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The function g is defined by g(x) = \frac{3\ln(x) - 7}{\ln(x) - 2} \quad x > 0 \quad x \neq k where k is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 2
Question 15
The function g is defined by g(x) = \frac{3\ln(x) - 7}{\ln(x) - 2} \quad x > 0 \quad x \neq k where k is a constant. (a) Deduce the value of k. (b) Prove that \(... show full transcript
Worked Solution & Example Answer:The function g is defined by g(x) = \frac{3\ln(x) - 7}{\ln(x) - 2} \quad x > 0 \quad x \neq k where k is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 2
Step 1
Deduce the value of k.
Answer
To find the value of k, we note that for the function g(x) to be well-defined, the denominator ( \ln(x) - 2 \neq 0 ), hence ( \ln(x) \neq 2 ) which means ( x \neq e^2 ). Therefore, we can set ( k = e^2 ). This gives us the condition that g(x) is defined for all x > 0 except x = e^2.
Step 2
Prove that g'(x) > 0 for all values of x in the domain of g.
Answer
To prove that ( g^{\prime}(x) > 0 ), we need to differentiate g(x) using the quotient rule:
Simplifying this:
- The numerator is: ( (\ln(x) - 2) \cdot \frac{3}{x} - (3\ln(x) - 7) \cdot \frac{1}{x} )
- Combine like terms:
- This leads to: ( \frac{3 \ln(x) - 6 - 3\ln(x) + 7}{x(\ln(x) - 2)^2} )
- Thus: ( g'(x) = \frac{1}{x(\ln(x) - 2)^2} )
Since x > 0 and ( \ln(x) - 2 > 0 ) when x > e^2, it follows that ( g'(x) > 0 ) in the defined domain.
Step 3