalous properties of lithium: differences and similarities with other alkali metals, diagonal relationship; similarities between lithium and magnesium.
• 9.4 Some important compounds of sodium- Sodium Carbonate; Sodium Chloride; Sodium Hydroxide; Sodium Hydrogen Carbonate.
• 9.5 Biological importance of Sodium and Potassium.

Group 2 Elements:

• 9.6 Alkaline earth elements; electronic configuration; ionization enthalpy; hydration enthalpy; physical properties; chemical properties; uses.
• 9.7 General. characteristics of compounds of the alkaline earth metals: oxides, hydroxides, halides, salts of oxyacids (carbonates; sulphates, and nitrates).
• 9.8 Anomalous behavior of beryllium; its diagonal relationship with aluminium.
• 9.9 Some important compounds of calcium: Preparation and uses of calcium oxide; calcium hydroxide; calcium carbonate; plaster of Paris; cement.
• 9.10 Biological importance of calcium and magnesium.

Chapter 10 p-Block Elements Group 13 (8 Periods)
Boron Family

• 10.1 General introduction – electronic configuration, atomic radii, ionization enthalpy, electronegativity; physical & chemical properties.
• 10.2 Important trends and anomalous properties of boron.
• 10.3 Some important compounds of boron-borax, orthoboric acid, and diborane.
• 10.4 Use of boron, aluminium, and their compounds.

Chapter 11 p-Block Elements Group 14 (8 Periods)
Carbon Family

• 11.1 General introduction – electronic configuration, atomic radii, ionization enthalpy, electronegativity; physical & chemical properties.
• 11.2 Important trends and anomalous properties of carbon.
• 11.3 Allotropes of carbon
• 11.4 Uses of carbon
• 11.5 Some important compounds of carbon and silicon-carbon monoxide, carbon dioxide, Silica, silicones, silicates, and zeolites.

Chapter 12 Environmental Chemistry (8 Periods)

• 12.1 Definition of terms: Air, Water, and Soil Pollutions
• 12.2 Environmental pollution
• 12.3 Atmospheric pollution; tropospheric pollution; gaseous air pollutants (oxides of sulphur; oxides of nitrogen; hydrocarbons; oxides of carbon (CO; CO₂), Global warming and greenhouse effect.
• 12.4 Acid Rain-particulate pollutants-smog.
• 12.5 Stratospheric Pollution: formation and breakdown of ozone-ozone hole-effects of depletion of the ozone layer.
• 12.6 Water Pollution: causes of water pollution; international standards for drinking water.
• 12.7 Soil Pollution: pesticides, industrial wastes.
• 12.8 Strategies to control environmental pollution – waste management- collection and disposal
• 12.9 Green chemistry: green chemistry in day-to-day life; dry cleaning of clothes; bleaching of paper; synthesis of chemicals.

Chapter 13 Organic Chemistry-Some Basic Principles and Techniques and Hydrocarbons (25 Periods)

• 13.1 General introduction
• 13.2 Tetravalency of carbon: shapes of organic compounds
• 13.3 Structural representations of organic compounds
• 13.4 Classification of organic compounds
• 13.5 Nomenclature of organic compounds
• 13.6 Isomerism
• 13.7 Fundamental concepts in organic reaction mechanisms
  • 13.7.1 Fission of covalent bond
  • 13.7.2 Nucleophiles and electrophiles
  • 13.7.3 Electron movements in organic reactions
  • 13.7.4 Electron displacement effects in covalent bonds
  • 13.7.5 Types of Organic reactions
• 13.8 Methods of purification of organic compounds
• 13.9 Qualitative elemental analysis of organic compounds
• 13.10 Quantitative elemental analysis of organic compounds

Hydrocarbons

• 13.11 Classification of Hydrocarbons.
• 13.12 Alkanes- nomenclature, isomerism (structural and conformations of ethane only)
  • 13.12.1 Preparation of alkanes
  • 13.12.2 Properties-physical properties and chemical reactivity, substitution reactions- halogenation (a free radical mechanism), combustion, controlled oxidation, isomerization, aromatization, reaction with steam, and Pyrolysis.
• 13.13 Alkenes- Nomenclature, the structure of ethane, Isomerism (structural and geometrical).
  • 13.13.1 Methods of preparation.
  • 13.13.2 Properties: Physical and chemical reactions, the addition of hydrogen, halogen, water, sulphuric acid, Hydrogen halides Mechanism-ionic and peroxide effect, Markovnikov’s anti-Markovnikov’s or Kharasch effect). Oxidation, ozonolysis, and polymerization.
• 13.14 Alkynes- nomenclature and isomerism, the structure of acetylene. Methods of preparation of acetylene.
  • 13.14.1 Physical properties, chemical reactions- the acidic character of acetylene, addition reactions of hydrogen, halogen, hydrogen halides, and water. Polymerization.
• 13.15 Aromatic hydrocarbons: Nomenclature and isomerism. Structure of benzene, resonance, and aromaticity.
  • 13.15.1 Preparation of benzene, Physical properties, Chemical properties: Mechanism of electrophilic substitution. Electrophilic substitution reactions- nitration, sulphonation, halogenation, Friedel-Craft’s alkylation, and acylation.
  • 13.15.2 Directive influence of functional groups in mono substituted benzene, Carcinogenicity and toxicity.

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Chemistry Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Jr Inter 1st Year Chemistry Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(h)

I.

Question 1.
Find the points of local extrema (if any) and local extrema of the following functions each of whose domain is shown against the function.
i) f(x) – x², ∀ x ∈ R.
Solution:
f(x) = x²
f'(x) = 2x ⇒ f”(x) = 2
For maximum or minimum f'(x) = 0
2x = 0
x = 0
Now f”(x) = 2 > 0 ,
∴ f(x) has minimum at x = 0
Point of local minimum x = 0
Local minimum = 0.

ii) f(x) = sin x, [0, 4π)
Solution:
Given f(x) = sinx
⇒ f'(x) = cosx
⇒ f”(x) = -sinx
For max on min,
f'(x) = 0
cos x = 0
⇒ x = \(\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}\)

i) f”(\(\frac{\pi}{2}\)) = -sin\(\frac{\pi}{2}\) = -1 < 0
f(x) = sin \(\frac{\pi}{2}\) = 1
∴ Point of local maximum x = \(\frac{\pi}{2}\)
local maximum – 1

ii) f”(\(\frac{3\pi}{2}\)) = – sin \(\frac{3\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{3\pi}{2}\) = -1
∴ Point of local minimum x = \(\frac{3\pi}{2}\)
local minimum x = -1

iii) f”(\(\frac{5\pi}{2}\)) = – sin \(\frac{5\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{5\pi}{2}\) = 1
∴ Point of local maximum x = \(\frac{5\pi}{2}\)
local maximum = 1

iv) f”(\(\frac{7\pi}{2}\)) = – sin \(\frac{7\pi}{2}\) = -1 > 0
f(x) = sin \(\frac{7\pi}{2}\) = -1
∴ Point of local maximum x = \(\frac{7\pi}{2}\)
local maximum = 1

iii) f (x) = x³ – 6x² + 9x + 15 ∀ x ∈ R.
Solution:
f (x) = 3x² – 12x + 9 and f'(x) = 6x – 12
For maximum or minimum f(x) = 0
⇒ 3x² – 12x + 9 = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1) (x – 3) = 0
⇒ x = 1 or 3
Now f”(1) = 6(1) – 12 = – 6 < 0
∴ f(x) has maximum value at x = 1
Max. valueis f(1)= 1³ – 6(1)² + 9(1) + 15
= 1 – 6 + 9 + 15 = 19

f”(3) = 6(3) – 12 = 18 – 12 = 6 > 0
∴ f(x) has minimum value at x = 3
Min. value is f(3) = 33 – 6.32 + 9.3 + 15
= 27 – 54 + 27+ 15
= 15

iv) f(x) = \(\sqrt{1-x}\) ∀ x ∈ (0, 1)
Solution:

For max. or min. f'(x) = 0

v) f(x) = 1/x² + 2 ∀ x ∈ R
Solution:

