PSEB 11th Class Economics Book Solutions Guide in Punjabi English Medium

Punjab State Board Syllabus PSEB 11th Class Economics Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Economics Guide | Economics Guide for Class 11 PSEB

Economics Guide for Class 11 PSEB | PSEB 11th Class Economics Book Solutions

PSEB 11th Class Economics Book Solutions in Hindi Medium

PSEB 11th Class Economics Book Solutions in English Medium

  • Chapter 1 What is Economics?
  • Chapter 2 Micro and Macro Economics
  • Chapter 3 Central Problems of an Economy
  • Chapter 4 Consumer’s Equilibrium
  • Chapter 5 Demand
  • Chapter 6 Price Elasticity of Demand
  • Chapter 7 Meaning of Production
  • Chapter 8 Concepts of Costs
  • Chapter 9 Concepts of Revenue
  • Chapter 10 Producer’s Equilibrium
  • Chapter 11 Supply & Price Elasticity of Supply
  • Chapter 12 Forms of Market
  • Chapter 13 Price Determination Under Perfect Competition: Equilibrium Price
  • Chapter 14 Meaning, Scope and Importance of Statistics in Economics
  • Chapter 15 Primary and Secondary Data
  • Chapter 16 Census and Sample Methods
  • Chapter 17 Organisaiton of Data
  • Chapter 18 Presentation of Data With Tables – Tabulation
  • Chapter 19 Diagrammatic Presentation
  • Chapter 20 Graphic Presentation
  • Chapter 21 Arithmetic Line Graphs : Time Series
  • Chapter 22 Measures of Central Tendency – Arithmetic Mean
  • Chapter 23 Measures of Central Tendency – Median
  • Chapter 24 Measures of Central Tendency – Mode
  • Chapter 25 Measures of Dispersion
  • Chapter 26 Man Power and Physical Resources of Punjab
  • Chapter 27 Agriculture Development of Punjab Since 1966
  • Chapter 28 Industrial Development of Punjab Since 1966
  • Chapter 29 Financial Position of Punjab Government

PSEB 11th Class Economics Syllabus

Class – XI (Pb.)
Economics
Time Allowed: 3 Hours

Theory: 80 Marks
Internal Assessment: 20 Marks
Marks Total: 100 Marks

Part – A
Introductory Micro Economics

Unit 1 Introduction
What is Economics? Definitions of Economics (Wealth, Material Welfare, Scarcity, and Growth Definitions). Economic Activities. Nature of Economics, Economic Policies, Economic Systems. Positive and Normative economics. Meaning of Microeconomics and Macroeconomics, Difference and interdependence between Micro and Macro Economics. Scope, Importance, subject matter, and limitations of Micro Economics.

What is an economy? Central problems of an economy: what, how, and for whom to produce. Production Possibility Curve, the slope of production possibility curve, the concept of opportunity cost, and marginal opportunity cost. Shifts and rotations of production possibility curve. Solution of various central problems with production possibility curve.

Unit 2 Consumer’s Equilibrium and Theory of Demand
Consumer’s equilibrium – meaning of consumer’s equilibrium, the meaning of utility, and various types of utility and their inter relationship. Law of diminishing marginal utility and Law of Equi-Marginal utility. Conditions of consumer’s equilibrium using marginal utility analysis in case of one and two commodities.

Theory of Demand: Meaning, types of demand, Demand schedule, Demand Curve and its slope, Law of Demand-its assumptions and exceptions. Determinants of demand. Movement along and shifts in the demand curve.

Price elasticity of demand – Meaning, degrees of price elasticity of demand, factors affecting price elasticity of demand; measurement of price elasticity of demand with percentage method and numerical.

Unit 3 Producer Behaviour and Supply
Theory of Production: Meaning of Production and production function. concepts of total product, Average Product, and Marginal Product. Concept of short Run and long run in production and laws of return to a variable factor and return to scale. Economies and Diseconomies of scale.

Theory of Cost: Meaning and types of cost. Short-run costs-total cost, total fixed cost, total variable cost; Average cost; Average fixed cost, average variable cost, and marginal cost. Relationship between various types of cost. Long-run cost curves.

Theory of Supply: Meaning and types. Supply schedule and Supply Curve and its slope. Determinants of supply. Movements along and shifts in the supply curve. Price elasticity of supply; measurement of price elasticity of supply with percentage method. Factors affecting price elasticity of supply.

Concepts of Revenue – Meaning and types of revenue. Total, average and marginal revenue – meaning and their relationship. Revenue curves under different market situations.

Producer’s equilibrium meaning and its conditions in terms of marginal revenue and the marginal cost approach. Concept of Gross profits and Net profits.

Unit 4 Forms of Market and Price Determination under Perfect Competition
Forms of Market: Meaning and features of a Market. Forms of Market: Perfect Competition, Monopoly and Monopolistic Competition and their features. Price Determination under Perfect competition-Determination of equilibrium through market forces and effect of shifts in demand and supply curves on equilibrium price and equilibrium quantity.

Part – B
Statistics for Economics

In this course, the learners are expected to acquire skills in the collection, organization, and presentation of quantitative and qualitative information pertaining to various simple economic aspects systematically. It also intends to provide some basic statistical tools to analyze, and interpret any economic information and draw appropriate inferences. In this process, the learners are also expected to understand the behaviour of various economic data.

Unit 5 Introduction
Statistics in Economics: Meaning, scope, functions, nature, limitations, and importance of statistics in Economics. Concept of statistics in singular and plural sense with their characteristics.

Unit 6 Collection, Organisation, and Presentation of Data
Collection of data sources of data primary and secondary data: their meaning, difference between primary and secondary data, methods for collection of primary and secondary data along with their suitability, advantages, and limitations. Some important sources of secondary data: are the Census of India and the National Sample Survey Organisation.

Theory of Census and Sampling: Meaning of census and sample method along with their suitability, merits, and demerits. Method of sampling: Random sampling, Stratified sampling, Systematic sampling, Quota sampling, Convenient sampling, and Purposive sampling. Sampling and Non-Sampling errors.

Organization of Data: Meaning and types of variables; Meaning and types of series: Individual, Discrete and Continuous series (various types of continuous series). Various concepts are related to the formation of series.

Presentation of Data: Tabular Presentation and Diagrammatic Presentation of Data:
(i) Geometric forms (bar diagrams and pie diagrams), (ii) Frequency diagrams (histogram, polygon, and ogive), and (iii) Arithmetic line graphs (time series graph).

Unit 7 Measures of Central Tendency and Dispersion
Measures of Central Tendency: Meaning of Central Tendency, Features, Arithmetic Mean (simple), Median and other positional averages (Quartile, Decile, and Percentile) and Mode (by inspection method only). Relationship between various measures of central tendency and choice of the best measure of central tendency.

Measures of Dispersion: Meaning, Absolute measures of Dispersion (Range, Quartile’s Range, Quartile Deviation, Mean Deviation, Standard Deviation). Relative measures of Dispersion (Co-efficient of range, co-efficient of quartile deviation, Coefficient of Mean deviation, Coefficient of Standard Deviation, and Coefficient of variation). Lorenz Curve.

Part – C
Punjab Economy

Unit 8 An Overview of Punjab Economy
Physical Resources of Punjab, Manpower Resources of Punjab, Agriculture Development of Punjab since 1966, Industrial Development of Punjab, and Financial System of Punjab.

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

Punjab State Board PSEB 11th Class Economics Book Solutions Chapter 4 उपभोगी का सन्तुलन Textbook Exercise Questions, and Answers.

PSEB Solutions for Class 11 Economics Chapter 4 उपभोगी का सन्तुलन

PSEB 11th Class Economics उपभोगी का सन्तुलन Textbook Questions and Answers

I. वस्तुनिष्ठ प्रश्न (Objective Type Questions)

प्रश्न 1.
कुल उपयोगिता की परिभाषा लिखें।
उत्तर-
एक वस्तु की सभी इकाइयों का उपभोग करने से प्राप्त होने वाली उपयोगिता के जोड़ को कुल उपयोगिता कहते हैं।

प्रश्न 2.
सीमान्त उपयोगिता की परिभाषा लिखें।
उत्तर-
एक वस्तु की एक और इकाई का उपयोग करने से प्राप्त होने वाली अतिरिक्त उपयोगिता को सीमान्त उपयोगिता कहते हैं।
अथवा
(MU = TUn TUn -1)

प्रश्न 3.
उपभोग से क्या अभिप्राय है ?
उत्तर-
उपभोग वह प्रक्रिया है जिस द्वारा वस्तुओं तथा सेवाओं के प्रयोग से आवश्यकताओं की सन्तुष्टि की जाती है।

प्रश्न 4.
उपयोगिता से क्या अभिप्राय है ?
उत्तर-
वस्तु या सेवा का वह गुण जिस द्वारा आवश्यकताएँ सन्तुष्ट होती हैं, उस गुण को उपयोगिता कहा जाता है।

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 5.
सीमान्त उपयोगिता से कुल उपयोगिता का आंकलन कैसे किया जाता है ?
उत्तर-
सीमान्त उपयोगिता के जोड़ से कुल उपयोगिता प्राप्त हो जाती है।

प्रश्न 6.
जब कुल उपयोगिता अधिकतम होती है तो सीमान्त उपयोगिता कितनी होती है ?
उत्तर-
जब कुल उपयोगिता अधिकतम होती है तो सीमान्त उपयोगिता शून्य (Zero) होती है।

प्रश्न 7.
जब सीमान्त उपयोगिता ऋणात्मक होती है तो कुल उपयोगिता पर क्या प्रभाव पड़ता है ?
उत्तर-
जब सीमान्त उपयोगिता ऋणात्मक होती है तो कुल उपयोगिता गिरना शुरू हो जाती है।

प्रश्न 8.
जब सीमान्त उपयोगिता शून्य होती है तो कुल उपयोगिता की क्या स्थिति होती है ?
उत्तर-
जब सीमान्त उपयोगिता शून्य होती है तो कुल उपयोगिता अधिकतम होती है।

प्रश्न 9.
उपभोक्ता से क्या अभिप्राय है ?
उत्तर-
उपभोक्ता एक व्यक्ति, परिवार या सरकार हो सकती है जिस द्वारा आवश्यकताओं की पूर्ति के लिए वस्तुओं तथा सेवाओं का प्रयोग किया जाता है।

प्रश्न 10.
उपभोक्ता सन्तुलन क्या है ?
उत्तर-
उपभोक्ता सन्तुलन वह अवस्था है जब वह अपने व्यवहार को वर्तमान परिस्थितियों में सबसे अच्छा मानता है और उसमें कोई परिवर्तन नहीं करना चाहता।

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 11.
घटती सीमान्त उपयोगिता के नियम से क्या अभिप्राय है ?
उत्तर-
घटती सीमान्त उपयोगिता के नियम अनुसार जब किसी वस्तु की इकाइयों का लगातार अधिक उपभोग किया जाता है तो प्रत्येक इकाई से प्राप्त होने वाली सीमान्त उपयोगिता घटती जाती है।

प्रश्न 12.
उपभोक्ता सन्तुलन की शर्त क्या होती है ?
उत्तर-
उपभोक्ता सन्तुलन = MUx = Price of X. MU of Money.

