IMAT Mathematics in One Shot | Every Topic Covered

[00:00:00]a one short video that will cover everything related to math and physics sections of IMAT. I am going to introduce you the Shouba who will be our math and physics tutor throughout this one short video in which we will cover everything regarding math and physics for IMAT including PYQs as well which stands for previous year questions. So Shoubam I would like to invite you to give your formal introduction to our students. Hello and welcome everyone.

[00:00:25]Myself Sham Raj outed my bachelor's in maths and physics and currently I'm undertaking a research project at IIT Madras under professor Narayan Narayan.

[00:00:34]It's a mathematics research project in the field or generalization of exact power graphs. I'm thrilled to be guiding you through the IMAT preparation for mathematics and physics part. That's it.

[00:00:44]>> Okay. It's wonderful that we have someone from IIT who's teaching us physics and mathematics for IMAT. I wish when I was preparing I had someone like that. So uh with regards to this I have some pointers for you in both physics and math. The essential point is to understand where the formula is coming from and then try to implement it when you're actually solving questions. This is what our whole approach would be throughout the whole oneshot video and hence we have also introduced the past year questions as well. Right? So let's go forward with the whole video and sham you must have had some pointers to add up our students. I want you to uh help them out with that. So whenever you want to start the lecture a water bottle with you all right don't memorize the formula understand the understand where it comes from and I would like your full attention throughout the lecture okay understand the concepts try and implement the formulas that you learn in practical ways right so yeah >> okay thank you so much Shubam and I wish the lecture goes very well thank you so much for helping all the students who are preparing for IMAT it really means a And this video will hopefully create a history among the IMAT space and I hope it helps a lot of students coming forward. Thank you so much.

[00:01:59]>> Welcome to this module. Today's discourse is designed to establish a robust theoretical foundation for your upcoming examinations.

[00:02:07]Before we can conclude the broader IMAT mathematics syllabus today, we need to anchor our understanding in the very foundation of quantitative systems. The number system. The number system is broadly divided into specific hardware concepts of natural numbers, whole numbers which include zero, natural numbers starting from one, integers which include the whole numbers and the negative of natural numbers, real numbers which can be plotted on the real number line and imaginary numbers which include iota and all that. Let us now direct our attention to the properties of real numbers. A real number is classified into rational and irrational numbers. Rational numbers are those numbers which can be written in the form of P upon Q where P and Q are inteious.

[00:02:55]All right? And irrational numbers are those numbers which cannot be written in the form of fractions. For instance, 3.141 this continuous forever this is an irrational number and pi is also an irrational number. Building on what we have already established, let us turn our attention to power and exponents. An exponent dictates how many times a base is multiplied by itself. The fundamental law governs the manipulation of these expressions. The first of its kind rule is the multiplication rule. A ^ m into a ^ n equals a ^ sum of the powers. In case of division rule you subtract the powers and if e to the power n and this entire thing is operated to the power n in this case the powers will be multiplied. If a into b they share the same power m it can be distributed among them in this form and if they are in the division form then it can be distributed in the division form like this b power m. The zero rule a power 0 any number power zero is always one and any number power one is always the number itself.

[00:04:04]Let us turn our attention to negative and fractional exponents. In case of negative exponents the number shifts in the denominator with a positive exponent. All right. So a power minus n= 1 upon a power m. The number shifts in the denominator with a positive exponent. In case of one power one upon n power 1 upon n this could also be written as under root of a where n here represents the nth power one upon nth power of a and if this is a power n upon n this could be written as a power m to the power 1 by n and this could further be written as a power m under root n years. Turning our attention to exponential equations, we see if ax equals a y that is the base is same then the powers has to be same x= y. In case of different bases however we can always write a power x= b and then when we'll further learn about logarithms we will see that x log of a equals log of b or we can write x n equals ln b with our preliminary definitions locked in let's hit the ground running with some of the past IMAT questions evaluate 8 into 10 ^ minus 5 upon under root of 1.6 into 10 ^ 7 * 1.2 into 10 ^ 3 squared. All right, let's separate the terms of 10 powers of powers of 10 and the normal numbers. So 8 into 10 ^ minus 5 upon under root of 1.6. The under root could be divided between the numbers 1.6 and 10 ^ 7 as 1.6 under root into 10 ^ 7 upon 2. This this is 2. So 1 upon 2 it would be in the power. So 7 upon 2 1.2 into 10 ^ If you further distribute the powers of 2 among 1.2 and ^ 3 it will come out to be this.

[00:06:10]So 8 into 1.2 ^ 2 into 10 ^ 6 and 5.

[00:06:16]This will be 6 - 5 1 upon under root of 1.6. We can write this as under root of 16 upon under root of 10 into 10 ^ 7 upon 2. Further simplify this as 8 into 1.2² will be 1.44 into 10 ^ upon in the denominator we will have <unk>16 is 4 m2. This is 10 ^ 1x2 into 10 ^ 7x2.

[00:06:46]So we will have 8 into 1.44 4 4 into 10 upon 4 into 10 ^ 7x2 minus of 1x2 so that will be this and when you subtract this you will get 7 - 1 6 upon 2 which is 3. So this will be simply three this boils down to 2 and you finally reduce this it will come 2 into 1.44 into 10 ^ -2 this is 1 1 - 3 will be -2 this boils down to 2.88 88 into 10 ^ -2.

[00:07:20]The correct option is option D. Let us tackle another IMAT past year question.

[00:07:26]Evaluate 27^ 2 - 23² plus of 14 2 - 6 square. We have to keep in mind the formula here which is a 2 - b² = a - b into a + b. Now we start with the solution. 27^ 2 - 23 2 + 14^ 2 - 6 2 This will be 27 - 23 27 + 23 + 14 - 6 into 14 + 6 so this will be 4 into 50 + 8 into 20 this simplifies to 200 + 160 this is 360 so the correct option is B 306. Let us look at another IT past here question. X is given to be 3 into 10 ^ m. Y is 5 into 10 ^ n. We are required to find the value for XY that in scientific notation. So X into Y will be 3 into 10 ^ N * 5 into 10 ^ N. This will be 15 * 10 ^ M into 10 ^ N. base is same different powers. So we'll add that 15 into 10 ^ m + n. What does it mean to be in scientific notation? In scientific notation we write expressions as n into 10 ^ n where n is a number less than 10.

