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23 Dec 2025, 00:44
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k^4+2k < 2k^3+k^2
k^4-2k^3-k^2+2k < 0
k(k^3-2k^2-k+2) < 0
Using the Wavy Curve Method for solving Inqualities:
+ve -1 -ve 0 +ve 1 -ve 2 +ve (Basically the LHS of the inequality is +ve for any value < -1, -ve for any value between -1 and 0 (exclusive) and so on...)
Since we want k(k^3-2k^2-k+2) to be lesser than 0, we check for the -ve regions:
1. Between -1 and 0 (exclusive) -> No such integral value of k between -1 and 0 (exclusive)
2. Between 1 and 2 (exclusive) -> No such integral value of k between 1 and 2 (exclusive)
Since there are no such possible integral values of k in either of the two negative regions, there does not exist any such value of k.
Hence, answer = 0 (Option A)
k^4-2k^3-k^2+2k < 0
k(k^3-2k^2-k+2) < 0
Using the Wavy Curve Method for solving Inqualities:
+ve -1 -ve 0 +ve 1 -ve 2 +ve (Basically the LHS of the inequality is +ve for any value < -1, -ve for any value between -1 and 0 (exclusive) and so on...)
Since we want k(k^3-2k^2-k+2) to be lesser than 0, we check for the -ve regions:
1. Between -1 and 0 (exclusive) -> No such integral value of k between -1 and 0 (exclusive)
2. Between 1 and 2 (exclusive) -> No such integral value of k between 1 and 2 (exclusive)
Since there are no such possible integral values of k in either of the two negative regions, there does not exist any such value of k.
Hence, answer = 0 (Option A)
23 Dec 2025, 00:56
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K^4 + 2k < 2k^3 + k^2
= k^4 - 2k^3 + 2k - k^2 < 0
= k^3(k - 2) - k(k - 2) < 0
= (k - 2)(k^3 - k) < 0
= k(k - 2)(k^2 - 1) < 0
= (k - 2)(k - 1)k(k + 1) < 0
Inflection points -1, 0, 1, 2
The expression is negative between {-1, 0} and {1, 2}. Rest of the values between other inflection boundaries give a positive outcome for the expression.
We only have non-integer values between these integer boundaries hence no integer values satisfy the solution
Option A
= k^4 - 2k^3 + 2k - k^2 < 0
= k^3(k - 2) - k(k - 2) < 0
= (k - 2)(k^3 - k) < 0
= k(k - 2)(k^2 - 1) < 0
= (k - 2)(k - 1)k(k + 1) < 0
Inflection points -1, 0, 1, 2
The expression is negative between {-1, 0} and {1, 2}. Rest of the values between other inflection boundaries give a positive outcome for the expression.
We only have non-integer values between these integer boundaries hence no integer values satisfy the solution
Option A
Bunuel
23 Dec 2025, 01:07
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\(\\
k^4+2k<2k^3+k^2\\
\\
\)
\(\\
k^4-k^2+2k-2k^3<0\\
\\
\)
\(\\
k^2(k^2-1)+2k(1-k2)<0\\
\\
\)
\(\\
k^2(k2-1)-2k(k2-1)<0\\
\\
\)
\(\\
(k^2-2k)(k^2-1)<0\\
\\
\)
\(=> k(k-2)(k-1)(k+1)<0\), or in a more convenient order: \((k-2)(k-1)k(k+1)<0\)
We can apply the interval method here (attached) and check which intervals provide the desired negative value. It turns out that the only suitable intervals are (-1;0) and (1;2), but since they're not incusive of the integer values (because the inequality cannot be equal to zero), none of these integers are included. Therefore, there are no integer values for K, and the answer is A.
k^4+2k<2k^3+k^2\\
\\
\)
\(\\
k^4-k^2+2k-2k^3<0\\
\\
\)
\(\\
k^2(k^2-1)+2k(1-k2)<0\\
\\
\)
\(\\
k^2(k2-1)-2k(k2-1)<0\\
\\
\)
\(\\
(k^2-2k)(k^2-1)<0\\
\\
\)
\(=> k(k-2)(k-1)(k+1)<0\), or in a more convenient order: \((k-2)(k-1)k(k+1)<0\)
We can apply the interval method here (attached) and check which intervals provide the desired negative value. It turns out that the only suitable intervals are (-1;0) and (1;2), but since they're not incusive of the integer values (because the inequality cannot be equal to zero), none of these integers are included. Therefore, there are no integer values for K, and the answer is A.
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