∴ For maximum or minimum f (x) = 0
⇒ \(\frac{-2x}{(x^{2}+2)^{2}}\) = 0 ⇒ x = 0
f”(0) = \(\frac{2(0-2)}{(0+2^{3})}=\frac{-4}{8}=\frac{-1}{2}\) < 0
∴ f(x) has max. value at x = 0
Max. value is f(0) = \(\frac{1}{0+2}=\frac{1}{2}\)

vi) f(x) = x²- 3x ∀ x ∈ R
Solution:
f(x) = 3x² – 3 and f”(x) = 6x
∴ For maximum or minimum f(x) = 0
⇒ 3x² – 3 = 0
⇒ x² – 1 = 0
⇒ x = ± 1
Now f”(1) = 6(1) = 6 > 0
∴ f(x) has minimum at x = 1
Minimum value is f(1) = 1³ – 3(1) = -2
f”(-1) = 6(-1) = -6 < 0
∴ f(x) has maximum value at x = -1
Maximum value is f(-1) = (-1)³ – 3(-1)
= -1 + 3 = 2

vii) f(x) = (x -1) (x + 2)² ∀ x ∈ R
Solution:
f(x) = (x – 1) (x + 2)²
f(x) = (x – 1) 2(x + 2) + (x + 2)²
= 2(x – 1) (x + 2) + (x + 2)²
f”(x) = 2(x – 1) + 2(x + 2) + 2(x + 2)
= 2(3x + 3) = 6(x + 1)
∴ For maximum or minimum f'(x) = 0
2(x – 1) (x + 2) + (x + 2)² = 0
(x + 2) [2(x – 1) + (x + 2)] = 0
⇒ (x + 2) (3x) = 0
⇒ x = 0, x = -2
Now f”(0) = 6(0 + 1) = 6 > 0
∴ f(x) has min. value at x = 0
Min. value is f(0) = (0 – 1) (0 + 2)² = -4
f'(-2) = 6 (-2 + 1) = -6 < 0
∴ f(x) has max. value at x = -2
Max. value is f(-2) = (-2 -1) (-2 + 2)² = 0

viii) f(x) = \(\frac{x}{2}+\frac{2}{x}\) ∀ x ∈ (0, ∞)
Solution:
f'(x) = \(\frac{1}{2}-\frac{2}{x^{2}}\) and f”(x) = \(\frac{4}{x^{3}}\)
∴ For max. or min. f'(x) = 0
⇒ \(\frac{1}{2}-\frac{2}{x^{2}}\) = 0 ⇒ x² – 4 = 0 ⇒ x = ± 2
f”(2) = \(\frac{4}{2^{3}}\) = \(\frac{1}{2}\) > 0 (Since x > 0)
∴ f(x) has min. value at x = 2
Min. value is f(2) = \(\frac{2}{2}+\frac{2}{2}\) = 1 + 1 = 2.

ix) f(x) = – (x – 1)³ (x + 1)² ∀ x ∈ R
Solution:
f(x) = -(x – 1)³ (x + 1)² = (1 – x)³ (x + 1)²
f”(x) = (1 – x)³ 2(x + 1) + 3(1 – x)² (-1) (x + 1)²
= (1 – x)² (x + 1) {2(1 – x) – 3(x + 1)}
= (1 – x)² (x + 1) {2 – 2x – 3x – 3}
= (1 – x)² (x + 1) (-1 – 5x)
f”(x) = (1 – x)² (x + 1) (-5) + (1 – x)² (-1 – 5x) + (x + 1) (-1 -5x) 2(1 – x) (-1)
= -5 (1 -x)² (x+ 1) – (1 + 5x) (1 – x)² + (x + 1) (1 + 5x) 2(1 – x)
∴ For maximum or minimum f(x) = 0
(1 – x)² (x + 1) (-1 – 5x) = 0
⇒ x = ± 1 or -1/5
f”(1) = 0 – 0 + 0 ⇒ critical value at x = 0
f”(1 +1)² (-1) = 0 – (1 – 5) + 0 = 16 > 0
∴ f(x) has min. value at x = -1
Min. value = f(-1) = (1 + 1)³ (-1 + 1)² = 0
f”(- \(\frac{1}{5}\)) < o
⇒ f(x) has max. value at x = – \(\frac{1}{5}\)
Min. value is f (-\(\frac{1}{5}\)) = \(\frac{3456}{3125}\)

x) f(x) = x² e^3x ∀ x ∈ R
Solution:
f'(x) = x² e^3x .3 + e^3x. 2x
For maximum or minimum f'(x) = 0
3x² e^3x + 2e^3x. x = 0
x² e^3x (3x + 2) = 0
x = 0, x = \(\frac{-2}{3}\) and e = 0 is not possible
Now f”(x) = 3(x² e^3x. 3 + e^3x 2x)
+ e^3x 2 + 2x e^3x
f”(x) = 9x²e^3x + 6x
e^3x+ 2 e^3x + 6xe^3x
= 9x²e^3x + 12xe^3x + 2e^3x
f”(0) = 2 > 0
∴ Point of local minimum = 0
local minimum = 0
f”(\(\frac{-2}{3}\)) = \(\frac{-2}{e^{2}}\) < 0
∴ point of local maximum = \(\frac{-2}{3}\)
local maximum = \(\frac{4}{9e^{2}}\)

Question 2.
Prove that the following functions do not have absoute maximum and absolute minimum.
i) e^x in R
Solution:
f(x) = e^x and f”(x) = e^x
∴ For maxima or minima f'(x) = 0 ⇒ e^x = 0
⇒ x value is not defined
Hence it has no maxima or minima.

ii) log x in (0, ∞)
Solution:
f(x) = \(\frac{1}{x}\) and f”(x) = – \(\frac{1}{x^{2}}\)
f (x) = 0 ⇒ x value is not defined
⇒ f(x) has no maxima or minima.

iii) x³ + x² + x + 1 in R
Solution:
f (x) = 3x² + 2x + 1 gives imaginary values.
⇒ It has no maximum or minimum values.

II.

Question 1.
Find the absolute maximum value and absoulte minimum value of the following functions on the domain specified against the function.
Solution:
f(x) = x³ on (-2, 2)
f(x) = 3x² and f'(x) – 6x
the value f(-2) = (-2)³ = – 8
Max Value f(2) = 2³ = 8

ii) f(x) = (x – 1)²+ 3 on [-3, 1]
Solution:
f(x) = 2(x – 1) and f'(x) = 2
Max. value f(-3) = (-3 – 1)² + 3 = 16 + 3 = 19
Min. value f(l) = 0 + 3 = 3

iii) f(x) = 2|x| on [-1, 6]
Solution:
f'(x) = \(\frac{2|x|}{x}\)
For max. or min., f'(x) = 0
\(\frac{2|x|}{x}\) = 0 ⇒ x = 0
f(0) = 0
f(-1) – 2|-1| = 2
f(6) = 2(6) = 12
Absolute minimum = 0
Absolute maximum = 12

iv) f(x) = sin x + cos x on [0, π]
Solution:
f'(x) cos x – sin x which exists at all x ∈ (0, π)
Now f'(x) = 0 ⇒ cos x – sin x = 0
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\) ∈ (0, π)
Now f(0) = sin 0 + cos 0 = 1
f(\(\frac{\pi}{4}\)) = sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
f(π) = sin π + cos π = 0 – 1 = -1
∴ Minimum value : -1
Maximum value: √2

v) f(x) = x + sin 2x on [0, π]
Solution:
f(x) = x + sin 2x
f(x) = 1 + 2 cos 2x
f (x) = 0 ⇒ 2 cos 2x + 1 = 0
⇒ cos 2x = –\(\frac{1}{2}\) = cos \(\frac{2 \pi}{3}\)
⇒ 2x = \(\frac{2 \pi}{3}\)
⇒ x = \(\frac{\pi}{3}\) ∈ (0, 2π)
Now f(0) = 0 + sin 2(0) = 0
f(\(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\) + sin 2.\(\frac{\pi}{3}\) = \(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)
f(2π) = 2π + sin 2. 2π = 2π + 0 = 2π
Minimum value = 0
Maximum value is = 2π

Question 2.
Use the first derivative test to find local extrema of f(x) = x³ – 12x on R.
Solution:
f(x) = x³ – 12x
f'(x) = 3x² – 12
f”(x) = 6x
For maximum or minimum, f'(x) = 0
3x² – 12 = 0
3x² = 12
x = ± 2
f”(2) = 12 > 0
Point of local minimum at x = 2
Local minimum = – 16
f”(-2) = -12 < 0
Point of local maximum at x = -1
Local maximum =16

Question 3.
Use the first derivative test to find local extrema of f(x) = x² – 6x + 8 on R.
Solution:
f(x) = x² – 6x + 8
f’(x) = 2x -6 ⇒ f”(x) = 2
For maximum or minimum f'(x) = 0
2x – 6 = 0
⇒ x = 3
f”(3) = 2 > 0
∴ Point of local minimum x = 3
Local minimum = -1