[00:09:00]So we have to keep it less than 10. For that we will divide this number by 10 and multiply this by 10. This part will make it 1.5. So this boils down to 10 into 1.5 into 10 ^ n + n. There is a 10 here. So this will join this. So this will become 1.5 into 10 ^ n + l. So the correct option is here 1.5 into 10 ^ m + n + 1. Before we wrap up this chapter, let's look at another IMAT question.

[00:09:33]Evaluate 3 into 10 ^ 3 ^ 3 into 10 ^ 2 into 10 ^ - 5. So this will be 3 into 10 ^ 3 cube and 2 into 10 ^ - 5. You can distribute the exponent among the numbers on the base. So this will be 3 into uh 3 ^ 3 into 10 ^ 3 into 3 * 2 into 10 ^ - 5 3^ 3 is 27 27 into 2 will be 54 and just when you divide it by 10 and multiply it by 10 you will get 10 into 5.4 into 10 ^ 4 this will come out to be 5.4 into 10 ^ 5 so the correct option is B 5.4 4 into 10 ^ 5. The natural progression from our previous discussion dates us to logarithms. A logarithm is the inverse operation to exponentiation. If you have a to the^ y = x, it could be represented as y= log of x base a. All right? And you have two conditions here. First of all, the base it has to be greater than one and not equal to one. And the argument has to be always positive. Uh talking about types of logarithm we have log base 10 or we could have log to the base e and there will be arguments. Uh the value of e is 2.718.

[00:11:00]Let us now direct our attention to the properties of logarithms and solve the rules. Uh first one is identity properties. Log of argument one and any base a is always zero. All right. Log of the base is one. Okay. where argument is also a and base is also a power n reciprocal rules. If you have log of base a and argument x ^ n in that case we can take out the exponent and write log of x base a log of the reciprocal.

[00:11:32]If you have something one upon x in the arent, you can always take negative sign out and put the argument x. Change of base formula. You can always change base of any logarithm. Let's suppose we have log of 2 base 3. In this case, we can always change the base to let's say five by putting log of 2 b 5 upon log of 3 b 5. Operational rules, the product rule and the quotient rule. If you have log of argument x y to the base a in this case you can always separate these x and y into sum of log sum of their log which is log of x base a plus log of y b a and similarly if you have x upon y in the argument in this case you can separate them with a negative sign and their logs all right so this will be log of x base a minus of log of y base a now we look at some of the past year versions of iMac let's look at this part it was asking IMAC 2024 f(x)= log of argument x² + 2 upon and the base is 2. What is the reciprocal of the value of f of 2?

[00:12:37]Just have to calculate the value of f of 2. f of 2 is log of 2 and this will be 2 ^ 2 + 12. All right. And this will yield 2 and 4 + 12 16. You can write this as log of two base 2 ^ 4 upon two and using the power of exponents you can always take four out log 2 base 2 log of the base is always 1 so this is 1 so this will be 4 into 1 4 f of 2 is 4 our option is 4 option B now let's look at another basio question of IMAC uh given log of 7 b 10 is x 2 b 10 is y 3 b 10 is E what is log of 14 upon 3 base 10 expressed in terms of x y and z. So this log of 14 base 14 upon 3 base 10. This 14 upon 3 could be written as 7 into 2 upon 3. This must give you an idea of what we should be doing. All right. If you have two logs and you sum it up, you get their product in the argument. And if you subtract, you get division in the argument. So you must have realized that these two will be multiplied and z will be divided. All right. So plus minus of log 3 base 10. So in this case you will get log of 7 into 2 base 10 minus of log of 3 base 10. This will be log of 7 into 2 upon 3 base 10. All right. So this is x. This is y - z. We have x + y - z. The third option is option C. Let's look at another past year question. Which one of the following is equivalent to ln x² y - 2 ln xy + 3 ln y x² into y? This is in product. We can always separate them with with some of their logs. This will be ln x 2 plus of ln y minus again we can separate them. 2 ln x plus of n y plus of 3 y. If we further simplify this to ln y - 2 ln x - 2 ln y + of 3 ln y - 2 ln y + 3 ln y this will reduce to + ln y + ln y and plus ln y + ln y this will become ln x² - 2 ln x + 2 ln y if you look at this x² you can always take out the exponent this It will be 2 ln x - 2 ln x plus of 2 ln y. This will get cancelled and you're left with 2 ln of y which is ln y² y. This is the final answer. Option C. Next objective is to define polomial functions. Also study about some of the properties of quadratic equations. All right. A polomial is an algebraic expression consisting of variables and coefficients involving operations of addition and subtraction, multiplication and strictly non- negative integer exponents where n here represents the non- negative integer. The value of the maximum n will be the degree of the function of the polomial ratio. The polomial operations could be addition or subtraction, multiplication or division. You can multiply two functions. You can divide two functions. You can subtract two functions. So you can perform any operations on polomial functions. All right? And still they will be polomials.

[00:16:13]Uh the crucial caveat here is if you divide certain function let's suppose x upon x² in that case this is not a polomial function because p of x will be 1 upon x = x of minus1. And here n is equal to minus1. But we have a strict conditions of n being a non- negative integer. So keep this in mind that you can always perform addition, subtraction, multiplication. But under division you have to keep in mind the degree does not become a negative addition. With that framework in place, we now move on to classification and the degree of polomial functions. We have already talked about the degree where n the greatest non- integer value that appears in the polomial function is the degree. Classification could be based on the number of terms in the equation. If the polomial function or equation has exactly one term in that case it is monomial. If it has exactly two terms it's binomial. If it has three terms it is tromial and multinnomial for four or more terms. All right. zeros of a function over the values of x for which fp of x equals zero. This is set to be roots of the polom. On to the next quadratic equations. Quadratic equation standard form and definition. A quadratic equation is a fundamental secondderee polomial equation in a single variable. A number of variables in the equation has to be single. And this is a second degree polomial meaning the largest value of the non-gative integer appearing in the equation has to be 2. So and you could find x² + 2 = 0.

[00:17:57]This is also a quadratic equation. x² + 3x this is also a quadratic equation and this is the standard form of quadratic equation where a cannot be zero and abc are the real numbers. If a is zero then the condition of having the maximum non- negative integer as two it collapses. Now we move on to the next methods of solution of the quadratic equation. There are three methods. First one is factorization.

