Standard +0.3 Part (i) requires applying the quotient rule to find dy/dx, setting it equal to 1/4, and solving a straightforward quadratic equation. Part (ii) involves splitting the integrand, integrating to get logarithms, and solving ln(2a/a) = ln 7. Both parts are standard applications of C3/C4 techniques with no novel insight required, making this slightly easier than average.
5. (i) Find the \(x\) coordinate of each point on the curve \(y = \frac { x } { x + 1 } , x \neq - 1\), at which the gradient is \(\frac { 1 } { 4 }\)
(ii) Given that
$$\int _ { a } ^ { 2 a } \frac { t + 1 } { t } \mathrm {~d} t = \ln 7 \quad a > 0$$
find the exact value of the constant \(a\).
\(\int\frac{t+1}{t}\,dt=(t+1)\ln t - \int \ln t\,dt\)
M1
\(=(t+1)\ln t - t\ln t + t\)
A1
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(x+1)-x}{(x+1)^2} = \frac{1}{(x+1)^2}$ or $y = 1 - \frac{1}{x+1} \Rightarrow \frac{dy}{dx} = \frac{1}{(x+1)^2}$ | M1 | Correct differentiation method |
| $\frac{1}{(x+1)^2} = \frac{1}{4}$ or $(x+1)^2 = 4$ or $x^2+2x+1=4$ | A1 | |
| $x = 1, -3$ | M1A1 | |
**(4 marks)**
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\frac{t+1}{t}\,dt = \int 1+\frac{1}{t}\,dt = t + \ln t \ (+c)$ | M1A1 | See notes for integration by parts method |
| $\left[t + \ln t\right]_a^{2a} = \ln 7 \Rightarrow 2a + \ln 2a - a - \ln a = \ln 7$ | | |
| $a + \ln\left(\frac{2a}{a}\right) = \ln 7 \Rightarrow a = \ln\left(\frac{7}{2}\right)$ or $a = \ln 7 - \ln 2$ | dM1A1 | Dependent on previous M1 |
**(4 marks)**
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct derivative e.g. $-(−1)(x+1)^{-2}$ (chain rule) | M1 | Correct use of quotient, product, implicit or chain rule – may not be simplified |
| $(x+1)^{-1} + (-1)x(x+1)^{-2}$ (product rule) | | |
| $y(x+1)=x$ gives $(x+1)\frac{dy}{dx}+y=1$ so $\frac{dy}{dx}=\frac{1-y}{x+1}$ or $\frac{1}{(x+1)^2}$ (implicit) | | |
| $\frac{dy}{dx}=\frac{1}{4}$ and proceeds to correct equation in $x$ only | A1 | |
| i.e. $\frac{1}{(x+1)^2}=\frac{1}{4}$ or $(x+1)^2=4$ or $x^2+2x+1=4$ | | |
| Solves quadratic to obtain two values for $x$ | M1 | By usual methods |
| Both correct answers for $x$, no working implies method | A1 | Ignore $y$ values if given |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes as sum of $1$ and $t^{-1}$, integrates giving one correct term e.g. $t + t^{-2}$ | M1 | May use parts (see variants below) |
| Both terms correct e.g. $t + \ln t$ (ignore arbitrary constant) | A1 | From parts may obtain $t+\ln t+1$ |
| Uses limits correctly, sets $I(2a)-I(a)=\ln 7$ and uses/states $\ln 2a - \ln a = \ln 2$ leading to $a=..$ | dM1 | Dependent on previous M |
| Correct answer e.g. $\ln 3.5$ | A1 | $\ln 7 - \ln 2$ is A1 but if followed by $\ln 5$ this is A0 |
### Parts in integration variants:
| Working | Mark |
|---|---|
| $\int\frac{t+1}{t}\,dt = \frac{1}{t}\left(\frac{t^2}{2}+t\right)-\int-\frac{1}{t^2}\left(\frac{t^2}{2}+t\right)dt = \frac{t}{2}+1+\int\frac{1}{2}+\frac{1}{t}\,dt$ | M1 |
| $=\frac{t}{2}+1+\frac{t}{2}+\ln t$ | A1 |
| $\int\frac{t+1}{t}\,dt=(t+1)\ln t - \int \ln t\,dt$ | M1 |
| $=(t+1)\ln t - t\ln t + t$ | A1 |
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5. (i) Find the $x$ coordinate of each point on the curve $y = \frac { x } { x + 1 } , x \neq - 1$, at which the gradient is $\frac { 1 } { 4 }$\\
(ii) Given that
$$\int _ { a } ^ { 2 a } \frac { t + 1 } { t } \mathrm {~d} t = \ln 7 \quad a > 0$$
find the exact value of the constant $a$.\\
\hfill \mbox{\textit{Edexcel C34 2015 Q5 [8]}}