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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Differentiation แพ็กฝึกเฉพาะหัวข้อ · จุดนิ่งและโจทย์ประยุกต์

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Stationary points & applied max/min, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Given that $y = 2x^3 - 5x^2 + 4x - 1$, find $\dfrac{dy}{dx}$.

Mark schemeshow ▾
M1 apply the power rule $x^n \to nx^{n-1}$ to each term  A1 at least two terms correct, e.g. $6x^2$ and $-10x$  A1 $\dfrac{dy}{dx}=6x^2-10x+4$.
Q2. 4 marks

The curve $C$ has equation $y = 3x^2 + \dfrac{16}{x}$, $x \ne 0$. Find $\dfrac{dy}{dx}$ and hence the gradient of $C$ at the point where $x=2$.

Mark schemeshow ▾
M1 write $\dfrac{16}{x}=16x^{-1}$ before differentiating  A1 $\dfrac{dy}{dx}=6x-16x^{-2}$  M1 substitute $x=2$: $12-\dfrac{16}{4}$  A1 gradient $=8$.
Q3. 4 marks

$f(x) = 6\sqrt{x} - \dfrac{4}{x}$, $x > 0$. Find $f'(x)$, and evaluate $f'(4)$.

Mark schemeshow ▾
M1 rewrite as $f(x)=6x^{1/2}-4x^{-1}$  A1 $f'(x)=3x^{-1/2}+4x^{-2}$  M1 substitute $x=4$: $\dfrac{3}{2}+\dfrac{4}{16}$  A1 $f'(4)=\dfrac{7}{4}$.
Q4. 5 marks

Find the equation of the tangent to the curve $y = x^3 - 4x + 2$ at the point where $x = 2$. Give your answer in the form $y = mx + c$.

Mark schemeshow ▾
M1 $\dfrac{dy}{dx}=3x^2-4$  A1 gradient at $x=2$ is $3(4)-4=8$  B1 $y=8-8+2=2$, so the point is $(2,2)$  M1 $y-2=8(x-2)$  A1 $y=8x-14$.
Q5. 5 marks

The point $P(1,-2)$ lies on the curve $y = x^2 - 6x + 3$. Find the equation of the normal to the curve at $P$, giving your answer in the form $ax + by + c = 0$ where $a$, $b$, $c$ are integers.

Mark schemeshow ▾
M1 $\dfrac{dy}{dx}=2x-6$  A1 tangent gradient at $P$ is $-4$  M1 normal gradient $=-\dfrac{1}{-4}=\dfrac14$  M1 $y+2=\dfrac14(x-1)$  A1 $x-4y-9=0$ (o.e. with integer coefficients).
Q6. 6 marks

Find the coordinates of the stationary points of the curve $y = 2x^3 - 9x^2 + 12x + 1$, and use the second derivative to determine the nature of each.

Mark schemeshow ▾
M1 $\dfrac{dy}{dx}=6x^2-18x+12$  M1 set $=0$: $6(x-1)(x-2)=0$, so $x=1,\ 2$  A1 $(1,6)$  A1 $(2,5)$  M1 $\dfrac{d^2y}{dx^2}=12x-18$, evaluated at both $x$ values  A1 at $x=1$: $-6<0$ → $(1,6)$ maximum; at $x=2$: $6>0$ → $(2,5)$ minimum.
Q7. 6 marks

$f(x) = x^3 + 3x^2 - 9x + 2$. (a) Find $f'(x)$, giving your answer in factorised form. (b) Find the set of values of $x$ for which $f$ is decreasing. (c) State the set of values of $x$ for which $f$ is increasing.

Mark schemeshow ▾
(a) M1 $f'(x)=3x^2+6x-9$  A1 $=3(x+3)(x-1)$.
(b) M1 decreasing where $f'(x)<0$, i.e. between the roots $-3$ and $1$  A1 $-3<x<1$.
(c) A1 $x<-3$  A1 $x>1$.
Q8. 7 marks

The curve $C$ has equation $y = x + \dfrac{16}{x}$, $x > 0$. (a) Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$. (b) Find the coordinates of the stationary point of $C$. (c) Determine the nature of this stationary point.

Mark schemeshow ▾
(a) M1 write $y=x+16x^{-1}$ and differentiate  A1 $\dfrac{dy}{dx}=1-16x^{-2}$  A1 $\dfrac{d^2y}{dx^2}=32x^{-3}$.
(b) M1 $1-\dfrac{16}{x^2}=0 \Rightarrow x^2=16$  A1 $x=4$ (reject $x=-4$ since $x>0$)  A1 $y=4+4=8$, point $(4,8)$.
(c) B1 $\dfrac{d^2y}{dx^2}=\dfrac{32}{64}=\dfrac12>0$ at $x=4$ → minimum.
Q9. 7 marks

An open-topped box has a square base of side $x$ cm and height $h$ cm. The volume of the box is $500$ cm³. (a) Show that the total external surface area, $S$ cm², is given by $S = x^2 + \dfrac{2000}{x}$. (b) Use calculus to find the value of $x$ for which $S$ is a minimum, and find this minimum value of $S$.

Mark schemeshow ▾
(a) B1 $x^2h=500$, so $h=\dfrac{500}{x^2}$  M1 $S=x^2+4xh=x^2+4x\cdot\dfrac{500}{x^2}$  A1 $S=x^2+\dfrac{2000}{x}$ ∎.
(b) M1 $\dfrac{dS}{dx}=2x-2000x^{-2}$  M1 set $=0$: $x^3=1000$  A1 $x=10$  A1 $S_{\min}=100+200=300$ cm².
Q10. 8 marks

A closed cylindrical can has base radius $r$ cm and height $h$ cm. The volume of the can is $250\pi$ cm³. (a) Show that the total surface area, $A$ cm², is given by $A = 2\pi r^2 + \dfrac{500\pi}{r}$. (b) Use calculus to find the value of $r$ for which $A$ is stationary. (c) Show that this value of $r$ gives a minimum, and find the minimum surface area.

Mark schemeshow ▾
(a) M1 $\pi r^2 h=250\pi$, so $h=\dfrac{250}{r^2}$  A1 $A=2\pi r^2+2\pi r h=2\pi r^2+2\pi r\cdot\dfrac{250}{r^2}=2\pi r^2+\dfrac{500\pi}{r}$ ∎.
(b) M1 differentiate: $\dfrac{dA}{dr}=4\pi r-500\pi r^{-2}$  A1 correct  M1 set $=0$: $4\pi r=\dfrac{500\pi}{r^2} \Rightarrow r^3=125$  A1 $r=5$.
(c) B1 $\dfrac{d^2A}{dr^2}=4\pi+\dfrac{1000\pi}{r^3}=12\pi>0$ at $r=5$ → minimum  A1 $A_{\min}=50\pi+100\pi=150\pi$ cm².