[00:18:24]Second is completing the square and the third is quadratic formula. Uh best one is the quadic formula. You can always employ this method in any circumstances.

[00:18:34]First one is little ingenious method but you can always use this. We will see first and second in some of the examples. The discussion on the methods of solution naturally leads us to the relationship between the roots and the coefficients. All right. There are two relationships for quadratic equation often referred to as retas formulas. Sum of roots is denoted by minus b upon a and product of roots is denoted by c upon a. All right. And this is the general form of quadratic equations. It could be represented as this where x² minus sum of the roots into x plus the product of volts into x product of those are simply if suppose and there is a coefficient attached to x² then you will have x² minus s x upon a plus of product upon a equals z. All right. Discriminant and the nature of roots discriminant often represented as d is the value of b ^ 2 - 4 d. If this is negative then you have no roots. If this is greater than equal to0 then you have roots. And if it is specifically d equals z in that case you have two equal roots. So yeah keep this in mind. Drop of a quadrating equation the continuous mapping of the function visually forms symmetric u-shaped curve known as a parabol. You try to graph any quadratic equation you will get this function. this graph either it could be upward facing or it could be downward facing and the location where it hits the x-axis it depends on the values of a b and c if a is positive then the geometric curvation opens upward if a is negative it will open downwards okay and this is the vertex vertex is this point the absolute maxima or the minima in this case it is maxima so this is the vertex and here it is another vertex for this function for this quadratic equation. So yeah, now we move on to some of the past year questions. Let us look at this question.

[00:20:35]It was asked in 2011.

[00:20:37]Which of the following is a simplification of this function? If you try and pay close attention to this function, the denominator can be broken down into x^2 + 2x - xus of 2 and then you can rearrange take out something common here. It will be x x + 2 - of x + 2 and you will get x -1 x + 2. So you're getting x -1 into x + 2 in the denominator. And so this will be a simple x + 2 upon x -1 x + 2. This will cut out and the answer will be x + 2 upon x - 1. Now we move on to the next question. Which of the following is a simplification of this 2 upon x² -1 - 1 upon x -1 and since x² - 1 could be written as 2 upon x + 1 and x -1 - 1 upon x -1 you could take out this common common factor is 1 upon x - 1. This will be 2 upon x + 1 - of 1. This could be further written as 2 upon x + 1 minus of x + 1. All right, we took LCM and multiplied the denominator with 1. So this will be x and 1 upon x - 1 2 - x - of 1. So this will be 1 - x upon x + 1.

[00:22:05]This 1 - x could be written as x - 1 with a negative coming out. So this would be - 1 upon x + 1. So our final answer will be option A. Go on to the next question.

[00:22:17]Which of the following is an expression for the mean X upon 3 X and X + 6? You have to calculate the mean. Mean of a function. Mean of any given variable is a sum of variable upon the number of variable which is three here. So the mean will be X upon 3 + this will be 2X + 6 upon 3. If you take the LCM, it will be 3 x + 6 x + 18 upon 3 and this 3 will be x + 6 x. This will be 7 x + 18 upon 9. So our option that matches this d.

[00:22:57]Let's solve another past question. The equation below has two roots. What is the sum of the roots? So this can be you have to rearrange the equation such that you see a quadratic equation. Whenever an equation is set to have two roots, it is sure that the equation is quadratic.

[00:23:16]And if it has three roots, then the equation is cubic. If it has four roots, then quadric. So, and so and so forth.

[00:23:23]So, here it is a quadratic equation. You can find the quadratic by cross multiplying the x² + x. And this will be x² - 4 = 0. This will be x = 4. x²= 4.

[00:23:37]And x = + -2. And this the sum of the equations of the roots will be -2 + 2 and this will be sum. So our final answer is option E. Let us shift gears.

[00:23:49]We have already established the basic rules. So let's dive into the nuts and bolts of linear equations. Linear equations in one variable are represented by ax + b=0 and it solution will be x= - b upon a. There are three conditions to it. If a is not equal to z. If a is not equal to zero, in that case the solution is uniquely determining. If a is zero, b is also zero. In that case, the solution is indeterminate because any value of x will satisfy the condition a x + b equals 0. So it has it will have infinitely many solutions in that case and the solution is impossible if a equals 0 and b is not equal to z. In this case, the solution is impossible.

[00:24:33]So we have three distinct solution ways.

[00:24:36]With that framework in place, we are perfectly positioned to tackle linear equations in two variables. All right.

[00:24:44]Linear equation in two variable is represented by two variables x and y and three coefficients a, b and c. All right. The x intercept meaning the value of the function. The linear equation in two variable is associated with an x intercept. This is the point where the line crosses the x-axis or where y = 0.

[00:25:06]Whenever you have let's suppose 3x + 2 2 y = 1 and if you put y = z you will get certain x and this x is the x intercept.

[00:25:18]Do it otherwise and you will get y intercept. The point where line crosses the x and yaxis. This is a point when you this is point you can always find by putting x equals z. So yeah you will get y intercept slope of this line is minus a upon b while linear equation represents one line a pair of linear equations represent two lines. All right if you have two lines you can talk about their intersection whether they in whether they intersect each other or they remain parall to each other. The nature of these lines can be further given as intersecting lines if a1 upon a2 is not equal to b1 upon b1.

[00:25:57]In this case the lines intersect. If you have a1 upon a2 equals b1 upon b2 and does is not equal to c1 upon c2. In this case the line remains equidistant and never intersect. Therefore no solution.

[00:26:09]If a1 upon a2 equals b1 upon b2 equals c1 upon c2. In this case the line completely overlap. This is one line.

[00:26:17]Another line completely overlaps it. And in this case you will have infinite solutions because they intersect each other infinitely many times. Now we solve some of the past year questions and look at the pattern or question asked regarding this chapter. Let's look at one of the past year questions of bi.

[00:26:34]The graph below shows the line joining A and B and it perpendicular bis sector is given by the dashed line. We have to find the equation of this dashed line.

