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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?
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Updated on: 05 Aug 2019, 03:25
3
chondro48 wrote:
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
A. \(-\frac{1}{{\sqrt{2}}}\)
B. \(-\frac{1}{2}\)
C. 1
D. \(\frac{1}{2}\)
E. \(\frac{1}{{\sqrt{2}}}\)
+1 kudo if this question is worthy of 700-level
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\) \(= .75 + t^2 - t^4\) where t = 1/x \(= .75 + t^2(1-t^2)\) Sum of \(t^2 + (1-t^2) =1\) =.75 + .5 (1 -.5) for expression to be max since product of 2 numbers whose sum is fixed is max when they are equal. =1
IMO C
Originally posted by Kinshook on 31 Jul 2019, 22:15.
Last edited by Kinshook on 05 Aug 2019, 03:25, edited 1 time in total.
Putting in values for x starting with option B & D
(b) when x= \(-\frac{1}{2}\)........answer is -ve value (d) when x= \(\frac{1}{2}\)..........again answer is -ve
(c) x=1 then the soln would be \(\frac{3}{4}\)
option a and e (extremes than b and d) should be cancelled out as it would give -ve number (check it out if you wish). Hence the maximum value of the equation is when x=1. Option (C) P.S: Before testing answers, make sure you simplify such complicated equations to the simplest form that you feel comfortable with. It makes testing the answers much easier and quicker.
Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?
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05 Aug 2019, 02:54
Kinshook wrote:
chondro48 wrote:
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
A. \(-\frac{1}{{\sqrt{2}}}\)
B. \(-\frac{1}{2}\)
C. 1
D. \(\frac{1}{2}\)
E. \(\frac{1}{{\sqrt{2}}}\)
+1 kudo if this question is worthy of 700-level
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\) \(= .75 + t^2 - t^4\) where t = 1/x \(= .75 + t^2(1-t^2)\) Sum of \(t^2 + 1-t^2 =1\) =.75 + .5 (1 -.5) for expression to be max since product of 2 numbers whose sum is fixed is max when they are equal. =1
IMO C
Hi Kinshook, Can you elaborate \(= .75 + t^2(1-t^2)\) Sum of \(t^2 + 1-t^2 =1\) =.75 + .5 (1 -.5)
How did you get \(t^2 =\) .5 and what is the significance of : Sum of \(t^2 + 1-t^2 =1\)
_________________
What is the maximum value of 3/4 + 1/x^2 - 1/x^4?
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Updated on: 05 Aug 2019, 04:48
stne wrote:
Kinshook wrote:
chondro48 wrote:
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
A. \(-\frac{1}{{\sqrt{2}}}\)
B. \(-\frac{1}{2}\)
C. 1
D. \(\frac{1}{2}\)
E. \(\frac{1}{{\sqrt{2}}}\)
+1 kudo if this question is worthy of 700-level
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\) \(= .75 + t^2 - t^4\) where t = 1/x \(= .75 + t^2(1-t^2)\) Sum of \(t^2 + 1-t^2 =1\) =.75 + .5 (1 -.5) for expression to be max since product of 2 numbers whose sum is fixed is max when they are equal. =1
IMO C
Hi Kinshook, Can you elaborate \(= .75 + t^2(1-t^2)\) Sum of \(t^2 + 1-t^2 =1\) =.75 + .5 (1 -.5)
How did you get \(t^2 =\) .5 and what is the significance of : Sum of \(t^2 + 1-t^2 =1\)
Here \(x=t^2;y=1-t^2 =>x+y=1\)
Suppose x+y = 6 and you have to find maximum value of xy If x=0 y=6 => xy=0 If x=1 y=5 => xy=5 If x=2 y=4 => xy=8 If x=3 y=3 => xy=9 If x=4 y=2 => xy=8 If x=5 y=1 => xy=5 If x=6 y=0 => xy=0
xy is max when x=3=y=\(\frac{6}{2}\)
In general if it is given that x+y=k where k is a positive number xy will be max when x=y=k/2 => xy max will be \(xy=\frac{k^2}{4}\)
Another way to solve it is calculus based:- \(E = .75 + t^2(1-t^2) = .75 + t^2 - t^4\) \(dE/dt =0 + 2t - 4t^3\) This gives condition for extremes (max or min) \(2t = 4t^3\) t=0 or \(t=1/\sqrt{2}\) or\(t^2=.5=\frac{1}{2}\) If t=0 => E=.75 If \(t^2\)=1/2 => E=.75+1/2(1-1/2) =.75+.5*.5 = .75 + .25 = 1 Since E=1 > .75 Emax = 1
Originally posted by Kinshook on 05 Aug 2019, 03:16.
Last edited by Kinshook on 05 Aug 2019, 04:48, edited 1 time in total.
Putting in values for x starting with option B & D
(b) when x= \(-\frac{1}{2}\)........answer is -ve value (d) when x= \(\frac{1}{2}\)..........again answer is -ve
(c) x=1 then the soln would be \(\frac{3}{4}\)
option a and e (extremes than b and d) should be cancelled out as it would give -ve number (check it out if you wish). Hence the maximum value of the equation is when x=1. Option (C) P.S: Before testing answers, make sure you simplify such complicated equations to the simplest form that you feel comfortable with. It makes testing the answers much easier and quicker.
Question is asking maximum value of the expression \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number, and NOT the value of x at which expression is maximum.
Please correct the explanation.
Originally posted by Kinshook on 05 Aug 2019, 03:24.
Last edited by Kinshook on 05 Aug 2019, 03:44, edited 1 time in total.
Expression depended on value of \(t^2(1-t^2)\) and I saw that \(t^2 + (1-t^2) = 1\) (Sum of \(x=t^2\) & \(y=1-t^2\) is 1 => x+y=1) If the sum is constant then product is max when x=y=\(\frac{1}{2}\)=.5 => \(x=t^2=.5\) and \(y=1-t^2=.5\)
Sometimes you have to see some relation to make the solution easy.
Originally posted by Kinshook on 05 Aug 2019, 03:38.
Last edited by Kinshook on 05 Aug 2019, 09:35, edited 2 times in total.
Expression depended on value of \(t^2(1-t^2)\) and I saw that \(t^2 + (1-t^2) = 1\) (Sum of \(x=t^2\) & \(y=1-t^2\) is 1 => x+y=1) If the sum is fix then product is max when x=y=\(\frac{1}{2}\)=.5 => \(x=t^2=.5\) and \(y=1-t^2=.5\)
Sometimes you have to see some relation to ease the solution, which I did.
Right!
Finally got it, learnt something new, if the sum of two expressions x and y is a constant then max value of the product xy will lie at x=y.
_________________
Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?
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05 Aug 2019, 09:28
chondro48 wrote:
What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?
A. \(-\frac{1}{{\sqrt{2}}}\)
B. \(-\frac{1}{2}\)
C. 1
D. \(\frac{1}{2}\)
E. \(\frac{1}{{\sqrt{2}}}\)
+1 kudo if you think it is good question
I am sure there is a very good theory you are trying to test, but you may want to re-evaluate the answer choices. Just plugging in 1 makes 3/4 so c is the only viable answer and the question takes about 10 sec.
Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?
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05 Aug 2019, 10:30
Maybe I could have missed something, but It can easily be solved by analyzing the answer choices: since both exponents are even numbers, choices A and E hold the same result as well choices B and D.