Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one quick statement before we start, and you tell me: true or false? If f prime of c equals zero, then f has a local maximum or a local minimum at c. The tangent is flat there, so it must be a peak or a valley, right? Most students say true. It sounds so obvious. Hold on to your answer. Today we do Part C of the unit problem bank: one line statements on limits, continuity and derivatives. Statements like these come up often in JEE, and one small word can hide a trap. So, let's begin.

0:38Part C: true or false, in one line

So, how does Part C work? Each item is one statement, and you decide: true or false. But a guess is not enough, okay? You give a one line reason, or a counterexample: one single case where the statement fails. Now think. Suppose a statement claims that something always happens. To prove it false, how many failing cases do you need? Just one! Yes, exactly. If it fails even once, the statement is false. That's it. But true needs more. Checking two or three nice functions proves nothing, okay? For true, you need a reason that works in every case.

1:19Statement 1: greatest integer of x at x = 1.5

Clear? Statement one. These square brackets around x mean the greatest integer less than or equal to x. So, true or false: the greatest integer of x is continuous at x equals one point five? Let's see. For x from one up to two, one included and two left out, the greatest integer of x is just one. This sign means less than or equal to, and this one means less than. So near one point five, the function is a constant. Okay? And a constant has no break. The left hand limit, LHL, and the right hand limit, RHL, are both one, and they equal the greatest integer of one point five, which is one. So it is continuous there. True! But look, here is the trap. Students remember that this function is discontinuous, and say false. It breaks only at integers. At an integer n, the left hand limit is n minus one, and the right hand limit is n. They do not agree, so the limit does not exist there, and the graph jumps. Got it?

2:30Statements 2 and 3: slopes and continuity

Statement two. If f is differentiable at a, then f is continuous at a. The double arrow means implies. True or false? True, and here is the one line reason. Write f of a plus h, minus f of a, as h times the same difference over h. Here h is a small step, and the dot means times. As h tends to zero, the fraction tends to f prime of a, the derivative; the small dash is called prime, and this single arrow means tends to. So the product tends to zero times f prime of a, which is zero: f of a plus h tends to f of a. Continuous! With me? Now statement three flips it. Continuous implies differentiable. True or false?

3:19Statements 3 and 4: corners and minima

False! Look at the left graph, y equals mod x; the two vertical bars mean the absolute value. No break at zero, so it is continuous. But L f prime of zero, the left hand derivative, is minus one, and R f prime of zero, the right hand derivative, is one. You can see both slopes on the graph. They differ, so there is no derivative: a corner. The right graph, y equals the cube root of x, with a small three on the root sign, fails the other way: a vertical tangent. So differentiable implies continuous, but not the reverse. Okay? Statement four, same left graph. Mod x has a minimum at zero, although f prime of zero does not exist. No derivative, so no minimum, right? True or false? True! Mod x is greater than or equal to zero, which is f of zero, for every x; this sign means greater than or equal to. So the lowest point is at zero. See, a minimum can sit where f prime does not exist. Clear?

4:27Example 1: is the limit of sin 1/x at 0 equal to 0?

Now Example one, another statement from Part C, okay? True or false: the limit as x tends to zero of sin of one over x equals zero. The arrow under lim, as before, means tends to. Pause the video, and try it. Done? Okay, let's check. Many students say: x goes to zero, and sin of zero is zero, so the limit is zero, right? No, no, careful! Inside the sine we have one over x, not x. As x tends to zero, mod of one over x tends to infinity; that sideways eight is infinity. So sin of one over x keeps oscillating between minus one and one. With me? In fact, sin of one over x equals one at x equals two over, four k plus one, times pi. And it equals minus one at x equals two over, four k plus three, times pi. Pi is the Greek letter you see, and k is zero, one, two, where the three dots mean and so on. Okay? As k grows, these points crowd in towards zero, from the right. So near zero the function hits one, then minus one, again and again. It never settles on a single value. The limit does not exist, so the statement is false. Got it? Now compare x times sin of one over x. Its absolute value is less than or equal to mod x, which tends to zero. So by the sandwich theorem, that limit is zero. One extra x in front changes everything!

6:07Example 2: f'(c) = 0 but no extremum, x cubed at 0

Now Example two, and it is our puzzle from the start. True or false: if f prime of c equals zero, then f has a local extremum at c. Local extremum means a local maximum or a local minimum. Pause the video, and try it. Done? Okay, let's check. Remember, false needs just one counterexample. So we want a function with a flat tangent at some point, but no peak and no valley there. Can you think of one? x cubed! Yes, exactly. The small three up top means cubed. f prime of x is three x squared, so f prime of zero is zero; the small two means squared. The tangent at zero is flat. Okay? But look at the values. For x less than zero, x cubed is negative, below f of zero. And for x greater than zero, the same sign turned the other way, x cubed is positive, above f of zero. Right? So at zero the graph just keeps climbing. Neither a maximum nor a minimum! One counterexample, so the statement is false. Said true at the start? No problem, most students do. That was the trap. The first derivative test tells us why. f prime is three x squared, greater than or equal to zero on both sides of zero. No change of sign means no extremum. So zero is an inflection point, with a horizontal tangent. Clear? So keep the direction straight. Every local extremum inside an interval is at a critical point, where f prime is zero or does not exist. But not every critical point is an extremum. Mod x had a minimum with no derivative; x cubed has a zero derivative with no extremum.

7:59Quick revision: mod x, x cubed, greatest integer, sin 1/x

So, quick revision. One: differentiable implies continuous, but not the reverse; mod x at zero is the counterexample. Two: f prime of c equal to zero does not force an extremum; think of x cubed at zero. Three: the greatest integer of x breaks only at integers, and sin of one over x has no limit at zero. Now go and try the practice problems on this page. One line each, with a reason. All the best!