प्रश्न 13.
एक वस्तु के लिए उपभोक्ता सन्तुलन कैसे प्राप्त होता है ?
उत्तर–
प्राप्त उपयोगिता (MU) = त्यागी गई उपयोगिता (Price).

प्रश्न 14.
दो वस्तुओं के लिए उपभोक्ता सन्तुलन कैसे प्राप्त होता है ?
उत्तर-
X वस्तु की सीमान्त उपयोगिता (MU.) = Y वस्तु की सीमान्त उपयोगिता (MUy).

प्रश्न 15.
सीमान्त उपयोगिता (MU) = ………
उत्तर-
सीमान्त उपयोगिता (MU) = (TUn – TUn-1) Or \(\frac{\Delta \mathrm{TU}}{\Delta \mathrm{Q}}\)

प्रश्न 16.
MU1 + MU2 + MU3 = ……………….. + MUn = ……..
उत्तर-
Total Utility (T.U.)

प्रश्न 17.
जब कुल उपयोगिता अधिकतम होती है तो सीमान्त उपयोगिता ……. होती है।
(a) 1
(b) 2
(c) 3
(d) 0 (शून्य)
उत्तर-
(d) 0 (शून्य)।

प्रश्न 18.
एक वस्तु की स्थिति में उपभोगी का सन्तुलन तब होता है जब
(a) MUm
(b) MUr
(c) MUp
(d) इनमें से कोई नहीं।
उत्तर-
(a) MUm

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 19.
दो वस्तुओं की खरीद में उपभोक्ता सन्तुलन उस समय होता है जब \(\frac{\mathbf{M U _ { x }}}{\mathbf{P}_{x}}=\frac{\mathbf{M U _ { y }}}{\mathbf{P}_{y}}\) ……………………… होती है।
(a) \(\frac{\mathrm{P}_{x}}{\mathrm{P}_{y}}\)
(b) \(\frac{P_{y}}{P_{x}}\)
(c) \(\frac{\mathrm{MU}_{x}}{\mathrm{MU}_{y}}\)
(d) MUm.
उत्तर-
(d) MUm.

प्रश्न 20.
किसी वस्तु की एक और इकाई का उपभोग करने से प्राप्त होने वाली अतिरिक्त उपयोगिता को …………….. कहते हैं।
उत्तर-
सीमान्त उपयोगिता।

प्रश्न 21.
जब सीमान्त उपयोगिता ऋणात्मक होती है तो कुल उपयोगिता अधिकतम होती है।
उत्तर-
सही।

प्रश्न 22.
एक वस्तु की स्थिति में उपभोगी को प्राप्त होने वाली सीमान्त उपयोगिता वस्तु की कीमत के बराबर हो जाती है तो इस को उपभोगी का सन्तुलन कहते हैं।
उत्तर-
सही।

प्रश्न 23.
जब एक मनुष्य के वस्तु भण्डार में वृद्धि होती है तो प्रत्येक वृद्धि के साथ प्राप्त होने वाला अतिरिक्त लाभ घटता जाता है। इस को घटती सीमान्त उपयोगिता का नियम कहते हैं।
उत्तर-
सही।

प्रश्न 24.
निम्नलिखित सारणी को पूरा करें-

उपभोग की गई इकाइयाँ: 1 2 3 4 5 6 7
सीमान्त उपयोगिता : 8 10 8 6 4 2 0
कुल उपयोगिता :

उत्तर-

उपभोग की गई इकाइयाँ : 1 2 3 4 5 6 7
सीमान्त उपयोगिता : 8 10 8 6 4 2 0
कुल उपयोगिता : 8 18 26 32 36 38 38

प्रश्न 25.
घटती सीमान्त उपयोगिता का नियम ……. ने दिया।
(a) मार्शल
(b) एड्म स्मिथ
(c) गोसन
(d) कोई भी नहीं।
उत्तर-
(c) गोसन।

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 26.
घटती सीमान्त उपयोगिता का नियम ………..से सम्बन्धित है।
(a) उपभोग
(b) विनिमय
(c) उत्पादन
(a) कोई भी नहीं।
उत्तर-
(a) उपभोग।

प्रश्न 27.
घटती सीमान्त उपयोगिता के नियम को ……….. भी कहा जाता है।
(a) अधिकतम सन्तुष्टि का नियम
(b) सन्तुष्टि का नियम
(c) गोसन का पहला नियम
(d) कोई भी नहीं।
उत्तर-
(c) गोसन का पहला नियम।

II. अति लघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
उपभोगी से क्या अभिप्राय है ?
उत्तर-
उपभोगी उस व्यक्ति को कहा जाता है जोकि अपनी आवश्यकताओं की सन्तुष्टि के लिए वस्तुओं तथा सेवाओं का प्रयोग करता है। उपभोगी परिवार, सरकार अथवा फ़र्म भी हो सकती है। सरकार लोगों के लिए वस्तुओं तथा सेवाओं की खरीद करती है ताकि सामाजिक भलाई में वृद्धि हो सके।

प्रश्न 2.
उपभोगी के सन्तुलन से क्या अभिप्राय है ? ‘
उत्तर-
उपभोगी का सन्तुलन एक ऐसी अवस्था होती है, जब वह अपने वर्तमान व्यवहार को कुछ विशेष स्थितियों में सबसे अच्छा अनुभव करता है तथा उसमें कोई परिवर्तन नहीं करता, जब तक स्थितियों में कोई परिवर्तन नहीं होता।

प्रश्न 3.
उपयोगिता से क्या अभिप्राय है ?
उत्तर-
उपयोगिता का अर्थ वस्तु अथवा सेवा में उस गुण से होता है जिस द्वारा मानवीय आवश्यकताएं पूरी होती हैं अथवा हम यह कह सकते हैं कि किसी वस्तु में वह शक्ति जो आवश्यकताओं की पूर्ति करती है, उसको उपयोगिता कहा जाता है।(“’Utility is the want satisfying power of a good.”)

प्रश्न 4.
सीमान्त उपयोगिता को पारिभाषित करो।
उत्तर-
सीमान्त उपयोगिता किसी वस्तु की एक अन्य इकाई का उपभोग करने से जो अधिक तुष्टिगुण प्राप्त होता है उसको सीमान्त उपयोगिता कहा जाता है। उदाहरणस्वरूप एक वस्तु की पाँच इकाइयों का उपभोग करने से 100 युटिलज़ सन्तुष्टि प्राप्त होती है तथा छः इकाइयों का उपभोग करने से 110 युटिलज़ सन्तुष्टि मिलती है तो छठी इकाई का उपभोग करने से 110 – 100 = 10 युटिलज़ को सीमान्त उपयोगिता कहा जाता है।
MU = TUn – TUn-1) = 110 – 100 = 10 Utils.

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 5.
कुल उपयोगिता से क्या अभिप्राय है ?
उत्तर-
किसी वस्तु की सभी इकाइयों के उपभोग से प्राप्त होने वाली उपयोगिता को कुल उपयोगिता कहा जाता है। मान लो एक उपभोगी तीन सेबों का उपभोग करता है। प्रथम सेब से 10 युटिल, द्वितीय सेब से 8 युटिल तथा तीसरे सेब से 6 युटिल उपयोगिता प्राप्त होती है तो कुल उपयोगिता 10 + 8 + 6 = 24 युटिल प्राप्त होगी।

प्रश्न 6.
टेबल की सहायता से बताइए कि जब सीमान्त तुष्टिगुण शून्य होता है तो कुल तुष्टिगुण अधिकतम होता है ?
उत्तर-
जब सीमान्त तुष्टिगुण शून्य होता है तो कुल तुष्टिगुण अधिकतम होता है। जैसा कि निम्नलिखित टेबल से स्पष्ट है –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 1
चार सेबों का उपभोग करने से सीमान्त तुष्टिगुण शून्य है और कुल तुष्टिगुण 12 अधिकतम है।

प्रश्न 7.
घटते सीमान्त तुष्टिगुण के नियम से क्या अभिप्राय है ?
उत्तर-
घटते सीमान्त तुष्टिगुण का नियम प्रोफैसर मार्शल की देन है। उनके अनुसार, “यदि अन्य बातें समान रहें जब एक निश्चित समय में एक उपभोगी किसी वस्तु का अधिक उपभोग करता है उस वस्तु की सीमान्त उपयोगिता घटती जाती है।” इसको घटते सीमान्त तुष्टिगुण का नियम कहते हैं।

III. लघु उत्तरीय प्रश्न (Short Answer Type Questions)

प्रश्न 1.
घटते सीमान्त उपयोगिता के नियम का संक्षेप में वर्णन करो।
उत्तर-
गोसन पहले अर्थशास्त्री थे जिन्होंने यह नियम दिया। इसलिए इस नियम को गोसन का पहला नियम भी कहते हैं। यह नियम लोगों की व्यावहारिक तथा मनोवैज्ञानिक स्थिति को ध्यान में रखकर बनाया गया है। प्रो० मार्शल के अनुसार, “जब किसी मनुष्य के वस्तु के भण्डार में वृद्धि होती है तो प्रत्येक वृद्धि से प्राप्त होने वाला लाभ घटता जाता है।” (“The additional benefit which a person derives from an increase of a stock of a thing diminishes with every increase in the stock that he already has.”) इस नियम को उपभोग का आधारपूर्वक नियम (Fundamental Law of Consumption) अथवा आधारपूर्वक मनोवैज्ञानिक नियम (Fundamental Psychological Law) भी कहा जाता है।

इस नियम अनुसार जब एक मनुष्य एक वस्तु की अधिक इकाइयों का उपभोग करता है तो उसको वस्तु से प्राप्त होने वाली सीमान्त उपयोगिता घटती जाती है अथवा कुल उपयोगिता में वृद्धि तो होती है परन्तु जिस अनुपात पर वृद्धि होती है, वह अनुपात घटता जाता है। उदाहरण-एक उपभोगी सेबों का उपभोग करता है, उसको प्राप्त M.U. तथा T.U. इस प्रकार है-
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 2
सूची-पत्र में तथा रेखाचित्र 1 में-

  • सेबों का उपभोग करने से सीमान्त उपयोगिता 4, 3, 2, 1, 0 युटिलज़ घटता जाता है।
  • जब सीमान्त उपयोगिता शून्य है तो. कुल उपयोगिता अधिकतम है।
  • जब सीमान्त उपयोगिता ऋणात्मक है तो कुल उपयोगिता घटने लगता है, इसको घटते सीमान्त उपयोगिता का नियम कहते हैं।