Question 4.
Use the second derivative test to find local extrema of the function
f(x) =x³ – 9x² – 48x + 72 on R.
Solution:
f(x) =x³ – 9x² – 48x + 72.
⇒ f'(x) = 3x² – 18x – 48
= 3(x – 8) (x + 2)
Thus the stationary point are – 2 & 8
f”(x) = 6x – 18 = 6(x – 3)
At x= 8, f”(8) = 30 > 0
∴ f (8) = (8)³ – 9(8)² – 48(8) + 72
= 512 – 576 – 384 + 72
= – 376
At x= -2, f”(-2) = – 30 < 0
f(-2) = (-2)³ – 9(-2)² – 48(-2)+72
= -8 – 36 + 96 + 72
= 124
Local minimum = -376
Local maximum = 124

Question 5.
Use the second derivative test to find local extrema of the function. f(x) = -x³ + 12x² – 5 on R.
Solution:
f(x) = -x³ + 12x² – 5
⇒ f'(x) = -3x² + 24x
= -3x(x – 8)
Thus the stationary point are 0, 8
f”(x) = – 6x + 24
At x= 0, f”(0) = 24 > 0.
f(0) = -5
At x= 8, f”(8) = -24 < 0
f(8) = -8³ + 12(8)² – 5
= -512 + 768 – 5 = 251
Local minimum = -5
Local maximum = 251

Question 6.
Find local maximum or local minimum of f(x) = -sin 2x – x defined on [-π/2, π/2].
Solution:
f(x) =-sin 2x – x
f'(x) = -2cos 2x – 1
f”(x) = 4 sin 2x
Thus the starting point are x = \(\frac{\pi}{3},\frac{-\pi}{3}\)

Local minimum = –\(\frac{\sqrt{3}}{2}-\frac{\pi}{3}\)
Local maximum = \(\frac{\sqrt{3}}{2}+\frac{\pi}{3}\)

Question 7.
Find the absolute maximum and absolute minimum of f (x) = 2x³ – 3x² – 36x + 2 on the interval [0, 5].
Solution:
f(x) = 2x³ – 3x² – 36x + 2
f(x) = 6x² – 6x – 36
f(x) = 12x – 6
for maxima or minima, f'(x) = 0
6x² – 6x – 36 = 0
x² – x – 6 = 0
x² -3x + 2x – 6 = 0
x(x – 3) + 2(x – 3) = 0
(x + 2) (x – 3) = 0
x = 3, -2
f”(3) = 30 > 0
f(x) has max/min value at x = 3
f(3) = 2(3)³ – 3(3)² – 36(3) + 2
= 54 – 27 – 108 + 2
= -79
Absolute minimum = – 79
Since 0 ≤ x ≤ 5
∴ f(0) = 0 – 0 – 0 + 2
= 2
∴ Absolute maximum = 2.

Question 8.
Find the absolute extremum of f(x) = 4x – \(\frac{x^{2}}{2}\) on [-2, \(\frac{9}{2}\)]
Solution:
f(x) = 4x – \(\frac{x^{2}}{2}\)
f'(x) =4 – x
f”(x) = – 1
for maxima or minima f'(x) = 0
4 – x = 0
x = 4
f”(4) = -1 < 0
∴ f has maximum value at x = 4
f(4) = 16 – \(\frac{16}{2}\) = 8
Since -2 ≤ x ≤ \(\frac{9}{2}\)
∴ f(-2) = -8 – \(\frac{4}{2}\)
= -8- 2 = -10
∴ Absolute minimum = -10
Absolute maximum = 8

Question 9.
Find the maximum profit that a company can make, if the profit function is given by P(x) = -41 + 72x – 18x²
Solution:
P(x) = – 41 + 72 x – 18x².
\(\frac{dp(x)}{dx}\) = 72 – 36x
for maxima or minima, \(\frac{dp}{dx}\) = 0
72 – 36x = 0
x = 2
\(\frac{d^{2}p}{dx^{2}}\) = -36 < 0
∴ The profit f(x) is maximum for x = 2
The maximum profit will be P(2) =
= -41 + 72(2) – 18(4)
= 31

Question 10.
The profit function P(x) of a company selling x items is given by P(x) = -x³ + 9x² – 15x – 13 where x represents thousands of units. Find the absolute maximum profits if the company can manufacture a maximum of 6000 units.
Solution:
P(x) = -x³ + 9x² – 15x – 13
\(\frac{dp(x)}{dx}\) = -3x² + 18x – 15
for maximum or minimum \(\frac{dp}{dx}\) = 0
-3x² + 18x – 15 = 0
x² – 6x + 5 = 0
x² – 5x – x + 5 = 0
x(x- 5) – 1(x- 5) = 0
(x – 1) (x – 5) = 0
x = 1, 5
P(1) = -1 + 9 – 15 – 13 = -10
P(5) = -125 + 225 – 75 – 13 = 12
∴ Maximum profit = 12.

III.

Question 1.
The profit function P(x) of a company selling x items per day is given by P(x) = (150 – x) x – 1000. Find the number of items that the company should manufacture to get maximum profit. Also find the maximum profit.
Solution:
Given that tbe profit function
P(x) = (150 – x)x -1000
for maximum or minimum \(\frac{dp}{dx}\) = 0
(150 – x(1) -x (-1) = 0
150 – 2x = 0
x = 75
Now \(\frac{d^{2}p}{dx^{2}}\) = -2 < 0
∴ The profit P(x) is maximum for x = 75
The company should sell 75 tems a day the maxma profit will be P (75) = 4625.

Question 2.
Find the absolute maximum and absolute minimum of f(x) = 8x³ + 81x² – 42x – 8 on [-8, 2].
Solution:
f(x) = 8x³ + 81x² – 42x – 8
f'(x) = 24x²+ 162x – 42
For maximum or minimum, f'(x) = 0
24x² + 162x – 42 = 0
4x² + 27x – 7 = 0
4x² + 28x – x – 7 = 0
4x(x + 7) – 1(x + 7) = 0
(x + 7) (4x – 1) = 0
x = – 7 or \(\frac{1}{4}\)

f(- 8) = 8(-8)³ + 81(-8)² – 42(-8) -8
= – 8(512) + 81(64) + 336 – 8
= -4096 + 5184 + 336 – 8
= 5520 – 4104
= 1416

f(2) = 8(2)³ + 81 (2)² – 42(2) – 8
= 64 + 324 – 84 – 8
= 296

f(-7) = 1246
Absolute maximum = 1416
Absolute minimum = \(\frac{-216}{16}\)

Question 3.
Find two positive integers whose sum is 16 and the sum of whose squares is minimum.
Solution:
Suppose x and y are the sum value
x + y =16
⇒ y = 16 – x
f(x) = x² + y² = x² + (16 – x)²
= x² + 256 + x² – 32x
f'(x) = 4x – 32
for maximum or minimum f'(x) = 0
⇒ 4x – 32 = 0
4x = 32 x = 8
f”(x) = 4 > 0
∴ f(x) is minimum when x = 8
y = 16 – x = 16 – 8 = 9
∴ Required number are 8, 8.

Question 4.
Find two positive integers x and y such that x + y = 60 and xy3 is maximum.
Solution:
x + y = 60 ⇒ y = 60 – x ………… (1)
Let p = xy³ = x(60 – x)³
\(\frac{dp}{dx}\) = x³(60 – x)²(-1) + (60-x)³
= -3x (60 – x)² + (60 – x)³
= (60 – x)² [-3x + 60 – x]
= (60 – x)² (60 – 4x) = 4(60 – x)² (15 – x)

\(\frac{d^{2}p}{dx^{2}}\) = 4[(60-x)² (-1) + (15-x) 2(60-x) (-1)]
= 4(60 – x) [-60 + x – 30 + 2x]
= 4(60 – x) (3x – 90)
= 12(60 – x) (x – 30)
For maximum or minimum \(\frac{dp}{dx}\) = 0
⇒ 4(60 – x)2 (15 -x) = 0
⇒ x = 60 or x = 15 ; x cannot be 60
∴ x = 15 ⇒ y = 60 – 15 = 45.
(\(\frac{d^{2}p}{dx^{2}}\))_{x = 15} = 12(60 – 15) (15 – 30) < 0
⇒ p is maximum
∴ Required numbers are 15, 45.

Question 5.
From a rectangular sheet of dimensions 30 cm x 80 cm four equal squares of side x cm are removed at the comers and the sides are then turned up so as to form an open rectangular box. Find the value of x, so that the volume of the box is the greatest?
Solution:
Length of the box = 80 – 2x = l
Breadth of the box = 30 – 2x = b

Height of the box = x = h
Volume = lbh = (80 – 2x) (30 – 2x). x
= x (2400 – 220 x + 4x²)
f(x) = 4x³ – 220 x² + 2400x
f'(x) = 12x² – 440x + 2400
= 4 [3x² – 110 x + 600]
f'(x) = 0
⇒ 3x² – 110 x +600 = 0

f(x) is maximum when x = \(\frac{20}{3}\)
Volume of the box is maximum when x = \(\frac{20}{3}\) cm

Question 6.
A window is in the shape of a rectangle surmounted by a semi-circle. If the peri-meter of the window is 20 feet, find the maximum area.
Solution:
Let length of the rectangle be 2x and breadth be y so that radius of the semi-circle is x.

