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.
A square with area equal to 256 has its center at the origin
[#permalink]
18 Nov 2020, 10:20
18 Nov 2020, 10:20
1
Expert Reply
6
Bookmarks
00:00
Question Stats:
63% (01:25) correct
36% (02:03) wrong based on 41 sessions
Hide Show timer Statistics
A square with area equal to 256 has its center at the origin, (0, 0), in the xy-coordinate plane. The graph of a function f defined by \(f(x) = cx^3\), for some positive constant c, intersects the square at two opposite vertices of the square. What is the value of c? (Enter your answer as a fraction.)
Show: ::
\(\frac{1}{64}\)
A square with area equal to 256 has its center at the origin
[#permalink]
21 Feb 2021, 08:54
21 Feb 2021, 08:54
6
void wrote:
plz explain 2nd half of the problem....
Hi!
Solution:
Firstly, the area of the square=256, Thus, the side of the square=16.
Next we are said that the origin (0,0) is the centre i.e. all the sides are equidistant. Hence, the vertices of the square are (8,8), (-8,-8), (8,-8) & (-8,8)
Now, the second half of the question says f(x)=c\(x^3\)
Whenever you see a f(x) in a coordinate geometry question simply substitute f(x) with y
So, we get, y=c\(x^3\)
The question then says this function passes through opposite vertices of the square i.e. it passes diagonally from point (8,8)to (-8,-8) or from point (8,-8) to (-8,8)
Lets substitute the value of x &y to get c
Considering (8,8)
8=c*\(8^3\)
c=\(\frac{1}{64}\)
Similarly, when you consider (-8,-8) you get C as \(\frac{1}{64}\)
And when we substitute the other two values we get \( \frac{-1}{64 } \). As c is a positive value we can negate the other two values
Thus, C=\(\frac{1}{64}\)
Hope this helps
General Discussion
Re: A square with area equal to 256 has its center at the origin
[#permalink]
18 Nov 2020, 23:00
18 Nov 2020, 23:00
1
here area of the square is 256=> the length of each side is 16 unit
as the center of this square is at the origin (0,0), so we can easily
determine the vertices of the square: (8,8);(-8,8); (-8,-8) and (8,-8)
in I, II, III, and IV quadrants respectively. vertices locate in I and III
quadrants are opposite, similarly, vertices locate in II and IV are opposit
to each other.
Here y= c*x^3 passes through the opposite vertices points. We get
c= 1/64, -1/64
as the c is positive, so we take c=1/64 (answer)
as the center of this square is at the origin (0,0), so we can easily
determine the vertices of the square: (8,8);(-8,8); (-8,-8) and (8,-8)
in I, II, III, and IV quadrants respectively. vertices locate in I and III
quadrants are opposite, similarly, vertices locate in II and IV are opposit
to each other.
Here y= c*x^3 passes through the opposite vertices points. We get
c= 1/64, -1/64
as the c is positive, so we take c=1/64 (answer)
