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

Key Idea: Collapse the tower using $a^{\log_a t} = t$ and reciprocal log identity.

Step 1: Domain: $\log_x 2012 > 0$ requires $$x > 1$$ .

Step 2: $f_3(x) = \left(\frac{1}{2013}\right)^{\log_{2013}(\log_x 2012)} = \frac{1}{\log_x 2012} = \log_{2012} x$

Step 3: $f_2(x) = 2012^{\log_{2012} x} = x$

Step 4: $$f_1(x) = 2^x$$ , $$x > 1$$ , so range is $(2,\infty)$

Answer: $(2,\infty)$

Trap: Domain is critical —

\log_x 2012

must also be positive (not just defined) because it is an argument of another log.

Key Concept: Nested exponent-log telescoping using

a^{\log_a t}=t

Question:

Step-by-Step Solution

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