[00:26:42]Okay. Since the dashed line is a perpendicular bis sector, it is a bis sector of A and B. We can find this midpoint as + 2 upon 2 since it is a midpoint of this point and this point. So we can take the mean 3 + 1 upon 2 and this will be 4 comma 2. We have already found the point on the line. If we could find a way to find a slope or another point on this line, we would have the equation of the line. So let's try finding the slope because we have been given another useful information which is the dash line is perpendicular to AB and if you don't know keep in mind that m_sub_1 m_sub_2 product of the slopes of two perpendicular lines is always -1. You can find the slope of this line as 3 -1 upon 6 - 2 which is y2 - y1 upon x2 - x1. And this will come out to be 2 of 4 = 1 upon 2. Because m1 m2= -1, we can find m_sub_2 the slope of this dashed line. It will be minus2. We know the slope of the line the dashed line and we know a point on the dashed line. We can find the equation of the dashed line by y - y1 = x - x. And the equation of the best line would be y - 2 = slope which is - 2 x - 4 is x1. All right. So our equation of the line will be y - 2 = - 2x + 8. And this will reduce to y + 2x = 10. So this is 10 - 2x. Option C will be our correct option. In this question you have to keep in mind we have done few things. The product of the slope of two perpendicular line is always negative -1. And slope can be represented by y2us y1 upon x2 - x1. x1 is this. And the equation of line if you have slope and a point on the line can be given by this equation. Keep these three things in mind and we'll move on to the next question. Let's re this question.

[00:28:58]Question 55. What is the equation of the straight line which passes through - 6 x 2 and is perpendicular to 4 y + 3x = 8?

[00:29:07]So you have to find the equation of line which passes through this that means you have point on the line. If you could find another point let's suppose x1 y1 you will have the equation of the line.

[00:29:21]If you could find a slope for this line you will also have the equation of the line. And in this case we have already discussed the equation of the line is given as y1 y - y1 = x - x. All right.

[00:29:34]So how do we find the slope of this line? It is given that this is perpendicular to this line. The slope of this line will be minus the upon b. So this will be -4 upon 3. Therefore m_sub_1 m_sub_2 will be -1.

[00:29:52]Since mqua is -4 upon 3, m2 m_sub_2 gets -1. m_sub_2 will be 3 upon 4. Therefore, y - y1 = 3 upon 4. x1 is - 6. So, this will be + 6. And this y1 is 2.

[00:30:09]Therefore, this will come out to be 4 y - 8 = 3x + 18. This will come out to be 4 y - 3x = 26. 4 y - 3x = 26 option B is the correct option. I will move on to the next topic. Having covered up all the theoretical formations, we now move on to our next core module, conscience.

[00:30:32]A function is a specific mathematical rule that maps every valid input to exactly one unique output. All right, crucial caveat here is the input has to be valid and the output has to be unique. All right, we'll come back to this later. The notation of a function is f ofx. x here represents the input value and f ofx is the output. For instance, if fx= 3x - 2, you put x = 4 and you will get 3 into 4 - 2 = 10. The domain and range of the functions.

[00:31:05]Domain is a set of all the permissible values that a function can acquire. For instance, f(x) let's suppose it is 3x - 2. then its domain will be all real numbers. Domain is a set upon permissible values at which a function is defined. All right. Let's suppose if it's f(x)= 3x -2 then for all x belongs to real number this function will be defined. Okay. But if we take fx= 1 upon x then for x=0 this function will approach infinity and therefore it is not defined. So x=0 will not be in the domain of this function.

[00:31:45]Range is a set of all the values for which the function gives certain output.

[00:31:51]Let's suppose for every input there is certain output and the set of all those output values it's the range of the function. A function could be of many types. It could be even, it could be odd, a function could be inverse, fractional part of a function, integer greatest integer function. Here we are specifically interested in the type of functions um which are polomials. So a function could be constant if the power of x and the greatest power of x in the equation is zero. It's linear if the highest power of x in the equation is one. It's quadratic if it's two and cubic if it's three. All right. So this this is the type of function purely based on the polomial. Sorry. And then we have a vertical line test. Vertical line test ensures that a function is defined. If a graph fails a vertical line test, that means it is not a function because at that point, it will have two values. Let's suppose a circle and you draw a vertical line. At these points, this and this. For a unique input value, you are getting two output.

[00:32:57]Let's call it y1 and y2. For one input value, you are getting two output.

[00:33:01]Therefore, this graph, the circle graph fails a vertical line test. Therefore, equation of a circle is not allowed. So, for a graph to be defined as a function, it must pass the vertical line test.

[00:33:14]That is the vertical line must intersect the graph only once. Building on what we have just established, our next logical step is to look at the restrictions that is there on the domain approach. There are three restrictions that we'll study.

[00:33:27]First one is division by zip. If you have a function, we already talked about this. If you have a function defined as f(x)= 1 upon(x), you cannot put x=0 here because then f(x) will tend to infinity.

[00:33:41]Okay? Therefore, function will be undefined at this point. All right? So, you cannot divide by zero. Square root of a function. Let's suppose f(x)= under root of x - x - strictly negative. Okay. So this is x - a and x - a is strictly negative. This is called negative. Okay. If we put less than equal to z and this is called strictly negative. If I say x - a is less than z. If it's strictly negative then under root will have a negative term and in that case the under root is not defined in real numbers. All right.

[00:34:21]If x is considered to be real numbered.

[00:34:23]Okay. So the function will not be defined in that case. Always employ this condition that uh if f(x) equals under root of any function. Let's suppose there is t ² + tq + y. So you put b ² + tq + 1 greater than equal to 0. Always use this condition for square roots. For logarithms you have to keep in mind that x is always positive. The x here let's switch gears for a moment and turn our attention to the composite functions. A composite function passes the output of one function directly into the another as an input. You have f of gx where gx is the input for this function f ofx. We take a concrete example here. Let's suppose fx is defined as x - 3 and gx is defined as x - 5x². Let's suppose it's 5x² and you are required to find g of f of 1. So what you will calculate? First you will calculate the input of g. You are working on function g. You have to calculate the input of this function f of 1. So you evaluate f of 1 as 1 - 3 = -2. And now you will put minus2 in here.

[00:35:38]So g of -2 will be -2 minus of f of -2².

[00:35:45]So this is uh the basic thing about composite functions. Inverse function.

