3 Given that \(y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }\).
Show mark scheme
Show mark scheme source
Question 3:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \frac{1}{2}(\ln(2x-1) - \ln(2x+1))\) M1
Use of \(\ln(a/b) = \ln a - \ln b\)
M1 Use of \(\ln\sqrt{c} = \frac{1}{2}\ln c\)
\(\frac{dy}{dx} = \frac{1}{2}\!\left(\frac{2}{2x-1} - \frac{2}{2x+1}\right)\) A1
o.e.; correct expression (if this line of working is missing, M1M1A0A0)
\(= \frac{1}{2x-1} - \frac{1}{2x+1}\) A1
NB AG [4] For alternative methods, see additional solutions
Copy
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln\!\left(\sqrt{\frac{2x-1}{2x+1}}\right) = \frac{1}{2}(\ln(2x-1) - \ln(2x+1))$ | M1 | Use of $\ln(a/b) = \ln a - \ln b$ |
| | M1 | Use of $\ln\sqrt{c} = \frac{1}{2}\ln c$ |
| $\frac{dy}{dx} = \frac{1}{2}\!\left(\frac{2}{2x-1} - \frac{2}{2x+1}\right)$ | A1 | o.e.; correct expression (if this line of working is missing, M1M1A0A0) |
| $= \frac{1}{2x-1} - \frac{1}{2x+1}$ | A1 | **NB AG** |
| **[4]** | | For alternative methods, see additional solutions |
---
Show LaTeX source
Copy
3 Given that $y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [4]}}