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# 11M01_Fundamentals of Mathematics_Avanti Module

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M1 – Fundamentals of Mathematics 1 Sixth Edition M1. Fundamentals of Mathematics TABLE OF CONTENTS M1. Fundamentals of Mathematics 1 M1.1 Logarithms........................................................................................................................................................................................2 M1.2 Differentiation.................................................................................................................................................................................8 M1.3 Chain Rule and Maxima Minima...........................................................................................................................................16 M1.4 Anti Derivatives and Integration .........................................................................................................................................22 Test Practice Problems .........................................................................................................................................................................27 Answer Key ................................................................................................................................................................................................34

M1 – Fundamentals of Mathematics 2 M1.1 Logarithms CONCEPTS 1. Revise the properties and formulae of Indices studied up to Grade X. 2. Revise the properties and rationalization method of surds studied up to Grade X. 3. Define the logarithm operation and basic terms such as base, argument and value of logarithms. 4. List the conditions on the base and argument for which logarithms are applicable. 5. Define two standard logarithms – common and natural logarithms. 6. Practice the inter - conversion of logarithmic form to exponential form, and find the logarithmic value of different argument base combinations. PRE-READING  Recap – Indices The use of exponents, also called powers or indices, allows us to write products of factors and also to write very large or very small numbers quickly. We have seen previously that 2 × 2 × 2 × 2 × 2, can be written as 25. 25 reads “two to the power of five” or “two with index five.” In this case 2 is the base and 5 is the exponent, power or index. We say that 25 is written in exponent or index notation. We shall first recap the basic properties of Indices which you have studied previously, as they will form the base for the understanding of Logarithms and Exponentials. Note: The below rules are always valid for ������ > 0, but not always valid when ������ < 0 Rule Property Example 1 2 ������������ × ������������ = ������������+������ 23 × 24 = (2 × 2 × 2) × (2 × 2 × 2 × 2) = 27 = 23+4 3 ������������ = ������������−������ 25 = (2×2×2×2×2) = 23 = 25−2 4 ������������ 22 (2×2) 5 (������������)������ = ������������×������ (23)2 = (2 × 2 × 2) × (2 × 2 × 2) = 26 = 23×2 6 (������������)������ = ������������������������ (3 × 4)2 = (3 × 4) × (3 × 4) = (3 × 3) × (4 × 4) = 32 × 42 7 ������0 = 1 (Put ������ = ������ in Rule 2) ⟹ 22 = 22−2 ⟹ 1 = 20 22 ������−������ = 1 (Put ������ = 0 in Rule 2) ⟹ 20 = 20−2 ⟹ 1 = 2−2 ������������ 22 22 1 ���√��� ������ (Put ������ = 1 in Rule 3) ⟹ 1 ������ = ���������1���×������ ⟹ 1 ������ = ������ ������ ������������ = (������������) (������������) 1 ���√��� ������ Taking ������th root on both sides, we get ������������ =  Introduction to Logarithms So far we have mainly encountered exponential equations of the type ������������ = ������ where either ������ or ������ are unknowns, and ������ is given. For instance, ������2 = 4 or 32 = ������, which are relatively simpler to calculate. M1.1

M1 – Fundamentals of Mathematics 3 However, we have no easy approach of finding ������ or the exponent, if it is unknown. For example, try solving 2������ = 5 to find the value of ������. You will realize that this is not as straightforward. Logarithms is a mathematical operator that helps us to easily calculate the value of the exponent which satisfies such equations. You will notice several other applications of logarithms in simplifying complex calculations as we learn more about the topic. Definition: If ������������ = ������, we define ������������������������ ������ = ������. This is read as “log of c to the base a is equal to b”. Notation: log������ ������ = ������ base (a) value or number (c) exponent (b) Mathematically, this means in the equation log������ ������ = ������ , the value ������ is the answer to the question \"To what power must ������ be raised, in order to yield ������?\" For example, we know that 102 = 100. Hence, we can say that log10 100 = 2. The reverse also applies, which means if log������ ������ = ������, then ������������ = ������ must be true. This notation is simply a new concept, not a difficult one, and by the time you complete this chapter you will be able to interrelate the logarithmic equation with the exponential form. Conditions and special cases: Condition / Special Case Rationale ������ > ������ i.e. Base must always be positive Negative numbers cannot be raised to all exponents. For example, 1 is not defined. Hence log−1 ������ = 1 cannot be defined 2 (−1)2 ������ ≠ ������ i.e. Base must not be equal to 1 1 raised to any value will always be equal to 1. For example, log1 2 = ������ cannot be defined. ������ > ������ i.e. Number must always be positive Since, ������ > 0, ������������ will always be positive, which means ������ must always ������������������������ ������ = ������, for all allowed values of ������ be positive. ������������������������ ������ = ������, for all allowed values of ������ ������0 = 1 for all values of ������ > 0 ������1 = ������ for all values of ������ > 0  Logarithmic Scale: One of the reasons logarithms was introduced was to ‘scale down’ large numbers to numbers of 1 or at most 2 digits. To understand this, fill the table below and notice the logarithmic scale in comparison to a linear scale. Logarithm Value Logarithm Value Logarithm Value Logarithm Value log10 1 log10 10 log10 100 log10 1,000 log10 10,000 log10 1,00,000 log10 10,00,000 log10 1,00,00,000

M1 – Fundamentals of Mathematics 4 You should have noted that as numbers become 10 times bigger, the value of the logarithm increases only by 1. Hence logarithm is used to ‘scale down’ large numbers. Linear Scale (number increases 10 times) → Linear Value ������ 1 10 100 1,000 10,000 100,000 Logarithmic 0 1 2 3 4 5 Value log10 ������ Logarithmic Scale (value increases by 1 unit) →  Standard bases in logarithms: 1. Common Logarithms: Logarithms to the base 10 ⟹ written as ������������������ ������ 2. Natural Logarithms: Logarithms to the base e ⟹ written as ������������ ������ Here ������ is an irrational number whose approximate value is 2.71828…. (also called Euler’s number) Choosing the appropriate base is important from the point of view of simplifying the numbers for calculation purposes. Please note that different sections of Science use different bases depending on the ease of calculation, hence you must always confirm the base when using logarithms.  Laws of Logarithms: \"The logarithm of a product is equal to the sum of the logarithms of each 1. ������������������������ ������������ = ������������������������ ������ + ������������������������ ������ factor with the same base.\" Let log������ ������ = ������ and log������ ������ = ������. ������ = ������������ ⟺ log������ ������ = ������ Thus, by the definition of logarithm, ������������ = ������ and ������������ = ������. from laws of indices ∴ ������������ = ������������ × ������������ ⇒ ������������ = ������������+������ ������ = ������������ ⟺ log������ ������ = ������ Thus, log������ ������������ = ������ + ������ Re-substituting the values of ������ and ������, we get ������������������������ ������������ = ������������������������ ������ + ������������������������ ������, which is the above mentioned law. 2. ������������������������ ������ = ������������������������ ������ − ������������������������ ������ \"The logarithm of a quotient is equal to difference of the logarithm of the ������ numerator and the denominator with the same base.\" Let log������ ������ = ������ and log������ ������ = ������. Thus, by the definition of logarithm, ������������ = ������ and ������������ = ������. ∴ ������ = ������������ or ������ = ������������−������ from laws of indices ������ ������������ ������ ������ = ������������ ⟺ log������ ������ = ������ ������ Thus, log������ ������ = ������ − ������ Re-substituting the values of ������ and ������, we get ������������������������ ������ = ������������������������ ������ − ������������������������ ������, which is the above mentioned law. ������ 3. ������������������������ ������������ = ������ ������������������������ ������ \"The logarithm of a number raised to an exponent is equal to the exponent times the logarithm with the same base.\" M1.1