[00:35:51]An inverse function completely reverses the action of the original function. All right. It is denoted as f of inverse of x. Let's turn our attention to our next topic. Inverse functions. An inverse function completely reverses the action of the original function. It is denoted as f of inverse x. The condition on f of inverse x to exist is that f ofx must be 1 one as well as on bit. So 1 one or 1 functions are those functions for which if fx1 equals fx2 then x1 equals x2 onto functions are those functions for which co- doain and range they are equal. Co- doain is the domain co- doain for which a function is defined. We define a function f of x as part to this is the co- doain. Okay. And range is a set of output values that a function outputs for a given input. So this is all about inverse functions. Graphically the curve of an inverse function is a perfect reflection of the original function across the diagonal y= x. If you have any constant let's suppose suppose an equation like this then if you take its inverse if its inverse is possible first of all if it satisfies the two conditions of being one to one and on to in that case if you take its reflection about this line y= x then you will get a perfect reflection of this graph if it's in if it's inverse exist in that case so this is all about inverse functions now I move on to some of the past tier questions of IMAT before tattling some of the past year questions. Let me quickly summarize some of the important points here uh in decreasing functions. Okay. So if you have two points let's suppose x1 and x2 and on increasing x the value of the function let's suppose this is y2 and corresponding to x1 this is y1 and y2 is greater than y1 then you see that on increasing the value of x you're getting a greater value of y. In this case, the function is set to be increasing function. If you're getting a smaller value, the function is set to be decreasing. All right? Maxima and minima of a function. Now, you have to keep in mind that for any function, if it bends smoothly, all right, in the opposite direction, in that case, this bending has a slope zero. FX is zero. And it could be a minima or it could be a maximum. For minima, you have this condition. f dx x is positive for a maximum you have this condition f d- x is negative all right so these are the two important points about maxima and minimum now we tackle some of the positive expressions let's bring inverse functions to life by crunching some numbers together uh this one's from 2023 48 consider the function defined for every x greater than 1 where ln integrates the natural logarithm f of x is defined as 2 ln x - 2 ln nx - 1.

[00:38:53]Which of these is expression for its inverse function f of inverse y? So we have f ofx = 2 ln x - 2 ln of x -1. You can replace f of x by y and this will be 2 bracket ln x - ln of x -1. You could move to the left. Then ln x there is minus. So you will use the formula length a - ln b= ln of a upon b. So that will be x upon x - 1 and this will be e / e ^ y by 2 = x /x -1 and rearranging you will get so e ^ y by 2 x - e ^ y by 2 = x and so you will get the final answer as e ^ y by 2 upon ^ y by 2 - 5.

[00:39:46]So our final answer is in option A. One crucial caveat in the previous question we solved was that we have to find the range of this function. Since the function is only defined for x greater than 1. So we can find the range at 1 + and at infinity. If we try to find the value of y at these points then we get the range. So at x ts to 1 + you will get y = 2 ln 1 upon 1 - 1 and this becomes infinity. So ln infinity will be infinity and at infinity y = 2 ln infinity upon infinity minus 1 you have something x upon x -1 where limit x tends to infinity in that case this value is one always remember that therefore this will be 2 of ln 1 and this will be zero so you have y= 0 all the way to y= infinity in that case you can safely assume that your function is greater and zero.

[00:40:49]>> If you want me and my team to help you out with the academics and also the documentation processes for the IMAT exam, you can fill the form below and my team will get in touch with you through WhatsApp.

[00:41:00]>> Let's set the stage for our next major milestone. Inequalities, fundamentals of inequalities, definitions and symbols. We have to keep track of all these symbols. All right.

[00:41:11]This represents less than or equal to.

[00:41:13]This is greater than or equal to. This is strictly less than. And this is strictly greater back. One important rule that you have to remember is whenever you try multiplying a negative across an inequality equation in that case the sign of inequality reverses right if you multiply it minus1 into this equation the sign reverses so you will get x less than equal so this will be 3x less than= minus of 12. So yeah always keep that in mind. Now we look at the interval notation and the number line. An open interval with suppose 2a 3. This means that you are including all the values between two and three excluding two and you are excluding two and three. If you're writing two closed three open this means you are including two but excluding three and including all the values between two and three. If you write 2, 3 closed in that case you are including two and three and obviously it also includes all the values between two and so yeah this is about all about open interval closed interval and mixed intervals. Now let's look at the combination and absolute inequalities combined inequalities could be and or it could be or. In case of and it must be satisfied simultaneously. For example, if we have a less than x and x less than b, in this case, all the values of x must be such that they satisfy both these inequalities simultaneous. In another case, we see if you have x less than a or x less x greater than b. In this case, you will have all the values of x, which could satisfy either of these inequalities at once. it does not need to satisfy both of them simultaneous. So this is all about combined inequalities or absolute inequalities. If you have mod x less than 1 in this case this transforms to this minus of a less than x less than a.

[00:43:19]And if we have x mod of x greater than 1 in this case this transforms to x less than minus a or x greater than a this is by and this is and all right now we move on to the next advanced solving methods.

[00:43:35]So there are two methods to solving quadratic inequalities. The first one is a x² + b x + c. You have this equation given and you try finding the roots of this. So let's suppose you have x - alpha x - beta is greater than zero.

[00:43:52]Then you represent the graph alpha and beta. Now you find all the values for which the quadratic equation graph is positive. In this case it is for x less than alpha and for x greater than beta.

[00:44:07]So this is one method which is quadratic inequalities. Another method is wavy curve method. To elicitate the WB curve method, let us take a representative case. All right. So let's suppose we have x -1 into x greater than 0. So you first find points at which this equals 0. So this will be x= 1 and x= 0. Now you plot these points on a line. So this is one and it would be zero. And try drawing the curve as like this. Now you see this is positive here and positive here and negative here. Below the number line it is assumed to be negative and above it it is assumed to be positive.

[00:44:48]So now you can see the function is positive at x less than zero and at x greater than one. And these conditions can be satisfied individually. So we have four. Now let's look at some of the pastions. Some of the least pastions.

[00:45:03]What is the solution to the inequality root 2x less than 1 + x? In this case we can square. Obviously we have to spread.

[00:45:12]So 2x^2 less than 1 + x. This will be 2x less than 1 2 + x^2 + of 2x 2x will get cancelled and we have x² + 1. This is greater than 0. And for all values of x this will always be the left hand side will always be a positive number will always be greater than zero. Or if you want to write like this for all values of x x² will be greater than minus1.

[00:45:41]Therefore option a will be our answer.

