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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 02 Dec 2021, 05:45
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C
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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. \(\frac{1}{{\sqrt{2}}}\)
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C
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 05 Dec 2021, 14:28
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Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)

Breaking Down the Info:

We can set \(y = x^{-2}\) and rewrite the expression as \(\frac{3}{4} + y - y^2 = -y^2 + y + \frac{3}{4}\).

The "vertex" of this expression is \(y = -\frac{b}{2a} = \frac{1}{2}\). Then we may plug that in for an extreme value.

\(-\frac{1}{4} + \frac{1}{2} + \frac{3}{4} = 1\) is the maximum since the leading coefficient is negative, giving us a parabola opening downwards.

Answer: C
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 04 Dec 2021, 09:33
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we can get maximum value when value of 1/X^2 & 1/x^4 is minimum. Assuming it is close to zero, that gives us 3/4=0.75.

Now checking option, only E is close to 0.75 hence E.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 13 Dec 2021, 10:02
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Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)

We can see that, if X = 0, the answer is 3/4. Since this is bigger than any of the other options, we are left with C:1 as the correct solution.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 27 Jan 2022, 20:01
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Hello CrackverbalGMAT, could you give us your insight on this? I feel option E comes the closest, if we assume x=0.

Correct me if I'm wrong, please.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 27 Jan 2022, 21:56
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Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)


I. Options
Now, \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}>\frac{3}{4}\)
Only C is greater than 3/4.

II. Algebraic
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)

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

Now for maximum value, we have to maximise \( \frac{1}{{x^2}}(1-\frac{1}{{x^2}})\)
This will happen when we multiply two equal terms as we are looking at two terms adding up to 1.
This, \(\frac{1}{{x^2}}=(1-\frac{1}{{x^2}})\)
\(\frac{2}{{x^2}}=1\)
\(\frac{1}{{x^2}}=\frac{1}{2}\)

So max value = \(\frac{3}{4}+\frac{1}{{2}}*\frac{1}{{2}}\)
\(\frac{3}{4}+\frac{1}{4}=1\)


III. Differentiation
However, this is not in scope of GMAT.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 25 Dec 2024, 01:30
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Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)
IMO easiest way to solve such a question will be using the options
check the answer of first or last two options, since square of a negative sign wont make any difference, you will get the same answer for A and E or B and D. Compare the values you get here with the value you get using option C and you will get your answer


check the answer of first or last two options, since square of a negative sign wont make any difference, you will get the same answer for A and E or B and D
Solely based on this you can mark C as the answer since two same results can't get you an answer.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 28 Dec 2024, 16:43
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chetan2u
Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)


I. Options
Now, \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}>\frac{3}{4}\)
Only C is greater than 3/4.

II. Algebraic
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)

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

Now for maximum value, we have to maximise \( \frac{1}{{x^2}}(1-\frac{1}{{x^2}})\)
This will happen when we multiply two equal terms as we are looking at two terms adding up to 1.
This, \(\frac{1}{{x^2}}=(1-\frac{1}{{x^2}})\)
\(\frac{2}{{x^2}}=1\)
\(\frac{1}{{x^2}}=\frac{1}{2}\)

So max value = \(\frac{3}{4}+\frac{1}{{2}}-\frac{1}{{2}}\)
\(\frac{3}{4}+\frac{1}{4}=1\)


III. Differentiation
However, this is not in scope of GMAT.
your algebra calculation seems wrong
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 28 Dec 2024, 18:13
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Thank you so much for this explanation. It makes so much sense and kinda also brings in practical implementation of the vertex formula.


JerryAtDreamScore
Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)

Breaking Down the Info:

We can set \(y = x^{-2}\) and rewrite the expression as \(\frac{3}{4} + y - y^2 = -y^2 + y + \frac{3}{4}\).

The "vertex" of this expression is \(y = -\frac{b}{2a} = \frac{1}{2}\). Then we may plug that in for an extreme value.