प्रश्न 2.
उपयोगिता, सीमान्त उपयोगिता तथा कुल उपयोगिता के सम्बन्ध को स्पष्ट करें। कुल उपयोगिता तथा सीमान्त उपयोगिता के सम्बन्ध को स्पष्ट करें।
उत्तर-
उपयोगिता का अर्थ (Meaning of Utility)-अर्थशास्त्र में ‘उपयोगिता’ एक महत्त्वपूर्ण धारणा है। उपयोगिता का अर्थ किसी वस्तु या सेवा में उस गुण से होता है जो हमारी ज़रूरतों की पूर्ति करता है। इसीलिए किसी वस्तु में उस गुण या शक्ति को उपयोगिता कहा जाता है जिसके द्वारा मानवीय ज़रूरतों की सन्तुष्टि होती है। (Utility is the want satisfying power of a commodity.) उपयोगिता का माप संख्याओं 1, 2, 3, 4, 5 आदि द्वारा किया जाता है जिनको युटिल कहा जाता है।

सीमान्त उपयोगिता (Marginal Utility)-किसी वस्तु का एक बार और उपभोग करने से जिन कुल उपयोगिता में वृद्धि होती है उस वृद्धि को सीमान्त उपयोगिता कहा जाता है। उदाहरण के तौर पर एक वस्तु की 5 इकाइयों का उपभोग करने के साथ 100 युटिल प्राप्त होती है और छठी इकाई का उपभोग करने के साथ कुल उपयोगिता में वृद्धि 110 युटिल हो जाती है। तो छठी इकाई के उपभोग के साथ 110 -100 = 10 युटिल उपयोगिता की वृद्धि हुई है। इसीलिए सीमान्त उपयोगिता 10 युटिल होगी। इसका माप निम्नलिखित अनुसार किया जाता है।
MU = TUn – TUn-1
= TU6th Unit – TU5th Unit (110 – 100 = 10 युटिलज़)
कल उपयोगिता (Total Utility) किसी वस्तु की सभी इकाइयों के उपभोग से प्राप्त होने वाले उपयोगिता को कुल उपयोगिता कहा जाता है। मान लो एक आदमी ने तीन सेबों का उपभोग किया। पहले सेब से 10 युटिल, दूसरे सेब से 8 युटिल और तीसरे सेब से 6 युटिल उपयोगिता प्राप्त हुआ तो कुल उपयोगिता 10 + 8 + 6 = 24 युटिलज़ हुआ।

कुल उपयोगिता और सीमान्त उपयोगिता का सम्बन्ध (Relationship Between T.U. & M.U.)-
कुल उपयोगिता (T.U.) और सीमान्त उपयोगिता (M.U.) के सम्बन्ध को सूची-पत्र और रेखाचित्र द्वारा स्पष्ट करते हैं –
कुल उपयोगिता तथा सीमान्त उपयोगिता का सम्बन्ध –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 4
सूची-पत्र तथा रेखाचित्र 2 अनुसार –

  • जब सेबों की इकाइयों का अधिक प्रयोग किया जाता है तो सीमान्त तुष्टिकरण (M.U.) घटता जाता है तथा कुल उपयोगिता में वृद्धि घटती दर पर होती है।
  • जब सीमान्त उपयोगिता शून्य (zero) हो जाती है तो कुल उपयोगिता अधिकतम होती है, जैसे चौथे सेब के उपभोग में दिखाया गया है।
  • जब पांचवें सेब का उपभोग किया जाता है तो सीमान्त उपयोगिता ऋणात्मक हो जाती है तो कुल उपयोगिता घटने लगती है, जैसे Mu द्वारा दिखाया गया है।

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 5

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

IV. दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

प्रश्न 1.
घटती सीमान्त उपयोगिता के नियम की व्याख्या करो। इस नियम के अपवाद तथा महत्त्व को स्पष्ट करो।
(Explain the Law of Diminishing Marginals Utility. Give its Exceptions and Importance.)
उत्तर-
घटती सीमान्त उपयोगिता का नियम, उपयोगिता विश्लेषण का आधारपूर्वक नियम है। इस नियम को सबसे पहले एच० एच० गोसन (H.H. Gossen) ने 1854 में दिया था, परन्तु इस नियम की ठीक रूप में व्याख्या प्रो० मार्शल ने 1890 में अपनी पुस्तक “अर्थशास्त्र के सिद्धान्त” में की। प्रो० मार्शल के शब्दों में, “मनुष्य के पास किसी वस्तु की जितनी मात्रा होती है, उस वस्तु की मात्रा में जैसे-जैसे वृद्धि होती है, उसकी सीमान्त उपयोगिता घटती जाती है।” (“The additional benefit which a person dervies from a given stock of a thing, diminishes, with every increase in the stock that he already has.”-Marshall)

प्रो० चैपमैन के अनुसार, “जितनी कोई वस्तु हमारे पास अधिक मात्रा में होती है, उतनी ही कम मात्रा में हम अन्य प्राप्त करना चाहते हैं।” (“The more we have of a thing the less we want additional increament of it.”-Chapman) -उदाहरणस्वरूप एक मनुष्य को प्यास लगी है। पानी का पहला गिलास उस मनुष्य को बहुत उपयोगिता देगा। परन्तु दूसरे तथा तीसरे गिलास से प्राप्त होने वाली उपयोगिता घटती जाएगी। अन्य पानी पीने से उसको तकलीफ भी हो सकती है, अर्थात् वस्तु की वृद्धि से उपयोगिता घटती जाती है, परन्तु एक सीमा के पश्चात् यह उपयोगिता शून्य अथवा ऋणात्मक हो जाती है। इसको घटती उपयोगिता का नियम कहा जाता है।

मान्यताएं (Assumptions) –

  1. उपयोगिता का गणनावाचक माप हो सकता है।
  2. मुद्रा की सीमान्त उपयोगिता स्थिर रहती है।
  3. वस्तुओं की इकाइयां समान आकार तथा गुण वाली हैं।
  4. वस्तु का उपभोग एक समय तथा निरन्तर किया जाता है।
  5. उपभोगी की आय दी हुई है तथा यह स्थिर रहती है।
  6. वस्तु की कीमत तथा स्थानान्तरण वस्तुओं की कीमत स्थिर रहती है।
  7. उपभोक्ता की रुचि, फैशन, आदतें, रीति-रिवाज इत्यादि में कोई परिवर्तन नहीं होता।

नियम की व्याख्या (Explanation of the Law)-
सीमान्त उपभोक्ता का नियम उपभोग का महत्त्वपूर्ण नियम है। एक मनुष्य के पास ₹ 7 हैं। इन पैसों से उपभोक्ता केले खरीदता है। जब यह पैसे केलों पर व्यय किए जाते हैं तो प्राप्त होने वाली सीमान्त उपयोगिता घटती जाती है। इसको सूचीपत्र तथा रेखाचित्र द्वारा दिखाया जा सकता है।
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 7
सूचीपत्र के अनुसार जब उपभोक्ता केलों पर व्यय करता है, तो प्राप्त होने वाली सीमान्त उपयोगिता 5, 4, 3, 2, 1 घटती जाती है। छठे केले से शून्य तथा सातवें केले से ऋणात्मक उपयोगिता प्राप्त होती है। रेखाचित्र 3 में दिखाया है कि उपभोक्ता को केलों से प्राप्त होने वाली सीमान्त उपयोगिता कम होती जाती है। छठी इकाई से शून्य तथा. सातवीं इकाई से ऋणात्मक उपयोगिता प्राप्त होती है। जिसको MU द्वारा दिखाया गया है। इसको घटती सीमान्त उपयोगिता का नियम कहा जाता है।

नियम के अपवाद (Exceptions of the Law)-घटती सीमान्त उपयोगिता का नियम निम्नलिखित स्थितियों में लागू नहीं होता है-

  1. दुर्लभ तथा अद्भुत वस्तुएं-दुर्लभ तथा अद्भुत वस्तुओं की मात्रा के बढ़ने से उपभोक्ता की उपयोगिता घटने की जगह पर बढ़ती जाती है।
  2. कंजूस मनुष्य-कंजूस मनुष्य के पास किसी वस्तु की मात्रा बढ़ने से वह मनुष्य उस वस्तु की अन्य मात्रा प्राप्त करना चाहता है।
  3. नशीले पदार्थ-यह नियम नशीले पदार्थों शराब, अफीम, तम्बाकू इत्यादि पर लागू नहीं होता। शराबी मनुष्य अधिक शराब पीकर अधिक उपयोगिता महसूस करता है।
  4. आरम्भिक इकाइयां-किसी वस्तु की आरम्भिक इकाइयों पर भी नियम लागू नहीं होता।
  5. अच्छी पुस्तकें-यह नियम अच्छी पुस्तकों, कविताओं, पुराने गानों के भण्डार पर भी लागू नहीं होता। .

नियम का महत्त्व (Importance of the Laws)-प्रो० टॉज़िग अनुसार, “घटती सीमान्त उपभोक्ता का नियम इतना व्यापक है कि इसको सर्वव्यापक कहना गलत नहीं होगा।” इसका महत्त्व इस प्रकार है –
1. नियमों का आधार-उपभोग के बहुत से नियम जैसे कि सम-सीमान्त उपयोगिता का नियम, मांग का नियम, उपभोक्ता की बचत का नियम इत्यादि इस नियम पर आधारित हैं।

2. उपभोक्ता के लिए लाभदायक-यह नियम उपभोक्ता के लिए महत्त्वपूर्ण है। एक उपभोक्ता अपनी सारी आय एक वस्तु पर ही व्यय नहीं करता; बल्कि विभिन्न प्रकार की वस्तुओं की खरीद करता है। इससे उस उपभोक्ता को अधिक उपभोक्ता प्राप्त होती है।

3. उत्पादकों के लिए लाभदायक-यह नियम उत्पादकों के लिए भी सहायक है। उत्पादक जब वस्तुओं का उत्पादन नहीं करते हैं तो एक प्रकार की वस्तुओं का ही उत्पादन नहीं किया जाता, बल्कि विभिन्न प्रकार की वस्तुएं विभिन्न वर्ग के लोगों के लिए बनाई जाती हैं। इससे उत्पादक को अधिकतम लाभ प्राप्त होता है।

4. मूल्य निर्धारण-यह नियम मूल्य निर्धारण में भी लाभदायक है। किसी वस्तु की मांग उस वस्तु से प्राप्त होने वाली सी :न्ति उपयोगिता पर निर्भर करती है, जब वस्तु की अधिक मात्रा की मांग की जाती है तो सीमान्त उपयोगिता कम होने के कारण वस्तु की कीमत कम दी जाती है। परन्तु कीमत के घटने से वस्तु की कम मात्रा बेची जाएगी। इसलिए कीमत निर्धारण उस बिन्दु पर होगा, जहां वस्तु की मांग, वस्तु की पूर्ति के समान होती है।

