Doubt in maths


Q6, option B doubt
Q3

For Q3 you can use Newton sums, or you can write it as product of two matrices.
For Q6, M^{2}+M N^{2}=N^{4}+N^{2} M=N^{2}\left(N^{2}+M\right).
We know that
0=N^{4}-M^{2}=\left(N^{2}+M\right)\left(N^{2}-M\right) (given in question).
So multiply both sides of first equation by N^{2}-M and it will come out to be zero matrix. So one possible value of the matrix “U” is N^{2}-M

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bhai ye kya hota hai? Theory bhej sakte ho?

https://artofproblemsolving.com/wiki/index.php/Newton's_Sums

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Ques 25 without putting values…
@Rade_Woosh @Dexter27 @Deadpool @LaveshGupta @Rudra_d @anon30992111

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@Achyut_Singh

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The number of permutations of n different objects, taken all at a time, when m specified
objects always come together is m! × (n – m + 1)!.
Koi iska proof batado by example…
@Dexter27 @Rade_Woosh @Anorak @VictoryGod and anyone

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proof toh yaad nahi try taking n=3 differently coloured objects where m=2 specified colours e.g red and blue together and subtitue in formula and also try same ques normal method you’ll get same answer

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dekho…ham n me se m ko hamesha saath rakhna chahte hai…so what we do is take the m objects together always and consider it as a single unit instead…so we get (n-m+1) from here…uske baad the internal permutation amongst that single unit is taken account by m! term…
@Achyut_Singh

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Ok samjh gya Thanks @Dexter27 @Anorak :slightly_smiling_face:

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mujhe number of solutions ko accurately count krne me bahaut dikkat aati hai…
pls help :weary:

@Anorak @Achyut_Singh @LaveshGupta @VictoryGod @Deadpool and everyone…

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@Dexter27
Try some methods like:-
1)Graphical methods
2)Equating Solutions
3)Comparing Domains …etc.

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bro like in 2nd question its evident that rhs will always assume negative value…so i get range for x…but what to after that?? :face_with_head_bandage:

1)take e^modsinx=t and solve using determinant D and sinx range condition
2.)only possible by graph plotting
3.)since they didnt give range of x i assumd 0,2pi and solved using general solution of sinpix^2/3=1
4.) we can solve by graph of tan{x}=1 but that is harder to visualise so so change {x}=x-[x] and expand and simplify then draw graph

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ln(sinx) ka graph kaise banega???

i took aproximation
we know that modsinx is oscilaatory and ranges between (0,1]
and since we have to find points of intersection my graph wasnt accurate
also we can further simplify to write x-1=(1-lnsinx)^1/2 and take condition from here as well

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Chamka?

but this we will not be able to know without graph?!

i am still struggling bro…

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ig this question will be easier if we write in form of e…because it will be e^(1-(x+1)^2) and the power is negative…so will be like e^(-x) form graph…and will cut |sinx| at 2 points…
@Anorak

yeah that is easier lol

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