Differential calculus? Honestly, idk.
Q.1 and Q.15 had me stumped.
I forgot the method for depressed cubics so I couldn't even factorise the function in 1.
As for 15, I don't understand how we solve the DE with an inequality? If it was an equation, then characteristic quadratic gives us the solution, but with a ā„, I have no clue.
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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Q.16 is also something I couldn't do, but that's a different topic, so, different thread?
For the 15th one, ig writing it as e^(-3x)(f''-3f')-e^(-3x)(2f'-6f)>=0 helps.
This would give e^(-3x)(f'-2f) is increasing. Since e^(-3x) is decreasing this means f'-2f must be increasing (monotonic).
1 seems okay enough?
graph banalo
or is it tough š±
f(f(x)) ka graph?
yeah just calculate f(f(x)) then plot the graph
6 degree equation banegi
If roots are weird then idk what to do
f(f(x)) ke derivative ke roots nikalne honge
šæ tedious work
This is doable
Because f'(x) is factorable easily
yeah
actually
lemme tru
try
Oh ok
f'(f(x)) will change signs at -ā3 0 ā3 and 2
if f(fx) = 0 wala case lu to
yeah just calculate f(fx) then check consecutive points jahan dy/dx = 0 hai that f(fx1) f(fx2) is positive or negative
7 roots aajaenge
you have to check f'(x) also
there will be 2 more critial pts -1 and +1
then you will get 7 roots
question no 15 assume the function is Ae^2x+Be^3x (we take 2 and 3 because they are the roots of x^2-5x+6) and then satisfy f(0)=1 and f'(0)=0 conditions and then you will get f(x) to be 3e^2x-2e^3x so a =3 b =2 and h(0)=1
so ans is 5
I solved for f, but it's not equations, it's all inequalities. That's what confused me.
there is equality also right so it doesnt matter
16 one all i got is the degree should be 4 or lower i am not getting anywhere after this
Ok, that's just weird.
Yeah
It's going to be the no.of integral solutions of
625a+125b+25c+5d+e = 1500
Such that
0ā¤a,b,c,d,eā¤24
In Q1, the minima of f(x) acquires a value of -1 rather than -3, so as for beta, there exists only one value of x such that f(x)=beta (Since beta<-1) [I made a slight error in the solution attached below.]
Ig (c) and (d) are by curve sketching of f(x)
Ah 16 would be a caseworking nightmare but here is something I thought of: Convert 1500 to base 25. Then 5a+b=(1500 in base 25) and we have info. about the number of digits in a and b. Does this logic work? :thinker:
I'm not sure if the above works, but just something I thought of.
Just a moment, why 5a+b? Wouldn't it be 25a+b?
Converting to base 25 covers positions of 1, 25 (the x^2 term), and so on but leaves out 5 (the x term), 125 and so on. Hence the multiplication by 5.
Nvm i understood
1500 = (2A0)_25
Where A denotes 10
2(625)+10(25)+0(1)
Ig we know that b must be divisible by 5, so considering it 5k would allow us to use multinomial ig.
b would be a 3-digit number while a would be a 2-digit one.
Rather messy, mb
@SirLancelotDuLac
When using multinomial ig we would have to do intermediate calculations in base 25 right?
To be on the safer side? Although that turns pretty hectic.
Um..Is the answer 1 for 16th?
It's given 3
Ah then I messed up :/
Wait, I hope the solutions copy is out.
I just have the answer key
Ig there are exactly 301 such polynomials.
Does this work?
Unfortunately, I don't know how to do maths in base 25. Let me try and decode it.
Neither do I lmao. I just tried to deconstruct and reconstruct the numbers from base 25 to 10 in my head.
Ig we dont need base 25 in the last 3 lines
Ok, where did you get the choice expression? (a+k)C1 ?
Beggar's method is not dependant on base (due to its derivation) so the expression remains same.
Ab ye beggar's method kya hota hai? :sweaty:
As for the value of the choice expression, that would also remain same due to its independancy from base.
The (n+r-1)C(r-1) thing. I heard someone call it beggar's method so I used it.
oh wait,
(n+r-1)C(r-1)?
I see.
Idk what anything's called honestly.
Oh, so we reduced it to no.of integral sols of a + k = 300
Yeah.
Well, if you have the solutions, do check if the value of k comes out to be 301.
Sorry, I'm still a bit confused. We got this expression, but how did we figure out the set of allowed values for a,k?
I think it'll be out around 7:30-8pm
The coeffecients of 1,x^2,x^4 are the digits of a. Since these can acquire values from 0 to 24, all numbers of upto 3 digits are valid.
Similar logic for b.
@SirLancelotDuLac
Yoo the method works š
I mean, this one is more intuitive, but the base changing thing really shortens stuff out.
I still didnt understand the base change stuff š¤¦āāļø
Gimme a bit.
Yeah, I didn't explain it nicely. This is not a universal method, just seemed to work here. We consider a number of base 25 because the digits will give us the coefficient of the polynomial.
We have 625a+125b+25c+5d+e=1500
which can be written as (625a+25b+e)+5(25b+d)=1500
Let m be a base 25 number of the first bracket, and n be that of the second bracket.
Then we have m+5n=(1500 in base 25), since a,b,c,d and e can acquire values from 0 to 24, m and n can acquire values of upto 3 digit and 2 digit numbers.
Then let m=5k and apply beggar's method.
Again, not a general method 'cuz coeffecients might not always vary from 0 to 24 or like-wise, but still in this particular case shortens stuff out.
I see.
+solved @SirLancelotDuLac @hithav @Dexter
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