5. वित्त मंत्री के लिए लाभदायक-यह नियम वित्त मंत्री के लिए भी लाभदायक है। कर लगाते समय वित्त मंत्री इस नियम की सहायता लेता है। जब एक मनुष्य की आय बढ़ जाती है तो मुद्रा का सीमान्त उपयोगिता घटता जाता है, इस कारण अमीर मनुष्यों पर कर की अधिक मात्रा लगाई जाती है।

6. समाजवाद का आधार-समाजवाद वह आर्थिक प्रणाली है, जिसमें देश की सरकार आय तथा धन का समान विभाजन करना चाहती है, जबकि अमीर लोगों के पास अधिक धन होता है तो उन पर अधिक कर लगाया जाता है, जिससे समाजवाद आर्थिक प्रणाली जन्म लेती है। देश में साधनों का समान विभाजन होने के कारण प्रत्येक मनुष्य को लाभ प्राप्त होता है।

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 2.
उपभोगी के सन्तुलन से क्या अभिप्राय है ?.एक वस्तु की स्थिति में उपभोगी सन्तुलन कैसे प्राप्त करता है ? (What is Consumer’s Equilibrium ? How does a consumer reach state of equilibrium in case of one commodity and two commodities ?)
उत्तर-
उपभोगी का सन्तुलन (Consumer’s Equilibrium)-उपभोगी का सन्तुलन उस स्थिति को कहा जाता है, जब एक उपभोगी अपनी आय तथा वस्तुओं की कीमतों अनुसार, अपना व्यय इस ढंग से करता है, जिससे उसको प्राप्त होने वाली सन्तुष्टि अधिक-से-अधिक होती है तथा इसमें वह कोई परिवर्तन नहीं करना चाहता, जब तक उपभोगी की आय, वस्तुओं की कीमतों, देश में फैशन, रीति-रिवाज इत्यादि में कोई परिवर्तन नहीं होता।

मान्यताएं (Assumptions) –

  1. विचारशील उपभोगी (Rational Consumer)-उपभोगी विचारशील है तथा अपनी आय से अधिक-सेअधिक सन्तुष्टि प्राप्त करना चाहता है।
  2. संख्यावाचक उपयोगिता (Cardinal Utility) -उपयोगिता का माप संख्यावाचक संख्याओं 1, 2, 3, 4 इत्यादि रूप में किया जा सकता है, जिनको युटिलज़ (Utils) कहा जाता है।
  3. मुद्रा का सीमान्त उपयोगिता स्थिर रहता है (Constant M.U. of money) – जब उपभोगी अपनी आय को व्यय करता है तो मुद्रा के सीमान्त उपयोगिता में कोई परिवर्तन नहीं होता।
  4. उपयोगिता स्वतन्त्र होती है (Utility is Independent)-वस्तु की उपयोगिता स्वतन्त्र होती है तथा अन्य वस्तुओं की उपयोगिता का इस पर कोई प्रभाव नहीं पड़ता।
  5. अन्य बातें समान रहती हैं (Other things being equal) – देश में फैशन, रीति-रिवाज, आदतें, दूसरी वस्तुओं की कीमतें स्थिर रहती हैं।

एक वस्तु की स्थिति में उपभोगी का सन्तुलन (Consumer’s Equilibrium in case of one Commodity)-मान लो एक उपभोगी ऐसी वस्तु की खरीद करता है, जिसका इस तरह से प्रयोग किया जा सकता है जब उपभोगी ऐसी वस्तु का उपभोग करता है तो घटते सीमान्त उपयोगिता का नियम (Law of Diminishing Marginal Utility) लागू होता है। इस नियम अनुसार जब एक उपभोगी एक वस्तु का निरन्तर उपभोग करता है तो अधिक उपभोग करने से प्राप्त होने वाली सीमान्त उपयोगिता घटती जाती है।

इसलिए विचारशील मनुष्य ऐसी वस्तु का उपभोग उस सीमा तक करेगा, जहां कि उस वस्तु की अन्तिम इकाई से प्राप्त सीमान्त उपयोगिता तथा उस इकाई के लिए दी जाने वाली कीमत के रूप में मुद्रा की उपयोगिता एक-दूसरे के समान हों अर्थात् उपभोगी का सन्तुलन निम्नलिखित स्थिति में होगा –
\(\frac{\mathrm{MU}_{x}}{\mathrm{MU}_{m}}=\mathrm{P}_{x}\)
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 8
इसको सूची-पत्र द्वारा स्पष्ट किया जा सकता है –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 9
मान लो X वस्तु की कीमत 1 रुपया प्रति वस्तु दी हुई है। मुद्रा ₹ 1 की सीमान्त उपभोक्ता 10 युटिल के समान है (मुद्रा की सीमान्त उपभोक्ता का कोई विशेष मापदण्ड नहीं होता। यह तो लोगों की वस्तु सम्बन्धी पसन्द पर निर्भर करती है, जो स्थिर रहती है।) उपभोगी चार इकाइयों की खरीद करता है तो वस्तु की सीमान्त उपयोगिता तथा मुद्रा कीमत की सीमान्त उपयोगिता एक-दूसरे के समान है। इसको उपभोगी का सन्तुलन कहा जाता है।

रेखाचित्र 4 में दिखाया है कि जब उपभोगी X वस्तु की अधिक इकाइयों का उपभोग करता है तो वस्तु से प्राप्त सीमान्त उपयोगिता (M.U.) घटती जाती है। ₹ 1 वस्तु की कीमत समान रहती है। एक रुपए की सीमान्त उपयोगिता 10 युटिल मानी गई है। सन्तुलन E बिन्दु पर स्थापित होगा, जिसको सन्तुलन का बिन्दु कहा जाता है।
उपभोगी का सन्तुलन = वस्तु से प्राप्त उपयोगिता = वस्तु की कीमत के रूप में उपयोगिता का त्याग।
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 10

V. संख्यात्मक प्रश्न (Numericals)

प्रश्न 1.
एक मनुष्य A की कुल उपयोगिता अनुसूची दी गई है। सीमान्त उपयोगिता अनुसूची ज्ञात करो।
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 11
उत्तर-
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 12

प्रश्न 2.
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 13
उत्तर
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 14

प्रश्न 3.
निम्नलिखित सारणी पूरी कीजिए :

प्रयोग की गई इकाइयां : 1 2 3 4
कुल उपयोगिता : 7 17 25 31
सीमान्त उपयोगिता :

उत्तर –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 15

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 4.
निम्नलिखित सारणी पूरी कीजिए :

प्रयोग की गई इकाइयां : 1 2 3 4 5 6 7
कुल उपयोगिता 10 25 38 48 55 60 63
सीमान्त उपयोगिता :

उत्तर –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 16

प्रश्न 5.
निम्नलिखित सारणी पूरी कीजिए :

प्रयोग की गई इकाइयां : 1 2 3 4 5 6 7
कुल उपयोगिता : 12 22 30 36 40 42 42
सीमान्त उपयोगिता :

उत्तर –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 17

प्रश्न 6.
निम्नलिखित सारणी को पूरा करें :

प्रयोग की गई इकाइयां : 0 1 2 3 4 5
कुल उपयोगिता : 1 15 30 42 52 61
सीमान्त उपयोगिता :

उत्तर-
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 18

प्रश्न 7.
निम्नलिखित सारणी को पूरा करें :

उपयोग की गई इकाइयां : 0 1 2 3 4 5
कुल उपयोगिता : 0 10 25 38 48 55
सीमान्त उपयोगिता :

उत्तर –
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 19

प्रश्न 8.
निम्नलिखित सारणी को पूरा करें :

उपयोग की गई इकाइयां : 0 1 2 3 4 5
कुल उपयोगिता : 0 20 35 47 57 65
सीमान्त उपयोगिता :

उत्तर-
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 20

PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन

प्रश्न 9.
नीचे दिये हुए आंकड़ों से सीमान्त उपयोगिता मालूम करें :

उपभोग की गई इकाइयां: 1 2 3 4 5
कुल उपयोगिता : 10 25 38 48 55

उत्तर-
PSEB 11th Class Economics Solutions Chapter 4 उपभोगी का सन्तुलन 21

PSEB 11th Class Maths Solutions Chapter 16 Probability Miscellaneous Exercise

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.3 Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Miscellaneous Exercise

Question 1.
A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that
(i) all will be blue?
(ii) atleast one will be green?
Answer.
The box contains 10 (red) + 20 (blue) + 30 (green)
= 60 marbles of which 5 marbles are drawn.
∴ Total number of outcomes, n(S) = \({ }^{60} C_{5}\)

(i) 5 blue marbles can be drawn from 20 blue marbles in \({ }^{20} C_{5}\) ways.
i.e., n(E1) = \({ }^{20} C_{5}\)
∴ P (all marbles blue) = \(\frac{n\left(E_{1}\right)}{n(S)}=\frac{{ }^{20} C_{5}}{{ }^{60} C_{5}}\)

= \(\frac{20 \times 19 \times 18 \times 17 \times 16}{60 \times 59 \times 58 \times 57 \times 56}=\frac{34}{11977}\)

(ii) 5 green marbles can be drawn from 30 green marbles in \({ }^{30} C_{5}\) ways i. e.,
n(E2) = \({ }^{20} C_{5}\)

∴ P (all marbles blue) = \(\frac{n\left(E_{2}\right)}{n(S)}=\frac{{ }^{30} C_{5}}{{ }^{60} C_{5}}\)

P(ar.least one will be green) = 1 – P (5 balls other than green)

= 1 – \(\frac{{ }^{30} C_{5}}{{ }^{60} C_{5}}\)

= 1 – \(\frac{30 \times 29 \times 28 \times 27 \times 26}{60 \times 59 \times 58 \times 57 \times 56}\)

= 1 – \(\frac{117}{4484}\) = \(\frac{4367}{4484}\)

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 2.
4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Answer.
Number of ways of drawing 4 cards from 52 cards = \({ }^{52} C_{4}\)
In a deck of 52 cards, there are 13 diamonds and 13 spades.
Number of ways of drawing 3 diamonds and one spade = \({ }^{13} C_{3} \times{ }^{13} C_{1}\)

Thus, the probability of obtaining 3 diamonds and one spade = \(\frac{{ }^{13} C_{3} \times{ }^{13} C_{1}}{{ }^{52} C_{4}}\)

Question 3.
A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine
(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3).
Answer.
Total number of faces = 6
(i) Number faces with number ‘2’ = 3
∴ P(2) = \(\frac{3}{6}=\frac{1}{2}\)

(ii) P (1 or 3) = P (not 2)
= 1 – P(2)
= 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

(iii) Number of faces with number ‘3’ = 1
∴ P(3) = \(\frac{1}{6}\)
Thus, P(not3) = 1 – P(3)
= 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\).