M1 – Fundamentals of Mathematics 5 This is easily derived from Law 1 i.e. log������ ������������ = log������ ������ + log������ ������ Let ������ = ������. Then, the above equation becomes log������ ������2 = 2log������ ������ Also, we can extend the law to log������ ������������������ = log������ ������ + log������ ������ + log������ ������ This would give log������ ������3 = 3log������ ������ on putting ������ = ������ = ������ This can be generalized to ������������������������ ������������ = ������ ������������������������ ������ for any exponent ������. 4. ������������������������ ������ = ������������������������ ������ \"The logarithm of one number with respect to another is equal to the ratio ������������������������ ������ of their logarithms with respect to another base\" Let log������ ������ = ������, log������ ������ = ������, log������ ������ = ������. ∴ ������ = ������������; ������ = ������������; ������ = ������������ Thus, ������ = ������������ = ������������ Substituting the value of ������ = ������������, we get (������������)������ = ������������ ⟹ ������������������ = ������������ Comparing the exponents, we have ������������ = ������ ⟹ ������ = ������ ������ Re-substituting the values of ������, ������ and ������, we have ������������������������ ������ = ������������������������ ������, which is the required law. ������������������������ ������ 5. ������������������������ ������ = ������ ������ \"The logarithm of one number with another is equal to the reciprocal of the ������������������������ logarithm of the second number with the first\" Replace ������ = ������ in Law 4 and you will get an important property that is often used in problem solving. PRE-READING EXERCISE Q1. Find the value of: II. (2 1)−2 I. (1)−1 2 4 IV. 30 − 3−1 III. (9 × 104) ÷ (3 × 103) II. log10 1000 − log10 100 Q2. Find the value of: I. log2 8 Q3. log10 16 = ������ log10 2. Find ������. Q4. State whether true or false: log64 4 = 3 Q5. Convert to logarithmic form: 8−2 = 1 64 IN CLASS EXERCISE LEVEL 1 3 2 II. 92 = 27 Q1. Find the value of (−27)3 Q2. Express the following in logarithmic form: I. 63 = 216

M1 – Fundamentals of Mathematics 6 Q3. Find the value of II. log5 3√5 M1.1 I. log2 32 III. log10 0.0001 IV. log1 8 Q4. If (������)������−1 = (������)������−3, find the value of ������. (Assume ������ ≠ ������) 2 ������ ������ LEVEL 2 Q5. 100.35 = ������; 100.70 = ������; ������������ = ������2. Find the value of ������ A) 0.35 B) 0.70 C) 2 D) 4 Q6. Solve for ������: log(3������ + 2) − log(3������ − 2) = log 5 Q7. Show that log6 7 = log2 7 1+log2 3 Q8. If log2 ������ + log4 ������ + log16 ������ = 21, find ������. 4 LEVEL 3 Q9. ������ = 1 + log������ ������������ ; ������ = 1 + log������ ������������ ; ������ = 1 + log������ ������������ ; ������������������ ≠ 0 Prove that 1 + 1 + 1 = 1 ������ ������ ������ Q10. For what values of ������ will log6(������ + 2) + log6(������ + 3) = 1 be true? A) ������ = 0 and −5 B) ������ = 0 only C) ������ = −5 only D) none of these Q11. If log10 2 = 0.301, the number of digits in 264 is: B) 19 D) 32 A) 64 C) 20 HOMEWORK LEVEL 1 Q1. (0.04)−1.5 =? A) 25 B) 0.08 C) 125 D) 0.2 Q2. Identify whether the below statements are true or false: II. log2 4√2 = 2.5 IV. log2 0 = 1 I. log2 1 = −1 √2 2 III. 5log5 12 = 12 Q3. Prove that log 540 = 2 log 2 + 3 log 3 + log 5 Q4. Find ������ such that log������ ������ = log������ 4 + log������(������ − 1) Q5. Evaluate 3 log2 10 + log2 3 − log2 15 − 2 log2 5

M1 – Fundamentals of Mathematics 7 LEVEL 2 Paragraph For Questions 6 and 7 Chemists define the acidity or alkalinity of a substance according to the formula ������H = – log[H+] where [������+] is the amount of hydrogen ions in the solution. Solutions with a ������������ value less than 7 are acidic; solutions with a ������������ value greater than 7 are basic; solutions with a ������������ value equal to 7 (such as pure water) are neutral. Q6. Suppose that you test apple juice and find that the hydrogen ion concentration is [H+] = 0.0001M. Find the pH value and determine whether the juice is basic or acidic. Q7. Find the [������+] ion concentration of pure water. Q8. Solve for ������: log2(������2) = (log2 ������)2 Q9. Show that log������ ������5 × log������ ������3 × log������ ������7 = 105 Q10. 1 + 1 + 1 = ? log3 60 log4 60 log5 60 A) 0 B) 1 C) 60 D) 5 Q11. Solve log2 ������ + log2(������ − 2) = 3 LEVEL 3 Q12. log4(������ + 1) + log16(������ + 1) = log4 8. Find ������. Q13. (������������������������)������+������−������ (������������������������ ������+������−������ (������������������������)������+������−������ = ) A) ������������������������ B) ������������+������+������ C) 1 D) ������������2+������2+������2

M1 – Fundamentals of Mathematics 8 M1.2 Differentiation CONCEPTS 1. Dependent and independent variables 2. Dependent variable as a mathematical function of an independent variable 3. Average rate of change of functions 4. Instantaneous rate of change for functions 5. Average and instantaneous rate of change with the slope of a line 6. List of derivatives for standard mathematical functions 7. Rules of differentiation Calculus - an Introduction Calculus is a branch of mathematics that is concerned with comparing quantities that are continuously changing. It is used extensively in science and engineering since most of the quantities we measure (such as velocity, acceleration, current in a circuit etc.) are varying. Calculus was developed independently by the Englishman, Sir Isaac Newton, and by the German, Gottfried Leibniz. They were both working on problems of motion towards the end of the 17th century. There was a bitter dispute between the men over who developed Sir Isaac Newton calculus first. There are two main branches of calculus. The first is differentiation (or derivatives), which helps us to find a rate of change of one quantity compared to another. The second is integration, which is the reverse of differentiation. We may be given a rate of change and we need to work backwards to find the original relationship (or equation) between the two quantities. Gottfried Leibniz Before we dive into calculus, look at the situations below and answer the questions that follow: Situation 1: A graph depicting Akshay’s savings over the past 5 months after he started a new job has been shown below. Savings 10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0 0123456 Month M1.2

M1 – Fundamentals of Mathematics 9 Q1. What were Akshay’s savings just before he started his new job? Q2. How much did Akshay save in the first month at his job? Q3. How much did Akshay save in the second month at his job? Q4. What was Akshay’s average rate of savings per month? Q5. Was Akshay’s average rate of savings the same as his savings in each month? Situation 2: A graph depicting Chinmay’s Mumbai marathon run in 2015. Chinmay ran the full marathon which was a distance of 42 km. Distance covered (in km) 45 Distance covered vs time taken distance 40 42 38 35 32 25 25 30 14 25 7 20 1234567 Time (in hours) 15 10 50 0 0 Q1. How much time did Chinmay take to complete the marathon? Q2. What was Chinmay’s speed in the first hour? Q3. What was Chinmay’s speed in the last hour? Q4. How many kilometres did Chinmay run in the 4th hour? Q5. In which hour was Chinmay fastest in the entire marathon? Q6. What was his average speed throughout the marathon? Q7. Was his average speed the same as his speed at all instants? Q8. From the graph, which situation is most likely to have happened? I. He ran at a constant speed throughout the marathon II. His speed gradually decreased throughout the marathon III. He initially ran at a constant speed to conserve energy, then accelerated due to which he lost a lot of energy. He then slept for an hour, and continued running at a gradually decreasing speed till he finished the marathon.

M1 – Fundamentals of Mathematics 10 Situation 3: A slice of pizza is kept in a fridge. Bacteria begins to grow on the slice, and starts multiplying itself to create more bacteria over time. The graph below depicts the number of bacteria on the pizza over time. Number of bacteria 70 Number of Bacteria vs Time 60 50 64 7 40 30 32 20 16 10 0 248 0 123456 0 Time (in hours) Q1. What was the change in number of bacteria in the second hour? Q2. What was the change in number of bacteria in the fifth hour? Q3. What was the average increase in bacteria per hour? Q4. Was the average rate of change equal to the change in every hour? Q5. What was the rate of change of bacteria at ������ = 4 hours? (Detailed solutions are available in the answer key at the end of the chapter. Check your answers before reading ahead) Slope – Average Rate of Change The concept of slope is central to the understanding of differentiation. In each of the situations above, you knowingly or unknowingly used the concept of slope to find the average rate of change. Slope of a straight line that passes through two points (������������, ������������) and (������������ , ������������) is equal to ������������−������������. ������������−������������ Situation 1 Average Savings (������������. ������������������������ − ������������. ������������������������) (������ ������������������������������������) Rs. 1000 per month M1.2

M1 – Fundamentals of Mathematics 11 Situation 2 Average Speed Situation 3 ������������ − ������ = ������ − ������ = 6 kmph Average Rate of Change versus Instantaneous Rate of Change In situation 1 above, you must have observed that the average savings were equal to the savings of each month, where in situation 2 and 3, the average speed or the average increase in bacteria was not equal to the change at every step. For instance, in situation 2, Chinmay’s average speed or rate of change of distance is 6 kmph (42 kms divided by 7 hours), but this data does not tell us whether he constantly ran at 6 kmph, or whether he ran faster than 6 for some time interval and rested / stopped for an entire hour. Thus, we need to be able to measure the rate of change of variables at every instant (INSTANTANEOUS), as opposed to over the entire period (AVERAGE) to have a better understanding of what actually occurred.

