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Senior Manager  D
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GMAT 1: 720 Q49 V40 GPA: 3.8
What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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Question Stats: 65% (01:52) correct 35% (02:06) wrong based on 94 sessions

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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. $$\frac{1}{{\sqrt{2}}}$$

+1 kudo if you think it is good question

Originally posted by chondro48 on 31 Jul 2019, 21:48.
Last edited by chondro48 on 31 Jul 2019, 23:04, edited 1 time in total.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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.
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GMAT 1: 720 Q49 V40 Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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1
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.

Posted from my mobile device
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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1
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

$$\frac{3}{4}+\frac{1}{{x^2}}-\frac{1}{{x^4}}$$
=$$\frac{3}{4}+\frac{1}{{x^2}} * \frac{(x^2-1)}{{x^2}}$$
= $$\frac{3}{4}+\frac{1}{{x^4}}$$ * (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.
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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 $$t^2 + 1-t^2 =1$$
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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EncounterGMAT 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 you think it is good question

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

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.

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.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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chetansood wrote:
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.

Posted from my mobile device

Please see above the correct solution based on calculus (Engineer's way).
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Joined: 27 May 2012
Posts: 945
Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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Hi Kinshook,

Why did you write Sum of $$t^2+1−t^2=1$$?
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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stne wrote:
Hi Kinshook,

Why did you write Sum of $$t^2+1−t^2=1$$?

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.
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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Kinshook wrote:
stne wrote:
Hi Kinshook,

Why did you write Sum of $$t^2+1−t^2=1$$?

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.
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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SSwami wrote:
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) should be -
= 3/4 + 1/2 - 1/4
= 1/2 +1/2
=1

Here we are lucky that a is +ve, otherwise the solution is invalid for -ve a.
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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.
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Re: What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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.
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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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.

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
SVP  D
Joined: 03 Jun 2019
Posts: 1877
Location: India
What is the maximum value of 3/4 + 1/x^2 - 1/x^4?  [#permalink]

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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

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 What is the maximum value of 3/4 + 1/x^2 - 1/x^4?   [#permalink] 06 Aug 2019, 06:13
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