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IIT JEE differentiability and boundedness

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Let f be a real-valued function defined on the interval from 0 to infinity. It does not include 0. And f is defined as the natural log of x plus the definite integral from 0 to x of the square root of 1 plus sine of t, d of t. Then which of the following statements are true? And so more than one of these might be true. So let's figure them out. So A says, the second derivative-- there's two apostrophes here. The second derivative of f exists for all x between 0, or really, all x greater than 0. Then they say the first derivative exists for all x greater than 0, and the derivative is continuous on that interval, but not differentiable. OK, so everything they're doing, they're relating the function to the first derivative and the second derivative. So let's just figure those things out. So let's just write the function first. I'll just rewrite it here. The function f is equal to the natural log of x plus the definite integral from 0 to x of the square root of 1 plus sine of t dt. The derivative of f-- so let me write it over here. So f prime of x is going to be equal to the derivative of this first term. Derivative of natural log of x is 1/x. And then derivative of this, that's just the fundamental theorem of calculus. It's going to be just this inside expression with respect to x, as a function of x, so plus the square root of 1 plus sine of x. Let's get the second derivative now. The second derivative-- I'll do it over here-- f prime prime of x is going to be equal to the derivative of this term. This is x to the negative 1. So it's going to be negative x to the negative 2 or negative 1 over x squared. That's negative x to the negative 2. You could view it that way. And then plus-- so let's do the chain rule. What's the derivative of the inside? Derivative of 1 plus sine of x is cosine of x. So plus cosine of x times-- now, it's 1 plus sine of x to the 1/2. So it's going to be times 1/2 times 1 plus sine of x to the negative 1/2. So I took the derivatives. Now let's look at the choices. So the first choice says the second derivative of the function exists for all x greater than 0. If you're going between 0 and infinity and not including 0, this is essentially for all x greater than 0. Let's see if that is true. So over here, the one area-- so this would be undefined if x is equal to 0, but that's not in the interval over here. So let's just be clear. When you take something to the negative 1/2 power-- let me rewrite it like this. I could rewrite it as-- instead of writing it to the negative 1/2 power, I could write it in the denominator as the square root of 1 plus sine of x. So if this denominator becomes undefined, the second derivative won't be undefined. So let's think about when that denominator might be undefined. So it'll be undefined if it equals 0. So 1 plus sine of x is equal to 0. Well, it's going to be equal to 0 if sine of x is equal to negative 1. And sine of x can equal negative 1 at x is equal to-- let's see. If you go around the unit circle, sine is the y-coordinate, so it has to go around-- you can either go down. It would be negative 90 degrees or positive 270 degrees. Or if in radians, if x is equal to-- because we're assuming, or at least I'm assuming that we're dealing in radians here. If x is equal to 3 pi over 2, assuming we're in radians, that will make this undefined. So this f prime prime not defined at x is equal to 3 pi over 2, which is clearly in this interval. So choice A is false. Choice A does not work. Now let's see. Choice B says f prime of x exists for all x's greater than 0. Well, that seems to be the case. You can't put x equal 0, but that's not in the interval. And f prime is continuous on 0 to infinity. Yeah, it's continuous-- no obvious discontinuities here-- but not differentiable on 0 to infinity. Well, it's clearly not differentiable at 0 to infinity. The first derivative or the second derivative does not exist at x is equal to 3 pi over 2. We just saw that. f prime of x is not differentiable at 3 pi over 2. So everything in this is true. Everything is true. It's not differentiable there because the second derivative is not defined at 3 pi over 2. Let's do part C now, part C. There exists an alpha greater than 1 such that the absolute value of f prime of x is less than the absolute value of f of x for all x essentially greater than-- for all x greater than that alpha. So let's think about this. Let's think about this a little bit. What we want to show is that, beyond some threshold, the derivative is always going to be smaller than the original f of x. The derivative will always be smaller than f of x. So the first thing we might want to see-- so let's just compare the parts that are comparable. So let's compare, first of all, the natural log of x. Let's compare the natural log of x to 1/x. Let's just set a threshold here. So is there a number so at any number larger than that, the natural log of x is always going to be larger than 1/x? And the number that pops into my head, just because it's an easy number to work with with natural logs, is e. So for x is greater than e, it's pretty clear that the natural log of x is going to be greater than 1/x. Think about it. At e, it's going to evaluate to the natural log of e, which is 1. And 1 is clearly greater than 1/e. e is 2.71 and so on and so forth. So this is clearly true. And then as you get beyond e, as you get numbers larger than e, as this becomes e plus something, this number will become larger. And then this number over here will become smaller because we'll be dividing by a larger number. So it's clear