M1 – Fundamentals of Mathematics 12 Slope – Instantaneous Rate of Change? Just as we use slope to define the average rate of change, we will now see how we can find the rate of change at any instant from the slope of the curve at that instant. Revisit question 5 of the bacteria example i.e. “What was the rate of change of bacteria at ������ = 4 hours?” This is equivalent to the question: “What is the instantaneous rate of change of bacteria at ������ = 4?” The average change occurs between 2 distinct points, while the instantaneous change occurs at a single point. Zoom in to the graph around ������ = 4 and consider secant 1 in the figure below. Between ������ = 3 and ������ = 5, slope of the secant = average rate of change = ������������−������ = ������������ ������−������ Moving much closer to ������ = 4 in the graph, consider secant 2. Between ������ = 3.8 and ������ = 4.2, slope of the secant = average rate of change = ������������−������������ = ������������ ������.������−������.������ (approximate values taken from the graph) If we could choose 2 points extremely close to ������ = 4, and join the corresponding points on the graph, the slope of the secant would be as good as the rate of change at ������ = 4. You notice that as we get closer to ������ = 4, this secant converges to the tangent of the curve at ������ = 4. (A tangent is a line that touches a curve at only one point) Instantaneous rate of change = slope of tangent to curve at that point What is Differentiation? Differentiation is a tool in calculus that helps us find the instantaneous rate of change of any mathematical function (a relation between a dependent and independent variable) using the concepts explained above. In mathematics, we will often encounter 2 variables that depend on each other. Any change in one will bring about a change in the other, because of this dependence. We usually call the independent variable ������, and the variable that depends on ������ is called ������. M1.2

M1 – Fundamentals of Mathematics 13 Suppose we vary ������ adding to it a very small amount which we call ������������. We are thus causing ������ to become ������ + ������������. Then, because ������ has been changed, ������ would have changed to ������ + ������������. Here the bit ������������ may be positive, negative or zero depending on the relation between ������ and ������. The rate of change of ������ with respect to ������ in this case would be the ratio of the change in ������ to the change in ������. This is given by ������������ The process of finding ������������ is called differentiation. ������������. ������������ Now we shall look at differentiating some standard functions: 1. Linear Function ������ = ������������ + ������ We have seen earlier when we studied straight lines that as ������ changes, ������ changes too. Take a value of ������, for which ������ = ������������ + ������, and increase it to ������ + ������������. Then, ������ + ������������ = ������(������ + ������������) + ������ On rearranging, we get {������} + ������������ = {������������ + ������} + ������������������ Since {������} = {������������ + ������}, we can cancel them on both sides, to get the new equation ������������ = ������������������ ������������ This gives ������������ = ������ This means that the rate of change of ������ at any point ������ is equal to the slope of the straight line itself. We saw earlier that the instantaneous rate of change = slope of tangent to curve, and we know that the tangent to a straight line is the line itself, hence the rate of change will be equal to the slope of the line. 2. Quadratic Function ������ = ������2 Using the same approach as above, ������ + ������������ = (������ + ������������)2 Expanding, we get, ������ + ������������ = ������2 + 2������������������ + (������������)2 Cancelling ������ and ������2, we get, ������������ = 2������������������ + (������������)2 Since ������������ is very small as we take the other value to be very close to ������, we know that (������������)2 would be negligible. Hence, we get, ������������ = 2������������������ ⇒ ������������ = 2������ ������������ We shall study more on this in detail in the class. Differentiation Formulae for Standard Functions: Function type ������ ������������ ������������ Constant K 0 ������ Polynomial ������2 1 ������3 Power Rule (valid for all ������ except zero) ������������ 2������ Exponential ������ ������ 3������2 Logarithmic ������������ ������������������−1 ������ ������ Trigonometric ln ������ ������������ ln ������ sin ������ 1 cos ������ ������ tan ������ cos ������ − sin ������ sec2 ������

M1 – Fundamentals of Mathematics 14 IN CLASS EXERCISE LEVEL 1 Q1. Using the basic rules of differentiation, solve ������������ for each of the functions below ������������ Sr No ������ ������������ ������������ I 3������2 II 3������ III −2������ IV 2√������ V 5 sin ������ + 2 VI 1 3 ������2 LEVEL 2 Paragraph for Questions 2 – 5 Displacement of a particle is defined as the shortest path length covered by it during its motion. Its velocity is defined as the rate of change of its displacement. The acceleration of the particle is defined as the rate of change of its velocity. The displacement (in metres) of a particle varies with time (in seconds) as ������(������) = ������3 − 6������. This means that as time 2 increases, its displacement will change according to the formula above. Q2. What is the velocity of the particle as a function of time? Q3. At what instant will the particle stop? Q4. What is the value of its acceleration at ������ = 1 s? Q5. What is the velocity of the particle when its acceleration is 9 ������/������2? LEVEL 3 Q6. Which function will increase the most at ������ = 4? B) ������ = 5√������ D) ������ = 125 A) ������ = 3 ln ������ − ������2 C) ������ = 3������2 + 2������ Q7. Using the product and quotient rule, solve ������������ for each of the functions below ������������ Sr No ������ ������������ ������������ I cot ������ II ������3 cos ������ III 2������3 7������ + 5 IV ������������ tan ������ M1.2

M1 – Fundamentals of Mathematics 15 HOMEWORK LEVEL 1 Q1. Using the basic rules of differentiation, solve ������������ for each of the functions below ������������ Sr No ������ ������������ ������������ I 5 − tan ������ II sec ������ III 4������ + 3 IV ������3(������2 − 4) LEVEL 2 Q2. The displacement (in metres) of a particle varies with time (in seconds) as ������(������) = 3������2 − 5������ + 2 I. Find the velocity of the particle at all instances when its displacement becomes zero. II. Find the acceleration of the particle when the velocity is 1 ������/������. III. Find the displacement of the particle when it comes to rest momentarily. Q3. Using the basic rules of differentiation, solve ������������ for each of the functions below ������������ Sr No ������ ������������ I ������3 ln ������ ������������ II 6������ cos ������ 4 III 3√������2 3 sin ������ IV √������ LEVEL 3 Q4. Which function will increase the fastest at ������ = 0? B) ������ = ������������ − 4 D) ������ = 5 sin ������ A) ������ = 4������ − 2������2 C) ������ = 3������2 + 2������ − 7 Q5. For which of the following functions between ������ and ������ will the instantaneous rate of change always be equal to the average rate of change? A) ������ = ������2 + 2 B) ������ = 3������ C) ������ = sin ������ D) ������ = ln ������

M1 – Fundamentals of Mathematics 16 M1.3 Chain Rule and Maxima Minima CONCEPTS 1. Chain rule of differentiation 2. Increasing / decreasing function 3. Local maximum / local minimum 4. Maximum / minimum value of mathematical functions PRE-READING Note: There is no pre-class exercise for this topic. Please attempt the solved examples on the basis of the pre reading before reading the solutions. Recap:  Differentiation or ������ of a function ������(������) denotes the value of the instantaneous rate of change of the function at any ������������ point ������.  Differentiation of standard polynomial, logarithmic and trigonometric functions form the base of calculus, and must be committed to memory.  ������ ������������(������) = ������ ������ ������(������) constant rule ������������ ������������ The differential of k times a function is equal to k times the differential of the function  ������ [������(������) ± ������(������)] = ������ ������(������) ± ������ ������(������) sum / difference rule ������������ ������������ ������������ To differentiate the sum / difference of two functions, differentiate each function independently and add / subtract them together.  ������ (������������) = ������ ������ ������ + ������ ������ ������ product rule ������������ ������������ ������������ To differentiate the product of two functions, multiply each function by the differential of the other, and add together the two products obtained.  ������ (������) = ������������������������������−������������������������������ quotient rule ������������ ������2 ������ To differentiate the quotient of two functions, multiply the divisor by the differential of the dividend; then multiply the dividend by the differential of the divisor; and subtract. Lastly divide by the square of the divisor. Derivative of a function of a function: When we use a function like sin 2������ or ������������2 or √������2 + 1, we are in fact dealing with composite functions or functions of a function. sin 2������ is the sin function of 2������. Similarly ������������2 is the exponential function of ������2. √������2 + 1 is the square root function of the polynomial ������2 + 1, read as ‘square root of (������2 + 1)′ When we talk about functions of a function in a general setting we will use the notation ������(������(������)) where both ������ and ������ are functions. For example, in the above cases, ������(������) and ������(������) would be: a) ������(������(������)) = sin 2������ ������(������) = sin ������ and ������(������) = 2������ b) ������(������(������)) = ������������2 ������(������) = ������������ and ������(������) = ������2 c) ������(������(������)) = √������2 + 1 ������(������) = √������ and ������(������) = ������2 + 1 M1.3

