Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Here is a function: e to the power k x, minus one, over sin two x. At x equals zero, the top is zero and the bottom is zero, so the function has a hole there. Now we fill that hole with the value three. Which k makes the graph join up, with no break? Is it three? Six? Something else? Today: Set one of the mixed JEE Main pattern problems, where limits, continuity and derivatives meet. Questions like these come up often. So, let's begin.

0:36Set 1: eight problems, four kinds

So, here is Set one: eight problems in the style of JEE Main. Problems one to three are limits. For them you need expansions, rationalising, and the standard limits. Okay? Problem four is continuity at a point. Clear? Problems five and six: a derivative, and then differentiability. Right? And problems seven and eight are short applications: a tangent, and a maximum. Now look. In a mixed set, nobody tells you the chapter. So, what is the very first thing you do with a mixed problem? Name the type, exactly! Is it a limit, continuity, or a derivative? Once you name it, the tool is clear. So today: two tools, then problems four and five, the easier one first. With me?

1:24Tool 1: standard limits and continuity

Tool one: the standard limits. The limit as x tends to zero of sin x over x is one. And the limit as x tends to zero of e to the power x, minus one, over x, is also one. The arrow under lim means tends to, and x is in radians. Okay? Now tell me. In these limits, must it be x exactly? Or can x be replaced by two x, or by k x? It can, yes! Any expression u that tends to zero can take the place of x. So sin u over u tends to one, and e to the power u, minus one, over u, tends to one. The only rule: the same u on the top and at the bottom. Got it? So, a quick result. As x tends to zero, sin a x over sin b x tends to a over b, with b not equal to zero: the equals sign with a slash through it means not equal to. Each sine just needs its own matching u. Right? Now continuity. f is continuous at x equals a when the left hand limit, the right hand limit, and the value f of a are all equal. The small minus on a means x tends to a from the left, and the small plus means from the right. Remember, the limit exists only when the left and right limits agree, and then it must also equal f of a. Clear?

2:50Tool 2: a variable base and a variable power

Tool two: derivatives where the base and the power both change. y equals u to the power v, where u and v are functions of x, and u is greater than zero. So, which rule do we use here? The power rule, or the exponential rule? Neither! The power rule needs a fixed power, and the exponential rule needs a fixed base. So we take log first. These letters, l n, mean the natural log, and we read l n y as log y. Log y equals v times log u: the log brings the power down. Okay? Now differentiate both sides with respect to x. On the left, by the chain rule: one over y, times d y by d x. On the right, the product rule. The small dash is called prime, so v prime is the derivative of v. We get v prime log u, plus v u prime over u. With me? Last step: multiply by y, which is u to the power v. The d by d x in front means the derivative with respect to x. So the derivative of u to the power v is u to the power v, times the bracket. And the trap? Students find the bracket and forget to multiply by y. Don't do that!

4:06Example 1: find k for continuity of (e^(kx) - 1)/sin 2x at 0

Now, remember our puzzle? Let's finish it. This is problem four from the set. f of x equals e to the power k x, minus one, over sin two x, for x not equal to zero. Remember, the slashed equals sign means not equal to. And f of zero is three. Find k, so that f is continuous at zero. Pause the video, and try it first. Done? Okay, let's check. Continuous at zero means the limit as x tends to zero of f of x must equal f of zero, which is three. One formula holds on both sides of zero, so the left and right limits come out the same. Okay? Now, the top looks like our standard limit, with k x in place of x. So the limit is just k, and k equals three, right? No, no, careful! The bottom is sin two x, not k x. So make each piece standard: divide the top by k x, put two x over sin two x, and fix it with k over two outside. The dots mean times. And k is not zero here; if it were, the top would be zero, and the limit zero, not three. Clear? As x tends to zero, k x and two x both tend to zero, so both pieces tend to one. So the limit is one times one times k over two: just k over two. Right? Now make it equal to f of zero. k over two equals three, and the double arrow means implies: k equals six. So the puzzle answer is six, not three. Three was the trap answer. See how one missing two changes everything?

5:43Example 2: derivative of x^(sin x) at x = pi/2

Now a slightly harder one, problem five from the set. Find the derivative of x to the power sin x, at x equals pi over two. Pi is the Greek letter you see on screen. Pause and try. Hint: use tool two. Done? Okay, let's check. The base is x, and the power is sin x. Both change, so call it y, and take log, with x positive. Log y equals sin x times log x. Okay? Now differentiate the right side, sin x times log x. Which rule, and what do you get? The product rule, yes! Cos x times log x, plus sin x times one over x. And on the left, one over y, times d y by d x. Right? Multiply by y. d y by d x equals x to the power sin x, times the bracket: cos x log x, plus sin x over x. Exactly the formula from tool two. Got it? Now put x equals pi over two. Sin of pi over two is one, and cos of pi over two is zero. So x to the power sin x is pi over two, to the power one: just pi over two. In the bracket, cos x log x becomes zero, and sin x over x becomes two over pi. So the answer is two over pi, right? No, no, careful! Two over pi is only the bracket; that is d y by d x, over y. Multiply by y, which is pi over two. Pi over two, times zero plus two over pi: exactly one. Forget that last y, and you lose the question. That's it!

7:30Quick revision: name the type, standard limits, take log

So, quick revision. One: in a mixed problem, name the type first, then pick the tool. Two: for standard limits, make each piece sin u over u, or e to the power u, minus one, over u, with the same u. Three: with a variable base and a variable power, take log, differentiate, and multiply by y. Now go and try the practice problems on this page. All the best!