Inequality question (1 Viewer)

[Run hard@thehsc]

Pls help with this question... The previous part is attached in the second image below

 

[cossine]

So ii)

In this case there are two possibilities, x^p-1 is positive. If x^p-1 is positive then x^q-1 must be positive otherwise (x^p-1)/(x^q-1) will be negative. Similarly if x^p-1 is negative then x^q-1 is also negative.

Does this this help.

 

[Run hard@thehsc]

@cossine it does give an idea thanks

 

[idkkdi]

Pls help with this question... The previous part is attached in the second image belowView attachment 34629

1. (0,1)
Considering the boundaries, 0^k = 0, 1^k = 1.
Since f(x) is monotonically increasing between (0,1), x^k - 1 is negative where kEQ+.
Therefore, x^p-1/x^q-1 >0.

2. (1, inf)
Considering the boundaries, 1^k = 1, and as x-> infinity, x^k -> infinity.
Since f(x) is monotonically increasing between (1,inf), x^k - 1 is pos where kEQ+.
Therefore, x^p-1/x^q-1 >0.

 

[5uckerberg]

1. (0,1)
Considering the boundaries, 0^k = 0, 1^k = 1.
Since f(x) is monotonically increasing between (0,1), x^k - 1 is negative where kEQ+.
Therefore, x^p-1/x^q-1 >0.

2. (1, inf)
Considering the boundaries, 1^k = 1, and as x-> infinity, x^k -> infinity.
Since f(x) is monotonically increasing between (1,inf), x^k - 1 is pos where kEQ+.
Therefore, x^p-1/x^q-1 >0.

The first case is simply just two negative numbers divided by each another thus leading to a positive number.

 

[idkkdi]

The first case is simply just two negative numbers divided by each another thus leading to a positive number.

yep