| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Implicit differentiation |
| Difficulty | Standard +0.3 This is a straightforward integration problem requiring two successive integrations of a simple power function, using the minimum condition to find one constant and an arithmetic progression condition to find another. While it has multiple parts (11 marks total), each step follows standard A-level techniques without requiring novel insight or complex manipulation. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| 11(i) | f ' ( x )=[ ( 4x+1 )1/2 ÷½ ] [ ÷4] (+c ) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | B1 FT | Expect ½ ( 4x+1 )1/2 − 3 . FT on theirf′ ( x )=k ( 4x+1 )1/2 +c. (i.e. c = −3k) |
| Answer | Marks |
|---|---|
| Total: | 3 |
| Answer | Marks |
|---|---|
| 11(ii) | ( )=1 |
| f ″ 0 SOI | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| f ' 0 SOI | B1 FT | Substitute x = 0 into their f ′(x) but must not involve c otherwise B0B0 |
| f(0) = – 3 | B1 FT | FT for 3 terms in AP. FT for 3rd B1 dep on 1st B1. Award marks for the AP |
| Answer | Marks |
|---|---|
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 11(iii) | f ( x )=½ ( 4x+1 )3/2 ÷3/2÷4 ] − [ 1½x (+k ) | |
| | B1 FT | |
| B1 FT | Expect (1/12) ( 4x+1 )3/2 −1½x (+k ) . FT from their f ′(x) but c numerical. | |
| −3=1/12−0+k ⇒ k =−37/12 CAO | M1A1 | Sub x=0, y=their f ( 0 ) into their f(x). Dep on cx & k present (c numerical) |
| Answer | Marks |
|---|---|
| 12 12 6 | A1 |
| Total: | 5 |
Question 11:
--- 11(i) ---
11(i) | f ' ( x )=[ ( 4x+1 )1/2 ÷½ ] [ ÷4] (+c ) | B1 B1 | Expect ½ ( 4x+1 )1/2 (+c )
f ' ( 2 )=0 ⇒ 3 +c=0 ⇒ c=− 3 (Sufficient)
2 2 | B1 FT | Expect ½ ( 4x+1 )1/2 − 3 . FT on theirf′ ( x )=k ( 4x+1 )1/2 +c. (i.e. c = −3k)
2
Total: | 3
--- 11(ii) ---
11(ii) | ( )=1
f ″ 0 SOI | B1
( )=1/2−1½=−1
f ' 0 SOI | B1 FT | Substitute x = 0 into their f ′(x) but must not involve c otherwise B0B0
f(0) = – 3 | B1 FT | FT for 3 terms in AP. FT for 3rd B1 dep on 1st B1. Award marks for the AP
method only.
Total: | 3
--- 11(iii) ---
11(iii) | f ( x )=½ ( 4x+1 )3/2 ÷3/2÷4 ] − [ 1½x (+k )
| B1 FT
B1 FT | Expect (1/12) ( 4x+1 )3/2 −1½x (+k ) . FT from their f ′(x) but c numerical.
−3=1/12−0+k ⇒ k =−37/12 CAO | M1A1 | Sub x=0, y=their f ( 0 ) into their f(x). Dep on cx & k present (c numerical)
27 37 23
Minimum value = f(2) = −3− =− or−3.83
12 12 6 | A1
Total: | 5The function f is defined for $x \geqslant 0$. It is given that f has a minimum value when $x = 2$ and that $\text{f}''(x) = (4x + 1)^{-\frac{1}{2}}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\text{f}'(x)$. [3]
\end{enumerate}
It is now given that $\text{f}''(0)$, $\text{f}'(0)$ and $\text{f}(0)$ are the first three terms respectively of an arithmetic progression.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $\text{f}(0)$. [3]
\item Find $\text{f}(x)$, and hence find the minimum value of f. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q11 [11]}}