Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take one plus tan x, over one plus sin x, and raise it to the power cosec cubed x. Now let x tend to zero. Top and bottom both tend to one, so the base tends to one. And one to the power anything is one. So the answer is one, right? Hold that thought. Today, Set three of our mixed problems. Problems like these come up often in JEE Main. Let's go.

0:32Warm-up 1: the mean value theorem

First, a warm-up. The mean value theorem needs two conditions. f must be continuous on the closed interval a comma b; the square brackets include both ends. And f must be differentiable on the open interval a comma b; the round brackets leave the ends out. Okay? Then, for some c between a and b, f prime of c equals f of b minus f of a, over b minus a. The little dash after f is prime, for the derivative. Right? Now, find c for e to the power x, on zero to one. But first: does e to the power x meet both conditions here? Yes! It is continuous and differentiable everywhere, and its derivative is e to the power x itself. So e to the power c equals e minus one, over one minus zero, which is just e minus one. Clear? So c equals the log of, e minus one, the whole bracket; the l n on screen is log, the natural log. And e minus one lies between one and e, so c lies between zero and one; the less than signs say this. Perfect!

1:39Warm-up 2: a tangent from a point off the curve

Warm-up two. Find the tangent to y equals e to the power x that passes through the origin, the point zero comma zero. Quick check first: is the origin on the curve? No! At x equals zero, e to the power x is one, not zero. So call the unknown point of contact t comma e to the power t. The slope there is e to the power t, so the tangent is y minus e to the power t, equals e to the power t, times x minus t. With me? Now put x and y both zero. Minus e to the power t equals minus t times e to the power t. e to the power t is never zero, so cancel it; the double arrow means so: t equals one. Got it? So the tangent is y equals e times x, touching at one comma e. Point off the curve? Name the point of contact. Okay?

2:35The tool for the 1 to the power infinity form

Now, the tool for our puzzle. f of x tends to one; the arrow means tends to. And g of x tends to infinity, the sideways eight. Then f to the power g is called the one to the power infinity form. So, is the answer simply one? No, not always! The base is a tiny bit away from one, and a huge power makes that bit count. So, here is the rule. The l i m on screen is short for limit. The limit of f of x to the power g of x equals e to the power, the limit of g of x times, f of x minus one. Power times, base minus one, goes up into the power of e. Clear? And the trap: never replace the base by one too early. That is what we did in the puzzle, right? Also keep two standard limits ready, as x tends to zero. tan x over x tends to one, and one minus cos x, over x squared, tends to one half. Example two needs both. Got it?

3:38Example 1: where max(|x|, x squared) is not differentiable

Now Example one. f of x is the max of the absolute value of x, the two vertical bars, and x squared. The curly brackets hold the two, and max means the larger one. At how many points is f not differentiable? Pause the video, and try it first. Done? Okay, let's check. First, where do the two functions meet? The absolute value of x equals x squared at x equals zero, and at x equals plus or minus one. Okay? Between minus one and one, the absolute value of x is the larger; outside, x squared wins. So f is the absolute value of x when mod x, the same absolute value, is less than or equal to one, and x squared when mod x is greater than one. Right? Now the slopes. At x equals one, on the left f is x, slope one. On the right f is x squared, slope two x, and at x equals one that gives two. The slopes one and two do not match, so there is a corner. With me? At minus one, on the left f is x squared, slope two times minus one: minus two. On the right f is minus x, slope minus one. Another corner. Clear? So f switches between the two only at plus one and minus one. So the answer is two, right? No, no, careful! Look at zero. Near zero, f is just the absolute value of x, and that has its own corner there: slope minus one on the left, and one on the right. f does not switch at zero, but the corner is still there. Got it? So the corners are at minus one, zero and one: three points. Three, not two. Perfect!

5:27Example 2: 1 to the power infinity limit with tan x, sin x

Now Example two, our puzzle from the start. Find the limit as x tends to zero, written under lim, of one plus tan x, over one plus sin x, all to the power cosec cubed x. Pause the video and try it. Hint: check the form first. Done? Okay, let's check. The base tends to one. And cosec x is one over sin x, so cosec cubed x is one over sine cubed x. As x tends to zero, it blows up. So this is the one to the power infinity form, and the answer need not be one. Right? So, we need base minus one. Take the common denominator, and the ones cancel: tan x minus sin x, over one plus sin x. Okay? Now think. The wavy equals sign means approximately. tan x is about x, and sin x is about x. So tan x minus sin x is about zero, and the answer is e to the power zero, which is one, right? Oh no, careful! The difference is much smaller than x, but it is not zero. Take tan x out: tan x minus sin x equals tan x, times one minus cos x. By our two standard limits, tan x is about x, and one minus cos x is about x squared over two. So the difference is about x times x squared over two, that is x cubed over two; the dot means times. Clear? Now power times, base minus one; the cross means times. The power is about one over x cubed, and one plus sin x tends to one. So we get x cubed over two, over x cubed times one: one half. This holds from the right and from the left, so the limit exists. With me? So the limit is e to the power one half, which is root e, the square root of e. Not one at all! Puzzle solved!

7:22Quick revision: MVT, tangents, corners, 1 to power infinity

So, quick revision. One: the mean value theorem needs continuity on the closed interval, differentiability on the open one. Two: from a point off the curve, call the contact point t comma f of t. Three: for the max of two functions, check where they switch, and each one's own corners. Four: for one to the power infinity, use e to the power, the limit of g times, f minus one, and never replace the base by one too early. Now try the three practice problems on this page. All the best!