Perimeter = 2x + 2y + π. x = 20
2y = 20 – 2x – πx.
y = 10 – x – \(\frac{\pi}{2}\). x
Area = 2xy + \(\frac{\pi}{2}\). x²
= 2x(10 – x – \(\frac{\pi}{2}\)) + \(\frac{\pi}{2}\)x²
= 20x – 2x² – πx² + \(\frac{\pi}{2}\) x²
f(x) = 20x – 2x² – \(\frac{\pi}{2}\) x²
f'(x) = 0 ⇒ 20 – 4x – πx = 0
(π + 4)x = 20
x = \(\frac{20}{\pi+4}\)
f”(x) = -4 – π < 0
f(x) is a maximum when x = \(\frac{20}{\pi+4}\)

Question 7.
If the curved surface of right circular a cylinder inscribed in a sphere of radius r is maximum, show that the height of the cylinder is √2r.
Solution:
Suppose r is the radius and h be the height of the cylinder.

From ∆OAB, OA² + AB² = OB²
r² + \(\frac{h^{2}}{4}\) = R² ; r² = R² – \(\frac{h^{2}}{4}\)
Curved surface area = 2πrh

f(h) is greatest when h = √2 R
i.e., Height of the cylinder = √2 R.

Question 8.
A wire of length l is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of the pieces of the wire respe-ctively so that the sum of the areas is the least?
Solution:
Suppose x is the side of the square and r is the radius of the circle.

Given 4x + 2πr = l
4x = l – 2πr
r = \(\frac{1-2 \pi r}{4}\)
Sum of the area = x² + πr²

∴ f(r) is least when r = \(\frac{l}{ 2(\pi+4)}\)
Sum of the area is least when the wire is cut into pieces of length \(\frac{\pi l}{\pi+4}\) and \(\frac{4l}{\pi+4}\)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g)

I.

Question 1.
Without using the derivative, show that
i) The function f(x) = 3x + 7 is strictly increasing on R.
Solution:
Let x₁, x₂ ∈ R with x₁ < x₂
Then 3x₁ < 3x₂
Adding 7 on both sides
3x₁ + 7 < 3x₂ + 7
⇒ f(x₁) < f(x₂)
∴ x₁ < x₂ ⇒ f(x₁) < f(x₂) ∀ x₁ x₂ ∈ R
∴ The given function is strictly increasing on R

ii) The function f(x) = (\(\frac{1}{2}\))^x is strictly decreasing on R.
Solution:
f(x) = (\(\frac{1}{2}\))^x
Let x₁, x₂ ∈ R
Such that x₁ < x₂
⇒ (\(\frac{1}{2}\))^x₁ > (\(\frac{1}{2}\))^x₂
⇒ f(x₁) > f(x₂)
∴ f(x) is strictly decreasing on R.

iii) The function f(x) = e^3x is strictly increasing on R.
Solution:
f(x) = e^3x
Let x₁, x₂ ∈ R such that x₁ < x₂
We know that of a > b then e^a > e^b
Then e^3x < e^3x₂
⇒ f(x₁) < f(x₂)
∴ f is strictly increasing on R.

iv) The function f(x) = 5. – 7x is strictly decreasing on R.
Solution:
f(x) = 5 – 7x
Let x₁ x₂ ∈ R
Such that x₁ < x₂
Then 7x₁ < 7x₂
-7x₁ > -7x₂
Adding 5 on bothsides
5 – 7x₁ > 5 – 7x₂
f(x₁) > (x₂)
∴ x₁ < x₂ ⇒ f(x₁) > f(x₂) V x₁ x₂ ∈ R.
The given function f is strictly decreasing on R.

Question 2.
Show that the function f(x) = sin x. Define on R is neither increasing nor decreasing on (0, π).
Solution:
f(x) = sin x
Since 0 < x < n
Consider 0 < x
f(0) < f(x)
sin 0 < sin x
0 < sin x ……….. (1)
Consider x < n
f(x) < f(π)
sin x < sin π 0 > sin x ………….. (2)
From (1) & (2); f(x) is neither increasing nor decreasing.

II.

Question 1.
Find the intervals in which the following functions are strictly increasing or strictly decreasing.
i) x² + 2x – 5
Solution:
Let f(x) = x² + 2x – 5
f'(x) = 2x + 2
f(x) is increasing if f'(x) > 0
⇒ 2x + 2 < 0 ⇒ x+ 1 > 0
x > -1
f(x) is increasing If x ∈ (-1, ∞)
f(x) is decreasing If f'(x) < 0
⇒ 2x + 2 < 0
⇒ x + 1 < 0
⇒ x < -1 f(x) is decreasing if x ∈ (-∞, -1).

ii) 6 – 9x – x².
Solution:
Let f(x) = 6 – 9x – x²
f'(x) = -9 – 2x
f(x) is increasing if f'(x) > 0
⇒ -9 -2x > 0
⇒ 2x + 9 < 0
x < \(\frac{-9}{2}\)
f(x) is increasing if x ∈ (-∞, \(\frac{-9}{2}\))
f(x) is decreasing if f'(x) < 0
⇒ 2x + 9 > 0
⇒ x > \(\frac{-9}{2}\)
f(x) is decreasing of x ∈ (\(\frac{-9}{2}\), ∞)

iii) (x + 1)³ (x – 1)³.
Solution:
Let f(x) = (x + 1)³ (x – 1)³
= (x² – 1)³
x⁶ – 1 – 3x⁴ + 3x ²
f'(x) = 6x⁵ – 12x³ + 6x
= 6(x⁵ – 2x³ + x)
= 6x(x⁴ – 2x² +1)
= 6x(x² – 1)²
f'(x) ≤ 0
⇒ 6x(x² – 1)² < 0
f(x) is decreasing when (-∞, -1) ∪ (-1, 0)
f'(x) > 0
f(x) is increasing when (0, 1) ∪ (1, ∞)

iv) x³(x – 2)²
Solution:
f'(x) = x³. 2(x – 2) + (x – 2)².3x²
= x² (x – 2) [2x + 3 (x- 2)]
= x² (x – 2) (2x + 3x – 6)
= x² (x – 2) (5x – 6) ∀ x ∈ R, x² ≥ 0
For increasing, f'(x) = 0
x²(x – 2) (5x – 6) > 0
x ∈ (-∞, \(\frac{6}{5}\)) ∪ (2, ∞)
For decreasing, f'(x) < 0
x²(x – 2) (5x – 6) < 0
x ∈ (\(\frac{6}{5}\), 2)

v) xe^x
Solution:
f'(x) = x . e^x + e^x. 1 = e^x(x + 1)
e^x is positive for all real values of x
f'(x)>0 ⇒ x + 1 > 6 ⇒ x > – 1
f(x) is increasing when x > – 1
f(x) < 0 ⇒ x + 1 < 0 ⇒ x < -1
f(x) is decreasing when x < – 1

vi) \(\sqrt{(25-4x^{2})}\)
Solution:
f(x) is real only when 25 – 4x² > 0
-(4x² – 25) > 0
-(2x + 5) (2x – 5) > 0
∴ x lies between –\(\frac{5}{2}\) and \(\frac{5}{2}\)
Domain of f = (-\(\frac{5}{2}\), \(\frac{5}{2}\))
f'(x) = \(\frac{1}{2 \sqrt{25-4 x^{2}}}\) (-8x)
= –\(\frac{4x}{\sqrt{25-4 x^{2}}}\)
f(x) is increasing when f'(x) > 0
⇒ \(\frac{-4x}{\sqrt{25-4 x^{2}}}\) > 0
i.e., x < o
f(x) is increasing when (-\(\frac{5}{2}\), 0)
f(x) is decreasing when f'(x) < 0
⇒ –\(\frac{4x}{\sqrt{25-4 x^{2}}}\) < 0
∴ x > 0
f(x) is decreasing when (0, \(\frac{5}{2}\)).

vii) ln (ln(x)); x > 1.
Solution:
f'(x) = –\(\frac{1}{lnx}•\frac{1}{x}\)
f(x) is decreasing when f'(x) > 0
\(\frac{1}{x.ln x}\) >0
⇒ x. In x > 0
ln x is real only when x > 0
∴ ln x < 0 = ln 1
i.e., x > 1
f(x) is increasing when x > 1 i.e., in (1, ∞)
f(x) is decreasing when f'(x) < 0 ⇒ ln x > 0 = ln 1
i.e., x < 1
f(x) is decreasing in (0, 1)

viii) x³ + 3x² – 6x + 12.
Solution:
f(x) = x³ + 3x² – 6x + 12
f(x) = 3x² + 6x – 6
= 3(x² + 2x – 2)
= 3((x + 1)² – 3)
= 3[(x + 1) + √3] [(x + 1) – √3]
= 3(x + (1 + √3) (x + (1 – √3 )
f (x) < 0
⇒ x = -(1+ √3 ) or -(1 – √3 )
x = -1 – √3 or √3 – 1
f(x) is decreasing in (-1 -√3 , √3 -1)
f (x) > 0
f(x) increasing when (-∞, -1 – √3 ) ∪ (√3 – 1, ∞)

Question 2.
Show that f(x) = cos²x is strictly increasing on (0, π/2).
Solution:
f(x) = cos² X
⇒ f(x) = 2 cos x (-sin x)
= -2 sin x cos x
= -sin 2x
Since 0 < x < \(\frac{\pi}{2}\)
⇒ 0 < 2x < π
Since ‘sin x’ is +ve between 0 and π
∴ f(x) is clearly -ve.
∴ f'(x) < 0
∴ f(x) is strictly decreasing.