M1 – Fundamentals of Mathematics 17 Chain Rule The chain rule helps us differentiate ‘composite functions’, which are essentially standard functions mixed inside one another, and not simply multiplied / divided by each other. In mathematical terms, a composite function would be written as ℎ(������) = ������(������(������)), where ������(������) and ������(������) are standard functions, and ������(������) is ‘inside’ ������(������). Now, to differentiate ℎ(������) with respect to ������, i.e. find ������ℎ, we essentially need to calculate the change in ℎ(������) produced ������������ by changing ������. We know that if we change ������, first ������(������) will change, and then because ������(������) will change, ������(������(������)) will change. This is similar to a chain reaction, and to calculate ������ = change in ������(������(������)), we must know ������ = change in ������(������(������)) and ������ = change in ������(������) change in ������ change in ������(������) change in ������ separately. You will notice that if you multiply ������ and ������, change in ������(������) will cancel out and you will get ������. Mathematically, this is written as: ������ ������(������(������)) ������������(������(������)) ������������(������) ������������ = ������������(������) × ������������ This is the chain rule of differentiation. For example, let ℎ(������) = ln cos ������. You can see that ������(������) = ln ������ and ������(������) = cos ������ Thus, applying the chain rule, we see that: ������ ln cos ������ ������ ln cos ������ ������ cos ������ ������������ = ������ cos ������ × ������������ (������) (������������) i) Substitute cos ������ = ������, Thus, (i) becomes ������ ln ������ which we know is 1 i.e. 1 ������������ ������ cos ������ ii) We know that ������ cos ������ = − sin ������ ������������ Thus, using chain rule, ������ ln cos ������ 1 ������������ = cos ������ × − sin ������ = − tan ������ Important points to note: 1. Chain rule can be extended to more than 2 standard functions combined together. 2. While using the chain rule, it is important to identify the outermost function and the sequence of functions that are linked to one another. 3. The composite function must be broken down into standard functions whose differential is known using the basic properties and table of differentials. Solved Examples: ������������ Find ������������ for: 1. ������ = tan 2������ 2. ������ = sin2 ������ 3. ������ = √(������2 + 3)

M1 – Fundamentals of Mathematics 18 Solution: 1. ������ = tan 2������ The first step is to identify the outermost function ������(������) and the inner function ������(������) We can see that ������ is tan (something) where ‘something' is not equal to ������ ������ = ������(������(������)) where ������(������) = tan ������ and ������(������) = 2������ Applying chain rule, ������������ ������ ������(������(������)) ������������(������(������)) ������������(������) ������ tan 2������ ������ tan 2������ ������2������ ������������ = ������������ = ������������(������) × ������������ ⇒ ������������ = ������2������ × ������������ ������ tan 2������ ������ tan ������ ������2������ Substituting 2������ = ������ to convert it into the standard differential ⇒ ������������ = ������������ × ������������ From basic differentiation formulae ������ tan 2������ = sec2 ������ ×2 ⇒ ������������ ⇒ ������ tan 2������ = 2 sec2 2������ ������������ Note: Had we not applied chain rule, and directly differentiated tan 2������ using the differentiation formula for tan ������ ������ tan ������ = sec2 ������ , we would get sec2 2������ , which is not the correct answer. i. e. ������������ 2. ������ = sin2 ������ Similarly, we can see that ������ is something squared. Which means ������(������) would be ������2 and the something (in this case sin ������) would be ������(������). i.e. ������ = ������(������(������)) where ������(������) = ������2 and ������(������) = sin ������ ������������ ������ ������(������(������)) ������������(������(������)) ������������(������) ������������ = ������������ = ������������(������) × ������������ ������ sin2 ������ ������ sin2 ������ ������ sin ������ Substituting sin ������ = ������ to convert it into standard differential ⇒ ������������ = ������ sin ������ × ������������ Using standard differentiation formulae Re-substituting ������ = sin ������ ������ sin2 ������ ������������2 ������ sin ������ ⇒ ������������ = ������������ × ������������ ������ sin2 ������ ⇒ ������������ = 2������ × cos ������ ������ sin2 ������ ⇒ ������������ = 2 sin ������ cos ������ 3. ������ = √(������2 + 3) This is equivalent to ������ = √something. Thus, ������(������) = √������ and ������(������) = ������2 + 3 ������������ ������ ������(������(������)) ������������(������(������)) ������������(������) ������������ = ������������ = ������������(������) × ������������ ⇒ ������ √(������ 2 + 3) = ������√(������2 + 3) ������ (������2 + 3) ������������ ������ (������2 + 3) × ������������ ⇒ ������ √(������2 + 3) = ������√������ ������ (������2 + 3) Substituting (������2 + 3) as ������ ������������ ������ ������ × ������������ Using the standard differentials ⇒ ������ √(������ 2 + 3) = 1 × 2������ ������������ 2√������ ⇒ ������ √(������ 2 + 3) = 1 × 2������ ������������ 2√������2 + 3 Therefore, ������ √(������2 + 3) = ������ 3) ������������ √(������2 + M1.3

M1 – Fundamentals of Mathematics 19 Finding Maximum And Minimum Values Of Functions: One very interesting application of derivatives is to find maximum/minimum values of functions. The local maximum of a function is the highest value of a function in a neighborhood. This means that the function increases up to this point and then decreases. Similarly, the local minimum of a function is the lowest value of a function in a neighborhood, where the function is decreasing before and increasing after. Condition For Maximum Value: For instance, let ������ = ������(������) be a function of ������. If this function were to have a maximum value at a point ������0, then the graph of the function would resemble the image alongside. ������(������) ������ ������������ ������(������) Before ������0 Increasing Positive At ������0 Maximum 0 After ������0 Decreasing Negative ������������ Notice the change in ������������ around the maximum point. It is going from positive values to negative values, which means it is decreasing around ������ = ������0. If you assume ������������ to be ������(������), the ������������������������ ������������ ������������������������������������ of ������(������) is negative. ������������ This means that ������ ������(������) < 0 or ������ ������������ < 0 ������������ ������������ (������������) ������2������ This is referred to as the second differential of ������, or simply ������������2. ������ ������������ If ������(������) is maximum at ������ = ������0, ������������ ������(������) = ������ and ������������������ ������(������) < ������ ������������ ������ = ������0 Condition For Minimum Value: Similarly, if ������(������) is to have a minimum at ������ = ������0, then ������(������) will decrease till ������ = ������0 and then start increasing as shown in the graph alongside. Here, Before ������0 ������(������) ������ At ������0 ������������ ������(������) After ������0 Decreasing Negative Maximum 0 Increasing Positive ������ ������������ If ������(������) is minimum at ������ = ������0, ������������ ������(������) = ������ and ������������������ ������(������) > ������ ������������ ������ = ������0

M1 – Fundamentals of Mathematics 20 Solved Examples: 1. Prove that sin ������ is maximum at ������ = 90° Sol: For sin ������ to be maximum at ������ = 90° ������ ������(������)������=90° = 0 and ������2 ������(������)������=90° < 0 ������������ ������������2 ������ ������ ������2 ������ ⇒ ������������ ������(������) = ������������ sin ������ = cos ������ and ������������2 ������(������) = ������������ cos ������ = − sin ������ ������2 ⇒ ������ ������(������)������=90° = cos 900 = 0 and ������������2 ������(������)������=90° = − sin 90° = −1 < 0 ������������ Thus, sin ������ is maximum at ������ = 900 2. Find the maximum value of ������(������) = 2 + 3������ − 4������2 ������ ������2 Sol: f(x) will be maximum at ������0 if ������������ ������(������)������=������0 = 0 − (������) & ������������2 ������(������)������=������0 < 0 − (������������) Using differentiation rules, we get ������ ������(������) = ������������ = 3 − 8������ ������������ ������������ ������ 3 From (i), ������������ ������(������)������=������0 = 0 ⇒ 3 − 8������0 = 0 ⇒ ������0 = 8 ������2 ������ From (ii), ������������2 ������(������) = ������������ (3 − 8������) = −8 ������2 ⇒ ������������2 ������(������)������=38 = −8 < 0 3 Thus, both conditions are satisfied at ������0 = 8 3 ∴ ������(������) is maximum at ������0 = 8 3 3 9 41 Substituting ������0 in ������(������), we get (8) = 2 + 3 × 8 − 4 × 64 = 16 , which is its maximum value. 3 Note: 8 is not the maximum value of (������), it is the value of ������ at which ������(������) is maximum. IN CLASS EXERCISE IN CLASS EXERCISE 1 Differentiate the following functions with respect to ������ using chain rule of differentiation: Q1. ������ = sin(2������ + 1) Q2. tan 5������ Q3. ln 3������2 Q4. ������ = cos(sin ������) IN CLASS EXERCISE 2 Q5. Which of the functions will have no local maximum or minimum value? I. ������2 II. sin ������ III. 3������ + 4 D) All three A) and II) only B) II) and III) only C) III) only M1.3

