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
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An engineers way

1/x^2 = p

3/4 + p - p^2

Diffrentiate it and equate to 0 for maxima

2p -1 = 0
P = 1/2
X = 2^-1/2

Value of expression 1.

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\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)
=\(\frac{3}{4}+\frac{1}{{x^2}} * \frac{(x^2-1)}{{x^2}}\)
= \(\frac{3}{4}+\frac{1}{{x^4}}\) * (x+1) (x-1)

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.

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 \(t^2 + 1-t^2 =1\)
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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
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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.
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Please see above the correct solution based on calculus (Engineer's way).
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Hi Kinshook,

Why did you write Sum of \(t^2+1−t^2=1\)?
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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.
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Let, 1/x^2 be a
therefore the expression can be written as:
3/4 + a - a^2

To find the maxima ---> Differentiate the expression and equate with zero

0 + 1 - 2a = 0
a = 1/2

therefore 1/x^2 = a = 1/2
1/x^2 = 1/2
x^2 =2
x = 2^(1/2)

value of expression: 3/4 + 1/2 + 1/(2^2)
= 3/4 + 1/2 - 1/4
= 1/2 +1/2
=1

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.
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Here we are lucky that a is +ve, otherwise the solution is invalid for -ve a.
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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.

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.

We can notice that at x=1, \(\frac{1}{x^2}-\frac{1}{x^4}=0\)
Hence maximum possible value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)≥0.75

Hence only C option is possible.

Asked: What is the maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\), where x is any real number?

\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)
\(=\frac{3}{4} - \{\frac{1}{x^4} - 2*\frac{1}{x^2}*\frac{1}{2} + \frac{1}{2^2}\} +\frac{1}{2^2}\)
\(=\frac{3}{4} - \{\frac{1}{x^2} - \frac{1}{2}\}^2 +\frac{1}{4}\)
\(=1 - \{\frac{1}{x^2} - \frac{1}{2}\}^2\)

Expression is max when \(\frac{1}{x^2}-\frac{1}{2}=0\)
Maximum value of \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\) = 1

IMO C
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For such problems we should try to convert equation of X into a perfect square

Let Y = \(\frac{1}{x^2}\) , therefore the equation can be re-written as

- \((Y^2 -Y -\frac{3}{4})\)

-\((Y^2 -Y -\frac{3}{4} + \frac{4}{4} - \frac{4}{4})\)

-\((Y^2 -Y + \frac{1}{4} - \frac{4}{4} )\)

- \(((Y-\frac{1}{2})^2 - 1)\)

- \(( Y -\frac{1}{2})^2 +1\)

Therefore maximum value will be 1 when Y = \(\frac{1}{2}\)