Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 1912:00 PM PDT
-01:00 PM PDT
Scoring 329 on the GRE is not always about using more books, more courses, or a longer study plan. In this episode of GRE Success Talks, Ashutosh shares his GRE preparation strategy, study plan, and test-day experience, explaining how he kept his prep....
May 1501:00 PM IST
-11:00 AM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.
May 2008:00 AM PDT
-08:30 AM PDT
What’s in it for you- Live Profile Evaluation Chat Session with Jenifer Turtschanow, CEO, ARINGO. Come with your details prepared and ARINGO will share insights! Pre-MBA Role/Industry, YOE, Exam Score, C/GPA, ECs Post-MBA Role/ Industry & School List.- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
23 Dec 2025, 00:44
Kudos
Bookmarks
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
Kudos
Bookmarks
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
Kudos
Bookmarks
\(\\
k^4+2k<2k^3+k^2\\
\\
\)
\(\\
k^4-k^2+2k-2k^3<0\\
\\
\)
\(\\
k^2(k^2-1)+2k(1-k2)<0\\
\\
\)
\(\\
k^2(k2-1)-2k(k2-1)<0\\
\\
\)
\(\\
(k^2-2k)(k^2-1)<0\\
\\
\)
\(=> k(k-2)(k-1)(k+1)<0\), 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.
Attachments
Screenshot_2.jpg [ 56.61 KiB | Viewed 376 times ]
