oh god... n! = n* (n-1)! when n > 1... their is no good reason for 0! = 1, but it is a convention to take 0! = 1... many people also tell the reason as- 5! / 5 = 4!; 4!/4 = 3!... and they say that similarly 1!/1 = 0! = 1... but again this is not a valid reason

This helped in in my add math i am from zone 4 Bangladesh

Let me explain it further, when you generalized it it's n! = n*(n-1)! and when you plug in 1 it's gonna be 1! = 1*0! in this example its still not clear as to what 0! even means because if it equals to 0 then 1 = 1*0 which is 0 (which 1 does not equal 0) and if its equal to 1 its it would be 1! = 1 * 1 which is 1 and this proves that 0! equals 1 otherwise 1 = 0. Hope this helped you. :)

3! = 3 x 2 x 1, write out 0! In the same way. You can’t. You can’t get to n! = n(n-1)! using 0 as the starting point. The rule of factorial is that you multiple whole numbers descending to or including 1. So 1! is 1x1. 0! violates the rules of the operation. Also how can any multiplicative function including 0 as a factor ever be >0? Finally using the equation of n!= n(n-1)! , if one uses -3 as n then is it true that -3! = -3(-3-1)! … -3! = -3(-4)! …-3 = 12?! No, -3! Is an imaginary number. And so should 0! be seen as well.

Thank you so much! I was really wondering this, You have saved my day!😊

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Hi Hassan, Would you mind checking your email inbox?

So 0!=0(-1)!, therefore -1! can be anything.

Absolutely excellent video. Wish the writing was a bit clearer but the explanations were perfect. Thank you!

dont know why this isnt the default way to teach this

OMG U ARE A LIFESAVER THANK U

My curiosity makes me come here😅

dang okay i see how it is

ans is missing a square for k?

waittt is it u used new points because ur new graph has lnx. If the axis still stays the same then u dont need to change the points right?

Good explanation ❤️

Thank u so much😊😊❤

Thank you! This was very helpful. God bless you.

Around 1:09:50, it should not be minus but plus.

Should the last step be instead: scale by a factor of 1/2 parallel to the y-axis?

The solution for 6(iii) is incorrect. I provide 2 classes of examples to why 288 is incorrect. 1. Numbers such as 3468 (whose individual numbers do not add up to a multiple of 6) are not subtracted from 360 in both the VJC and your proposed solution. 2. On the other hand, numbers such as 3465 which is not divisible by 6 are subtracted from 360. You have inadvertently included such classes of (odd) numbers in the subtraction by taking 4!. The correct answer is 60. But I cannot provide a systematic solution for this question since I brute-forced my way to the solution by systematically checking numbers from 3456 to 8765. Will provide an update if I manage to figure one out. But I agree - this question is quite unfair to students in an examination setting.

paper in 8 hours 😭😭

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wow!!! This method of explaining the concept of range of composite function is actually so impressive!!! Save me one day before my math paper lol

10. part a has a mistake, its not 3x+3, should be 2x+3

Your a goooooood teacher. thank you so much.

0! = 1 is more of an axiomatic choice. You have to take it as 1; otherwise, so many different parts of mathematics fall apart. Using the integer definition, there is no way to actually calculate 0!.

Absolute Masterclass

More bullshit from mathematicians who have to come up with ways to handle zero when it's problematic.

thank you omg

Very good explanation!!

i really like how u actually teach us the proofs of the formulas for all ur videos instead of jus throwing it like our school teachers, i wish all teachers were like u bro

Quad root 4 is just square root 2

Sec 4 here

Is the presentation not wrong? Dy/dx=0 is for stationary points not increasing y. Should be Dy/dx more than 0

period of tan graph is 180/b

best h2 math lecture video ever

dude your actually saving me, i am learning the whole topics from basic knowledge/scratch for o lvls which is in abt 2 weeks , man god bless u

very helpful

thank you for sharing your experience😊 at least i know im not alone with strugging for vectors in j1. now i also know how to improve 😁wish i found this vid sooner tho

Thank you mr, your explanation is really helpful, i hope i can study in SG in the future

I think 3rd last working got careless mistake you forgot ²

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Best explanation ....but for always negative a< 0 .....😊.. printing mistake not yours ,sir .. From india 🙏

Invalid argument; the generalized case should be n(n-1)! where n is not equal 0. To prove this plug 0 in for n and then 'do the math'. 0! should be undefined.

U are saving my life

I Think there was a slight mistake. 2/y=-1/x + 4 should equal to 1/y = -(1/2)(1/x) + 2, since you're dividing the equation throughout by 2.

At 20:30, your U substitution seem not quite right =8(2x+l/x)? You let u=x^2+lnx in the first place? 😁

Its easy. Just solve the equation 2x²-3/x-2 = 1 2x²-3= x-2 2x²-x-1=0 (2x+1)(x-1)=0 x=1 and x= -1/2 and also the region x>2 So the answer is 1>x>-1/2 and x>2