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 4.
In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting prize if you buy
(a) one ticket
(b) two tickets
(c) 10 tickets?
Answer.
Total number of tickets sold = 10,000
Number of prizes awarded = 10
(i) If we buy one ticket, then
∴ P(getting a prize) = \(\frac{10}{10000}=\frac{1}{1000}\)
∴ P(not getting a prize) = 1 – \(\frac{1}{1000}\)
= \(\frac{999}{1000}\)

(ii) If we buy two tickets, then
Number of tickets, not awarded = 10,000 – 10 = 9990
P(not getting a prize) = \(\frac{{ }^{9990} C_{2}}{{ }^{10000} C_{2}}\)

(iii) If we buy 10 tickets, then
P(not getting a prize) = \(\frac{{ }^{9990} C_{10}}{{ }^{10000} C_{10}}\)

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 5.
Out of 100 students, two sections of 40 and 60 are formed If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?
Answer.

PSEB 11th Class Maths Solutions Chapter 16 Probability Miscellaneous Exercise 1

(a) Let there be two sections A and B of 40 and 60 students, respectively.
∴ 38 students are to selected out of 98, since two particular students are already in section A.
∴ Number of ways of selecting 38 students out of 98 = \({ }^{98} C_{38}\) ways
Number of exhaustive cases of selecting 40 students out of 100 = \({ }^{100} C_{40}\) ways
∴ P(both studensts enter section A) = \(\frac{{ }^{98} C_{38}}{{ }^{100} C_{40}}=\frac{98 !}{38 ! 60 !} \div \frac{100 !}{40 ! 60 !}\)

= \(\frac{98 !}{38 ! 60 !} \times \frac{40 ! 60 !}{100 !}=\frac{40 \times 39}{100 \times 99}=\frac{26}{165}\)

If both students enter the section B. Then, the number of ways of selecting
58 students out of 98 = \({ }^{98} C_{58}\) ways.
Total number of ways of selecting 60 students out of 100 = \({ }^{100} C_{60}\) ways.
∴ Probability that two students enter section
B = \(\frac{{ }^{98} C_{58}}{{ }^{100} C_{60}}=\frac{98 !}{58 ! 40 !} \div \frac{100 !}{60 ! 40 !}\)

= \(\frac{98 !}{58 ! 40 !} \times \frac{60 ! 40 !}{100 !}=\frac{60 \times 59}{100 \times 99}=\frac{59}{165}\)

[∵ \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)]

∴ P (that two particular students enter either section A or B) = \(\frac{26}{165}+\frac{59}{165}=\frac{85}{165}=\frac{17}{33}\)

(b) The probability that they enter different sections
= 1 – P (that two particular students enter either section A or B)

= \(\frac{1}{1}-\frac{17}{33}=\frac{16}{33}\)

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 6.
Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope
Answer..
Let L1, L2, L3 be three letters and E1, E2 and E3 be their corresponding envelopes respectively.
1 letter in correct envelope and 2 in wrong envelope may be put as

PSEB 11th Class Maths Solutions Chapter 16 Probability Miscellaneous Exercise 2

Two lettes or consequently all the letters are in correct envelope may be put in one way.
i.e., (E1L1, E2L2, E3L3)
∴ Number of cases = 3! = 6
Number of favourable cases = 4
Thus, the required probability is \(\frac{4}{6}=\frac{2}{3}\).

Question 7.
A andB are two events such that P(A) = 0.54, P(B) = 0.69 and P (A ∩ B) = 0.35.
Find (i) P (A uP)
(ii) P (A nB)
(iii) P(A nffj
(iv) P (B ∩ A)
Answer.
(i) P(A ∪ B) = P(A) + P(B) – P(A nB)
= 0.54 + 0.69 – 0.35
= 1.23 – 0.35 = 0.88

(ii) A’ ∩ B’ = (A ∪ B)’ By Demorgan’s law.
P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B)
= 1 – 0.88 = 0.12

(iii) A ∩ B’ = A – A ∩ B
P(A ∩ B’) = P(A) – P(A ∩ B )
= 0.54 – 0.35 = 0.19

(iv) P(B ∩ A’) = P(B) – P (A ∩ B)
= 0.69 – 0.35 = 0.34.

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 8.
From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

PSEB 11th Class Maths Solutions Chapter 16 Probability Miscellaneous Exercise 3

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will he either male or over 35 years?
Answer.
Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age.
Accordingly, P(E) = \(\frac{3}{5}\) and P(F) = \(\frac{2}{5}\)
Since there is only one male who is over 35 years of age, P(E ∩ F) = \(\frac{1}{5}\)
We know that P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
∴ P(E ∪ F) = \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}=\frac{4}{5}\)
Thus, the probability that the spokesperson will either be a male or over 4 35 years of age is \(\frac{4}{5}\).

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 9.
If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated?
(ii) the repetition of digits is not allowed?
Answer.
(i) When the digits are repeated Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.
The remaining 3 places can be filled by any of the digits 0, 1, 3, 5 or 7 as repetition of digits is allowed.
∴ Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 – 1.
= 250 – 1 = 249.

[In this case, 5000 can not be counted; so 1 is subtracted]
A number is divisible by 5 if the digit at its units place is either O orS.
∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 – 1 = 100 – 1 = 99.
Thus, the probability of forming a number divisible by 5 when the digits are repeated is = \(\frac{99}{249}=\frac{33}{83}\).

(ii) When repetition of digits is not allowed
The thousands place can be filled with either of the two digits 5 or 7.
The remaining 3 places can be filled with any of the remaining 4 digits.
∴ Total numbers of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2 = 48
When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴ Here, number of 4-digit numbers starting with 5 and divisible by 5 = 3 × 2 = 6
When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits.
∴ Here, number of 4-digit numbers starting with 7 and divisible by 5 = 1 × 2 × 3 × 2 = 12.
∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18.
Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is \(\frac{18}{48}=\frac{3}{8}\).

PSEB 11th Class Maths Solutions Chapter 16 Probability Ex Miscellaneous Exercise

Question 10.
The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
Answer. When the digits are not repeated, then first place may have one of 10 digits, the second 9, third 8 and fourth 7.
Number of 4-digit numbers, n(S) = 10 × 9 × 8 × 7 = 5040
Now, lock may be opened only in 1 way.
∴ n(E) = 1
∴ Probability of opening the lock = \(\frac{n(E)}{n(S)}=\frac{1}{5040}\).

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1

Question 1.
Evaluate the given limit: \(\lim _{x \rightarrow 3}\) x + 3.
Answer.
\(\lim _{x \rightarrow 3}\) x + 3 = 3 + 3 = 6.

Question 2.
Evaluate the given limit: \(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)).
Answer.
\(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)) = (π – \(\frac{22}{7}\)).

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 3.
Evaluate the given limit : \(\lim _{x \rightarrow 1}\) πr2.
Answer.
\(\lim _{x \rightarrow 1}\) πr2 = π (1)2 = π.

Question 4.
Evaluate the given limit : \(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}\).
Answer.
\(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}\)

Question 5.
Evaluate the given limit: \(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}\)
Answer.
\(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}\)

= \(\frac{1-1+1}{-2}=-\frac{1}{2}\)

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 6.
Evaluate the given limit : \(\lim _{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}\).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 1

Question 7.
Evaluate the given limit : \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}\).
Answer.
At x = 2, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{(x-2)(3 x+5)}{(x-2)(x+2)}\)

\(\lim _{x \rightarrow 2} \frac{3 x+5}{x+2}=\frac{3(2)+5}{2+2}=\frac{11}{4}\)

Question 8.
Evaluate the given limit: \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\).
Answer.
At x = 3, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\) = \(\lim _{x \rightarrow 3} \frac{(x-3)(x+3)\left(x^{2}+9\right)}{(x-3)(2 x+1)}\)

\(\lim _{x \rightarrow 3} \frac{(x+3)\left(x^{2}+9\right)}{2 x+1}\) = \(\frac{(3+3)\left(3^{2}+9\right)}{2(3)+1}\)

= \(\frac{6 \times 18}{7}=\frac{108}{7}\)

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 9.
Evaluate the given limit; \(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}\) = .

Question 10.
Evaluate the given limit: \(\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 2

Question 11.
Evaluate the given limit: \(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}\), a + b + c ≠ 0.
Answer.
\(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}=\frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}\)

= \(\frac{a+b+c}{a+b+c}\) = 1. [a + b + c ≠ 0]

Question 12.
Evaluate the given limit: \(\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 3

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 13.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}\).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 4

Question 14.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\), a, b ≠ 0.
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 5

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 15.
Evaluate the given limit: \(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
Answer.
\(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
put π – x = θ, As x → π, θ → 0 (zero)
\(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\pi \theta}=\lim _{\theta \rightarrow 0} \frac{1}{\pi} \frac{(\sin \theta)}{\theta}=\frac{1}{\pi}\)

Question 16.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}\)

Question 17.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}\).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 6

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 18.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}\)
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 7

Question 19.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) x sec x.
Answer.
\(\lim _{x \rightarrow 0}\) x sec x = \(\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}\) = 0.

Question 20.
Evaluate the given limit: \(\) a, b, a + b ≠ 0.
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 8

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 21.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) (cosec x – cot x).
Answer.
\(\lim _{x \rightarrow 0}\) (cosec x – cot x)
At x = 0, the value of the given function takes the form ∞ – ∞.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 9

Question 22.
Evaluate the given limit \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\).
Answer.
\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\)

At x = \(\frac{\pi}{2}\), the value of the given function takes the form \(\frac{0}{0}\).

