Overview

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Some function f is differentiable, and you are told only two numbers: f of one is one, and f prime of one is three. That's all. No formula for f at all. Now find a limit with f of x inside it. Impossible? Not at all, and you will see it today. Look, in this chapter, problems from the whole unit come mixed together, just like a real JEE paper. So today, we see how to use this problem bank well. Let's go.

0:36How this chapter works

So, look at how this chapter is built. First, Part A, at JEE Main level: eight solved problems from the whole unit, all mixed. Okay? Now tell me, when should you open Part A? Right after the chapter on limits? No! Only after Chapters one to eight, the whole unit. The little dash on screen means one to eight. And attempt Part A as a thirty minute test, with a clock running. Treat it like the real exam, right? Next comes Part B: longer problems, at JEE Advanced level. We will look inside it in a minute. Then comes Part C, which checks the ideas often tested in statement form, as true or false. Small words, big traps. Got it?

1:26Topper's trick: budget your 30 minutes

Now, a small trick that toppers use. Every problem here is marked easy, medium or hard. Count them first. Part A has seven easy problems and one medium problem. With me? So, eight problems in thirty minutes. On average, how many minutes does each problem get? Thirty divided by eight: three point seven five minutes each, on average. That is the division sign on screen. Clear? And that is only an average. The easy ones must go fast, so that the medium one gets the extra time it needs. Right?

2:05Part B: JEE Advanced level

Alright, now Part B, the JEE Advanced level. Six longer problems, every one marked hard. So give each one real time, okay? What kind of problems? Limits with n tending to plus infinity inside them, like a function that is itself defined by such a limit. The arrow means tends to, and the sideways eight is infinity. And limits built from f prime of a. The small dash is the prime: f prime of a is the derivative of f at a. Our second example is one of these. With me? And proofs: an identity with y double prime, the second derivative, with two dashes, an inequality, and the largest area of a rectangle in a semicircle. Here you show that something is true. Clear?

2:52Part C: is the statement exactly right?

Now, let's try Part C, okay? First, a true statement. If f is differentiable at a point a, then f is continuous at a. The double arrow means implies. Differentiable always gives continuous. Right? Now turn it around. Every continuous function is differentiable. True or false? False! Take mod x, the absolute value of x: the two vertical bars. At x equals zero it is continuous, but its graph has a sharp corner there, so it is not differentiable. So differentiable gives continuous, but continuous does not give differentiable. Got it? One more. If f prime of c is zero, then f has a local extremum at c, that means a local maximum or a local minimum. Looks perfectly natural, doesn't it? True or false? Oh no, careful! False. Take f of x equals x cubed. Its derivative at zero, f prime of zero, is zero. But x cubed keeps increasing right through zero, so there is no maximum and no minimum there. That is exactly the kind of slip Part C catches. With me?

4:09Example 1: limit of sin 2x/(e to the 3x - 1) as x tends to 0

Now, our first example, a Part A problem, marked easy. Find the limit as x tends to zero, of sine two x, over e to the power three x, minus one. The little arrow under lim means x tends to zero. Start your clock, pause the video, and try it. Done? Okay, let's check. First, what happens if we simply put x equal to zero? Sine of zero is zero, and e to the power zero, minus one, is also zero. So it is a zero over zero form. And careful: that does not mean the limit fails to exist. Zero over zero only says, do more work. Okay? Now, two standard limits from this unit. Theta, the Greek letter on screen, stands for any quantity that tends to zero. As theta tends to zero, sine theta over theta tends to one. And e to the power theta, minus one, over theta, also tends to one. Right? So, under sine two x we want two x, and under e to the power three x, minus one, we want three x. So we write: sine two x over two x, times three x over, e to the power three x minus one, times two x over three x. The dots mean times. Everything new cancels, so nothing has changed. With me? Now let x tend to zero. The first factor tends to one, the second also tends to one, and two x over three x is just two over three. So the limit is two over three. That's it! Now a quick check with L'Hôpital's rule. It is allowed here: a zero over zero form, the top and the bottom are both differentiable, and, as we will see, the new limit exists. Differentiate them separately: two cos two x, over three e to the power three x. At x equals zero, that is two over three again. Perfect!

6:10Example 2: limit with f unknown, using f(1) and f'(1)

Now, remember our challenge? Let's finish it. This one is from Part B, marked hard. f is differentiable, f of one is one, and f prime of one is three. Find the limit as x tends to one of, x squared times f of one, minus f of x, all over x minus one. Hard means more time. Pause the video, and try it. Done? Okay, let's check. We do not know f. We know only f of one and f prime of one. So tell me, f prime of one is the limit of what? Yes, exactly! The definition of the derivative. f prime of one is the limit as x tends to one, of f of x minus f of one, over x minus one. So our job is to make this exact piece appear. Clear? Now, a quick guess. The top has f of one minus f of x, so the answer is just minus f prime of one, that is minus three, right? No, no, wait! That is the trap: the x squared is still there. So, subtract and add f of one. x squared times f of one, minus f of x, equals x squared minus one, times f of one, minus the big bracket, f of x minus f of one. Open the bracket, and the two f of one terms cancel. Okay? Now divide each piece by x minus one. First piece: x squared minus one is x minus one, times x plus one. So we are left with x plus one, times f of one. As x tends to one, that goes to two times f of one, which is two. With me? And the second piece is exactly our definition. Because f is differentiable at one, it tends to f prime of one, which is three. Right? So the limit is two times f of one, minus f prime of one: two minus three, that is minus one! No formula for f at all. Challenge solved. Got it?

8:14Quick revision: test plan, derivative definition, continuity

So, quick revision. Part A: a thirty minute test, after Chapters one to eight. Part B: hard problems, and when f prime of a is given but not f itself, use its definition. Part C: differentiable gives continuous, but not the reverse. Now open Part A, start your clock, and begin. Then try the practice problems on this page. All the best!