Question 3.
Show that x + \(\frac{1}{x}\) is increasing on [1, ∞)
Solution:
Let f(x) = x + \(\frac{1}{x}\)
f'(x) = 1 – \(\frac{1}{x^{2}}\) = \(\frac{x^{2}-1}{x^{2}}\)
Since x ∈ [1, ∞) = \(\frac{x^{2}-1}{x^{2}}\) > 0
∴ f'(x) > 0
∴ f(x) is increasing.

Question 4.
Show that \(\frac{x}{1+x}\) < ln (1 + x) < x ∀ x > 0
Solution:
Let f(x) = ln(1 + x)- \(\frac{x}{1+x}\)
= ln(1 + x) – \(\frac{1+x-1}{1+x}\)
= ln(1 + x) – 1 + \(\frac{1}{1+x}\)
f'(x) = \(\frac{1}{1+x}\) – \(\frac{1}{(1+x)^{2}}\)
= \(\frac{1+x-1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\) > 0 since x > 0
f(x) is increasing when x > 0
∴ f(x) > f(0)
f(0) = ln 1 – \(\frac{0}{1+0}\) = 0 – 0 = 0
Since xe [1, ∞) =
ln (1 + x) – \(\frac{x}{1+x}\) > 0
⇒ ln (1 + x) > \(\frac{x}{1+x}\) ……… (1)
Let g(x) = x – ln (1 + x)
g'(x) = 1 – \(\frac{x}{1+x}=\frac{1+x-1}{1+x}\)
= \(\frac{x}{1+x}\) > 0 since x > 0
g(x) is increasing when x > 0
i.e., g(x) > g(0)
g(0) = 0 – ln (1) = 0 – 0 = 0
∴ x – ln (1 + x) > 0
x > ln(1 + x) ………….. (2)
From (1), (2) we get
\(\frac{x}{1+x}\) < ln (1 + x) < x ∀ x > 0

III.

Question 1.
Show that \(\frac{x}{1+x^{2}}\) < tan^-1 x < x when x > 0.
Solution:
Let f(x) = tan^-1 x – \(\frac{x}{1+x^{2}}\)

f(x) is increasing when x > 0
f(x) > f(0)
But f(0) = tan^-1 0 – 0 = 0 – 0 = 0
i.e., f(x) > 0

g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – tan^-1 0 = 0 – 0 = 0
∴ x – tan^-1 x > 0
⇒ x > tan^-1 x ………. (2)
From (1), (2) we get
\(\frac{x}{1+x^{2}}\) <tan^-1 x< x for x > 0

Question 2.
Show that tan x > x for all (0, \(\frac{\pi}{2}\))
Solution:
Let f(x) = tan x – x
f'(x) = sec² x – 1 > 0 for every
x ∈ (0, \(\frac{\pi}{2}\))
f(x) is increasing for every x ∈ (0, \(\frac{\pi}{2}\))
i.e., f(x) > f(0)
f(0) = tan 0 – 0 = 0 – 0 = 0
∴ tan x – x > 0
⇒ tan x > x for every x ∈ (0, \(\frac{\pi}{2}\))

Question 3.
If x ∈ (0, \(\frac{\pi}{2}\)) then show that \(\frac{2x}{\pi}\) < sin x < x.
Solution:
Let f(x) = x – sin x
f(x) = 1- cos x > 0 for every x ∈ (0, \(\frac{\pi}{2}\))
f(x) is increasing for every x ∈ (0, \(\frac{\pi}{2}\))
⇒ f(x) > f(0)
f(0) = 0 – sin 0 = 0 – 0 = 0
∴ x – sin x > 0
⇒ x > sin x ………….. (1)
Let g(x) = sin x – \(\frac{2x}{\pi}\)
g'(x) = cos x – \(\frac{2}{\pi}\) > 0 for every x ∈ (0, \(\frac{\pi}{2}\))
g(x) is increasing in (0, \(\frac{\pi}{2}\))
g(x) > g(0)
g(0) = sin 0 – 0 = 0 – 0 = 0
∴ sin x – \(\frac{2x}{\pi}\) > 0
⇒ sin x > \(\frac{2x}{\pi}\) ………… (2)
From (1), (2) we get
\(\frac{2x}{\pi}\) < sin x < x for every x ∈ (0, \(\frac{\pi}{2}\))

Question 4.
If x e (0,1) then show that 2x < ln [latex]\frac{(1+x)}{(x-1)}[/latex] < 2x [1 + \(\frac{x^{2}}{2(1+x^{2})}\)] Solution:
Let f(x) = ln \(\frac{(1+x)}{1-x}\) – 2x
= ln (1 + x) – ln (1 – x) – 2x

f(x) is increasing ih (0, 1)
i.e., x > 0 ⇒ f(x) > f(0)
f(0) = ln 1 – 0 = 0 – 0 = 0


g(x) is increasing when x > 0
g(x) > g(0)
g(0) = 0 – ln 1 = 0 – 0 = 0

for x ∈ (0,1)

Question 5.
At what point the slopes of the tangents y = \(\frac{x^{3}}{6}-\frac{3x^{3}}{2}+\frac{11x}{2}\) + 12 increases?
Solution:
Equation of the curve is
y = \(\frac{x^{3}}{6}-\frac{3}{2}x^{2}+\frac{11x}{2}\) + 12

Slope = m = \(\frac{x^{2}}{c}-3x+\frac{11}{2}\)
\(\frac{dm}{dx}\) = \(\frac{2x}{2}\) -3 = x – 3
Slope increases ⇒ m > 0
x – 3 > 0
x > 3
The slope increases in (3, ∝)

Question 6.
Show that the functions ln \(\frac{(1+x)}{x}\) and \(\frac{x}{(1+x)ln(1+x)}\) are decrasing on (0, ∞).
Solution:
i)


∴ f(x) is decreasing for x ∈ (0, ∝)

ii) let f(x) = \(\frac{x}{(1+x)ln(1+x)}\)

∴ f(x) is decreasing for x ∈ (0, ∝)

Question 7.
Find the intervals in which the function f (x) = x3 – 3×2 + 4 is strictly increasing all x e R.
Solution:
f(x) = x³ – 3x² + 4
f'(x) = 3x² – 6x
f(x) is increasing if f'(x) > 0
3x² – 6x > 0
3x(x – 2) > 0
(x – 0)(x – 2) > 0
f(x) is increasing if x £ (-∞, 0) u (0, ∞)
f(x) is decreasing if f'(x) < 0
(x – 0) (x – 2) < 0
x ∈ (0, 2)

Question 8.
Find the intervals in which the function f(x) = sin⁴x + cos⁴x ∀ x ∈ [0, \(\frac{\pi}{2}\)] is increasing and decreasing.
Solution:
f(x) = sin⁴x + cos⁴x
f(x) = (sin²x)² + (cos²x)²
= (sin²x + cos²x)² – 2sin²x cos²x
= 1 – \(\frac{1}{2}\) sin² 2x
f'(x) = \(\frac{-1}{2}\) 2sin 2x. cos 2x(2)
= -2 sin 2x. cos 2x
= -sin 4x
Let 0 < x < \(\frac{\pi}{4}\)
∴ f(x) is decreasing if f'(x) < 0
⇒ -sinx < 0 ⇒ sinx > 0
∴ x ∈ (0, \(\frac{\pi}{4}\))
f(x) is increasing if f'(x) > 0
⇒ – sinx > 0
⇒ sinx < 0
∴ x ∈ (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

I.