M1 – Fundamentals of Mathematics 21 Q6. Which of the functions has a minimum value at ������ = 0? A) ������ = 5 − ������2 B) ������ = 3������ + 7 C) ������ = cos ������ D) ������ = ������2 Q7. Displacement of a particle is given by ������(������) = ������2 − 4������ + 5. At what instant is the displacement of the particle minimum? What is the velocity of the particle at this instant? Q8. Find two positive numbers whose sum is 16 and product is maximum. Q9. A wire of length 28 m is to be shaped into a rectangle. What should be the length of the sides such that its area is maximum? HOMEWORK LEVEL 1 Q1. Use chain rule to differentiate the functions below: I. ������(������) = cos 3������ II. ������(������) = ln 5������ III. ℎ(������) = ������−������ LEVEL 2 Q2. Find the maximum value of sin ������ + cos ������ and the value of ������ at which it occurs. Q3. Use chain rule to find the derivative of the functions below I. ������ = ������tan ������ II. (������2 + ������ + 1)4 III. √3������ + 2 Q4. Two numbers add up to 10. Find the minimum value of the sum of their squares. LEVEL 3 Q5. The height reached in time ������ by a particle thrown upward with a speed ������ is given by ℎ = ������������ − 1 ������ t2. Find the 2 time taken to reach the maximum height. (Assume ������ and ������ to be constant) Q6. Prove that the local minimum value of ������ + 1 is larger than its local maximum value. ������

M1 – Fundamentals of Mathematics 22 M1.4 Anti Derivatives and Integration CONCEPTS 1. Anti-derivative ������(������) of a function ������(������) 2. Anti-derivatives of standard functions 3. Indefinite and Definite Integration 4. Area under a curve PRE-READING 1. Introduction By now you will be familiar with differentiating common functions and will have had the opportunity to practice many techniques of differentiation. In this topic we will carry out the process of differentiation in reverse. That is, we will start with a given function, ������(������) and ask what function or functions, ������(������), would have ������(������) as their derivative. This leads us to the concepts of an anti-derivative and integration. 2. Anti-derivatives ⇒ differentiation in reverse. Consider the function ������(������) = ������2. Suppose we write its derivative as ������(������), that is ������(������) = ������������ . You already know ������������ ������������ how to find this derivative ������(������) = ������������ = 2������. Suppose now that we work back to front and ask ourselves which function or functions could possibly have 2������ as a derivative. Clearly, one answer to this question is the function ������2. We say that ������(������) = ������2 is an anti-derivative of ������(������) = 2������. This process is illustrated below: Derivative ������2 2������ Anti-Derivative There are however, other functions whose derivative would be 2������. Such as ������2 + 2, ������2 + 7, ������2 − 3, and so on. The reason why all of these functions have the same derivative is that the constant term disappears during differentiation. So, all of these are anti-derivatives of 2������. Given any anti-derivative of ������(������), all others can be obtained by simply adding a different constant. In other words, if ������(������) is an antiderivative of ������(������), then so is ������(������) + ������ for any constant C. Summary: A function ������(������) is an antiderivative of ������(������) if ������������ = ������(������). ������������ If ������(������) is an anti-derivative of ������(������) then so is ������(������) + ������ for any constant ������. M1.4

M1 – Fundamentals of Mathematics 23 3. Standard Values of Anti-Derivatives: Using the differentiation values of standard functions, we can find the anti-derivatives of a range of functions without having to memorize them. In order to complete the below table, all you need is to find ������(������) which on differentiation would give ������(������): ������(������) ������(������) ������(������) ������(������) 0 1 cos ������ 2������ − sin ������ 3������2 sec2 ������ 1 ������������������−1 ������ ������ ������ (check the answers at the end of the chapter in the answer keys before reading ahead) 4. Anti-Derivative as the area under the curve: One of the most important applications of finding the anti-derivatives of functions is to calculate the area under the curve for mathematical functions. Area under a curve Consider the graph of ������ = ������(������) that lies completely above the ������-axis as shown alongside. Let the area within the curve and the ������-axis between the points ������ = ������ and ������ = ������ be represented by ������. This area will depend on the curve, as well as the limits ������ and ������, hence ������ will depend on ������, i.e. ������ = ������(������). Using areas of standard shapes To measure the area of complex shapes, we break the shape into smaller shapes of standard figures such as squares, rectangles, circles etc whose area we can easily calculate, and then add all the areas to get our resultant area. Area of 2 rectangles If we divide the area into 2 rectangles, as shown alongside, we lose out on area outside the rectangles, which means this is not a good approximation.

M1 – Fundamentals of Mathematics 24 Area of 5 rectangles However, if we make each rectangle thinner, and increase the number of rectangles, the approximation improves. With just 5 rectangles, we have been able to get a significant overlap of the area. Area of many rectangles Now, if we could make the width of each rectangle very small, we could increase the number of rectangles, and get a closer approximation of the total area. In the adjoining figure, we have taken a very thin strip of width = ������������ (remember that ������������ is a small change in ������) Since both the points are very close to each other, the points on the curve can both be approximated to be at ������(������). Thus the area of this strip would be the area of the rectangle which is ������(������)������������. Since this area would be very small, we call it ������������ (a small area). Thus, ������������ = ������(������)������������ By re-arranging the terms, this converts to the standard differentiation format i.e. ������������ = ������(������) ������������ This says that ������(������) is the derivative of ������, the area under the curve of ������(������). From the concept of anti-derivative, we conclude that ������ must be the anti-derivative of ������(������), i.e. ������(������) = ������(������) + ������, where ������������ = ������(������). (i) ������������ The value of ������ can be obtained by noticing that at ������ = 0, the area enclosed in the figure is zero, i.e. 0 = ������(0) + ������ or ������ = −������(0) Suppose now we want to find the area under the graph between ������ = ������ and ������ = ������ as shown in the figure. Using (i), we can write: ������(������) = ������(������) + ������ (ii) and ������(������) = ������(������) + ������ (iii) Subtracting (iii) from (ii), we get ������(������) − ������(������) = ������(������) − ������(������) … (note that C cancels out) Summary: When ������(������) lies entirely above the ������- axis between ������ and ������, we can find the area under ������ = ������(������) between a and b by finding an anti-derivative ������(������), and evaluating this at ������ = ������ and at ������ = ������. The area is then ������ = ������(������) − ������(������) M1.4

M1 – Fundamentals of Mathematics 25 5. Integration and Anti-Derivatives Up until now we have been using the term anti-derivatives in this topic. In calculus, anti-derivatives are called Integrals, and the process of finding anti-derivatives is called Integration. Standard notations: ������������(������) = ������(������) ⇒ ������(������) = ∫ ������(������)������������ ������������ Anti-Derivative ������(������) Integration symbol ∫ Area under the curve of ������ = ������(������) between ������ and ������ is given by ∫������������ ������(������)������������ Illustrations: 1. Find the area under the curve for ������ = cos ������ between ������ = 00 and ������ = 900 Sol: We know from the standard tables that if ������������(������) = cos ������, then ������(������) = sin ������ ������������ 900 ∴ ∫ cos ������ ������������ = sin 900 − sin 00 = 1 − 0 = 1 unit. 00 2. Find the area under the curve for ������ = 2������ between ������ = 0 and ������ = 10 Sol: We know from the standard tables that if ������������(������) = 2������, then ������(������) = ������2 ������������ 10 ∴ ∫ 2������������������ = 102 − 02 = 100 units. 0 If you draw the graph of ������ = 2������, you will notice that this is the area of a triangle whose base is 10 units and height is 20 units. The area of this triangle is 100 units indeed! PRE-READING EXERCISE Q1. If ������ ������(������) = ℎ(������), then ∫ ℎ(������)������������ = __________ ������������ Q2. ∫ cos ������ ������������ = ___________ Q3. ∫ − 1 ������������ = __________ ������2 Q4. ∫24 2������ ������������ = __________ Q5. ∫04050 sec2 ������ ������������ = __________ IN CLASS EXERCISE IN CLASS EXERCISE 1 Find the anti-derivatives of each of the below functions, using the knowledge of the differentiation of standard functions and the chain rule: Q1. ������(������) = ������3 Q2. ������(������) = −5

