| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding stationary points after integration |
| Difficulty | Moderate -0.3 This question requires integration by substitution (or recognition of a standard form) and finding where f'(x) < 0, both routine A-level techniques. Part (i) involves solving a simple inequality with a square root, and part (ii) is straightforward integration with a constant of integration determined by a boundary condition. While it requires multiple steps, each step is standard and the question follows a familiar pattern for P1 level. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2x-1)^{\frac{1}{2}} < 2\) or \(3(2x-1)^{\frac{1}{2}} < 6\) | M1 | SOI |
| \(2x - 1 < 4\) | A1 | SOI |
| \(\frac{1}{2} < x < \frac{5}{2}\) | A1 A1 | Allow 2 separate statements |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = \left[3(2x-1)^{3/2} \div \left(\frac{3}{2}\right) \div (2)\right]\ [-6x]\ (+c)\) | B1 B1 | |
| Substitute \(x = 1,\ y = -3\) into an integrated expression | M1 | Dependent on \(c\) being present \((c = 2)\) |
| \(f(x) = (2x-1)^{\frac{3}{2}} - 6x + 2\) | A1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2x-1)^{\frac{1}{2}} < 2$ or $3(2x-1)^{\frac{1}{2}} < 6$ | M1 | SOI |
| $2x - 1 < 4$ | A1 | SOI |
| $\frac{1}{2} < x < \frac{5}{2}$ | A1 A1 | Allow 2 separate statements |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = \left[3(2x-1)^{3/2} \div \left(\frac{3}{2}\right) \div (2)\right]\ [-6x]\ (+c)$ | B1 B1 | |
| Substitute $x = 1,\ y = -3$ into an integrated expression | M1 | Dependent on $c$ being present $(c = 2)$ |
| $f(x) = (2x-1)^{\frac{3}{2}} - 6x + 2$ | A1 | |8 A function f is defined for $x > \frac { 1 } { 2 }$ and is such that $\mathrm { f } ^ { \prime } ( x ) = 3 ( 2 x - 1 ) ^ { \frac { 1 } { 2 } } - 6$.\\
(i) Find the set of values of $x$ for which f is decreasing.\\
(ii) It is now given that $\mathrm { f } ( 1 ) = - 3$. Find $\mathrm { f } ( x )$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q8 [8]}}