It's -1. We have a quadratic in terms of sin x, we are trying to find the minimum for that. But that is a parabola in (terms of sin x) that opens downward, the minimum value must be at the endpoints of h(x). Since sin has a range of [-1,1], the parabola has endpoints at sin x = -1 and sin x =1. The minimum will occur when sin x=-1, which means cos x = 0. So the minimum value is 3(0)^2+2(-1)+1=-1.
I remember doing this question in the exam. I got the answer really quickly as that was the only solution out of the multiple choices which was greater than 4. Taking cos(x) = 1 you know that the largest value must be at least 4.
From looking at the thumbnail without doing any work, I would use the Pythagorean identity to simplify into exponents of sin(x), then use the quadratic formula/complete the square to solve for sin(x), and lastly use some inverse trig to get values of x.
Yes sir that is the exact problem with me too...as I see these type of questions I just use calculus and don't even think of alternative non calculus solutions...but I was sooo happy to take the challenge and find the answer...hope to see such type of videos more often by you...Thank u...Love from India Edit: Thank you soo much for these many likes👍
@@HDitzzDH you don't even have to do second derivative here, you have 2 points (infinitely many but repeating 2) where the derivative is 0 (namely where cosx=0 or sinx=1/3) so you just straight up plug them in. First point gives us y=3 or -1, 2nd point gives us y=13/3 (at first point cosx=0 then sinx can either be 1 or -1, and at the second point we know sin so using the identity we can find cos^2)
we have f(x) = 3cosx^2 + 2sinx +1 f'(x) =-6sinxcosx + 2cosx to get min/max we need to see when it differential equal to zero when f'(x)=0 : 2cosx(1-3sinx)=0 { 2cosx=0 { 1-3sinx=0 -->sinx= 1/3 to get cosx^2 we need square both side of sinx=1/3 then we get: sinx^2 = 1/9 cosx^2 = 1 - sinx^2= 1-1/9 = 8/9 so now we get sinx = 1/3 and cosx^2 = 8/9 then apply to the function 3cosx^2+3sinx+1 we get: 3(8/9)+3(1/3)+1 = 13/3
i took the MAT this year and it didnt go so well, the 2020 edition of the MAT (where this question is from) was most certainly the easiest paper in the last 8 years, this question is perhaps one of the easiest in any multi choice section the mat has had
Can you make some videos about elliptic functions (and perhaps elliptic integrals too)? There's a way of looking at elliptic functions as a form of trigonometry on ellipses, and then elliptic integrals are sort of inverses of elliptic trig functions
@@jessebosker122 in a parabola of equation y=ax²+bx+c, the y value of the vertex, that is the maximum value of the function, is equal to -∆/4a, and since the given function is in the form of a parabola you can apply that formula
@@jessebosker122 If you have a parabola of the form y=ax^2+bx+c, then the quadratic equation finds the 2 points where y=0, that is of the form (-b +/- sqrt(b^2-4ac)) / (2a), which can be simplified to (-b +/- sqrt(∆))/(2a). Due to the symmetry of parabolas, this places the vertex is midway between these points at x=-b/2a Subbing that back into the equation you get the maximum (or minimum) at: y=a*(-b/2a)^2 + b*(-b/2a) + c =b^2/4a - b^2/2a + c Rewriting so everything is on 4a, and factoring out the 1/4a: (b^2 - 2b^2 + 4ac)/4a =(-b^2+4ac)/4a =-(b^2-4ac)/4a =-∆/4a A less formal, but faster, way would be after you have gotten your vertex, note that you can shift your axes so the vertex is at x=0, and that would make the equation just y=Ax^2+C, with the roots now at +/- sqrt(∆)/2a, and the max or min being C. You can sub that in and rearrange to get the above, or note that you need to go over in the x direction by sqrt(∆)/2a, so the change in y must be that squared. To determine if it is a maximum or minimum you need to look at the sign of a. If a is positive, it is a minimum, if as is negative it is a maximum. In this case a is negative to it is a maximum.
Used derivatives and got the same result. Actually seemed a bit easier. For instance, the derivative of the function is -6 sinx cosx +2cosx Then for critical points we should have -6 sinx cosx +2cosx=0 -2cosx (3sinx -1)=0 In the case the second factor is zero, we obtain the result you get. But yeah, I get the message that calculus will not always be the only way. Interesting video.
