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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
08 May 2013, 22:28

emmak wrote:

If n is a prime number greater than 2, is 1/x > 1? (1) x^n < x < x^{\frac{1}{n}} (2)x^{n-1} > x^{2n-2}

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n. Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

In such a case, 1/x>1 or x belongs to {0,1} -ve values again will not satisfy the equation.

Statement 2: x^(n-1) > x^2(n-1), for n being odd, n-1 will be even. Hence for the above equation the solution set will be {-1,1} As mentioned above, the negative values of x will not hold the 1/x>1 equation. Hence the range is again {0,1}.

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... x^{n-1} > x^{2n-2} ...

x^{n-1} > x^{2n-2}

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...x^2 > x^4

& when n = 5 .. we will get x^4 > x^8 ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation: 1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
08 May 2013, 23:45

Expert's post

arpanpatnaik wrote:

emmak wrote:

If n is a prime number greater than 2, is 1/x > 1? (1) x^n < x < x^{\frac{1}{n}} (2)x^{n-1} > x^{2n-2}

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n. Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As x>2 and is a prime number, x will always be an odd integer. For x = -8,x^{1/3} = -2. _________________

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... x^{n-1} > x^{2n-2} ...

x^{n-1} > x^{2n-2}

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...x^2 > x^4

& when n = 5 .. we will get x^4 > x^8 ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation: 1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards, Arpan

My answer is still A ... _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Last edited by manishuol on 09 May 2013, 00:34, edited 2 times in total.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
09 May 2013, 00:06

vinaymimani wrote:

arpanpatnaik wrote:

emmak wrote:

If n is a prime number greater than 2, is 1/x > 1? (1) x^n < x < x^{\frac{1}{n}} (2)x^{n-1} > x^{2n-2}

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n. Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As n>2 and is a prime number, will n always be an odd integer. For x = -8,x^{1/3} = -2.

My Bad!! I missed that... Thanks for correcting me Vinay! I got the answer now! [C] it is Can't believe I used the odd principle in Statement 2 and failed to use it in Statement 1. The range x < -1 will be valid and the solution set needs to be an intersection of Statement 1 and 2 i.e. x belongs to {0,1}.

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