Question 1.
Verify Rolle’s theorem for the following functions.
i) x² – 1 on [-1, 1]
Solution:
Let f(x) = x² – 1
f is continuous on [-1,1]
since f(-1) = f(1) = 0 and
f is differentiable on [-1, 1]
∴ By Rolle’s theorem ∃ c ∈ (-1, 1) such that f'(c) = 0
f'(x) = 2x = 0
∴ = f'(c) = 0
2c = 0
c = 0
The point c = 0 ∈ (-1, 1)
Then Rolle’s theorem is verified.

ii) sin x – sin 2x on [0, π].
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π]
since f(0) = f(π)= 0 and
f is differentiable on [0, π]
By Rolle’s theorem ∃ c ∈ (0, π)
such that f'(c) = 0
f'(x) = cos x – 2 cos 2x
f'(c) = 0 ⇒ cosc – 2 cos 2c = 0
⇒ cosc – 2(2cos²c – 1):
cosc – 4 cos²c + 2=0 2
4 cos² c – cosc – 2 = 0

iii) log (x² + 2) – log 3 on [-1, 1]
Solution:
Let f(x) = log (x² + 2) – log 3
f is continuous on [-1, 1]
Since f(-1) = f(1) = 0 and f is
Differentable on [-1, 1]
By Rolle’s theorem ∃ c ∈ (-1, 1)
Such that f'(c) = 0
f'(x) = \(\frac{1}{x^{2}+2}\) = (2x)
f'(x) = \(\frac{2c}{c^{2}+2}\) = 0
2c = 0
c = 0
c = 0 ∈ (-1, 1).

Question 2.
It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2t + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Solution:
Given f(x) = x³ + bx² + ax
f'(x) = 3x² + 2bx + a
∴ f'(c) = 0 6 ⇔ 3c² + 2bc + a = 0

⇔ b = 6 and b² – 3a = 3
36 – 3 = 3a
33 = 3a
a = 11
Hence a = 11 and b = -6.

Question 3.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1].
Solution:
Clearly f(0) = f(c)
0 – 0 + k = 1 – 3 + k
0 = -2
Which is not possible
∴ There is no real number K.

Question 4.
Find a point on the graph of the curve y = (x – 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).
Solution:
Given points (3, 0) and (4, 1)
The slope of chord = \(\frac{1-0}{4-3}\) = 1
Given y = (x- 3)²
\(\frac{dy}{dx}\) = 2(x – 3)
⇒ Slope = 2(x – 3)
1 = 2(x – 3)
\(\frac{1}{2}\) = x – 3
x = \(\frac{1}{2}\) + 3 = \(\frac{7}{2}\)
y = (x – 3)² = (\(\frac{7}{2}\) – 3)² = \(\frac{1}{4}\)
∴ The point on the curve is (\(\frac{7}{2}\), \(\frac{1}{4}\))

Question 5.
Find a point on the graph of the curve y = x³, where the tangent is parallel to the chord joining (1, 1) and (3, 27).
Solution:
Given points (1, 1) and (3, 27)
Slope of chord = \(\frac{27-1}{3-1}\) = 13
Given y = x³
\(\frac{dy}{dx}\) = 3x²
⇒ Slope = 3x²
13 = 3x²

∴ The point on the curve is (\(\frac{\sqrt{39}}{3}\), \(\frac{13\sqrt{39}}{9}\))

Question 6.
Find ‘c’, so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases.
i) f(x) = x² – 3x – 1, a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\).
Solution:

f'(x) = 2x – 3
f'(c) = 2c – 3
Given f'(c)= \(\frac{f(b)-f(a)}{b-a}\)

ii) f(x) = e^x ; a = 0, b = 1
Solution:
f(b) = f(1) = e’ = e
f(a) = f(0) = e° = 1
Given f(x) = e^x
f'(x) = e^x
Given condition f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

Question 7.
Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [-1, 2]. Find the point in the interval where the derivate vanishes.
Solution:
Let f(x) = (x² – 1) (x- 2) = x³ – 2x² – x + 2
f is continous on [-1, 2]
since f(-1) = f(2) = 0 and f is
Differentiable on [-1, 2]
By Rolle’s theorem ∃ C ∈ (-1, 2)
Let f'(c) = 0
f'(x) = 3x² – 4x – 1
3c² – 4c – 1 = 0

Question 8.
Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem
i) x² -1 on [2, 3]
Solution:
Solution:
Let f(x) = x² – 1
f is continous on [2, 3]
and f is differentiable
Given f(x) = x² – 1
f'(x) = 2x
By Lagrange’s mean value theorem ∃ C ∈ (2, 3) such there

ii) sin x – sin 2x on [0, π]
Solution:
Let f(x) = sin x – sin 2x
f is continuous on [0, π] and f is differentiable
Given f(x) = sin x – sin 2x
f'(x) = cos x – 2 cos 2x
By Lagrange’s mean value than ∃ C ∈ (0, π) such there
f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)
cosc – 2 cos 2c = 0
cosc 2(2cos² – 1) = 0
cosc – 4 cos²c + 2 = 0
4 cos² c – cos c – 2 = 0

iii) log x on [1, 2].
Solution:
Let f(x) = log x
f is continuous on [1, 2] and f is differentiable
Given f(x) = log x
f'(x) = \(\frac{1}{x}\)
By Lagrange’s mean value theorem ∃ c ∈ (1, 2) such that

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e)

I.

Question 1.
At time t, the distance s of a particle moving in a straight line is given by s = -4t² + 2t. Find the average velocity between t = 2 sec and t = 8 sec.
Solution:
s = -4t² + 2t ds
v = \(\frac{ds}{dt}\) = -8t + 2 dt
Velocity at t = 2 is v = (\(\frac{ds}{dt}\))_t=2
v = -16 + 2 = -14 units/sec.
Velocity at t = 8 is v = (\(\frac{ds}{dt}\))_t=8
v = -64 + 2 = -62
Average velocity = \(\frac{-62-14}{2}\) = -38 units/sec.

Question 2.
If y = x⁴ then find the average rate of change of y between x = 2 and x = 4.
Solution:
y = x⁴ ⇒ \(\frac{dy}{dt}\) = 4x³
(\(\frac{dy}{dt}\))_x=2 = 32
(\(\frac{dy}{dt}\))_x=4 = 256
Average rate of change = \(\frac{256+32}{2}\) = 144.

Question 3.
A particle moving along a straight line has the relation s = t³ + 2t + 3, connecting the distance s describe by the particle in time t. Find the velocity and acceleration of the particle of t = 4 sec.
Solution:
s = t³ + 2t + 3
\(\frac{ds}{dt}\) = 3t² + 2, velocity v = \(\frac{ds}{dt}\) = 3t² + 2
Velocity at t = 4
⇒ (\(\frac{ds}{dt}\))_t=4 = 48 + 2 = 50 units/sec
v = 3t² + 2
\(\frac{dv}{dt}\) = 6t ⇒ a = (\(\frac{dv}{dt}\))_t=4 = 24 units/sec².

Question 4.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and where the velocity is zero.
Solution:
Given s = t³ – 9t² + 24t – 18
v = \(\frac{ds}{dt}\) = 3t² – 18t + 24
v = 0 ⇒ 3(t² – 6t + 8) = 0
∴ (t – 2) (t – 4) = 0
∴ t = 2 or 4
The velocity is zero after 2 and 4 seconds.

Case (i):
t = 2
s = t³ – 9t² + 24t – 18
= 8 – 36 + 48 – 18 = 56 – 54 = 2

Case (ii) :
t = 4 ; s = t³ – 9t² + 24t – 18
= 64 – 144 + 96 – 18
= 160 – 162 = -2
The particle is at a distance of 2 units from the starting point ‘O’ on either side.

Question 5.
The displacement s of a particle travelling in a straight line in t seconds is given by s = 45t + 11t² – t³. Find the time when the particle comes to rest.
Solution:
s = 45t + 11t² – t³
v = \(\frac{ds}{dt}\) = 45 + 22t – 3t²
If a particle becomes to rest
⇒ v = 0 ⇒ 45 + 22t – 3t² = 0
⇒ 3t² – 22t – 45 = 0
⇒ 3t² – 27t + 5t – 45 = 0
⇒ (3t + 5) (t – 9) = 0
∴ t = 9 or t = – \(\frac{5}{3}\)
∴ t = 9
∴ The particle becomes to rest at t = 9 seconds.

II.

Question 1.
The volume of a cube is increasing at the rate of 8 cm³/sec. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Suppose ‘a’ is the edge of the cube and v be the volume of the cube.
v = a³ ……………….. (1)
\(\frac{dv}{dt}\) = 8 cm³/sec.
a = 12 cm
Surface Area of cube. S = 6a²
\(\frac{ds}{dt}\) = 12a\(\frac{da}{dt}\) ………….. (2)

Question 2.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm., how fast is the enclosed area increases?
Solution:
Suppose r is the value of the outer ripple and A be its area
Area of circle A = πr²
\(\frac{dA}{dt}\) = 2πr \(\frac{dr}{dt}\)
Given r = 8, \(\frac{dr}{dt}\) = 5
\(\frac{dA}{dt}\) = 2π (8) (5)
= 80π cm²/sec.