I must be missing something. This gives asin(1/3)=0.3398 which is also what I got but does not equal 13/3. Could you please clarify this for me? I think I'm being really dumb here
Exactly what I was thinking. The fact that we have a negative multiplied by the square of an expression meant that the maximum is achieved whenever sinx=1/3, because then nothing gets subtracted from 13/3. Otherwise we always subtract some nonzero value from 13/3. The only thing you have left to do is to check whether sin(x)=1/3 actually exists, which it obviously does in this case. But if the expression inside the square was something like sin(x)-2, we would always subtract something from 13/3, even at the maximum. In that case, the maximum would occur whenever (sin(x)-2)^2 would be at minimum (which happens at sin(x)=1 or x=pi/2 because that provides the smallest difference). In that example case, the maximum would then be 13/3 - 3 = 4/3. I think that BPRP missed this critical step, because the vertex formula for this problem works only sometimes. It's only a coincidence that it worked here, because the maximum happens to be at a real argument of the sine. But in the sin(x)-2 case, for example, the maximum would be at a complex argument. The true maximum would still be 13/3, but the way the question is formulated, we assume that they want from us the maximum over the *real number line* which would be smaller
The Slope of the Cosine Squared isn't linear, neither is the Slope of the Sine. So around 0.3Rad the Cosine Squared has decreased very slowly, while the Sine has increased very rapidly. In fact at 0.3Rad, the Cosine Squared has decreased by 0.3 while the Sine has increased by double that, 0.6. from there you know at 0.3Rad it would be 3*(1-0.3)+2(0+0.6)+1
Why do you complete the square like that? If you have x^2 + bx + c, complete the square you just do (x+b/2)^2 - (b/2)^2 + c... no need for "what is the magic number" or anything
We can easily get to 2nd solution provided by sir if we have ideas about parabola and can relate every quadratic equation to a parabola and thus to find minimum and maximum values we just need to get the vertex.
What if we directly put x= zero degrees.. We'll get 3(1) + 2(0) + 1 = 4 And if we look at the integer type ans., then 4 is correct.[bcz 4.33 = 4(ineger)] So am i right?
You can then find the vertex by factorising to 2*cos(x)*(-3*sin(x)+1) and setting it equal to 0, which then gives: 0=cos(x)*(3*sin(x)-1) cos(x)=0 or sin(x)=1/3 And noting that if cos(x)=0, cos^2(x)=0 & sin(x)=1 or sin(x)=-1, and if sin(x)=1/3, cos^2(x)=8/9 Then substituting that back in, if cos(x)=0 we get 3 or -1. If sin(x)=1/3, we get 8/3+2/3+1 = 13/3=4 +1/3. And combined with the knowledge that sin and cos are bounded to -1 and 1, so we can't have an infinite minimum or maximum, this gives us a global maximum of 4 & 1/3, a global minimum of -1, and a local minimum, maximum or inflection point at 3. By doing a bit more work by noting that a small change in x at this unknown point will keep sin(x) close to 1 and reducing x will make cos(x) negative and increasing it makes cos(x) positive, we can determine this point is a local minimum. But you don't need to do the differentiation there. You could do the substitution, get 4+2*x-3x^2 and use calc there.
very easy in fact, i wouldn’t use calculus to find the maxima of this equation. 1. translate cos^2(x) in terms of sin, gives the equation -3sin^2(x)+2sinx+2 2. redefine sinx as s or x giving 3s^2+2s+2 3. complete the square the find your a value to be 7/3 Edit: wrote problem down as +2 not +1 but works either way
I think using calculus is a lot easier than trying to remember a bunch of formulae. I had never even heard of the vertex formula before. Yet it is trivial to derive using calculus. For a general equation of y=ax^2+bx+c, the derivative is y'=2*a*x+b set y'=0, and solve for x to get x=-b/2a. Using calculus doesn't mean differentiating the first thing you see.
@@gamerdio2503 yes I'm aware, in the vid it looks like 1/3 is used in f to get the max value instead of arcsin(1/3) which just seems wrong was the point?