M1 – Fundamentals of Mathematics 26 Q3. ������(������) = ������ 3 Q4. ������(������) = √������ Q5. ������(������) = 4������2 − 3������ + 7 Q6. ������(������) = sin 2������ IN CLASS EXERCISE 2 Using the concept of definite integration, solve the following questions: Q7. The velocity of a particle is given by ������ = 2������. Find the displacement covered by the particle in the first ten seconds. Q8. Find the area bounded by the curve ������ = ������2 between the ������ − axis and the lines ������ = 1 and ������ = 2 Q9. Show that the area under the curve for the function ������ = ������������ for all negative values of ������ is 1 unit. Q10. Find the area under the curve for the function ������ = ������ between ������ = ������ and ������ = ������. Show that this is equal to the area of the trapezium enclosed between the line and the ������ −axis. (Area of a trapezium is equal to 1 × 2 (sum of length of parallel sides) × distance between them) HOMEWORK LEVEL 1 Q1. Find the anti-derivative ������ of ������ defined by ������(������) = 4������3 − 6 , where ������(0) = 3 Q2. Find the anti-derivative ������ of ������ defined by ������(������) = cos ������ , where ������(0) = 2 Q3. Find the area under the curve for the function ������ = sin ������ between ������ = 00 and ������ = 900 LEVEL 2 Q4. ∫13(2������2 + 3������ + 5)������������ = __________ Q5. ∫14 1 ������������ = __________ √������ Q6. Find the area bounded by the curve ������ = 2������3 between the ������ − axis and the lines ������ = 2 and ������ = 4 LEVEL 3 Q7. The velocity of a particle is given by ������ = 3������2 + 2 ������/������. Find the displacement covered by the particle in the first 5 seconds. Q8. Find the area under the curve ������ = ������−������ for all positive values of ������ M1.4

M1 – Fundamentals of Mathematics 27 Test Practice Problems Purpose: To practice a mixed bag of questions in a speed based format similar to what you will face in entrance examinations. In most entrance examinations, you will get not more than 3 minutes to attempt a question. Hence, you need to be able to attempt a question in less than 3 minutes, and at the end of 3 minutes skip the question and move to the next one. Approach:  Attempt the Test Practice Problems only when you have the stipulated time available at a stretch.  Start a timer and attempt the section as a test.  DO NOT look at the answer key / solutions after each question.  DO NOT guess a question if you do not know it. Competitive examinations have negative marking.  Fill the table at the end of the TPP and evaluate the number of attempts, and accuracy of attempts, which will help you evaluate your preparedness level for the chapter. No. of questions: 15 TEST PRACTICE PROBLEMS – 1 Time per question: 3 Minutes Total time: 45 Minutes Q1. A local minimum occurs when A) ������������ changes sign and ������ = 0 B) ������ changes sign and ������������ = 0 ������������ ������������ C) ������������ changes sign and ������2������ > 0 D) ������������ changes sign and ������2������ <0 ������������ ������������2 ������������2 ������������ Q2. ������ = ln ������ is an increasing function A) At ������ = 0 B) For all positive values of ������ C) For all values of ������ D) Only for ������ > 1 Q3. ������ (4 cos 3������) = B) 12 sin 3������ D) 4 sin 3������ ������������ A) −4 sin 3������ C) −12 sin 3������ Q4. log������ ������ = ������ is defined if B) ������ > 0, ������ > 0, ������ ≠ 1 D) ������ > 1, ������ > 0, ������ ≠ 1 A) ������ > 0, ������ > 0, ������ ≠ 1 C) ������ > 0, ������ > 0, ������ ≠ 1 Q5. ∫35(3������ − 2) ������������ = B) 20 C) 44 D) 48 A) 10 Q6. log������ √������ . log������ ������3 log������ 3√������2 = B) 6 C) 1 D) 2/3 A) 3 Q7. Displacement of a particle is given as ������ = 4������ − 2������2. Its velocity will be zero at ������ = A) 2 B) 1 C) 0 D) 4 Q8. The slope of the tangent to the curve ������ = 5������2 + 3 at ������ = 2 is A) 23 B) 3 C) 10 D) 20

M1 – Fundamentals of Mathematics 28 Q9. log3 log2 ������ = 1 then ������ = B) 9 C) 8 D) 6 A) 3 Q10. ������ (si������n2������) = ������������ A) ������ sin ������−2 cos ������ B) ������ cos ������+2 sin ������ C) −������ cos ������−2 sin ������ D) ������ cos ������−2 sin ������ ������3 ������3 ������3 ������3 Q11. When will the quadratic expression ������ = ������������2 + ������������ + ������ be minimum? A) At ������ = 0, if ������ > 0 B) At ������ = −������ , if ������ > 0 C) At ������ = −������ , if ������ < 0 2������ 2������ D) At ������ = 0, if ������ < 0 Q12. ∫ 3 cos 4������ ������������ = A) 3 sin 4������ + ������ B) −3 sin 4������ + ������ C) 12 sin 4������ + ������ D) 3 sin 4������ + ������ Comprehension for Q13 - 15: The velocity of a particle is given as ������ = ������2 − 3������ + 2 ������/������ 4 Q13. Find its displacement as a function of time, if at ������ = 0 seconds its displacement was zero. D) ������3 − 3������2 + 4������ A) ������3 − 3������2 + 2������ B) ������3 − 3������2 + 2������ C) ������3 − 3������2 + 3������ 32 32 34 32 Q14. When will its acceleration be zero? A) 3 ������ B) 3 ������ 4 2 C) 1 ������ D) None of these 2 Q15. How much displacement will it cover in the first 3 seconds? A) 3 ������ B) 1 ������ 4 2 C) 3 ������ D) None of these 2 TEST PRACTICE PROBLEMS – 2 No. of questions: 15 Total time: 45 Minutes Time per question: 3 Minutes Q16. log4 18 is B) An irrational number D) None of these. A) A rational number C) A prime number Q17. If ������, ������, ������ are consecutive positive integers and log(1 + ������������) = 2������, then the value of ������is A) log ������ B) log ������ C) 2 D) 1 Q18. If log4 5 = ������ and log5 6 = ������, then log3 2 is equal to A) 1 B) 1 C) 2������������ + 1 D) 1 2������+1 2������+1 C) 5 2������������−1 Q19. If ������4. ������5 = 1, then the value of log������(������5������4) equals B) 1 D) 8/5 D) None of these A) 9/5 B) 4 Q20. The value of 3log4 5 − 5log4 3 is A) 0 C) 2 T.P.P.

M1 – Fundamentals of Mathematics 29 Q21. The value of ������ satisfying √3(−4+2 log√5 ������) = 1/9 is B) 4 D) None of these A) 2 C) 1 B) Prime number D) Irrational number Q22. If √log2 ������ − 0.5 = log2 √������,then ������ is a/an C) 10 A) Odd integer C) Composite number C) 8 Q23. If (4)log9 3 + (9)log2 4 = (10)log������ 83, then ������ is equal to C) 2 A) 2 B) 3 D) 30 D) 16 Q24. If 2������log4 3 + 3log4 ������ = 27, then ������ is equal to D) ∞ D) 2 A) 2 B) 4 D) − tan ������ Q25. The number of solutions of ������log������(������+3)2 = 16 is log������ cos ������ A) 0 B) 1 D) 2 D) Not Defined Q26. If ������(������) = ������√3 ������ sin (������ + ������) then ������′(0) is equal to 3 A) ������ B) ������ C) 1 6 3 Q27. ������ = log������[log������ cos ������] then ������������ is ������������ A) 1 B) − sin ������ C) log������ cos ������ (log������ cos ������) log������ cos ������ cos ������ Q28. If ������ = ������−1 then ������������ is ������������ A) ������2 B) −1 C) −1 D) 1 ������2 ������2 Q29. If ������ = sec(������. ������������) then ������������ at ������ = 0 is ������������ A) 0 B) 1 C) −1 Q30. If ������(������) = tan √������ then ������′(0) is A) 0 B) 1 C) 1 2 TEST PRACTICE PROBLEMS – 3 No. of questions: 15 Total time: 45 Minutes Time per question: 3 Minutes Q31. ������ [sin (tan ������)] is B) [sin(cos2 ������)] D) None of these ������������ C) 2,2 A) [cos(tan ������)] C) [cos(tan ������)] sec2 ������ C) ������ cos ������ Q32. If ������(������) = tan 2������, then ������(������) and ������′(������) are respectively 180 A) 1, −2 B) 0,2 D) 1 , 1 Q33. The derivative of sin(������∘)is 2 A) cos ������∘ B) − cos ������∘ D) ������ cos ������������ 180 180

