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22 Dec 2025, 11:52
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given equation k^4 + 2k < 2k^3 + k^2, upon simplifying the equation:
k^4 - 2k^3 - k^2 + 2k < 0 => k(k^3 -2k^2 -k + 2) < 0 => k(k-2)(K+1)(K-1) < 0
on plotting on number line we see only possible range is (-1,0) & (1,2) -- no integers lie in-between hence 0.
k^4 - 2k^3 - k^2 + 2k < 0 => k(k^3 -2k^2 -k + 2) < 0 => k(k-2)(K+1)(K-1) < 0
on plotting on number line we see only possible range is (-1,0) & (1,2) -- no integers lie in-between hence 0.
22 Dec 2025, 12:08
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Bunuel
Factored, it becomes \(k(k^3+2) < k^2(2k+1)\)
We can't go any further with this equation because we don't know if k is positive or negative to divide. Use trial and error at this point, testing values close to 0 to look for trends.
0: \(0<0\) Invalid.
1: \(3<3\) Invalid.
2: \(20<20\) Invalid.
-1: \(-1 < -1\) Invalid.
2: \(-12 < -12\) Invalid.
Trend seems to be that each side is always equal. Thus we can confidently say that no values will ever satisfy this, so A is the answer.
22 Dec 2025, 12:15
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k^4+2k<2k^3+k^2
k^4-2k^3-k^2+2k<0
On solving we will get,
k(k-1)(k+1)(k-2)<0
So possible values of this as equality is 0, -1, 1 and 2
If we put it in equation, none of them got satisfied.
Hence the answer is (A)
k^4-2k^3-k^2+2k<0
On solving we will get,
k(k-1)(k+1)(k-2)<0
So possible values of this as equality is 0, -1, 1 and 2
If we put it in equation, none of them got satisfied.
Hence the answer is (A)
