Functions Question - None

Question

For the function

f(x) = \frac{e^{x} + 1}{e^{x} - 1}

, if

n(d)

denotes the number of integers which are not in its domain and

n(r)

denotes the number of integers which are not in its range, then

n(d) + n(r)

is equal to -

Accepted Answer

Step 1: Find n(d) - Integers not in the domain

The function f(x) = (e^x + 1)/(e^x - 1) is undefined when the denominator equals zero.

e^x - 1 = 0

e^x = 1

x = 0

Therefore, only x = 0 is excluded from the domain. Since 0 is an integer, n(d) = 1

Step 2: Find the range of f(x)

Let y = (e^x + 1)/(e^x - 1), where x ≠ 0

Solving for x in terms of y:

y(e^x - 1) = e^x + 1

ye^x - y = e^x + 1

ye^x - e^x = y + 1

e^x(y - 1) = y + 1

e^x = (y + 1)/(y - 1)

For e^x to be valid, we need e^x > 0, which means (y + 1)/(y - 1) > 0

This inequality holds when: y > 1 or y < -1

Step 3: Check behavior at boundaries

As x → 0^+: e^x → 1^+, so f(x) → +∞

As x → 0^-: e^x → 1^-, so f(x) → -∞

As x → +∞: e^x → ∞, so f(x) → 1^+

As x → -∞: e^x → 0, so f(x) → -1^-

Therefore, Range = (-∞, -1) ∪ (1, +∞)

Step 4: Find n(r) - Integers not in the range

Integers in the range must satisfy y < -1 or y > 1

Integers NOT in the range: {-1, 0, 1}

Therefore, n(r) = 3

Step 5: Calculate n(d) + n(r)

n(d) + n(r) = 1 + 3 = 4

∴ Answer: C

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