Question

Let $$f(x) = \frac{x^2}{x^2 - 1}.$$find the largest integer $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$

Answer 1

The question asks: "Let
`f(x) = ( x^(2) )/( x^(2) - 1)`. Find the largest integer n so that f(2) · f(3) · f(4) · ... · f(n-1) · f(n) < 1.98"

The answer is n = 98

Step-by-step explanation:

First thing, consider that the function can be written as:

`( x^(2) )/( x^(2) - 1) = (x \cdot x)/((x - 1)(x + 1))`

Now, let's expand the product, substituting the function with its equation for the requested values:

`(2 \cdot 2)/((1)(3)) \cdot (3 \cdot 3)/((2)(4)) \cdot (4 \cdot 4)/((3)(5)) \cdot ... \cdot ((n-1) \cdot (n-1))/((n-2))(n)) \cdot (n \cdot n)/((n - 1)(n + 1)) \ \textless \ 1.98`

As you can see, the intermediate terms cancel out with each other, leaving us with:

`(2 \cdot n)/(1 \cdot (n+1)) \ \textless \ 1.98`

This is a simple inequality that can be easily solved:

`(2n)/(n+1) \ \textless (198)/(100)`

200n < 198(n + 1)
200n < 198n + 198
2n < 198
n < 99

Hence, the greatest integer n < 99 (extremity excluded) is 98.