| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find second derivative |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring conversion to index form (x^{1/2} and x^{-1/2}), applying power rule twice, then algebraic substitution to verify an identity. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation in part (iii). |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}\) | M1 A2 |
| Answer | Marks |
|---|---|
| \(\frac{d^2y}{dx^2} = -\frac{1}{4}x^{-\frac{3}{2}} - 3x^{-\frac{5}{2}}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| LHS \(= 4x^2(-\frac{1}{4}x^{-\frac{3}{2}} - 3x^{-\frac{5}{2}}) + 4x(\frac{1}{2}x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}) - (x^{\frac{1}{2}} - 4x^{-\frac{1}{2}})\) | M1 A1 | |
| \(= -x^{\frac{1}{2}} - 12x^{-\frac{1}{2}} + 2x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\) | M1 A1 | |
| \(= 0\) | A1 | (8) |
# Question 7:
## Part (i):
$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}$ | M1 A2 |
## Part (ii):
$\frac{d^2y}{dx^2} = -\frac{1}{4}x^{-\frac{3}{2}} - 3x^{-\frac{5}{2}}$ | M1 A1 |
## Part (iii):
LHS $= 4x^2(-\frac{1}{4}x^{-\frac{3}{2}} - 3x^{-\frac{5}{2}}) + 4x(\frac{1}{2}x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}) - (x^{\frac{1}{2}} - 4x^{-\frac{1}{2}})$ | M1 A1 |
$= -x^{\frac{1}{2}} - 12x^{-\frac{1}{2}} + 2x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}$ | M1 A1 |
$= 0$ | A1 | **(8)**
---\begin{enumerate}
\item Given that
\end{enumerate}
$$y = \sqrt { x } - \frac { 4 } { \sqrt { x } }$$
(i) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$,\\
(ii) find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$,\\
(iii) show that
$$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = 0 .$$
\hfill \mbox{\textit{OCR C1 Q7 [8]}}