Question

Math, Please give an detailed answer.

Answer 1

Solution :

It is given that -

1/a + 1/b € I

0 < 1/a < 1/b

Assume the following lemma initially -

For any integer k ;

k² > 0

We have to find here which is not greater than 0

Option 1 -

> [ 1/a + 1/b ]²

> 1/a² + 1/b² + 2/ab

1/a² > 0, 1/b² > 0 , according to the lemma , and 1/ab also > 0

So , option 1 is always > 0

Option 2 -

> [ 1/a - 1/b ]²

It is mentioned that 1/a < 1/b

So 1/a - 1/b will be < 0 , but squaring will make it positive

Option 3 :

> b² - a²

If 1/a > 0

, a > 0

1/b > 0

> b > 0

But , 1/b > 1/a

So a < b

Hence , b² - a² < 0

The correct answer is option 3

As for option 4, as it is addition of cubes it is greater than 0

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Answer : Option (3), b² - a²

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Answer 2

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Given:

• a and b are integers

• a > b > 1

To find:

• Which of the given answer choices cannot be a multiple of either ’a’ or ‘b’

Approach and Working:

• For a number to be a multiple of ‘a’, it must be greater than ‘a’

• Similarly, for a number to be a multiple of ‘b’, it must be greater than ‘b’

• a > b > 1, and a, b are integers, implies that a and b are positive integers greater than 1.

• a – 1 is less than a, but it can be greater than b.

o Thus, it can be a multiple of b• b + 1 is always greater than b.

o Thus, it can be a multiple of b• b – 1 is less than both, a and b.

o Thus, it cannot be a multiple of both a and b• a + b is greater than both a and b.

o Thus, it can be a multiple of both a and b• ab is greater than both a and b.

o Thus, it can be a multiple of both a and b

Hence, the correct answer is option C