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minimum value of m is \(1\) so \(k_{min}=5\)

How can you get it?

lol i mean \((a+b+c+d)_{min}=5\)

not \(k_{min}\)

cuz \(m\) is a positive integer

yeah but m is given + integer

i didnt understand , answer given is option (b.) , btw m not clever

unlucky me, i didnt studied calculus yet

can a b c d repeat ?

Like a b c d can be literraly any integers?

yes \(\Large \{a,b,c,d\}\in \mathbb{Z}\)

So you just need to look for minimum value of \(k\)

yes i need to eliminate one option from 2

Now as you said, you just need to eliminate. You can do that by letting \(m = 1\).

\(a+b+c+d=5\), if \(m=1\)

So it of course cannot hold the value \(4m^2 - 2m + 1\), thus the answer is \(4m^2 + 2m + 1\).

Is there some way to use AM-GM-HM here?

but how can it be \(m+\dfrac14\) as all variables are given as integers

u maximized \(k\) and this is \(k_{max}\) right ?
\(k_{max} = (4m+1)^2 - 2 \cdot 6(m+ 1/4)^2\)

Yes! This is \(k_{max}\) for real numbers.

for m=1 option a-> 3 and option b->7 ?

option a doesn't makes sense , so option b is correct . nice

I literally said the same thing in the first reply... :|

lol yea ,sry

thnks all.