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 â€śUâ€ť is N^{2}-M

bhai ye kya hota hai? Theory bhej sakte ho?

Ques 25 without putting valuesâ€¦

@Rade_Woosh @Dexter27 @Deadpool @LaveshGupta @Rudra_d @anon30992111

The number of permutations of n different objects, taken all at a time, when m specified

objects always come together is m! Ă— (n â€“ m + 1)!.

Koi iska proof batado by exampleâ€¦

@Dexter27 @Rade_Woosh @Anorak @VictoryGod and anyone

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 youâ€™ll get same answer

dekhoâ€¦ham n me se m ko hamesha saath rakhna chahte haiâ€¦so what we do is take the m objects together always and consider it as a single unit insteadâ€¦so we get (n-m+1) from hereâ€¦uske baad the internal permutation amongst that single unit is taken account by m! termâ€¦

@Achyut_Singh

mujhe number of solutions ko accurately count krne me bahaut dikkat aati haiâ€¦

pls help

@Anorak @Achyut_Singh @LaveshGupta @VictoryGod @Deadpool and everyoneâ€¦

@Dexter27

Try some methods like:-

1)Graphical methods

2)Equating Solutions

3)Comparing Domains â€¦etc.

bro like in 2nd question its evident that rhs will always assume negative valueâ€¦so i get range for xâ€¦but what to after that??

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

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

Chamka?

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

i am still struggling broâ€¦

ig this question will be easier if we write in form of eâ€¦because it will be e^(1-(x+1)^2) and the power is negativeâ€¦so will be like e^(-x) form graphâ€¦and will cut |sinx| at 2 pointsâ€¦

@Anorak

yeah that is easier lol