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Which of the following represents the complete range of x over which
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07 Apr 2021, 00:09
07 Apr 2021, 00:09
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Which of the following represents the complete range of x over which \(x^3 – 4x^5 < 0\)?
A. \(0 < |x| < \frac{1}{2}\)
B. \(|x| >\frac{1}{2}\)
C. \(–\frac{1}{2} < x < 0\) or \(\frac{1}{2} < x\)
D. \(x < –\frac{1}{2}\) or \(0 < x < \frac{1}{2}\)
E. \(x < –\frac{1}{2}\) or \(x > 0\)
A. \(0 < |x| < \frac{1}{2}\)
B. \(|x| >\frac{1}{2}\)
C. \(–\frac{1}{2} < x < 0\) or \(\frac{1}{2} < x\)
D. \(x < –\frac{1}{2}\) or \(0 < x < \frac{1}{2}\)
E. \(x < –\frac{1}{2}\) or \(x > 0\)
Re: Which of the following represents the complete range of x over which
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07 Apr 2021, 04:30
07 Apr 2021, 04:30
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x^3–4x^5<0
taking x^3 common
x^3(1-4x^2)<0
x^3(1-2x)(1+2x)<0
a.b.c<0 implies either one is negative and two is positive OR all are negative
Hence,
case 1) All are negative
x^3<0 and 1-2x<0 and 1+2x<0
x<0 and x>1/2 and x<-1/2
No solutions
case 2) one is negative rest positive
case a) x^3<0 and (1-2x)>0 and (1+2x)>0
-> x<0 and x<1/2 and x>-1/2
-> x belongs to (-1/2,0)
case b) x^3>0 and (1-2x)<0 and (1+2x)>0
-> x>0 and x>1/2 and x>-1/2
-> x belongs to (1/2,inf)
case c) x^3>0 and (1-2x)>0 and (1+2x)<0
-> x>0 and x<1/2 and x<-1/2
-> no solution
Hence, the final answer is
(-1/2,0) and (1/2,inf)
Hence correct answer is C
taking x^3 common
x^3(1-4x^2)<0
x^3(1-2x)(1+2x)<0
a.b.c<0 implies either one is negative and two is positive OR all are negative
Hence,
case 1) All are negative
x^3<0 and 1-2x<0 and 1+2x<0
x<0 and x>1/2 and x<-1/2
No solutions
case 2) one is negative rest positive
case a) x^3<0 and (1-2x)>0 and (1+2x)>0
-> x<0 and x<1/2 and x>-1/2
-> x belongs to (-1/2,0)
case b) x^3>0 and (1-2x)<0 and (1+2x)>0
-> x>0 and x>1/2 and x>-1/2
-> x belongs to (1/2,inf)
case c) x^3>0 and (1-2x)>0 and (1+2x)<0
-> x>0 and x<1/2 and x<-1/2
-> no solution
Hence, the final answer is
(-1/2,0) and (1/2,inf)
Hence correct answer is C
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Re: Which of the following represents the complete range of x over which
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31 Dec 2021, 01:41
31 Dec 2021, 01:41
1
Given that \(x^3 – 4x^5 < 0\) and we need to find the range of values of x
\(x^3 – 4x^5 < 0\)
=> \(x^3 – 4x^3*x^2 < 0\)
=> \(x^3 (1 – 4x^2) < 0\)
=> \(x^3 (1 – (2x)^2) < 0\)
=> \(x*x^2 (1 – 2x) (1+2x) < 0\) [ Using \((a-b)^2 = a^2 - 2ab + b^2\) ]
Multiply both sides by -1 we get
=> \(x*x^2 (2x -1) (2x+1) > 0\)
Now we know that \(x^2\) is always >= 0 so we can ignore \(x^2\) rom above equation
=> \(x* (2x -1) (2x+1) > 0\)
Using Sine Wave method (Watch this video to learn about Sine Wave Method)
We will get the three points as \(\frac{-1}{2}\), 0, \(\frac{1}{2}\)
-0.5 to +0.5 Sine Wave.JPG [ 18.28 KiB | Viewed 2707 times ]
Since question is asking for > 0 so we will take the "+" range in the above figure
=> \(\frac{-1}{2}\) < x < 0 and x > \(\frac{1}{2}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Inequality Problems
\(x^3 – 4x^5 < 0\)
=> \(x^3 – 4x^3*x^2 < 0\)
=> \(x^3 (1 – 4x^2) < 0\)
=> \(x^3 (1 – (2x)^2) < 0\)
=> \(x*x^2 (1 – 2x) (1+2x) < 0\) [ Using \((a-b)^2 = a^2 - 2ab + b^2\) ]
Multiply both sides by -1 we get
=> \(x*x^2 (2x -1) (2x+1) > 0\)
Now we know that \(x^2\) is always >= 0 so we can ignore \(x^2\) rom above equation
=> \(x* (2x -1) (2x+1) > 0\)
Using Sine Wave method (Watch this video to learn about Sine Wave Method)
We will get the three points as \(\frac{-1}{2}\), 0, \(\frac{1}{2}\)
Attachment:
-0.5 to +0.5 Sine Wave.JPG [ 18.28 KiB | Viewed 2707 times ]
Since question is asking for > 0 so we will take the "+" range in the above figure
=> \(\frac{-1}{2}\) < x < 0 and x > \(\frac{1}{2}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Inequality Problems