@@wijo605 sin(x) = 1/3, so he puts in 1/3 for sin(x). If you plugged in arcsin(1/3) for x, you'd just get a bunch of sin(arcsin(1/3)) which is just 1/3
There is a result called the Pythagorean Identity which states that cos^2x + sin^2x = 1; subtract sin^2x from both sides to obtain cos^2x = 1 - sin^2x.
there's no reason to not differentiate. it's an upside down parabola in the variable y = sin x which ranges from -1 to 1. derivatives aren't hard, especially when they're just of quadratic expressions.
personally I would've used the double angle formula for cos, and then done some manipulation so we only get y = cos(ωx + φ) + k, then the maximum/minimum value would be easy to find edit: oops this doesn't work because we'd have cos2x and cosx, nevermind
The "difficult" way is doable and kinda short Notation: f(x) = 3*cos²(x) + 2*sin(x) + 1 Obs.: f is C^{\inf} Solution: f ' (x) = 2 * cos(x) * (3 - sin(x)) So extremum in two cases: CASE #1 => sin²(x) = 1 CASE #2 => sin(x) = 1/3 f '' (x) = -6 -2*sin(x) + 12*sin²(x) Subtituting in the cases: C#1 => f''£{4,8} => Minima C#2 => f'' Maxima From the observation follows max{f} = 3*8/9 + 2/3 + 1 = 13/3 for all x£R _____ Mr. Pens, you managed to bring us another bad example of a good solution. Again!
Like the math, love the poke ball ... do not like beard lol. And when I saw that I did think -- oh quadratic equation so that is simple to generally solve.
The second they starting throwing the alphabet into math I tuned out, history and English are easy math is the least used common core subject in most professions
Challenge: can you find the smallest value of this function without using calculus?
I think we can try to complete the square...
Take sinx =-1
ans=-1
3
It's -1. We have a quadratic in terms of sin x, we are trying to find the minimum for that. But that is a parabola in (terms of sin x) that opens downward, the minimum value must be at the endpoints of h(x). Since sin has a range of [-1,1], the parabola has endpoints at sin x = -1 and sin x =1. The minimum will occur when sin x=-1, which means cos x = 0. So the minimum value is 3(0)^2+2(-1)+1=-1.
Yes, just use desmos
"You don't have to differentiate yourself from everybody else."
BPRP 2021
I remember doing this question in the exam. I got the answer really quickly as that was the only solution out of the multiple choices which was greater than 4. Taking cos(x) = 1 you know that the largest value must be at least 4.
I took this approach as well, but was unable to determine whether values could go beyond.
@@qwerty6574 That's not correct.
sin x = cos x | x = π/4
cos(π/4) = √2/2
3 cos²(π/4) = 3/2
From looking at the thumbnail without doing any work, I would use the Pythagorean identity to simplify into exponents of sin(x), then use the quadratic formula/complete the square to solve for sin(x), and lastly use some inverse trig to get values of x.
Agh, going too fast. Instead of completing the square, just find the vertex with -b/(2a)
YOU DON'T EVEN NEED TO SOLVE FOR X, AH! I need to learn to read
Yes sir that is the exact problem with me too...as I see these type of questions I just use calculus and don't even think of alternative non calculus solutions...but I was sooo happy to take the challenge and find the answer...hope to see such type of videos more often by you...Thank u...Love from India
Edit: Thank you soo much for these many likes👍
True, setting it's derivative equal to 0 then using the second derivative test to see if it's a maximum was my immediate thought.
@@HDitzzDH you don't even have to do second derivative here, you have 2 points (infinitely many but repeating 2) where the derivative is 0 (namely where cosx=0 or sinx=1/3) so you just straight up plug them in. First point gives us y=3 or -1, 2nd point gives us y=13/3 (at first point cosx=0 then sinx can either be 1 or -1, and at the second point we know sin so using the identity we can find cos^2)
we have f(x) = 3cosx^2 + 2sinx +1
f'(x) =-6sinxcosx + 2cosx
to get min/max we need to see when it differential equal to zero
when f'(x)=0 : 2cosx(1-3sinx)=0
{ 2cosx=0
{ 1-3sinx=0 -->sinx= 1/3
to get cosx^2 we need square both side of sinx=1/3 then we get:
sinx^2 = 1/9
cosx^2 = 1 - sinx^2= 1-1/9 = 8/9
so now we get
sinx = 1/3 and cosx^2 = 8/9 then apply to the function 3cosx^2+3sinx+1 we get:
3(8/9)+3(1/3)+1 = 13/3
i took the MAT this year and it didnt go so well, the 2020 edition of the MAT (where this question is from) was most certainly the easiest paper in the last 8 years, this question is perhaps one of the easiest in any multi choice section the mat has had
Competing the square is my favorite cause you can intuitively see the maximum value when it's written in that form
Can you make some videos about elliptic functions (and perhaps elliptic integrals too)? There's a way of looking at elliptic functions as a form of trigonometry on ellipses, and then elliptic integrals are sort of inverses of elliptic trig functions
the max value can be found with -∆/4a where ∆ = b^2 - 4ac (the discriminant)
Proof it to me.