[00:45:44]So pron values of x x² will be greater than minus1. But you have to also keep in mind that <unk>2x is defined when 2x is greater than equal to 0. Therefore x is greater than equal to z. This is the domain of the function. All right? So you have to only find intervals inside this domain. When I say x² is greater than minus1 in this case I'm only strictly talking about the interval in the domain which is option a. So yeah we have a lot of grounds to cover today from probability to permitation and combination. So let's get started with probability first. Probability is the mathematical measure of the likelihood that a specific event will occur. It quantifies the concept of chance on a strict numerical scale from 0 to one with a zero being an absolute impossible to one being an absolute certain event and.5 is an even chance. Sum of probabilities of all the events always adds up to one. Empirical probability.

[00:46:50]Empirical probability is based on historical data, observations and experiments.

[00:46:57]And this is conducted like you have a coin that's suppose tails and heads and you throw it n number of times let's suppose 100 times and you get 48 heads and 52 bails then the probability of getting head will be 48 upon 100 and aes would be 52 upon 100. This is empirical pro and yeah there is law of large numbers that as you appro as you increase the number of trials you approach even all right like probability of getting head here we uh said it was 48 upon 100 if you increase the number of trials from 100 to let's suppose thousand or even more this will approach five critical probability and sample space the sample space is a complete exhaustive set of own possible equally likely outcomes of an experiment. For instance, when rolling a standard sixthsided die, the sample space is defined as 1 2 3 4 5 6. All the possible possibilities that the experiment holds, it forms the sample space. Theoretical calculation is number of favorable outcomes dividing by the total possible outcome. This is a theoretical calculation. We are not actually performing the experiment. We are just calculating the possibility that two will occur. So the total number of ways two can occur is just one total and the total number of outcomes are six. The probability that uh a die with phase two will be shown. It is one upon six. Now we take a look at complimentary events.

[00:48:33]The complement of an event E represents the exact scenario where that specific event does not occur. Okay. If E dash is the complement of an event key, then the sum of their probabilities will always equal to one. This complimentary property is highly useful for efficiently calculating at least one probability. Instead of adding up, we will come back to this with past your questions. All right? Where probability of calculating at least one will always be equal to probability of man subtracted from one. This topic naturally leads us to our next core concept of mutually exclusive and independent events. Mutually exclusive events are those events that cannot possibly occur at the exact same time.

[00:49:18]For instance, probability of getting one on throwing a dice and probability of getting two on throwing a dice. If you throw a single dice, you cannot get one and two at the same time. So these are mutually exclusive events and in this case you can always take their inter their intersection will always yield a probability of zero because obviously they are not possible and their union will have a probability of their sum. In case of independent events the occurrence of one event has absolutely no effect on the mathematical probability of the second event. For example, if you roll two dice separately, in this case, the probability of getting anything on this dice is independent of this t. In this case, you can always use the multiplication rule that probability of getting a on this die and b on this die is and its intersection and therefore its probability will be a product of probability of getting a into probability of getting b on the second die. All right, I move on to the next.

[00:50:15]Let's look at another concept combined events. Combined events involve calculating the probability of multiple distinct events happening either sequentially one after another or simultaneously at the exact same time.

[00:50:28]These combined events include with replacement questions and another set of questions which are asked without replacement. When you are doing replacement that suppose there is a bag containing the red and blue balls and you took one ball out and then you again replaced it. You put it back in the bag.

[00:50:46]In this case, the sample space remains completely identical for each subsequent event. If you are not putting it back, in this case the sample space shrinks after each event occurs. If let's suppose you are taking two balls out of the bag each after each event in this case the sample space shrinks by two after each event. All right. So this is the combined events with and without replacements. Now we move on to the next. Building on to what we have already covered. Now let's look at when diagrams and the probability formulas. A v diagram usually represents sample spaces and events as overlapping geometric reasons making complex combinations highly intuitive. All right. If for example we have a this is set a and this is set b then a union b it represents the entire set. All right.

[00:51:40]of A and B everything counted once because obviously in a set element is not counted twice. All right. So A union B equals the the probability of A union B equals probability of A plus of probability of B. And this reason is counted twice. All right? So you subtract this. So this is P A intersection B. A intersection B represents the common region between A and B. If you talk in terms of probability, this intersection represents they are not mutually exclusive. Okay, their intersection must be mathematically subtracted. All right, so yeah, that's it for when diagrams and probability formulas. Now we move on to some of the past questions of probability. Let's rect question form probability. In a bag there are three red balls, seven green balls.

[00:52:34]>> Indistinguishable by touch. Two extractions are made with the first ball being returned down to the back before the second extraction. That means this is with replacement. What is the probability of extracting two green balls? So probability of extracting first green ball. Let's look at the probability of extracting first green balls. This will be 7 upon 10. There are total seven green balls and the total sample space is 10. So the probability of extracting the first green ball is 7 upon 10. Now because the ball has been returned to the bag before the second extraction, therefore the sample space for the next event is also 10 and the probability of getting a green ball this time. Since we have already gotten this, so the sample space for this uh second event will also be 10. Since the number of green balls does not change.

[00:53:25]Therefore, for second event, the total number of green balls will also be seven. So the total probability of getting a green then a green will be 7 upon 10 into 7 upon 10. This will be 49 upon. So the correct option will be a 49 upon 100. Let's get our hands on another past year question. It was asked in 2023. Two standard six-sided dice numbered 1 to six are rolled. What is the probability that the product of the two numbers tained is the square of a crank. All right. So when two dice are drawn with each having a number 1 to 6 in this case the maximum product you can obtain is 6 into 6 which is 36. And the minimum you can obtain is 1 2 and this.

[00:54:15]And you have to find the probability that the product of the two numbers obtained is the square of a prime. You have primes 2 3 5 7.

[00:54:25]All right. The square of this will be four. This is contained in this. It could appear anywhere. Uh square of this number is 9. Square is 25. This is 49.

[00:54:36]Well, 49 will not be an outcome. All right. So 4 9 and 25. You can find four using four one if the outcomes of the two dice four and one. If it is one and four in this case as well if one appears on the first dice and the four appears on the second dice in this case also you are getting four as a product or if first if on the first dice you get two and the second dice you also get two in this case as one you will get the product of four. Nine could be found by three and three only. All right, you get three on the first die and three on the second die. And similarly, this can be found by only five five. If when you get five on the first eye and this five on the second die. So you have a total of five cases where you are getting values such that their product is a square of a prime. All right. So the total possibilities or the probability of getting this probability of X will be where X is this. This will be five upon total value total outcome 36. So our correct option is A. Having tied up all the loose ends we can turn our attention to the pivotal concept of statistics. An important concept in statistics is central tendency. There are three measures of central tendency. First one is mean. Second is median and third one is mode. Mean often referred to as arithmetic average is defined as a sum of all individual raw values in the data data set divided by the number of observations or the data points. It is mathematically written as sigma of all the math that refers to as summation of all the individual raw values in the data set divided by the number of observations or the data points.

