What is the maximum value of 3/4 + 1/x^2 - 1/x^4?

Question

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}}}

JerryAtDreamScore · Accepted Answer

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

chetan2u · Answer

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.

Gokha · Answer

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.

happy1900 · Answer

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.

Vidhi12 · Answer

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.

onlymalapink · Answer

your algebra calculation seems wrong

chetan2u · Answer

There is nothing wrong in calculations. There was a typo, * sign was written as -
Thanks

StacyArko · Answer

“This will happen when we multiply two equal terms”. Could you kindly elaborate on this a little for me? Thank you in advance.

Krunaal · Answer

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.

princevora · Answer

x can not be zero. put x=1 to get the value 3/4. putting x=0 makes the term undefined.

What is the maximum value of 3/4 + 1/x^2 - 1/x^4?

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What is the maximum value of 3/4 + 1/x^2 - 1/x^4?

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