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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.