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 10

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 23.
Find \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x), where f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 11.
Answer.
Given, f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 11
(i) Now, LHL = \(\lim _{x \rightarrow 0^{-}}\) f(x) = \(\lim _{x \rightarrow 0^{-}}\) (2x + 3)
= \(\lim _{h \rightarrow 0}\) [2 (0 – h) + 3] = 3
[putting x = 0 – h as x → 0, then h → 0]
RHL = \(\lim _{x \rightarrow 0^{+}}\) f(x) = \(\lim _{x \rightarrow 0^{+}}\) 3(x + 1)
= \(\lim _{h \rightarrow 0}\) [3(0 + h) + 1] = 3
[putting x = 0 + h as x → 0,then h → 0]J
Here, LHL = RHL
∴ \(\lim _{x \rightarrow 0}\) f(x) = 3

(ii) We have to find \(\lim _{x \rightarrow 1}\) f(x)
\(\lim _{x \rightarrow 1}\) f(x) = \(\lim _{x \rightarrow 1}\) 3 (x + 1)
= 3 (1 + 1) = 6

Question 24.
Find \(\lim _{x \rightarrow 1}\) f(x), where f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 12
Answer.
Given, f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 12
At x = 1,
RHL = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 + h)
= \(\) (1 + h)2 – 1 [put x = 1 + h]
= – (1 + 0)2 – 1
= – 1 – 1 = – 2
LHL = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 – h)
= \(\lim _{h \rightarrow 0}\) (1 – h)2 – 1 [put x = 1 – h]
= (1 – 0)2 – 1 = 1 – 1 = 0
RHL ≠ LHL
Hence, at x = 1 , limit does not exist.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 25.
Evaluate \(\lim _{x \rightarrow 0}\) f(x), where f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 13.
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 14

Question 26.
Find \(\lim _{x \rightarrow 0}\) f(x), where f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 15.
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 16

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 17

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 27.
Find \(\lim _{h \rightarrow 5}\) f(x), where f(x) = |x| – 5.
Answer.
The given function is f(x) = |x| – 5
when x > 5, put x = 5 + h, where h is small
|x| = |5 + h| = 5 + h
∴ \(\lim _{x \rightarrow 5^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) [(5 + h) – 5] = \(\lim _{h \rightarrow 0}\) h = 0
when x < 5, put x = 5 – h, where h is small
∴ |5 – h| = 5 – h
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) (5 – h – 5) = \(\lim _{h \rightarrow 0}\) (- h) = 0
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{x \rightarrow 5^{+}}\) f(x) = 0
∴ \(\lim _{h \rightarrow 5}\) f(x) = 0

Question 28.
Suppose f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 18 and if \(\lim _{h \rightarrow 1}\) f(x) = f(1) what are possible values of a and b ?
Answer.
The given function is f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 18
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (a + bx) = a + b
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (b – ax) = b – a
f(1) = 4
It is given that \(\lim _{x \rightarrow 1}\) f(x) = f(1).
∴ \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) f(x)
= \(\lim _{x \rightarrow 1}\) f(x) = f(1)
a + b = 4 and b – a = 4.
On solving these two equations, we obtain a = 0 and b = 4.
Thus, the respective possible values of a and b are 0 and 4.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 29.
Let a1, a2, ………. a, be fixed real numbers and define a function f(x) = (x – a1) (x – a2) ………….. (x – an). What is \(\lim _{x \rightarrow a_{1}}\) f(x)? For some a ≠ a1, a2, ………………. an compute \(\lim _{x \rightarrow a}\) f(x).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 19

Question 30.
If f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 20 For what value(s) of a does f(x) exists?
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 21

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 31.
If the function f(x) satisfies \(\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}\) = π, evaluate \(\lim _{x \rightarrow 1}\) f(x).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 22

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1

Question 32.
If f(x) = PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 23 For what integers m and n does \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x) exist?
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.1 24

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 1.
Find the derivative of x2 – 2 at x = 10.
Answer.
We have, f(x) = x2 – 2
By using first principle of derivative,
f'(10) = \(\lim _{h \rightarrow 0} \frac{f(10+h)-f(10)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{\left[(10+h)^{2}-2\right]-\left[(10)^{2}-2\right]}{h}\)

= \(\lim _{h \rightarrow 0} \frac{100+20 h+h^{2}-2-100+2}{h}\)

= \(\lim _{h \rightarrow 0} \frac{h^{2}+20 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (h + 20) = 20.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 2.
Find the derivative of 99x at x = 100.
Answer.
Let f(x) = 99x. Accordingly,
f'(100) = \(\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 \times 100+99 h-99 \times 100}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (99) = 99
Thus, the derivative of 99x at x = 100 is 99.

Question 3.
Find the derivative of x at x = 1.
Answer.
Let f(x) = x. Accordingly,
f(1) = \(\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{(1+h)-1}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)\) = 1.
Thus, the derivative of x at x = 1 is 1.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 4.
Find the derivative of the following functions from first principle.
(i) x3 – 27
(ii) (x – 1) (x – 2)
(iii) \(\frac{1}{x^{2}}\)
(iv) \(\frac{x+1}{x-1}\)
Answer.
(i) We have, f(x) = x3 – 27
By using first principle of derivative,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 1

(ii) We have, f(x) = (x – 1) (x – 2) = x2 – 3x + 2
By first principle of derivative, we have

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 2

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

(iii) Let f(x) = \(\frac{1}{x^{2}}\).
Accordingly, from the first, principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 3

(iv) Let f(x) = \(\frac{x+1}{x-1}\)
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 4

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 5.
For the function f(x) = \(\). Prove that f'(1) = 100 f'(0).
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 5

At x = 0, f'(0) = 1
At x = 1, f'(1) = 199 + 198 + ………….. + 1 + 1
= [1 + 1 + 1 + …………… + 1 + 1]100 terms
= 1 × 100 = 100.

Question 6.
Find the derivative of xn + axn – 1 + a2 xn – 2 + ………….. + an – 1 x + an for some fixed real number a.
Answer.
Let f(x) = xn + axn – 1 + a2 xn – 2 + ………….. + an – 1 x + an
On differentiating both sides, we get
f'(x) = nxn – 1 + a(n – 1)xn – 2 + a2 (n – 2)xn – 3 + …………. + an – 1 . 1 + 0
On putting x = a both sides, we get
f'(a) = nan – 1 + a (n – 1) xn – 2 + a2 (n – 2) an – 3 + ……………. + an – 1
= n an – 1 + (n – 1) an – 1 + (n – 2) an – 1 + ……………. + an – 1
= an – 1 [n + (n – 1) + (n – 2) + ………….. + 1]
[∵ sum of n natural numbers = \(\frac{n(n+1)}{2}\)]
= \(a^{n-1} \frac{n(n+1)}{2}\).

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 7.
For some constants a and b, find the derivative of
(i) (x – a) (x – b)
(ii)(ax +b)
(iii) \(\frac{x-a}{x-b}\)
Ans.
(i) Let f(x) = (x – a) (x – b)
f(x)= x2 – (a + b) x + ab
∴ f'(x) = \(\frac{d}{d x}\) [x2 – (a + b) x + ab]
= \(\frac{d}{d x}\) (x2) – (a + b) \(\frac{d}{d x}\)(x) + \(\frac{d}{d x}\) (ab)
On using theorem \(\frac{d}{d x}\) (xn) = nxn – 1, we obtain
f'(x) = 2x – (a + b) + 0 = 2x – a – b.

(ii) Let f(x) = (ax2 + b)2
f(x) = a2x4 + 2abx2 + b2
∴ f'(x) = \(\frac{d}{d x}\) (a2x4 + 2abx2 + b2)
= a2 \(\frac{d}{d x}\) (x4) + 2ab \(\frac{d}{d x}\) (x2) + \(\frac{d}{d x}\) (b2)
On using theorem \(\frac{d}{d x}\) (xn) = nxn – 1, we obtain
f’(x) = a2 (4x3) + 2ab (2x) + b2(O) = 4a2x3 + 4abx= 4ax (ax2 + b).

(iii) Let f(x) = \(\frac{x-a}{x-b}\)

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 6

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 8.
Find the derivative of \(\frac{x^{n}-a^{n}}{x-a}\) for some constant a.
Answer.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 7

Question 9.
Find the derivative of
(i) 2x – \(\frac{3}{4}\)
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x– 3 (5 + 3x)
(iv) x5 (3 – 6x– 9)
(v) x-4 (3 – 4x– 5)
(vi) \(\frac{2}{x+1}-\frac{x^{2}}{3 x-1}\)
Answer.
(i) Let f(x) = 2x – \(\frac{3}{4}\)
f'(x) = \(\frac{d}{d x}\left(2 x-\frac{3}{4}\right)\)

= \(2 \frac{d}{d x}(x)-\frac{d}{d x}\left(\frac{3}{4}\right)\)

= 2 – 0 = 2

(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)
By Leibnitz product rule,
f(x) = (5x3 + 3x – 1) \(\frac{d}{d x}\) (x -1) + (x – 1) \(\frac{d}{d x}\) (5x3 + 3x – 1)
= (5x3 + 3x – 1) (1) + (x – 1) (53x2 + 3 – 0)
= (5x3 +3x – 1) +(x – 1) (15x2 + 3)
= 5x3 + 3x – 1 + 15x33 + 3x – 15x2 – 3
= 20x3 – 15x2 + 6x – 4.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

(iii) Let f(x) = x– 3 (5 + 3x) .
By Leibnitz product rule,
f(x) = x– 3 \(\frac{d}{d x}\) (5 + 3x) + (5 + 3x) \(\frac{d}{d x}\) (x– 3)
= x– 3 (0 + 3) + (5 + 3x) (- 3x– 3 – 1)
= x– 3 (3) + (5 + 3x) (- 3x– 4)
= – 3x– 3 (2 + \(\frac{5}{x}\))
= \(\frac{-3 x^{-3}}{x}\) (2x + 5)
= \(=\frac{-3}{x^{4}}\) (5 + 2x)

(iv) Let f(x) = x5 (3 – 6x-9)
By Leibnitz product rule,
f'(x) = \(x^{5} \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x}\left(x^{5}\right)\)
= x5 {0 – 6 (- 9) x– 9 – 1} + (3 – 6x– 9) (5x4)
= x5 (54x– 10) + 15x4 – 30x– 5
= 54x– 5 + 15x4 – 30x– 5
= 24x– 5 + 15x4
= 15x4 + \(\frac{24}{x^{5}}\)

(v) Let f(x) = x– 4 (3 – 4x– 5)
By Leibnitz product rule,
f’(x) = x– 4 \(\frac{d}{d x}\) (3 – 4x– 5) + (3 – 4x– 5) \(\frac{d}{d x}\) (x– 4)
= x– 4 {0 – 4 (- 5) x– 5 – 1} + (3 – 4x– 5) (- 4)x– 4 – 1
= x– 4 (20 x– 6) + (3 – 4x– 5) (- 4 x– 5)
= 20x– 10 – 12x– 5 + 16x– 10
= 36x– 10 – 12x– 5
= \(-\frac{12}{x^{5}}+\frac{36}{x^{10}}\).

(vi) PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 8

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

Question 10.
Find the derivative of cos x from first principle.
Answer.
Let f(x) = cos x.
Accordingly, from the first principle, /'(*) = Um /(* +10-/00 = ^ COSUM 111 <mx

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 9

= – cos x (0) – sin x(1)
[∵ \(\lim _{h \rightarrow 0} \frac{1-\cos h}{h}\) = 0 and \(\lim _{h \rightarrow 0} \frac{\sin h}{h}\) = 1]
= – sin x
∴ f'(x) = – sin x

Question 11.
Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x+7
(vii) 2 tan x – 7 sec x
Answer.
(i) Let f(x) = sin x cos x.
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 10

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

(ii) Let (x) = sec x.
Accordingly, from the first principle.

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 11

= \(\frac{1}{\cos x} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)}\)

= \(\frac{1}{\cos x} \cdot 1 \cdot \frac{\sin x}{\cos x}\)

= sec x tan x.