M1 – Fundamentals of Mathematics 30 Q34. The derivative of log10 ������ w.r.t ������ is A) 1 B) 1 C) 1 D) 0 ������ 10������ ������������������������������10 Q35. If ������ = 3������−2, then ������������ is equal to 2������+3 ������������ A) 3 B) 3 C) 13 D) −13 (2������+3)2 (2������+3)2 2 2������+3 Q36. If ������ = |������| then ������������ is B) 0 C) −1 D) |������| ������������ ������ A) 1 Q37. If ������ = cos(log������ ������) , ������ = log������ cos ������, then A) ������������ = ������������ B) ������������ = log������ ������. ������������ ������������ ������������ ������������ ������������ D) ������������ = 1 . cot ������. sin(log ������) C) cos ������. ������������ = ������������ ������������ ������ ������������ ������������ Q38. If ������ = ������������ and ������ = ������ then ������������ is ������ ������������ A) 2 B) 2 C) − 1 D) − 2 ������������3 ������3 ������2 ������������3 Q39. The function ������ = sin ������ (1 + cos ������) is maximum for ������ = A) ������ B) ������ C) ������ D) ������ 2 4 3 6 Q40. The function ������5 has A) Both maximum and minimum B) Neither maximum nor minimum at ������ = 0 C) Maximum but not minimum D) Minimum but not maximum Q41. The two numbers ������ and ������ such that ������ + ������ = 12 and ������������2 is maximum are A) 4,8 B) 2,10 C) 6,6 D) 5,7 Q42. The function ������(������) = 2������3 − 3������2 − 12������ + 5 has a minimum at ������ = A) −1 B) 2 C) − 1 D) 3 2 2 Q43. The minimum value of the function ������(������) = 7 − 20������ + 11������2is A) 177 B) − 177 C) − 23 D) 23 11 11 11 11 Q44. The minimum value of 2������2 + ������ − 1 is A) − 1 B) 3 C) − 9 D) 9 4 2 8 4 Q45. The function ������(������) = 2������3 − 3������2 − 12������ + 4 has A) No maxima and minima B) One maximum and one minimum C) Two maxima D) Two minima TEST PRACTICE PROBLEMS – 4 No. of questions: 15 Total time: 45 Minutes Time per question: 3 Minutes D) − ������ Q46. The function ������ = 1 − cos ������ is maximum, when ������ = 6 A) 0 B) ������ C) ������ 2 T.P.P.

M1 – Fundamentals of Mathematics 31 Q47. The value of ∫ tan2 2������ ������������ is A) sec2 2������ − ������ + ������ B) tan 2������ + ������ + ������ 2 2 C) tan 2������ − ������ + ������ D) 1 cot 2������ − ������ + ������ 2 2 Q48. The value of ∫ cos 2������ ������������ is B) − (1) cot 2������ + ������ sin2 2������ 2 A) cot 2������ cosec 2������ + ������ D) cosec 2������ cot 2������ + ������ C) − (1) cosec 2������ + ������ 2 Q49. The value of ∫ ������(���3���) ������������ is A) (1) ������ + ������ B) 3������������/3 + ������ C) (3) ������������/3 + ������ D) ������������/3 + ������ 3 ������ 3 ������ Q50. The value of ∫ sin ������ + cos ������ ������������ is A) (sin ������ + cos ������)2 + ������ B) cos ������ − sin ������ + ������ C) ������������2 + ������ D) (cos ������ − sin ������)2 + ������ Q51. The value of ∫(2������ + 3)3 ������������ B) (2������+3)4 + ������ A) (2������+3)4 + ������ 6 8 D) (2������+3)4 + ������ C) (2������+3)4 + ������ 14 12 Q52. The value of ∫ √1 − ������ ������������ is A) (2) (1 − ������)3/2 + ������ B) − (2) (1 − 3 + ������ 3 3 ������)2 C) (3) (1 − ������)3/2 + ������ D) −3 (1 − ������)3/2 + ������ 2 2 Q53. ∫ 2������+3������ ������������ = 5������ A) ������ log (2) + ������ log (3) + ������ B) log (2) + log (3) + ������ 55 55 C) (25)������ + (53)������ + ������ D) (2)������ log (2) + (3)������ log (3) + ������ log(25) log(35) 5 55 5 Q54. The value of ∫ 1 ������������ is (������−5)2 A) 1 + ������ B) − 1 + ������ ������−5 ������−5 C) 2 + ������ D) −2(������ − 5)5 + ������ (������−5)3 Q55. If ������(������) = ������2������−31, then ∫ ������(������) ������������ ix A) 1 + 1 + ������ B) log ������ + 1 + ������ ������2 2������3 2������2 C) − 1 + 3 + ������ D) log ������ + 1 + ������ ������2 ������4 2������3 Q56. ∫ 3������3−2√������ ������������ = ������ A) ������3 − √������ + ������ B) ������3√������ + ������ C) ������3 − 2√������ + ������ D) ������3 − 4√������ + ������

M1 – Fundamentals of Mathematics 32 Q57. ∫ ������������−2+������������−1+������ ������������ = ������−3 A) 2������������2 + 3������������3 + 4������������4 + ������ B) 6������������2 + 4������������3 + 3������������4 + ������ D) 1 ������������2 + 1 ������������3 + 1 ������������4 + ������ C) ������ + ������ + ������������2 + ������ 234 Q58. ∫ (������ + 1)3 ������������ = ������ A) 1 (������ + 1) + ������ B) ������4 + 3������2 + 3 log ������ − 1 + ������ 4 2 2������2 4 ������ C) ������4 + 3������2 + 3 log ������ + 1 + ������ D) 4 (������ + 1) + ������ 4 2 ������2 ������ Q59. If ������′(������) = ������2 + 5 and ������(0) = −1, then ������(������) = A) ������3 + 5������ − 1 B) ������3 + 5������ + 1 C) 1 ������3 + 5������ − 1 D) 1 ������3 + 5������ + 1 3 3 Q60. If ������′(������) = 1 + ������ and ������(1) = 5, then ������(������) = ������ 2 B) log ������ + ������2 + 1 A) log ������ + ������2 + 2 2 2 D) log ������ − ������2 + 1 C) log ������ − ������2 + 2 2 2 T.P.P.

M1 – Fundamentals of Mathematics 33 DATA ANALYSIS Guide A # of questions Total problems in TPP B # Attempts Total attempts in OMR C # Correct Total questions correct D # Incorrect Out of the ones marked in OMR E # Unattempted ������ – ������ F Percentage attempts ������ G Percentage Accuracy ������ × 100 ������ ������ × 100 Question type # Correct (C) # Incorrect (I) # Unattempted (U) Easy Medium Hared Tip: To begin with, your accuracy must be high, typically > 60%. Percentage attempts should be > 50% As time progresses, your percentage attempts should increase without a reduction in accuracy. Additionally, you should be able to get > 80% Easy questions correct, as they involve basic recall of the concepts and formulae of the chapter.

M1 – Fundamentals of Mathematics 34 PRE-READING EXERCISE Answer Key Q1. I. 4 M1.1 LOGARITHMS II. 4 LEVEL 3 25 Q9. Proof Q10. B III. 30 Q11. C IV. 2 HOMEWORK 3 LEVEL 1 Q1. C Q2. I. 3 Q2. I. True II. 1 II. True Q3. 4 III. True IV. False Q4. False Q3. Proof Q4. ������ = 4 Q5. log8 1 = −2 64 3 IN CLASS EXERCISE Q5. 3 LEVEL 1 LEVEL 2 Q6. ������������ = 4 and Acidic Q1. 9 Q7. 10−7 Q8. ������ = 1 or ������ = 4 Q2. I. log6 216 = 3 Q9. Proof Q10. B II. log9 27 = 3 Q11. ������ = 4 2 LEVEL 3 Q3. I. 5 Q12. 3 II. 1 Q13. C 3 III. −4 IV. −3 Q4. 2 LEVEL 2 Q5. D Q6. 1 Q7. Proof Q8. 8 M1.2 DIFFERENTIATION Situation 1: Q2. Savings at the end of month 1 = Rs. 5000. Savings before he started his job = Rs. 4000. Q1. In the question it is mentioned that the graph ∴ Savings in the first month depicts Akshay’s savings just after he started a = ������������. 5000 − ������������. 4000 = ������������. 1000 new job. Which means that month 0 is when he started a new job. At this time he had ������������. 4000 Q3. Similarly as above, in his account. Therefore, Akshay’s savings Savings at the end of month 2 = Rs. 6000. when he started his new job is ������������. 4000. Savings at the end of month 1 = Rs. 5000. Ans.