@@jessebosker122 in a parabola of equation y=ax²+bx+c, the y value of the vertex, that is the maximum value of the function, is equal to -∆/4a, and since the given function is in the form of a parabola you can apply that formula
@@jessebosker122 If you have a parabola of the form y=ax^2+bx+c, then the quadratic equation finds the 2 points where y=0, that is of the form (-b +/- sqrt(b^2-4ac)) / (2a), which can be simplified to (-b +/- sqrt(∆))/(2a).
Due to the symmetry of parabolas, this places the vertex is midway between these points at x=-b/2a
Subbing that back into the equation you get the maximum (or minimum) at:
y=a*(-b/2a)^2 + b*(-b/2a) + c
=b^2/4a - b^2/2a + c
Rewriting so everything is on 4a, and factoring out the 1/4a:
(b^2 - 2b^2 + 4ac)/4a
=(-b^2+4ac)/4a
=-(b^2-4ac)/4a
=-∆/4a
A less formal, but faster, way would be after you have gotten your vertex, note that you can shift your axes so the vertex is at x=0, and that would make the equation just y=Ax^2+C, with the roots now at +/- sqrt(∆)/2a, and the max or min being C.
You can sub that in and rearrange to get the above, or note that you need to go over in the x direction by sqrt(∆)/2a, so the change in y must be that squared.
To determine if it is a maximum or minimum you need to look at the sign of a. If a is positive, it is a minimum, if as is negative it is a maximum.
In this case a is negative to it is a maximum.
Used derivatives and got the same result. Actually seemed a bit easier. For instance, the derivative of the function is
-6 sinx cosx +2cosx
Then for critical points we should have
-6 sinx cosx +2cosx=0
-2cosx (3sinx -1)=0
In the case the second factor is zero, we obtain the result you get. But yeah, I get the message that calculus will not always be the only way. Interesting video.
I must be missing something. This gives asin(1/3)=0.3398 which is also what I got but does not equal 13/3. Could you please clarify this for me? I think I'm being really dumb here
@@kingcrimson1631 you have x= 0.3398 (in radians btw), you plug this into the formula and then you get 13/3
And you find the relative minima at cos x = 0, f(pi/2) = 3 and f(3pi/2) = -1
@@lj_ljh ah cool. Duh. Thanks for that
@@kingcrimson1631 you know sinx=1/3 using identity cos^2x=8/9 plug them into the original eq and get 13/3
I thought about sharing this with my students, but I'm pretty sure they would just giggle at the graph the whole period...
"Life doesn't have to be that hard."
-BPRP
Completing the square was both nicer looking and more insightful.
Exactly what I was thinking. The fact that we have a negative multiplied by the square of an expression meant that the maximum is achieved whenever sinx=1/3, because then nothing gets subtracted from 13/3. Otherwise we always subtract some nonzero value from 13/3. The only thing you have left to do is to check whether sin(x)=1/3 actually exists, which it obviously does in this case. But if the expression inside the square was something like sin(x)-2, we would always subtract something from 13/3, even at the maximum. In that case, the maximum would occur whenever (sin(x)-2)^2 would be at minimum (which happens at sin(x)=1 or x=pi/2 because that provides the smallest difference). In that example case, the maximum would then be 13/3 - 3 = 4/3. I think that BPRP missed this critical step, because the vertex formula for this problem works only sometimes. It's only a coincidence that it worked here, because the maximum happens to be at a real argument of the sine. But in the sin(x)-2 case, for example, the maximum would be at a complex argument. The true maximum would still be 13/3, but the way the question is formulated, we assume that they want from us the maximum over the *real number line* which would be smaller
Wait, bprp, is this seriously enough to get into Oxford. It looks really easy...
It can't possible be _that_ trivial to get into ivy legue college.