(iii) Let f(x) = 5 sec x + 4 cos x.
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 12

= \(\frac{5}{\cos x}\left[\lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right]\) -4 sin x

= \(\frac{5}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1\) – 4 sin x

= 5 sec x tan x – 4 sin x

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

(iv) Let f(x) = cosec x
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 13

(v) Let f(x) = 3cot x – 5 cosec x
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 14

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 15

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2

(vi) Let f(x) = 5sin x – 6cos x + 7.
Accordingly, from the first principle,
f’(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 sin (x + h) – 6 cos (x + h) + 7 – 5 sin x + 6 cos x – 7]

= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 {sin (x + h) – sin x) – 6 {cos (x + h) – cos x}]

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [sin(x + h) – sin x] – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [cos(x + h) – cos x]

= \(5 \lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+x}{2}\right) \sin \left(\frac{x+h-x}{2}\right)\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{\cos x \cos h-\sin x \sin h-\cos x}{h}\)

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{(2 x+h)}{2}\right) \sin \frac{h}{2}\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}\right]\)

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 16

= 5 cos x . 1 – 6 [(- cos x) . (0) – sin x . 1]
= 5 cos x + 6 sin x

(vi) Let f(x) = 2 tan x – 7 sec x
Accordingly, from the first principle,

PSEB 11th Class Maths Solutions Chapter 13 Limits and Derivatives Ex 13.2 17

= \(\) \(\)

= \(\)

= 2 sec2 x – 7 sec x tan x.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Three vertices of a parallelogram ABCD are A (3,- 1, 2), B(1, 2, – 4) and C (- 1, 1, 2). Find the coordinates of the fourth vertex.
Answer.
The three vertices of a parallelogramABCD are given as A (3,- 1, 2), B (1, 2, – 4) and C (- 1, 1, 2).
Let the coordinates of the fourth vertex be D(x, y, z).

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise 1

We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
\(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)=\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
(1, 0, 2) = \(\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
\(\frac{x+1}{2}\) = 1, \(\frac{y+2}{2}\) = 0, and \(\frac{z-4}{2}\) = 2
x = 1 ,y = – 2, and z = 8
Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Question 2.
Find the lengths of the medians of the trisìngle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
Answer.
ABC is a thangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise 2

[∵ Coordinates of mid-points = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)]

Let points D, E and F are the mid-points of BC, AC and AB, respectively.
So, AD, BE and CF will be the medians of the triangle.
Coordinates of point D = \(\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)\) = (3, 2, 0)

Coordinates of point E = \(\left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right)\) = (3, 0, 3)

Coordinates of point F = \(\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)\) = (0, 2, 3)

Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Now, length of median
AD = Distance between points A and D
AD = \(\sqrt{(0-3)^{2}+(0-2)^{2}+(6-0)^{2}}\)
[∵ distance = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}\)]
= \(\sqrt{9+4+36}\)
= √49 = 7

similarly, BE = \(\sqrt{(0-3)^{2}+(4-0)^{2}+(0+3)^{2}}\)
= \(\sqrt{9+16+9}\) = √34

and CF = \(\sqrt{(6-0)^{2}+(0-2)^{2}+(0-3)^{2}}\)
= \(\sqrt{36+4+9}\) = √49 = 7
Hence, length of the median are 7, √34 and 7.

Question 3.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (- 4, 3b, -10) and R (8, 14, 2c), then find the values of a, b and c.
Answer.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise 3

It is known that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3),are \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)\)

Therefore, coordinates of the centroid of ∆PQR = \(\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)\)

= \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

It is given that the origin is the centroid of ∆PQR.
∴ (0, 0, 0) = \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

\(\frac{2 a+4}{3}\) = 0, \(\frac{3 b+16}{3}\) = 0 and \(\frac{2 c-4}{3}\) = 0

a = – 2, b = – \(\frac{16}{3}\) and c = 2.

Thus, the respective values of a, b and c are – 2, – \(\frac{16}{3}\) and 2.

Question 4.
Find the coorlinsites of a point on y-axis which are at a distance of 5√2 from point P(3, – 2, 5).
Answer.
If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.
Let A (0, b, 0) be the point on the y-axis at a distance of 5√2 from point P (3, – 2, 5).
Accordingly, AP = 5√2
∴ AP2 = 50
(3 – 0)2 + (- 2 – b)2 + (5 – 0)2 = 50
⇒ 9 + 4 + b2 + 4b + 25 = 50
⇒ b2 + 4b – 12 = 0
⇒ b2 + 6b – 2b – 12 = 0
⇒ (b + 6) (b – 2) = 0
⇒ b = – 6 o r2.
Thus, the coordinates of the required points are (0, 2, 0) and (0, – 6, 0).

Question 5.
A point R with x-coordinnte 4 lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10). Find the coordinates of the point R.
[Hint : Suppose R divides PQ in the ratio k : 1. The co-ordinates of the point R are given by, \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\)]
Answer.
Let the coordinates of R be (4, y, z) and R divides PQ in ratio k : 1
∴ Coordinate of R is \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\).
[∵ using internal ratio formula]

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise 4

But x – coordinate of R is 4.
So, \(\frac{8 k+2}{k+1}\) = 4
⇒ 8k + 2 = 4k + 4
⇒ 4k = 2
⇒ k = \(\frac{1}{2}\)

img 5

Hence, coordinates of R are (4, – 2, 6).

Question 6.
If A and B be the points (3, 4, 5) and (- 1, 3, – 7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.
Answer.
The coordinates of points A and B are given as (3, 4, 5) and Q (- 1, 3, – 7) respectively.
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain .
PA2 = (x – 3)2 + (y – 4)2 + (z – 5)2
= x2 + 9 – 6x + y2 +16 – 8y + z2 + 25 – 10z
= x2 – 6x + y2 – 8y + z2 – 10z + 50
PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2
= x2 + 2x +1 + y2 – 6y + 9 + z2 + 14z + 49
= x2 + 2x + y2 – 6y + z2 + 14z + 59
Now, if PA2 + PB2 = k2, then
(x2 – 6x + y2 – 8y + z2 – 10z + 50) + (x2 + 2x + y2 – 6y + z2 + 14z + 59) = k2
⇒ 2x2 + 2y2 + 2z2 – 4x – 14y + 4z +109 = k2
⇒ 2(x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109
⇒ x2 + y2 + z2 – 2x – 7y + 2z = \(\frac{k^{2}-109}{2}\)
Thus, the required equation is x2 + y2 + z2 – 2x + 7y + 2z = \(\frac{k^{2}-109}{2}\).

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3

Question 1.
Find the coordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally,
(ii) 2 : 3 externally.
Answer.
Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and B(1, – 4, 6) in the ratio 2 : 3 internally.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 1

Here, the ratio is 2 : 3
m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}, \frac{m_{1} z_{2}+m_{2} z_{1}}{m_{1}+m_{2}}\right)\right]\)

= \(\left[\frac{2 \times 1+3 \times(-2)}{2+3}, \frac{2(-4)+3(3)}{2+3}, \frac{2(6)+3(5)}{2+3}\right]\)

= \(\left(\frac{2-6}{5}, \frac{-8+9}{5}, \frac{12+15}{5}\right)=\left(\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)

(ii) Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and 8(1, – 4, 6) in the ratio 2 : 3 externally.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 2

Here, the ratio is 2 : 3
∴ m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}-m_{2} x_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}, \frac{m_{1} z_{2}-m_{2} z_{1}}{m_{1}-m_{2}}\right)\right]\)

= \(\left[\frac{2(1)+(-3)(-2)}{2+(-3)}, \frac{2(-4)+(-3) \times 3}{2+(-3)}, \frac{2(6)+(-3)(5)}{2+(-3)}\right]\)

= \(\left(\frac{2+6}{2-3}, \frac{-8-9}{2-3}, \frac{12-15}{2-3}\right)=\left(\frac{8}{-1}, \frac{-17}{-1}, \frac{-3}{-1}\right)\)

= (- 8, 17, 3).

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.3

Qiestion 2.
Given that P (3, 2, -4), Q (5, 4, – 6) and R (9, 8, – 10) are coimear. Find the ratio in which Q divides PR.
Answer.
Let point Q(5, 4, -6) divide the line segment joining point P (3, 2, – 4) and R (9, 8, – 10) in the ratio k : 1.
Therefore, by section formula:
(5, 4, – 6) = \(\left(\frac{k(9)+3}{k+1}, \frac{k(8)+2}{k+1}, \frac{k(-10)-4}{k+1}\right)\)

⇒ \(\frac{9 k+3}{k+1}\) = 5
⇒ 9k + 3 = 5k + 5
⇒ 4k = 2
⇒ k = \(\frac{2}{4}=\frac{1}{2}\).
Thus, point Q divides PR in the ratio 1 : 2.

Qiestion 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).
Answer.
Let the YZ plane divide the line segment joining points (- 2, 4, 7) and (3, – 5, 8) in the ratio k: 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{k(3)-2}{k+1}, \frac{k(-5)+4}{k+1}, \frac{k(8)+7}{k+1}\right)\)

On the YZ plane, the x-coordinate of any point is zero.
⇒ \(\frac{3 k-2}{k+1}\) = 0
⇒ 3k – 2 = 0
⇒ k = \(\frac{2}{3}\)
Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2 : 3.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.3

Qiestion 4.
Using section formula, show that the points A (2, – 3, 4), B(- 1, 2, 1) and C(0, \(\frac{1}{3}\), 2) are collinear.
Answer.
Let C(0, \(\frac{1}{3}\), 2) divides the join of A(2, – 3, 4) and B(- 1, 2,1) in the ratio k : 1.

Then, coordinates of C are \(\left(\frac{-k+2}{k+1}, \frac{2 k-3}{k+1}, \frac{k+4}{k+1}\right)\) ………………. (i)
[using internal ratio formula]

But coordinates of C are (0, \(\frac{1}{3}\), 2) ………………(ii) [given]

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 3

\(\frac{-k+2}{k+1}\) = 0
⇒ – k + 2 = 0
⇒ k = 2

\(\frac{2 k-3}{k+1}=\frac{1}{3}\)
⇒ 6k – 9 = k + 1
⇒ k = 2

\(\frac{k+4}{k+1}\) = 2
⇒ k + 4 = 2k + 2
⇒ k = 2

From each of these equations, we get k = 2
Since, each of these equations give same value of k.
Therefore, the given points are collinear and C divides AB internally in the ratio 2 : 1.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.3

Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4,2, -6) and Q (10, -16, 6).
Answer.
Let R and S be two points which trisect the joining of P and Q.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 4

Point R divides the join of PQ in the ratio 1 : 2.
∴ Coordinates of R are \(\left[\frac{1(10)+2(4)}{1+2}, \frac{1(-16)+2(2)}{1+2}, \frac{1(6)+2(-6)}{1+2}\right]\)

= \(\left(\frac{10+8}{3}, \frac{-16+4}{3}, \frac{6-12}{3}\right)\)

= \(\left(\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}\right)\)

= (6, – 4, – 2).