M1 – Fundamentals of Mathematics 35 ∴ Savings in the second month Q5. To answer this question we need to calculate = ������������. 6000 − ������������. 5000 = ������������. 1000 the speed for every hour. Q4. In order to calculate Akshay’s average savings Duration Distance covered Speed per month first we will find his savings in the 1st hour (in km) job and then divide that by the number of 2nd hour (km/hr) months he was in the job. 3rd hour ������ − ������ = ������ Savings in the job 4th hour ������������ − ������ = ������ ������ = Savings at end of month 5 – Savings at the 5th hour ������������ − ������������ = ������������ ������ = ������ start of Month 1 = 9000 − 4000 = ������������. 5000 6th hour ������������ − ������������ = ������ ������ Number of months in the job = 5 7th hour ������������ − ������������ = ������ ������ = ������ Therefore, the average savings per month ������������ − ������������ = ������ ������������ obtained is ������������ − ������������ = ������ ������ = ������������ 9000 − 4000 5000 ������ ������������. ( 5 ) = 5 = ������������. 1000. ������ = ������ ������ Q5. Yes, we notice that the answer to Q4 is equal to ������ = ������ the answer to Q2 and Q3. Hence Akshay’s ������ average savings are equal to the savings per ������ = ������ month, and you would get the same answer for ������ the savings in month 3,4 and 5. ������ = ������ Situation 2: Thus Chinmay ran fastest during the 3rd hour. Q1. From the graph, we see that Chinmay reached Q6. From the graph we can see that Chinmay ran a the 42 km mark at 7 hours. This means that total of 42 km in 7 hours. So his average speed Chinmay took 7 hours to complete the ������������ marathon. will be ������ = ������ ������������/ℎ������. Q2. In order to calculate Chinmay’s speed in first Q7. No, In question 6 we calculated Chinmay’s hour, we will have to see how much distance he average speed which is 6 km/hr. In question 5 covered in this duration. And from the graph it we saw Chinmay’s speed at different hours, is quite clear that he covered 7 km. Therefore, which differs every hour. Hence we can ������ conclude that Chinmay’s average speed is not Chinmay’s speed would be = ������ = ������ ������������/ℎ������ always the same as his speeds at different instants. Q3. To calculate Chinmay’s speed in the last hour we will look for the distance covered by Q8. From the graph and the earlier questions we Chinmay between the 6 hour and 7 hour mark know that options A and B are not true. Option and from the graph we can see that he covered C bears the closest resemblance to the graph. ������������ − ������������ = ������ ������������. To calculate his speed we will divide the distance covered by the time taken to Situation 3: cover this distance which is 1 hour. Hence, ������ Q1. Change in bacteria in second hour = ������ − ������ Chinmay’s speed in the last hour ������ = ������ ������������/ℎ������. = ������ bacteria Q4. Note that the 4th hour is the hour between the 3 Q2. Change in bacteria in fifth hour = ������������ − ������������ hour and 4 hour mark. In this duration the = ������������ bacteria distance covered does not change. It remains constant at 25 km. This means that Chinmay ran Q3. Average increase in bacteria per hour 0 ������������������ in the 4th hour. ������������������������������ ������������������������������������������������ ������������ ������������������������������������������������ ������������ = ������������������������������������ ������������ ������������������������������ = ������ = ������������. ������������ ������������������������������������������������/ℎ������������������ Q4. No, Since the answers to Q1, 2 and 3 are different, the average rate of change is not equal to the change in every hour.

M1 – Fundamentals of Mathematics 36 Q5. The concepts you have studied so far are not HOMEWORK ������ ������������ sufficient to answer this question, since you are LEVEL 1 ������������ required to calculate the rate of change at a Q1. 5 − tan ������ − sec 2������ particular instant. You will be able to answer sec ������ this after you have studied differentiation. Sr No 4������ + 3 sec ������ tan ������ ������3(������2 − 4) 4������ ln 4 IN CLASS EXERCISE I ������2(5������2 − 12) II LEVEL 1 III IV Q1. Sr No ������ ������������ ������������ I 3������2 6������ LEVEL 2 II 3������ 3������ ln 3 III −2������ Q2. I. At ������ = 1, ������ = 1 ������/������; At ������ = 2 , ������ = −1 ������/������ −2 IV 2√������ 1 3 √������ 5 cos ������ II. 6 ������/������2 III. −1 ������ 12 V 5 sin ������ + 2 Q3. ������ ������������ Sr No ������������ ������2(1 + 3 ln ������) LEVEL 2 I ������3 ln ������ Q2. velocity = 3������2 − 6 ������/������ II 6������ cos ������ −6������ sin ������ + 6 cos ������ 2 4 −8 III 3√������2 Q3. time = 2������ 5 Q4. acceleration = 3 ������/������2 3 sin ������ Q5. velocity = 7.5 ������/������ IV 3������3 6������ cos ������ − 3 sin ������ √������ 2������√������ LEVEL 3 LEVEL 3 Q6. C Q4. D Q5. B Q7. Sr No ������ ������������ ������������ I cot ������ −cosec2������ II ������3 cos ������ −������3 sin ������ + 3������2 cos ������ III 2������3 28������3 + 30������2 7������ + 5 49������2 + 70������ + 25 IV ������������ tan ������ ������������(sec 2������ + tan ������) IN CLASS EXERCISE 1 M1.3 CHAIN RULE AND MAXIMA MINIMA Q1. 2 cos(2������ + 1) Q4. − sin(sin ������) × cos ������ Q2. 5 sec2 5������ Q3. 2 IN CLASS EXERCISE 2 Q5. C ������ Q6. D Ans.

M1 – Fundamentals of Mathematics 37 Q7. Minimum displacement happens at ������ = 2������, and LEVEL 2 its velocity is zero at that instant. Q2. Maximum value = √2 at ������ = 45° Q8. 8 and 8 Q3. I. ������tan ������. sec 2������ Q9. Length and breadth both are 7 m II. 4(2������ + 1)(������2 + ������ + 1)3 HOMEWORK III. 3 LEVEL 1 2√3������+2 Q1. I. −3 sin 3������ Q4. 50 II. 1 LEVEL 3 ������ Q5. ������/������ Q6. Proof III. −������−������ M1.4 ANTI DERIVATIVES AND INTEGRATION PRE-READING TABLE Q6. − 1 cos 2������ + ������ ������(������) ������(������) ������(������) ������(������) 2 0 ������ sin ������ + ������ 1 ������ + ������ cos ������ cos ������ + ������ IN CLASS EXERCISE 2 2������ ������2 + ������ tan ������ + ������ Q7. 100 ������ − sin ������ Q8. 7 units 3������2 ������3 + ������ sec2 ������ ln ������ + ������ 1 3 ������������������−1 ������������ + ������ ������ ������������ + ������ ������ ������ Q9. Proof Q10. Area = 1 × (������ + ������)(������ − ������) 2 PRE-READING EXERCISE HOMEWORK Q1. ������(������) + ������ LEVEL 1 Q2. sin ������ + ������ Q1. ������(������) = ������4 − 6������ + 3 Q3. 1 + ������ Q2. ������(������) = sin ������ + 2 Q3. 1 ������ LEVEL 2 Q4. 12 Q4. 118 Q5. 1 3 IN CLASS EXERCISE 1 Q1. ������4 + ������ Q5. 2 Q6. 120 4 LEVEL 3 Q2. −5������ + ������ Q7. 135 ������ Q3. ������2 + ������ Q8. 1 6 Q4. 2 √������3 + ������ 3 Q5. 4������3 − 3������2 + 7������ + ������ 32

M1 – Fundamentals of Mathematics 38 TEST PRACTICE PROBLEMS Q. No. Ans. Level Mark (C) / (I) / (U) Q. No. Ans. Level Mark (C) / (I) / (U) as appropriate as appropriate Q1. C Easy Q31. C Medium Q2. B Medium Q32. B Medium Q3. C Medium Q33. D Medium Q4. C Easy Q34. C Hard Q5. B Medium Q35. C Medium Q6. C Hard Q36. D Hard Q7. B Medium Q37. D Hard Q8. D Medium Q38. C Medium Q9. C Medium Q39. C Medium Q10. D Hard Q40. B Medium Q11. B Hard Q41. A Medium Q12. D Medium Q42. B Medium Q13. A Hard Q43. C Medium Q14. B Medium Q44. C Medium Q15. C Hard Q45. B Medium Q16. B Hard Q46. C Medium Q17. A Hard Q47. C Medium Q18. D Medium Q48. C Medium Q19. A Hard Q49. B Easy Q20. A Easy Q50. B Easy Q21. C Medium Q51. A Medium Q22. B Hard Q52. B Medium Q23. C Hard Q53. C Medium Q24. D Hard Q54. B Medium Q25. A Hard Q55. B Medium Q26. D Hard Q56. D Medium Q27. D Medium Q57. D Medium Q28. C Easy Q58. B Hard Q29. A Medium Q59. C Medium Q30. D Medium Q60. A Medium Ans.