Back to basics. Nicely explained
The Slope of the Cosine Squared isn't linear, neither is the Slope of the Sine. So around 0.3Rad the Cosine Squared has decreased very slowly, while the Sine has increased very rapidly. In fact at 0.3Rad, the Cosine Squared has decreased by 0.3 while the Sine has increased by double that, 0.6. from there you know at 0.3Rad it would be 3*(1-0.3)+2(0+0.6)+1
I liked the completing of the square better. Great video. Thanks.
I thought the title says "Can you solve this question without crying?"
Why do you complete the square like that? If you have x^2 + bx + c, complete the square you just do (x+b/2)^2 - (b/2)^2 + c...
no need for "what is the magic number" or anything
Instead of taking - b/2a and then replacing couldn't we take -discriminant/4a?
please make one example for power 3 functions . to find the extreme value without derivative.
We can easily get to 2nd solution provided by sir if we have ideas about parabola and can relate every quadratic equation to a parabola and thus to find minimum and maximum values we just need to get the vertex.
> without using calculus
> channel is just calculus
Can we just take the derivative and equate to 0 and the find the x and put it in the eqn
What if we directly put x= zero degrees..
We'll get 3(1) + 2(0) + 1 = 4
And if we look at the integer type ans., then 4 is correct.[bcz 4.33 = 4(ineger)]
So am i right?
Why do you round 4.33?
We're looking for solutions in the real numbers, not just integers. Minus points for you
I can never remember the vertex formula so I have to use calculus to find the derivative of ax^2+bx+c to remember -b/2a anyway
Completing the square is harder than differentiating
AM-GM is also applicable to find the maximum though...
Isn't it six
-1
this is not correct, because sin x and cos x can never be both equal to 1 with the same x value.
@@Ninja20704 you're right, thnx
But in the equation -3 (sin x)^2 + 2 sin x + 4, we can choose max sin x =1 so the maxima global is -3 + 2 + 4 = 3. Where did I do wrong?
@@zakihasny there is -3 in front of (sinx)^2, which might reduce the value
@@zakihasny the maximum of this equation is not when sin = 1
d/dx 3 cos^2 x + 2 sin x + 1 = -6 cos x sin x + 2 cos x
I dont think it helps tho
It makes it quite messy, might work but will have to do work arounds where the cos x sin x might have to be simplified using the identities
@@johnnguyen1493 Is identities necessary? Cant you equate it to 0. You get cosx=0 or sinx=1/3. Substitute to find max value.
@@dhanish19 unfortunate as x=pi/2 isn’t a maxima
edit: oh wait i get what you mean now
That equation can be solved pretty easily. Just factor out cos x, so we get cosx(2-6sinx)= 0, which means cos x = 0 or sin x = 1/3.
You can then find the vertex by factorising to 2*cos(x)*(-3*sin(x)+1) and setting it equal to 0, which then gives:
0=cos(x)*(3*sin(x)-1)
cos(x)=0 or sin(x)=1/3
And noting that if cos(x)=0, cos^2(x)=0 & sin(x)=1 or sin(x)=-1, and if sin(x)=1/3, cos^2(x)=8/9
Then substituting that back in, if cos(x)=0 we get 3 or -1.
If sin(x)=1/3, we get 8/3+2/3+1 = 13/3=4 +1/3.
And combined with the knowledge that sin and cos are bounded to -1 and 1, so we can't have an infinite minimum or maximum,
this gives us a global maximum of 4 & 1/3, a global minimum of -1, and a local minimum, maximum or inflection point at 3.
By doing a bit more work by noting that a small change in x at this unknown point will keep sin(x) close to 1 and reducing x will make cos(x) negative and increasing it makes cos(x) positive, we can determine this point is a local minimum.
But you don't need to do the differentiation there. You could do the substitution, get 4+2*x-3x^2 and use calc there.
very easy in fact, i wouldn’t use calculus to find the maxima of this equation.
1. translate cos^2(x) in terms of sin,
gives the equation -3sin^2(x)+2sinx+2
2. redefine sinx as s or x
giving 3s^2+2s+2
3. complete the square the find your a value to be 7/3
Edit: wrote problem down as +2 not +1 but works either way
why not derivative =0? it isnt faster/simpler?
I think using calculus is a lot easier than trying to remember a bunch of formulae. I had never even heard of the vertex formula before.
Yet it is trivial to derive using calculus.
For a general equation of y=ax^2+bx+c, the derivative is y'=2*a*x+b
set y'=0, and solve for x to get x=-b/2a.
Using calculus doesn't mean differentiating the first thing you see.
