| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: general ax²+bx form |
| Difficulty | Standard +0.8 Part (a) is a standard first-principles differentiation of a quadratic, which is routine but requires careful algebraic manipulation (around 0.0 difficulty). Part (b) significantly elevates the difficulty: students must use their derivative result, apply given conditions about gradient and area to form simultaneous equations in a and b, integrate to find the area, then find the tangent line equation and its x-intercept. This multi-step problem requiring integration of several techniques (differentiation, integration, tangent lines, simultaneous equations) with abstract parameters makes it notably harder than average. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x+h) = a(x+h)^2 + b(x+h)\) \(= a(x^2+2xh+h^2)+b(x+h)\) | M1 | Considers \(f(x+h)\) and attempts to expand bracket squared |
| \(f(x+h)-f(x) = (ax^2+2ahx+ah^2+bx+bh)-(ax^2+bx) = 2xah+ah^2+bh\) | A1 | Correct simplified expression for \(f(x+h)-f(x)\) |
| \(\dfrac{f(x+h)-f(x)}{h} = 2ax+ah+b\) | A1 | Correct simplified expression |
| \(f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} = 2ax+b\) | A1 | cao – must be explicit that the limit (and not simply \(h=0\)) is considered |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int(ax^2+bx)\,dx = \frac{1}{3}ax^3+\frac{1}{2}bx^2\,(+c)\) | M1* | Attempt to integrate (with at least one term correct) |
| \(\int_1^4(ax^2+bx)\,dx = \left(\frac{64}{3}a+8b\right)-\left(\frac{1}{3}a+\frac{1}{2}b\right)\left(=21a+\frac{15}{2}b\right)\) | M1dep* | Correct use of limits \(x=1\) and \(x=4\) in their integrated expression (need not be simplified) |
| \(21a+\dfrac{15}{2}b=9\) | M1 | Dependent on both previous M marks – setting up an equation in \(a\) and \(b\) using the area of shaded region |
| \((f'(4)=)8a+b=-0.75\) | B1 | Correct equation in \(a\) and \(b\) |
| \(a=-0.375, b=2.25\) | A1 | BC (oe) |
| \(y=-0.375x^2+2.25x\) with \(x=4\) gives \(y=3\); Equation of tangent: \(y-3=-0.75(x-4)\) | M1 | Sets up the equation of the tangent at \(x=4\) using \(4\), \(-0.75\) and their \(y\) value at \(x=4\) (dependent on all previous M marks) or for \(-\dfrac{\text{their }y}{k-4}=-0.75\) |
| \(0-3=-0.75(k-4)\Rightarrow k=8\) | A1 | cao |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x+h) = a(x+h)^2 + b(x+h)$ $= a(x^2+2xh+h^2)+b(x+h)$ | M1 | Considers $f(x+h)$ and attempts to expand bracket squared |
| $f(x+h)-f(x) = (ax^2+2ahx+ah^2+bx+bh)-(ax^2+bx) = 2xah+ah^2+bh$ | A1 | Correct simplified expression for $f(x+h)-f(x)$ |
| $\dfrac{f(x+h)-f(x)}{h} = 2ax+ah+b$ | A1 | Correct simplified expression |
| $f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} = 2ax+b$ | A1 | cao – must be explicit that the limit (and not simply $h=0$) is considered |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int(ax^2+bx)\,dx = \frac{1}{3}ax^3+\frac{1}{2}bx^2\,(+c)$ | M1* | Attempt to integrate (with at least one term correct) |
| $\int_1^4(ax^2+bx)\,dx = \left(\frac{64}{3}a+8b\right)-\left(\frac{1}{3}a+\frac{1}{2}b\right)\left(=21a+\frac{15}{2}b\right)$ | M1dep* | Correct use of limits $x=1$ and $x=4$ in their integrated expression (need not be simplified) |
| $21a+\dfrac{15}{2}b=9$ | M1 | Dependent on both previous M marks – setting up an equation in $a$ and $b$ using the area of shaded region |
| $(f'(4)=)8a+b=-0.75$ | B1 | Correct equation in $a$ and $b$ |
| $a=-0.375, b=2.25$ | A1 | **BC** (oe) |
| $y=-0.375x^2+2.25x$ with $x=4$ gives $y=3$; Equation of tangent: $y-3=-0.75(x-4)$ | M1 | Sets up the equation of the tangent at $x=4$ using $4$, $-0.75$ and their $y$ value at $x=4$ (dependent on all previous M marks) or for $-\dfrac{\text{their }y}{k-4}=-0.75$ |
| $0-3=-0.75(k-4)\Rightarrow k=8$ | A1 | cao |
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\begin{enumerate}[label=(\alph*)]
\item The quadratic polynomial $a x ^ { 2 } + b x$, where $a$ and $b$ are constants, is denoted by $\mathrm { f } ( x )$.\\
Use differentiation from first principles to determine, in terms of $a , b$ and $x$, an expression for $\mathrm { f } ^ { \prime } ( x )$.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-07_565_1043_516_317}
$$y = a x ^ { 2 } + b x$$
The diagram shows the quadratic curve $y = a x ^ { 2 } + b x$, where $a$ and $b$ are constants. The shaded region is enclosed by the curve, the $x$-axis and the lines $x = 1$ and $x = 4$.
The tangent to the curve at $x = 4$ intersects the $x$-axis at the point with coordinates $( k , 0 )$.\\
Given that the area of the shaded region is 9 units ${ } ^ { 2 }$, and the gradient of this tangent is $- \frac { 3 } { 4 }$, determine the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q8 [11]}}