Question 3.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?
Solution:
\(\frac{dr}{dt}\) = 0.7 cm/sec
Circumference of a circle, c = 2πr
\(\frac{dc}{dt}\) = 2π \(\frac{dr}{dt}\)
= 2π(0.7)
= 1.4π cm/sec.

Question 4.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 15 cm.
Solution:
\(\frac{dv}{dt}\) = 900 c.c/sec
r = 15 cm
Volume of the sphere v = \(\frac{4}{3}\) πr³

Question 5.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm.?
Solution:
\(\frac{dr}{dt}=\frac{1}{2}\) cm/sec
radius r = 1 cm
Volume sphere v = \(\frac{4}{3}\) πr³
\(\frac{dv}{dt}\) = 4πr² \(\frac{dr}{dt}\)
= 4π(1)²\(\frac{1}{2}\)
= 2π cm³/sec.

Question 6.
Assume that an object is launched upward at 980 m/sec. Its position would be given by s = 4.9 t² + 980 t. Find the maximum height attained by the object.
Solution:
s = – 4.9 t² + 980 t
\(\frac{ds}{dt}\) = -9.8 t + 980
v = -9.8 t + 980
for max. height, v = 0
-9.8 t + 980 = 0
980 = 9.8 t
\(\frac{980}{9.8}\) = t
100 = t
s = -4.9(100)² +980(100)
s = -49000 + 98000
s = 49000 units.

Question 7.
Let a kind of bacteria grow in such a way that at time t sec. there are t^(3/2) bacteria. Find the rate of growth at time t = 4 hours.
Solution:
Let g be the amount of growth of bacteria at t then g(t) = t^3/2
The growth rate at time t is given by
g'(t) = \(\frac{3}{2}\)t^1/2
given t = 4hr
g'(t) = \(\frac{3}{2}\) (4 × 60 × 60)^1/2
= \(\frac{3}{2}\) (2 × 60) = 180

Question 8.
Suppose we have a rectangular aquarium with dimensions of length 8m, width 4 m and height 3 m. Suppose we are tilling the tank with water at the rate of 0.4 m³/sec. How fast is the height of water changing when the water level is 2.5 m?
Solution:
Length of aquarium l = 8 m
Width of aquarium b = 4 m
Height of aquarium h = 3
\(\frac{dv}{dt}\) = 0.4 m³/sec.
v = lbh
= 8(4)(3)
= 96
v = lbh
⇒ log v = log l + log b + log h

Note : Text book Ans. \(\frac{1}{80}\) will get when h = 3.

Question 9.
A container is in the shape of an inverted cone has height 8m and radius 6m at the top. If it is filled with water at the rate of 2m³/minute, how fast is the height of water changing when the level is 4m?
Solution:

h = 8m = OC
r = 6m = AB
\(\frac{dv}{dt}\) = 2 m³/minute
∆ OAB and OCD are similar angle then
\(\frac{CD}{AB}=\frac{OC}{OA}\)
\(\frac{r}{6}=\frac{h}{8}\)
r = h \(\frac{3}{4}\)
Volume of cone v = \(\frac{1}{3}\)πr²h

Question 10.
The total cost C(x) in rupees associated with the production of x units of an item is given by C(x) 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
Let m represents the marginal cost, then
M = \(\frac{dc}{dx}\)
Hence
M = \(\frac{d}{dx}\)(0.007x³ – 0.003x² + 15x + 4000)
= (0.007) (3x²) – (0.003) (2x) + 15 /.
∴ The marginal cost at x = 17 is
(M)_m=17 = (0.007) 867 – (0.003) (34) + 15
= 6.069-0.102+ 15
= 20.967.

Question 11.
The total revenue in rupees received from the sale of x units of a produce is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution:
Let m denotes the marginal revenue. Then
M = \(\frac{dR}{dx}\)
Similar R(x) = 13x² + 26x +15
∴ m = 26x + 26
The marginal revenue at x = 7
(M)_{x = 7} = 26(7) + 26
= 208.

Question 12.
A point P is moving on the curve y = 2x². The x co-ordinate of P is increasing at the rate of 4 units per second. Find the rate at which y co-ordinate is increasing when the point is (2, 8).
Solution:
Given y = 2x²
\(\frac{dy}{dx}\) = 4x. \(\frac{dx}{dt}\)
Given x = 2, \(\frac{dx}{dt}\) = 4.\(\frac{dy}{dt}\)
= 4(2).4 = 32
y co-ordinate is increasing at the rate of 32 units/sec.

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)

I. Find the angle between the curves given below.

Question 1.
x + y + 2 = 0 ; x² + y² – 10y = 0.
Solution:
x + y + 2 = 0 ⇒ x = -(y + 2)
x² + y² – 10y = 0
(y + 2)² + y² – 10y = 0
y² + 4y + 4 + y² – 10y = 0
2y² – 6y + 4 = 0
y² – 3y + 2 = 0
(y + 1) (y – 2) = 0
y = 1 or y – 2
x = – (y + 2)
y = 1 ⇒ x = -(1 + 2) = -3
y = 2 ⇒ x = -(2 + 2) = -4
The points of intersection are P(-3, 1) and Q(-4, 2), equation of the curve is
x² + y² – 10y = 0
Differentiating w.r.to x.
2x + 2y\(\frac{dy}{dx}\) – 10 \(\frac{dy}{dx}\) = 0
2\(\frac{dy}{dx}\)(y – 5) = -2x
\(\frac{dy}{dx}\) = –\(\frac{x}{y-5}\)
f'(x₁) = –\(\frac{x}{y-5}\)
Equation of the line is x + y + 2 = 0
1 + \(\frac{dy}{dx}\) = 0 ⇒ \(\frac{dy}{dx}\) = -1
g'(x) = -1

Case (i):
At P(-3, 1), f'(x₁) = \(\frac{3}{1-5}=-\frac{3}{4}\), g'(x₁) = -1

Case (ii):

Question 2.
y² = 4x, x² + y² = 5.
Solution:
Eliminating y; we get x² + 4x = 5
x² + 4x – 5 = 0
(x – 1) (x + 5) = 0
x – 1 = 0 or x + 5 = 0
x = 1 or -5
Now y² = 4x
x = 1 ⇒ y² = 4
y = ±2
x = -5 ⇒ y is not real.
∴ Points of interstection of P(1, 2) and Q(1, -2) equation of the first curve is y² = 4x
2y.\(\frac{dy}{dx}\) = 4
\(\frac{dy}{dx}\) = \(\frac{4}{2y}\)
f(x) = \(\frac{2}{y}\)
Equation of the second curve is x² + y² = 5 dy
2x + 2y \(\frac{dy}{dx}\) = 0 dx
2.\(\frac{dy}{dx}\) = -2x

Question 3.
x² + 3y = 3 ; x² – y² + 25 = 0.
Solution:
x² = 3 – 3y; x² – y² + 25 = 0
3 – 3y – y² + 25 = 0
y² + 3y – 28 = 0
(y – 4) (y + 7) = 0
y – 4 = 0 (or) y + 7 = 0
y = 4 or – 7
x² = 3 – 3y
y = 4 ⇒ x² = 3 – 12 = – 9
⇒ x is not real
y = -7 ⇒ x² = 3 + 21 = 24
⇒ x = ± √24 = ± 2√6
Points of intersection are
P(2√6, -7), Q(-2√6, -7)
Equation of the first cur ve is x² + 3y = 3
3y = 3 – x²
3.\(\frac{dy}{dx}\) = -2x
\(\frac{dy}{dx}\) = –\(\frac{2x}{3}\) i.e, f'(x₁) = –\(\frac{2x}{3}\)
Equation of the second curve is
x² – y² + 25 = 0
y² = x² + 25
2y.\(\frac{dy}{dx}\) = 2x ⇒ \(\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}\)

Question 4.
x² = 2(y + 1), y = \(\frac{8}{x^{2}+4}\).
Solution:
x² = 2(\(\frac{8}{x^{2}+4}\) + 1) = \(\frac{16+2x^{2}+8}{x^{2}+4}\)
x²(x² + 4) = 2x² + 24
x⁴ + 4x² – 2x² – 24 = 0
x⁴ + 2x² – 24 = 0
(x² + 6) (x² – 4) = 0
x² = -6 or x² = 4
x² = -6 ⇒ x is not real
x² = 4 ⇒ x = ±2
y = \(\frac{8}{x^{2}+4}=\frac{8}{4+4}=\frac{8}{8}\) = 1
∴ Points of intersection are P(2, 1) and Q(-2, 1)
Equation of the first curve is x² = 2(y + 1)
2x = 2.\(\frac{dy}{dx}\) ⇒ \(\frac{dy}{dx}\) = x
f'(x₁) = x₁
Equation of the second curve is y = \(\frac{8}{x^{2}+4}\)