Also, point S divides the join of PQ in the rano 2 : 1
∴ Coordinates of S are \(\left[\frac{2(10)+1(4)}{1+2}, \frac{2(-16)+1(2)}{1+2}, \frac{2(6)+1(-6)}{1+2}\right]\)

= \(\left(\frac{20+4}{3}, \frac{-32+2}{3}, \frac{12-6}{3}\right)\)

= \(\left(\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}\right)\)

= (8, – 10, 2).

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3, 5) and (4, 3, 1)
(ii) (- 3, 7, 2) and (2, 4, – 1)
(iii) (- 1, 3, – 4) and (1, – 3, 4)
(iv) (2, – 1, 3) and (- 2, 1, 3).
Answer.
The distance between point P(x1, y1, z1) and Q(x2, y2, z2) is given by
PQ = \(\)
(i) Distance between points (2, 3, 5) and (4, 3, 1)
= \(\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}\)
= \(\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\)

(ii) Distance between points (- 3, 7, 2) and (2, 4, – 1)
= \(\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{25+9+9}=\sqrt{43}\)

(iii) Distance between points ( – 1, 3, – 4) and (1, – 3, 4)
= \(\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}\)
= \(\sqrt{(2)^{2}+(-6)^{2}+(8)^{2}}\)
= \(\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}\)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= \(\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}\)
= \(\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}\)
= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.2

Question 2.
Show that the point (- 2, 3, 5), (1, 2, 3) and (7,0, – 1) are collinear.
Answer.
Let points (- 2, 3, 5), (1, 2, 3) and (7, 0, – 1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 1

Here, PQ + QP = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P (- 2, 3, 5), Q(1, 2, 3), and P (7, 0, – 1) are collinear.

Question 3.
Verify the following:
(i) (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled triangle.
(iii) (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) are the vertices of a parallelogram.
Answer.
(i) Let points (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) be denoted by A, B and C respectively.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 2

Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.2

(ii) Let (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) be denoted by A, B, and C respectively.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 3

Now, AB2 + BC2 = (3√2)2 + (3√2)2
= 18 + 18 = 36 = AC2
Hence, the given points are the vertices of a right-angled triangle.

(iii) Let points (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) be denoted by A, B, C and D respectively.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 4

Now, AB = CD = 6, BC = AD = √43.
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1).
Answer.
Let P(x, y, z) be the point that is equidistant from points A(l, 2, 3) and B(3, 2, -1).
Accordingly, PA = PB
⇒ PA2 = PB2
⇒ (x – 1)2 + (y – 2)2 + (z – 3)2 = (x – 3)2 + (y – 2)2 + (z + 1)2
x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z +1
⇒ – 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 5.
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (- 4, 0, 0) is equal to 10.
Answer.
We have, a point P(x, y, z) such that, PA + PB = 0

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 5

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 +6

On squaring both sides, we get
x2 + y2 + z2 – 8x + 16 = 100 + x2 + y2 + z2 + 8x + 16 – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ – 16x – 100 = – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ 4x + 25 = 5 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\) [dividing both sides by -4]
Again squaring on both sides, we get
16x2 + 200x + 625 = 25 (x2 + y2 + z2 + 8x +16)
⇒ 16x2 + 200x + 625 = 25x2 + 25y2 + 25z2 +200x + 400
⇒ 9x2 + 25y2 + 25z2 – 225 = 0.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1

Question 1.
A point is on the x-axis. What are its y – coordinates and z – coordinates?
Answer.
Coordinates of any point on the X-axis is (x, 0, 0).
Because at X-axis, both y and z – coordinates are zero.
So, its y and z – coordinates are zero.

Question 2.
A point is in the XZ – plane. What can you say about its y-coordinate?
Answer.
Any point on the XZ – plane will have the coordinate (x, 0, z), so its y – coordinate is 0.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.1

Question 3.
Name the octants in which the following points lie : (1, 2, 3), (4, – 2, 3), (4, – 2, – 5), (4, 2, – 5), (- 4, 2, – 5), (- 4, 2, 5), (- 3, – 1, 6), (2, – 4, – 7).
Answer.
The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, – 5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, – 5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, – 5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate of point ( – 3, – 1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The x-coordinate, y-coordinate, and z-coordinate of point (2, – 4, – 7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

PSEB 11th Class Maths Solutions Chapter 12 Introduction to three Dimensional Geometry Ex 12.1

Question 4.
Fill in the blanks :
(i) The x-axis and y-axis taken together determine a plane known as
(ii) The coordinates of points in the XY-plane are of the form
(iii) Coordinate planes divide the space into octants.
Answer.
(i) XY plane
(ii) (x, y, 0)
(iii) eight

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise

Question 1.
If a parabolic refletor is 20 cm in diameter and 5 cm deep, find the focus.
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 1

Taking vertex of the parabolic reflector at origin, X-axis along the axis of parabola.
The equation of the parabola is of the fonn y2 = 4ax.
Given, depth is 5 cm and diameter is 20 cm.
∵ Point P(5, 10) lies on parabola.
∴ (10)2 = 4a(5) = a = 5
Clearly, focus is at the mid-point of given diameter i.e., S(5, 0).

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 2.
An arch is in the form of a parabola with its Rxi vertical. The arch is 10 m high and 5 m wide at the base. How wide Is it 2m from the vertex of the parabola?
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 2

Let the vertex of the parabola be at the origin and axis be along OY.
Then, the equation of the parabola is
x2 = 4ay
The coordinates of end A of the (2.5, 10) and it lies on the eq. (i).
∴ (25)2 = 4a × 10
⇒ a = \(\frac{6.25}{40}=\frac{5}{32}\) ……………(ii)

On putting the value of a from eq. (ii) in eq.(i), we get
x2 = 4(\(\frac{5}{32}\))y ………………..(iii)

On substituting y = 2 in eq. (iii), we get
x2 = \(\frac{5}{8}\) × 2
⇒ x2 = \(\frac{5}{4}\)
⇒ x = \(\frac{\sqrt{5}}{2}\) m.
Hence, the width of the arc at a height of 2 m from vertex is 2 × \(\frac{\sqrt{5}}{2}\) i.e., √5 m.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 3.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 3

The cable is in the form of a parabola x2 = 4ay.
Focus is at the middle of the cable, the shortest and longest vertical supports are 6 m and 30 m and roadway is 100 m long.
Since, point A(50, 24) lies on parabola x2 = 4ay.
(50)2 = 4a(24)
= 4a = 625
Equation of parabola is x2 = \(\frac{625}{6}\) y
[Put 4a = \(\frac{625}{6}\)]
Let the support at 18 m from middle be l m, then B(18, l – 6) lies on the parabola.
∴ (18)2 = \(\frac{625}{6}\) (l – 6)
l = \(\frac{18 \times 18 \times 6}{625}\) + 6
= 3.11 + 6 = 9.11 (approx)
Hence, the length of supporting wire is 9.11 m.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 4.
An arch is in the form of a semi-ellipse. It is 8 m wide and 2m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 4

Clearly, equation of ellipse is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 ……………(i)
Here, it is given that, 2a = 8 and b = 2.
⇒ a = 4 and b =2
On putting the values of a and b in eq. (i), we get
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1
which is required equation.
Given, AP = 1.5 m
∴ OP = OA – AP = 4 – 15
OP = 25m
Let PQ = k
∴ Coordinate of Q (2.5, k) will satisfy the equation of ellipse.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 5

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 5.
A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 6

Let l be the length of the rod and at any position meet X-axis at A(a, 0) and Y-axis at B(0, b), so that
l2 = a2 + b2
⇒ (12)2 = a2 + b2 …………….(i) [∵ l = 12]
Let P be the point on AB which is 3 cm from A and hence 9 cm from B.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 7

This means that P divides AB in ratio 3 : 9 i.e., 1 : 3.
If P = (x, y), then by section formula, we have

(x, y) = \(\left(\frac{1 \times 0+3 \times a}{1+3}, \frac{1 \times b+3 \times 0}{1+3}\right)\)

(x, y) = \(\left(\frac{3 a}{4}, \frac{b}{4}\right)\)

⇒ x = \(\frac{3 a}{4}\)
⇒ a = \(\frac{4 x}{3}\) and b = 4y
On putting the values of a and b in eq. (i), we get
144 = (\(\frac{4 x}{3}\))2 + (4 y)2
⇒ 1 = \(\frac{x^{2}}{9 \times 9}+\frac{y^{2}}{9}\)

⇒ \(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1
which is required equation.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 6.
Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.
Answer.
The given parabola is x2 = 12y.
On comparing this equation with x2 = 4ay, we obtain
4a = 12
⇒ a = 3.
The coordinates of foci are S (0, a) = S (0, 3)

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 8

Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At y = 3, x2 = 12 (3)
x2 = 36
⇒ x = ± 6.
The coordinates of A are (- 6, 3), while the coordinates of B are (6, 3).
Therefore, the vertices of ∆OAB are 0 (0, 0), A (- 6, 3), and B (6, 3).
Area of ∆OAB = \(\frac{1}{2}\) |0 (3 – 3) + (- 6) (3 – 0) + 6 (0 – 3)|
= \(\frac{1}{2}\) |(- 6)(3) + 6(- 3)| unit2
= \(\frac{1}{2}\) |- 18 – 18| unit2
= \(\frac{1}{2}\) |- 36| unit2
= \(\frac{1}{2}\) × 36 unit2
= 18 unit2
Thus, the required area of the triangle is 18 unit2.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise

Question 7.
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Answer.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 9

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as.
The equation of the ellipse will be of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 , where a is the semi-major axis.
Accordingly, 2a = 10
⇒ a = 5.
Distance between the foci, 2c = 8
⇒ c = 4
On using the relation c = \(\sqrt{a^{2}-b^{2}}\) we obtain 4 = \(\sqrt{25-b^{2}}\)
⇒ 16 = 25 – b2
⇒ b = 25 – 16 = 9
⇒ b = 3
Thus, the equation of the path traced by the man is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

Question 8.
An equilateral triangle is inscribed in the parabola y =4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Answer.
Given, equation of parabola is y2 = 4ax.
Let p be the side of equilateral ∆OAB, whose one vertex is the vertex of parabola.
Then, by symmetry, AB is perpendicular to the axis ON of parabola.
Let ON = x, then BN = \(\frac{p}{2}\)
Since, B(x, \(\frac{p}{2}\)) lies on parabola.

PSEB 11th Class Maths Solutions Chapter 11 Conic Sections Miscellaneous Exercise 10

⇒ p2 = 64 × 3 × a2
⇒ p = 8a√3
Hence, side of an equilateral triangle is 8a√3.