I don't think I understand what you're saying, I was talking about the last step
$$\lim_{h\to 0} \frac{h-\tan h}{h^2 \cdot \tan h}=\lim_{h\to 0} \frac{h-\tan h}{h^3 \cdot \frac{\tan h}{h}}=\frac{\lim_{h \to 0} h-\tan h}{\lim_{h\to 0} h^3 \cdot \frac{\tan h}{h}}=\frac{\lim_{h \to 0} h-\tan...
I didn't knew that!
But in the original solution also they used the quotient rule even though both the numerator and denominator is zero. Is that also incorrect?
This was the question,
The above solution is the one that I got originally by the question setters,
Below are my attempts (I don't know why is the size of image automatically reduced but hope that its clear enough to understand),
As you can see that both these methods give different answers...
Okay, yes know I get it!
yes I would end up with some positive average marks!
But why this isn't as simple as it looks is because it assumes that there is an equal distribution in the answer key, i.e., these are 15 questions whose answer is A, 15 whose answer is B and so on.
But in real life...
Let's say if the situation was, that I knew 40 out 60 question, then what is the average marks now if I attempt all?
Also this way I'm getting a feeling that even if I knew 0 out of 60 questions, I would still score some positive marks?
What if I told you that the test is not for grades but for ranks.
That means you don't even know whether 180 is a good score or bad. That depends on how others perform as well.
So, in this case the more you score, the better.
If you followed my solution,
you'll see that if I attempted all those remaining questions, then there is a 76% that I will score 180 or more. Of course I can also score less than 180 with a chance of 24%, But, if you look closely, you'll see that, 76% is the chance for you to score anywhere...
Think of this as the student in the middle of this test.
You did all you could and you're sure that you got 45 of these questions definitely correct.
But now your thinking, what should I do with the remaining question,
Should I attempt some and leave some?
Should I attempt all and leave none...
You don't necessarily need to guess all 15. I mentioned that.
I assumed that that student guessed all 15 because then I knew what to do for this case.
It is not the desired solution but one can reach an answer if they want with this method.
As I said,
Originally the intent of this question was to find whether its better to guess or to leave no response for the remaining questions. Thus, while framing the question I always had in my mind that the student will guess even though the net profit after that is zero.
So, yeah the question should...
To approach this, I first assumed the case when the students attempts all the remaining questions.
Probability that they gain 4 marks for a guess = ##\frac 1 4##
Probability that they lose 1 for a guess = ##\frac 3 4##
Now let us say the number of correct guesses = ##r##
Now we should have at...
I'm so sorry but this is beyond me. I thought that there must be some geometric way to see this through graphs, like the one in the first two responses by @BvU. I appreciate your efforts but I think this question is not for me (atleast not for now), But I do hope to understand this someday...
I actually didn't read that because when I started with section 7.5, a lot of stuff went over my head, I thought that I might have to read the whole thing from beginning or maybe my knowledge isn't enough to understand all of that, so I just saved that document for now.
I can still give it...
That is very interesting, in fact this video which I shared earlier also talks about how this has something to do with slopes.
But I can't quite get why is it so?
It somewhat feels like the concept of stable and unstable equilibrium in physics, where the root at which this function converges...
I can obviously see that when you get close to an asymptote (x=0 in the first graph and x=1 in the second), it will not converge. Now what are the possible inputs that will reach to an asymptote, that I'm not sure about.
Obviously starting on an asymptote surely doesn't converge, but there are...
It looks like,
for the first graph only the input (1-√5)/2 will get us to the alternate solution,
and similarly for the second graph only (1+√5)/2 gives us exactly the intersection between the lines y=x and y=1/(x-1),
anything other than that seems to slowly-slowly move away from (1+√5)/2...
On simplifying the given equation we get, x^2-x-1=0 and using the quadratic formula we get x=(1+√5)/2 and x=(1-√5)/2
Now, as the formula suggests, there are two possible values for x which satisfies the given equation.
But now, if we follow a process in any general calculator by entering...
If we are putting the numbers 1,2,3... in a fixed order, then it could be either ascending or descending 15,14,13... is that why we could get two different selection of five A's?
That means these 21 elements comprise of the ones we won't even choose and yet we will end up with a set of 5 non-consecutive numbers from this set?
let me try to understand this step by step,
STEP 1: Take the 10 elements which will not end up in the final selection and label them as B.
STEP...
But why take 21 slots in the first place? Is it because like in the original post we first took away 5 elements then we were left with 10 elements and 11 respective gaps between them? But then the 5 elements which took initially didn't even end up in our final selection so what was the thinking...