GRE Prep 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.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
[#permalink]
07 Sep 2021, 09:41
07 Sep 2021, 09:41
4
Bookmarks
00:00
Question Stats:
50% (01:49) correct
50% (01:55) wrong based on 18 sessions
Hide Show timer Statistics
A function, \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(t\) |
\(2\sqrt{15}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
[#permalink]
Updated on: 09 Sep 2021, 04:10
Updated on: 09 Sep 2021, 04:10
1
1
Given : \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = +/-2\sqrt{15}\)
Answer is D
Thanks to gernico for correcting.
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = +/-2\sqrt{15}\)
Answer is D
Thanks to gernico for correcting.
A function, f(x) = 3x^2 - tx + 5 is tangent to x-axis.
[#permalink]
08 Sep 2021, 19:02
08 Sep 2021, 19:02
2
eskay1981 wrote:
Given : \(f(x) = 3x^2 - tx + 5\) is tangent to x-axis
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = 2\sqrt{15}\)
Answer is C
so two observation from the above statement
1. at the point of tangent intersection at x, y=0
2. and \(f'(x) = 0\)
so first take the derivative and equate to 0 to find the point of minimum
\( f' = 6x - t = 0 => x = t/6\) when x = t/6, the value for f(x) is minimum that's 0
so substitute \(x =t/6 \) in f(x) and this will give \(t = 2\sqrt{15}\)
Answer is C
When subbing in \(\frac{t}{6}\) for \(x\), don't forget that when you get to:
\(t^2 = 60\)
\(t = 2\sqrt{15}\) OR \(-2\sqrt{15}\)
Which is why the answer is D.