You don't have to use the vertex formula ( I haven't heard of it before either). If you complete the square, it's pretty obvious 13/3 is the max value
@@two697 And I never said you did.
All my point was is that it isn't magically easier to not use calculus.
the max value of the sine thing isn't at 1/3 tho, it's at acsin of 1/3 right? Desmos agrees with that...
like if the polynomial is f(x) then max of f is at x = 1/3 but the og equation is f(sin(x)) so the max is at x where sin(x)=1/3 so x = arcsin(1/3).
That's not the maximum value, that's just where the maximum value occurs. We don't want x, we want f(x)
@@gamerdio2503 yes I'm aware, in the vid it looks like 1/3 is used in f to get the max value instead of arcsin(1/3) which just seems wrong was the point?
@@wijo605 sin(x) = 1/3, so he puts in 1/3 for sin(x). If you plugged in arcsin(1/3) for x, you'd just get a bunch of sin(arcsin(1/3)) which is just 1/3
@@gamerdio2503 oooh right yeah sry :D
*looking at the video title. looking at the channel name*
I feel betrayed!
Lol
max / min value = ( 4ac - b^2 ) / 4a
Hi ! could someone explain me why can we replace 3cos^2X by 3(1-sin^2X) ? Thanks a lot if someone do !!
There is a result called the Pythagorean Identity which states that cos^2x + sin^2x = 1; subtract sin^2x from both sides to obtain cos^2x = 1 - sin^2x.
@@sameersingh9895 Thank you a lot for explaining the Pythagorean identity to me !!
I feel an expansion of cerebral capacity after solving this problem… then i realize it’s just an admission question ;-;
there's no reason to not differentiate. it's an upside down parabola in the variable y = sin x which ranges from -1 to 1. derivatives aren't hard, especially when they're just of quadratic expressions.
this is the first time i think i've done one of these correctly. don't even do calc anymore lol
I couldn't understand that ,vertex?
Firstly I thought it's 6, but forgot about sin^x + cos^x = 1 with respect to the same x, so answer is 13/6
Not really
It's actually (sin(x))^2 + (cos(x))^2 = 1.
personally I would've used the double angle formula for cos, and then done some manipulation so we only get y = cos(ωx + φ) + k, then the maximum/minimum value would be easy to find
edit: oops this doesn't work because we'd have cos2x and cosx, nevermind
The "difficult" way is doable and kinda short
Notation: f(x) = 3*cos²(x) + 2*sin(x) + 1
Obs.: f is C^{\inf}
Solution:
f ' (x) = 2 * cos(x) * (3 - sin(x))
So extremum in two cases:
CASE #1 => sin²(x) = 1
CASE #2 => sin(x) = 1/3
f '' (x) = -6 -2*sin(x) + 12*sin²(x)
Subtituting in the cases:
C#1 => f''£{4,8} => Minima
C#2 => f'' Maxima
From the observation follows
max{f} = 3*8/9 + 2/3 + 1 = 13/3
for all x£R
_____
Mr. Pens, you managed to bring us another bad example of a good solution. Again!
Dunno why this got recommended to me but why is he holding a Pokéball?
42?
-delta/4a
5/2 + sqrt2
Just use AM-GM inequality
Shalom sir. Happy Hanukkah
Dude, you really look like the Shaolin Monk of Calculus. Because you are. 🤠
Like the math, love the poke ball ... do not like beard lol. And when I saw that I did think -- oh quadratic equation so that is simple to generally solve.
I didn't even watch video, but I'm sure that it's about differentiation
Vertex I like 👍🏻
Definitely the vertex formula
It was so easy to solve
no idea how i ended up here
But but but-
This is just calculus
I got the answer in under 2 seconds.
No, I cannot solve this problem without, or with calculus.
Im in year 11 and I got it right!!! (year 11 is 10th grade, we havent started calculus yet even though i know a lot about it lol)
I just impose that sinx=t and It become a second grade equations😢
You look better without a beard
PLS DERIVATIVE, PLS
Wait did people not know that you can just use trig identity’s to basically make it a quadratic
Lol this question can be done by head
easy af btw
Lol it's from ncert 😂
L
The second they starting throwing the alphabet into math I tuned out, history and English are easy math is the least used common core subject in most professions
Too easy
Sir i am from India.can you teach me mathematics. Please sir
Pagal hai kya bhai
He's already teaching you. Learn from his videos!