that it exists just for this part, for the first part, the natural log of x beyond e. And we just have to find some threshold. That might not be the first threshold. But the natural log of x is definitely greater than 1/x when x is greater than e. Now let's see if we can find a threshold for the second part. And if we can, then we can literally just pick the larger of the two thresholds, and then that will be a threshold above which f of x will always be larger than f prime of x. And to think about it, we just have to see that essentially this part of the function-- let me write it this way. If I define this as h of x, this right here is h prime of x. We've already shown that for x is greater than e, h of x is greater than h prime of x. We've seen that already. Now let me define this over here. Let me define that as g of x. Let me call that g of x. And then this is its derivative. This is g prime of x. That is g prime of x. Now, let's just draw g of x. Actually, let me draw g prime of x right over here. Let me draw it. It's actually a pretty straightforward graph to draw. So let me draw it over here. So when x is equal to 0, sine of 0 is 0 plus 1. You take the square root. So 0, 1, that's the first point. That is f of x equal to 1 when x is equal to 0. And then the next interesting point, maybe we'll take x is equal to pi over 2. So when x is pi over 2, sine of pi over 2 is 1. 1 plus 1 is 2, take the square root. So it's going to be the square root of 2, which is someplace around here, so the square root of 2. And then let's just take when x is pi, sine of pi is 0. This becomes the square root of 1 again, so we're back over here. Then let's go to 3 pi over 2. Sine of 3 pi over 2 is negative 1. Negative 1 plus 1 is going to be 0, so we're going to be down here. And then we go again to 2 pi. Sine of 2 pi is 0. 0 plus 1 is 1. Take the square root, you're back over here. So this graph, g prime of x we're calling it. This graph over here looks like this. It looks something like. This is my best attempt at drawing it. It just keeps oscillating. If its maximum point is square root of 2, its minimum point is 0, and it keeps oscillating. So this is g prime of x. G prime of x is equal to the square root of 1 plus sine of x. Now, what is g of x? What is this thing over here? Well, this is the definite integral from 0 to x of this thing. So you take any x-value, this thing over here, g of x, is the area over here. So what's happening here? So let's think about this. As we get to larger and larger x-values, the area gets bigger because this line never drops below the x-axis. So as we get larger and larger x-values, this area just keeps getting bigger and bigger and bigger. So we can write g of x unbounded as x goes to infinity. G of x approaches infinity. What about g prime of x? g prime of x, it maxes out. It just keeps oscillating between 0 and square root of 2. So g prime of x max-- or maybe I should write it this way. The max of g prime of x is equal to the square root of 2. So it's pretty clear that at some point this area is going to be either some x-value or this area is going to be more than the square root of 2. Let's call that point alpha. And after that alpha, where maybe it's over here, I mean, clearly we could say, at x equals 10, we definitely have a lot of area under this curve. The area is definitely bigger than the square root of 2. So let me just write it there because we just need to find some boundary. Let's just say at x greater than-- I'm just picking an arbitrary number. At x is greater than 10, it's clear that the integral from 0 to x of the square root of 1 plus sine of x, or I should say 1 plus sine of t dt, it's clear that that is going to be greater than the square root of 1 plus sine of x, because this is the area. At any value above 10, this thing at most can ever get to square root of 2. This thing's just going to keep growing and growing. At 10, it's already going to be larger than the square root of 2. So we know that there's some boundary. So for x is greater than 10-- so let's write it this place. So for x greater than 10, well, 10 is definitely greater than e. We know that h of x is greater than h prime of x. We know that's true. 10 is clearly greater than e, so we can state that. And we also know that g of x is greater than g prime of x. So we know that f of x, which is equal to h of x plus g of x, is going to be greater than f prime of x, which was the derivative, the sum of the derivative, which is equal to h prime of x plus g prime of x. So we know that this is true. We can make alpha equal 10 and this thing is going to be true. So this is true. Now let's do part D. There exists a beta greater than 0 such that the absolute value of f of x plus the absolute value of the derivative of f is going to be less than or equal to beta for all x. Well, we already know that f of x is unbounded, because this integral part of it right here just keeps increasing as x gets larger and larger and larger. It never dips below. We never have negative area. So this area just keeps growing. So this is unbounded. And actually, so is the natural log of x. The natural log of x is also going to be unbounded. So you cannot say that it's going to be less than some value because it's going to approach infinity. So it's not going to be less than some value for all x from 0 to infinity. So you can't bound it. This is an unbounded function. Just this part alone is an unbounded function. f prime of x is-- well, even there you have to be careful. But f of x is clearly unbounded, so this can't be true. There's not any kind of roof you can put on the-- well, I think you get the general idea. So this is definitely not true. So the answer here is B and C.