[00:56:26]Similarly, median is defined as the exact positional center of a data set when strictly arranged in descending or ascending order. When you arrange a data set or data points into strict ascending or descending order, then the middle value is referred to as median. If the number of data points in the set is odd, then median is the exact middle value given by n + 1 by. If the number of data sets or the data points is even, in that case there is no single middle number.

[00:56:55]The median is calculated as the arithmetic mean of the true central overlapping values. Coming to the next measure of central tendency, the mode.

[00:57:03]It is defined as the most frequent value in the data set. Mode identifies as the most commonly occurring value in the data set. It is determined strictly by the direct counting of the raw data. All right. If the data set has only one single most frequent random, then it is unimodern. If it has two distinct values that share the highest frequency exactly then it is biodern and similarly multiodern and no more when every single value in the data data set appears with the exact same frequency then that data set has no more preliminary definitions unlocked. Let's set the ground running with some of the past time at questions.

[00:57:40]Which one of the following is an expression for the mean of this? All right. So you have to calculate the mean of this data set x upon 3 plus of x plus of x + 6. So we have three data data points. We will defer we will divide this by three the summation by three and this will be mean. This comes out to be x upon 3 plus of 2x + 6 upon 3. This will be x + 6 x + 18 upon 9. This comes out to be 7 x + 18 upon 9. Therefore, the correct option is D. Let's look at another question. The mean of five non-positive integers is 20. The median is 24. What is the maximum possible value of the largest number in the data set? So, the mean of five positive numbers is 20. That means let's suppose the five numbers are x1, x2, x3, x4, and f x5. Then their sum is 20. So the sum of the numbers will be 100. What is median? We already defined median of an odd number will be n + 1 by 2. So 5 + 1x2 the third number. So x1, x2, x3, x4 and x5 assuming they are used in the sequence of increasing order. All right?

[00:59:04]In the ascending order. In this case x3 will be medium. All right. We have to find a data point in this set such that it is largest. Okay, we have to maximize the data point. So in that case we can always put x1 as one because it has to be non-zero positive number. So we can put x1 as 1. We can put x2 as 1. Median is 24. We can put x4 as 24 because x4 has to be greater than equal to x3.

[00:59:35]Therefore we can put it as 24. We cannot put x5 as 24 obviously and the fact that we are putting x1 as 1 and x2 as 1 is to maximize one data point in the set.

[00:59:48]Let's say x4 and x5. We want to maximize. So we can maximize x5. So how far can we reach? So that's why we have assumed x4 to be 24. It could not be 23 because that would violate the rules for medium. So x5 will be 1 + 1 + 24 + 24 + x5 = 100. So x5 comes out to be 48 into 50. 100 minus 50 this will be 50. So the largest value of the data point in this set would be a 50. The natural progression from our previous discussion leads us to permutation and combination.

[01:00:28]Permutation and combination is merely a counting method and we approach this problem through some formulas like combinations and formulas of permutation. There are two basic principle multiplication and addition principle. If you need to make one choice and then another choice, you multiply these options. You want three shirts and four pair of pants. Then you have 14 or 12 different outfits in this case. So you multiply the total number of options. But if you have two bus routes or three train routes to reach to a particular destination that suppose Dean in this case you add the totem number which is five total travel option you cannot multiply and say you have six. Now we look at the basic definition of factorial is defined as continuous product of numbers. This is factorial factor all the way up to one and we have four factorial permutation and combination we have two different this is factoriation and combination we have two different we can arrange the letters in permutation any fation let's suppose different numbers let's suppose we have five letters A A D A B C okay D in this case we are not these letters for any fashion that is the ordering of the number of these letters if we take into account the ordering of these letters we will in this case we are not taking into account the formula for NP or only about the combination to the combinations of any given formula is We take n factorial combination factorial combinations have given n factori in this case we take as we will see the boundaration and combination factorial upon r factorial how many ways are there to order the letters let's look at this with question let us ground the conation and combination with Last year the question already how many ways clarified to order the letters matters.

[01:03:05]Therefore C we take for example B A total number of ways to order these letters the question already matters matters permutation. Therefore now we take as total number of repeated letters letters in this case A is repeating so divide by two. I will use permutation again. BB is repeating. Now one question you will divide the answer by two. When you have if it was repeating letters BB in this case then you would have divided it by so you will divide the answer by 2 factorial. So this is again BB is repeated 5.

[01:03:43]If it was and this will be factorial upon then you would have divided it by 3 factorial upon 0 factori. So this is final answer we have already discussed as factorial factor. This is one and this will be factorial upon 2 factorial this will be 30 y we have already discussed that 0 factorial is one and two factorial 2 factorial this is 1 t upon 4 this is 30 so our correct option is option our fourth command the scores will be anchored entirely in the field of geometry all right without further ado let's type the distance formula gives the distance between two points x1 and x2 m m is given as d = under root of x1 - x2 - x1 square plus of y2 - y1 squ and is on the entire term. The section formula gives the coordinates of a point which divides the line into two ratios m is to n. All right. Let's suppose this is a point a b and we want to find the coordinate of this point which divides a into mf to n ratio. MS to n ratio. In that case, you can find the coordinate of this point using this formula. Now you move on to the next slope or gradient formula. Let's turn our attention to the slope or the gradient formula. Provided you have two points x1 x y1 and x2 y2. You can find the slope of the line connecting these two points using this formula which where m= y2 - y1 upon x2 - x1. Equation of a line can be represented in various forms. First of its kind is slope intercept form where y = mx + c. m is the slope, c is the constant and point slope form where you have x1 y1 as the point and m is the slope. Then you can use the formula y - y1 = mx - x1 and you'll get the equation of the line. General form of the equation of line is given as a x + b y + c= z. All right? Now move on to the next. That naturally leads us to our next core concept of parallel and perpendicular line and their conditions.

