2015 THREE(a)

Achievement
Question
For what value(s) of xx does the tangent to the graph of the function f(x)=5ln(2x3)f(x)=5\ln(2x-3) have a
gradient of 44?

*You must use calculus and show any derivatives that you need to find when solving this
problem.*
Official Answer
f(x)=52x3×2=102x3f'(x)=\frac{5}{2x-3}\times 2=\frac{10}{2x-3}
102x3=4\frac{10}{2x-3}=4
8x12=108x-12=10
x=2.75x=2.75
Grading Criteria

Achievement (u)

  • Correct solution with correct derivative.

Merit (r)

-

Excellence T1

-

Excellence T2

-