∴ The given curves cut orthogonally
i.e., θ = \(\frac{\pi}{2}\)
At Q (-2, -1),f'(x₁) = -2, g'(x₁) = \(\frac{32}{64}=\frac{1}{2}\)
f'(x₁) g'(x₁) = -2 × \(\frac{1}{2}\) = -1
∴ The given curves cut orthogonally
⇒ θ = \(\frac{\pi}{2}\)

Question 5.
2y² – 9x = 0, 3x² + 4y = 0 (in the 4^th quadrant).
Solution:
2y² – 9x = 0 ⇒ 9x = 2y²
x = \(\frac{2}{9}\)y²
3x² + 4y = 0
⇒ 3.\(\frac{4}{8}\) y⁴ + 4y = 0
\(\frac{4y^{2}+108y}{27}\) = 0
4y(y³ + 27) = 0
y = 0 or y³ = -27 ⇒ y = -3
9x = 2y² 2 × 9 ⇒ x = 2
Point of intersection (in 4^th quadrant) is P(2, -3)
Equation of the first curve is 2y² = 9x
4y\(\frac{dy}{dx}\) = 9 ⇒ \(\frac{dy}{dx}=\frac{9}{4y}\)
f'(x₁) = \(\frac{9}{4y}\)
At P(2, -3), f'(x₁) = \(\frac{9}{-12}=-\frac{3}{4}\)
Equation of the second curve is
3x² + 4y = 0
4y = -3x²
4.\(\frac{dy}{dx}\) = -6x

Question 6.
y² = 8x, 4x² + y = 32
Solution:
4x² + 8x = 32 ⇒ x² + 2x = 8
x² + 2x – 8 = 0
(x – 2) (x + 4) = 0
x = 2 or -4
y² = 8x
x = -4 ⇒ y² is not real
x = 2 ⇒ y² = 16 ⇒ y = ±4
Point of intersection are P(2, 4), Q(2, -4)
Equation of the first curve is y² = 8x
2y.\(\frac{dy}{dx}\) = 8 ⇒ \(\frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y}\)
f'(x₁) = \(\frac{4}{y}\)
Equation of the second curve is
4x² + y² = 32
8x + 2y.\(\frac{dy}{dx}\) = 0

Question 7.
x²y = 4, y(x² + 4) = 8.
Solution:
x²y = 4 ⇒ x = \(\frac{4}{y}\)
y(x² + 4) = 8
y(\(\frac{4}{y}\) + 4y) = 8
y\(\frac{(4+4y)}{y}\) = 8
4y – 4 ⇒ y = 1
x² = 4 ⇒ x = ±2
Points of intersection are P(2, 1), Q(-2, 1)
x²y = 4 ⇒ y = \(\frac{4}{x^{2}}\)
\(\frac{dy}{dx}\) = –\(\frac{8}{x^{3}}\) ⇒ f'(x₁) = –\(\frac{8}{x^{3}}\)
y(x² + 4) = 8 ⇒ y = \(\frac{8}{x^{2}+4}\)

Question 8.
Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at (\(\frac{1}{2}\), \(\frac{1}{2}\))
Solution:
Equation of the first curve is
6x² – 5x + 2y = 0
2y = 5x – 6x²
2.\(\frac{dy}{dx}\) = 5 – 12x
\(\frac{dy}{dx}=\frac{5-12x}{2}\)

Equation of the second curve is 4x² + 8y² = 3

∴ f'(x₁) = g'(x₁)
The given curves touch each other at P(\(\frac{1}{2}\), \(\frac{1}{2}\)).

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)

I.

Question 1.
Find the lengths of subtangent and sub-normal at a point of the curve
Solution:
Equation of the curve is y = b. sin \(\frac{x}{a}\)
\(\frac{dy}{dx}\) = b. cos \(\frac{x}{a}.\frac{1}{a}=\frac{b}{a}\). cos \(\frac{x}{a}\)
Length of the sub-tangent = |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)|
= \(\frac{b \cdot \sin \frac{x}{a}}{\frac{b}{a} \cdot \cos \frac{x}{a}}=\left|a \cdot \tan \frac{x}{a}\right|\)
Length of the sub-normal = |y₁, f'(x₁)|

Question 2.
Show that the length of sub normal at any point on the curve xy = a2 varies as the cube of the ordinate of the point.
Solution:
Equation of the curve is xy = a²
y = \(\frac{a^{2}}{x}\)
\(\frac{dy}{dx}=\frac{-a^{2}}{x^{2}}\)
Length of the sub-normal = |y₁, f’ (x₁)|

\(\frac{y_{1}^{3}}{a^{2}}\) ∝ y³₁ = cube of the ordinate.

Question 3.
Show that at any point (x, y) on the curve y = be^x/a, the length of the sub¬tangent is a constant and the length of the sub normal is \(\frac{y^{2}}{a}\).
Solution:
Equation of the curve is y = b.e^x/a
\(\frac{a^{2}}{x}\) = b.e^x/a.\(\frac{1}{a}=\frac{y}{a}\)
Length of the sub tangent
= |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)| = \(\frac{y_{1}}{\left(\frac{y_{1}}{a}\right)}\) = a = constant
Length of the sub-normal
= |y₁f'(x₁)| = |y₁\(\frac{y_{1}}{a}\)| = \(\frac{y_{1}^{2}}{a}\).

II.

Question 1.
Find the value of k so that the length of the sub normal at any point on the curve xy^k = a^{k + 1} is a constant.
Solution:
Equation of the curve is x.y^k = a^{k + 1}
Differentiating w.r.to x
x.k.y^k-1 \(\frac{dy}{dx}\) + y^k.1 = 0
x.k.y^k-1 \(\frac{dy}{dx}\) = -y^k
\(\frac{dy}{dx}=\frac{-y^{k}}{k \cdot x \cdot y^{k-1}}=-\frac{y}{kx}\)
Length of the sub-normal = |y₁f'(x₁|

Length of the sub-normal is constant at any point on the curve is independent of x₁ and y₁
∴ \(\frac{y_{1}^{k+2}}{k \cdot a^{k+1}}\) is independent of x₁, y₁
⇒ k + 2 = 0 ⇒ k = -2

Question 2.
At any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, sub tangent and sub
normal.
Solution:
Equation of the curve is x = a (t + sin t), y = a (1 – cost)
\(\frac{dx}{dt}\) = a (1 + cos t), \(\frac{dy}{dt}\) = a. sin t

Length of the tangent = |y₁|\(\sqrt{1+\frac{1}{\left[f^{\prime}\left(x_{1}\right)\right]^{2}}}\)

Length of the normal = |y₁|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)

Length of the sub tangent.

= |a (2 sin t/2. cos t/2|
= |a. sin t|
Length of the sub normal
= |y₁.f'(x₁)| = |a(1 – cos t).\(\frac{\sin t / 2}{\cos t / 2}\)|
= |2a sin² t/2. tan t/2|
= |2a sin² t/2. tan t/2|

Question 3.
Find the lengths of normal and sub normal at a point on the curve y = \(\frac{a}{2}\) (e^x/a + e^-x/a).
Solution:
Equation of the curve is y = \(\frac{a}{2}\) (e^x/a + e^-x/a)
= a. cosh (\(\frac{x}{a}\))
Length of the normal = |y₁|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)
= |a.cosh \(\frac{x}{a}\)|\(\sqrt{1+\sinh ^{2} \frac{x}{a}}\)
= a. cosh \(\frac{x}{a}\). cosh \(\frac{x}{a}\) = a. cosh² \(\frac{x}{a}\)
Length of the sub normal = |y₁ f'(x₁)|

Question 4.
Find the lengths of subtangent, sub normal at a point t on the curve x = a (cos t + t sin t), y=a (sin t – t cos t)
Solution:
Equations of the curve are x = a (cos t + t sin t)
\(\frac{dx}{dt}\) = a (- sin t + sin t +1 cost) = at cos t
y = a (sin t – t cos t)
\(\frac{dy}{dt}\) = a (cos t – cos t +1 sin t) = a t sin t

Length of the sub tangent

= |a cot t (sin t – t cos t)|
Length of the sub-normal = |y₁.f'(x₁)|
= |a (sin t – t cos t) tan t|
= |a tan t (sin t – t cos t)|