[01:05:53]For a parallel line, the slope will be equal and for a pair of perpendicular lines, the product of their slopes will be -1. All right. To find the distance of a point from a given line, you can use this formula. D equ= A into X1 B into Y1 + C upon in the denominator you have under root of A square + B square where you where X1 Y1 is the point and Ax + B Y + C = Z is the line and want to find the distance between X1 Y1 and the line. If you have two points x1 y1 and x2 y2 then the line passing through both of them is uniquely determined by using this equation. This is the equation of the n. If you put all the constants here y1 y2 x2 x1 you will get an equation of the form a x + b y + t = z. And this will be the line that will pass through both these points. Let us now direct our attention to areas of the geometric figures. For a triangle, area is given by half into base into height. For rectangle squared, length into width.

[01:07:07]For trapezium, it is half into a + b into h where a and b are the parall bases and h is the perpendicular height.

[01:07:15]For a circle, this is p<unk> r².

[01:07:18]Important thing to here is for a sector subending theta angle at the center.

[01:07:23]Area of a sector is given as s=<unk> r² theta upon 360 or we could write a theta upon 360 where a is the area of the circle. Here a represents area of the circle theta upon 360 that will give you area of the sector subending theta angle at the center. Building on to what we have already established let's talk more about the circle specifically the equation and some properties of the circle. The equation of a circle with its center at origin is given by x² + y^2 = r² where r represents the radius of the circle. If the center is at a specific point h, k in that case the equation of a circle is represented by x - h^ 2 where h represents the xcoordinate of the center. y - k^ 2 where k represents the ycoordinate of the center equals r² where r is the radius of the circle. General form of the equation of circle is given by x^2 + y^2 + 2 gx + 2 f_sub y + c = z where minus g minus f is the center of the circle and radius is given by under root of g² + f² minus c. All right. So the basic properties of circle you already are aware about diameter is 2 * rs circumference is 2 pi r the tangent rule that radius and tangent are always perpendicular to each other and arc angles angle subended by exact same ark at the circumference are mathematically equal to each other. All right so now we move on to some of the past year questions of him. Let's switch the gears for a moment to turn our attention to some of the basic properties of triangles. First one is pythogous theorem. We are already well aware of this fact that given this is perpendicular, this is base and this is hypotenus. Then h² equ= p ^² + b² perpendicular bis sector theorem which states any point placed directly on the perpendicular bis sector of a line segment is perfectly equidistant from the end points of that lines. It bicts the line segment into two halves.

[01:09:31]Angry bis sector theorem. It states an angle bis sector of a triangle divides the opposite side into two distinct segments that are mathematically proportional to the length of the other two sides of the triangle. All right, let's suppose this is a bis sector of this angle A. AB is a bis sector of the angle A. Then AB upon AC equals BD upon DZ. Let us turn our attention to our most important theorem midpoint theorem.

[01:10:01]It states the line segment connecting the midpoints of any two sides of a triangle is strictly parallel to the third side first point and its length is exactly half of that third side the second point. Consider a triangle ABC such that there is a line which runs half which bisects P A B as well as AC.

[01:10:24]Then this line T will be parallel to BC.

[01:10:29]Thus the length of B will be half of BC.

[01:10:33]Equal sides means equal angles. Consider a triangle where two sides are equal in benth then the angle corresponding to them will also be equal in depth. Theta and this will also be theta. Conversely, if angles are of the same size, then the sides will also be corresponding sides will also be of equal length. the side opposite to that angle. Finally, we conclude this session with basic proportionality theorem which states if a continuous line is drawn parall to the third side of a triangle so that it intersects the other two sides, it strictly divides the two sides in the exact same proportion. All right. So let's suppose this is a b and c and this is d e. that we have a b upon bb which is theta upon bb equals a e upon e right so this will be equal to a upon east let me elucidate the concepts we have already studied with some of the past IMAT questions this one's from 2022 59th question the diagram shows two concentric circles both with center O the angle of the shaded sector is theta radius of the larger circle is 50% and greater than the radius of the smaller circle. Total area of the smaller circle is 36 pi and area of the shaded region is given to be 27<unk> upon 8 cm². What is the value of theta? Let's try and find the radius first. The relationship between the radius of these two circles.

[01:12:07]The radius of the larger circle is given as r + 50% of r. 50% of r is r upon 2. So this will be r + r upon 2 which is 3 r upon 2. All right. So this is 3 r upon 2. This is r.

[01:12:24]We already have the area of the smaller circle which is 36 pi. Area of the smaller circle will be p<unk> r² and this is given to be 36 pi. Pi will be cancelled and r comes out to be + - 6. - 6 is redundant obviously. So r = 6 cm.

[01:12:45]that is the radius of the smaller circle. So radius of the larger sunken will be 3 r 2. This comes out to be 3 into 6 upon 2 which is 9 cm. Now we can find the area of shaded region which will be pi r² area of the bigger circle into theta upon 360. We already discussed this formula earlier. Area of the shaded region will be area of the bigger circle into theta upon 360. Area of the bigger circle will be p<unk> r² which is 3 r² by 2. Radius of the bigger circle is 9. So 9² is 81 into theta upon 360. This is equal to 27<unk>i upon 8.

[01:13:28]Pi will be cancelled. Let me try solving this further. 27 upon 8 = 81 theta upon 360. This will come out to be 93 93.

[01:13:41]Let's take it all one side. 360 upon 24 that this will go in line 49 4 6 15 so theta comes out to be 15 right let's check option here will get the correct answer okay so we have theta equ= 15° move through it once again and try and stack what I did let us look at another past iMac question a right angle triangle has an area of 18 cm square one of the two shorter sides is twice the length of the other one. What is the length of the hypotenus? All right. So the length of one suppose this is a shorter side of length L. So this will be 2 of L and hypotenus will be okay. So the area of the triangle is given to be 18 cm squared. Area of a right angle triangle is given as half into perpendicular into base. All right. So this will be half into L into 2 L that includes a D. So two two will be cancelled. L comes out to B or under root AB. All right. Now what is the length of the hypotenus on the triangle?

[01:14:49]Hypotenus will be hypotenus squ + L² + 4 L² using by theorem. So H comes out to be under root of 5 L² which is under root of 5 into L² 18.

[01:15:05]This comes out to be 9 which is 9 into 10 9 under root 9 will be 3. So three in root of 10.