Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Jan 2410:00 AM EST
-12:00 PM EST
The Target Test Prep team is proud to introduce GMAT OnDemand, a video course created by Scott Woodbury-Stewart, founder and CEO, designed to provide you with everything you need to master the GMAT and achieve your score goals.
Jan 2110:00 AM PST
-11:59 PM PST
For a limited time, get 20% off our GMAT courses, including private tutoring! Work with an expert who will help you ace the GMAT and achieve your dreams. Use code 20NEW.
Jan 2511:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Jan 26
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease.
16 Sep 2014, 00:49
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (01:54) correct
60%
(02:20)
wrong
based on 141
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(a\) positive?
(1) \(x^2 - 2x + a\) is positive for all \(x\).
(2) \(ax^2 + 1\) is positive for all \(x\).
(1) \(x^2 - 2x + a\) is positive for all \(x\).
(2) \(ax^2 + 1\) is positive for all \(x\).
16 Sep 2014, 00:49
Kudos
Bookmarks
Official Solution:
Is \(a\) positive?
(1) \(x^2-2x+a\) is positive for all \(x\):
It's crucial to notice that we are given that the expression is positive not just for some value of \(x\) but for every value of \(x\).
Next, since the coefficient of \(x^2\) is positive, \(f(x)=x^2-2x+a\) represents an upward parabola. Since \(f(x)\) is positive for all \(x\), it means that the parabola lies entirely above the x-axis and has no intersections with it. Therefore, the equation \(x^2-2x+a=0\) has no real roots.
For a quadratic equation not to have real roots, its discriminant must be negative. Thus, we have \(discriminant = 2^2 - 4a = 4 - 4a \lt 0\), which simplifies to \(1-a \lt 0\) and finally to \(a \gt 1\). Sufficient.
(2) \(ax^2+1\) is positive for all \(x\):
Notice that, if \(a > 0\), then \(ax^2+1\) is positive for every \(x\). But the trick here is that \(ax^2+1\) is also positive for every \(x\) when \(a = 0\), giving a NO answer to the question! Not sufficient.
Answer: A
Is \(a\) positive?
(1) \(x^2-2x+a\) is positive for all \(x\):
It's crucial to notice that we are given that the expression is positive not just for some value of \(x\) but for every value of \(x\).
Next, since the coefficient of \(x^2\) is positive, \(f(x)=x^2-2x+a\) represents an upward parabola. Since \(f(x)\) is positive for all \(x\), it means that the parabola lies entirely above the x-axis and has no intersections with it. Therefore, the equation \(x^2-2x+a=0\) has no real roots.
For a quadratic equation not to have real roots, its discriminant must be negative. Thus, we have \(discriminant = 2^2 - 4a = 4 - 4a \lt 0\), which simplifies to \(1-a \lt 0\) and finally to \(a \gt 1\). Sufficient.
(2) \(ax^2+1\) is positive for all \(x\):
Notice that, if \(a > 0\), then \(ax^2+1\) is positive for every \(x\). But the trick here is that \(ax^2+1\) is also positive for every \(x\) when \(a = 0\), giving a NO answer to the question! Not sufficient.
Answer: A
General Discussion
27 Sep 2014, 12:00
Kudos
Bookmarks
I used another approach for S1 and got inequality a>4. Is it correct?
x^2-2x+4-4+a>0
(x-2)^2-4+a>0 for all x. Min of (x-2)^2 if x=2.
Therefore a>4
Sufficient
x^2-2x+4-4+a>0
(x-2)^2-4+a>0 for all x. Min of (x-2)^2 if x=2.
Therefore a>4
Sufficient