\(-\frac{1}{4} + \frac{1}{2} + \frac{3}{4} = 1\) is the maximum since the leading coefficient is negative, giving us a parabola opening downwards.

Answer: C
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 28 Dec 2024, 18:57
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chetan2u
Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)


I. Options
Now, \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}>\frac{3}{4}\)
Only C is greater than 3/4.

II. Algebraic
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)

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

Now for maximum value, we have to maximise \( \frac{1}{{x^2}}(1-\frac{1}{{x^2}})\)
This will happen when we multiply two equal terms as we are looking at two terms adding up to 1.
This, \(\frac{1}{{x^2}}=(1-\frac{1}{{x^2}})\)
\(\frac{2}{{x^2}}=1\)
\(\frac{1}{{x^2}}=\frac{1}{2}\)

So max value = \(\frac{3}{4}+\frac{1}{{2}}-\frac{1}{{2}}\)
\(\frac{3}{4}+\frac{1}{4}=1\)


III. Differentiation
However, this is not in scope of GMAT.
your algebra calculation seems wrong
There is nothing wrong in calculations. There was a typo, * sign was written as -
Thanks
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 28 Feb 2025, 08:07
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“This will happen when we multiply two equal terms”. Could you kindly elaborate on this a little for me? Thank you in advance.

chetan2u
Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)


I. Options
Now, \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}>\frac{3}{4}\)
Only C is greater than 3/4.

II. Algebraic
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)

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

Now for maximum value, we have to maximise \( \frac{1}{{x^2}}(1-\frac{1}{{x^2}})\)
This will happen when we multiply two equal terms as we are looking at two terms adding up to 1.
This, \(\frac{1}{{x^2}}=(1-\frac{1}{{x^2}})\)
\(\frac{2}{{x^2}}=1\)
\(\frac{1}{{x^2}}=\frac{1}{2}\)

So max value = \(\frac{3}{4}+\frac{1}{{2}}*\frac{1}{{2}}\)
\(\frac{3}{4}+\frac{1}{4}=1\)


III. Differentiation
However, this is not in scope of GMAT.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 01 Mar 2025, 10:39
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StacyArko
“This will happen when we multiply two equal terms”. Could you kindly elaborate on this a little for me? Thank you in advance.

chetan2u
Bunuel
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. \(\frac{1}{{\sqrt{2}}}\)


I. Options
Now, \(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}>\frac{3}{4}\)
Only C is greater than 3/4.

II. Algebraic
\(\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}\)

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

Now for maximum value, we have to maximise
This will happen when we multiply two equal terms as we are looking at two terms adding up to 1.
This, \(\frac{1}{{x^2}}=(1-\frac{1}{{x^2}})\)
\(\frac{2}{{x^2}}=1\)
\(\frac{1}{{x^2}}=\frac{1}{2}\)

So max value = \(\frac{3}{4}+\frac{1}{{2}}*\frac{1}{{2}}\)
\(\frac{3}{4}+\frac{1}{4}=1\)


III. Differentiation
However, this is not in scope of GMAT.
If \(a + b = 1\)

The value of \(a * b\) is maximum when \(a = b\) => \(a = b = 0.5\)

Now there's a mathematical way to prove this, but if you want to understand intuitively why this works then think if \(a\) is \(0.9\) and \(b\) is \(0.1\), \(a*b = 0.09\); here \(a*b\) is small because \(a\) and \(b\) are far apart, the closer they get their products value will increase and will be maximum when \(a = b\)

In the given question, \(\frac{1}{{x^2}}\) + \((1-\frac{1}{{x^2}})\) = \(1\)

Value of \(\frac{1}{{x^2}}*(1-\frac{1}{{x^2}})\) will be maximum when, \(\frac{1}{{x^2}}\) = \((1-\frac{1}{{x^2}})\)

Hope it helps.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4? 02 Mar 2025, 04:52
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This is very helpful. Thank you.

[quote="Krunaal"][quote="StacyArko"]“This will happen when we multiply two equal terms”. Could you kindly elaborate on this a little for me? Thank you in advance.
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