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

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

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

Please see above the correct solution based on calculus (Engineer's way).

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.

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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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.

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.