Differentiation (first principles, tangents, normals, stationary points) – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

Given that $$y=4x^{3}-\frac{6}{x^{2}}+5\sqrt{x},\qquad x>0,$$ find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, giving each term in its simplest form. [3]

for $x^{n}\to x^{n-1}$ for at least one term (after writing as powers of $x$)M1
two terms correctA1
$12x^{2}+12x^{-3}+\tfrac52x^{-\frac12}$ all correctA1
Answer: $12x^{2}+\dfrac{12}{x^{3}}+\dfrac{5}{2\sqrt{x}}$
Examiner tip: Rewrite $-\frac{6}{x^{2}}$ as $-6x^{-2}$ first; its derivative is $+12x^{-3}$ – the two negatives make a positive.

2StandardCalculator4 marks

$$f(x)=x^{3}-5x$$

Prove, from first principles, that $f'(x)=3x^{2}-5$. [4]

$f(x+h)=(x+h)^{3}-5(x+h)=x^{3}+3x^{2}h+3xh^{2}+h^{3}-5x-5h$ (correct expansion seen anywhere; allow $\delta x$ for $h$)B1
for forming the gradient of the chord $\dfrac{f(x+h)-f(x)}{h}$ and cancelling the $x^{3}$ and $5x$ terms, e.g. $\dfrac{3x^{2}h+3xh^{2}+h^{3}-5h}{h}$M1
$=3x^{2}+3xh+h^{2}-5$A1
as $h\to0$, $3xh+h^{2}\to0$, so $f'(x)=\lim_{h\to0}\left(3x^{2}+3xh+h^{2}-5\right)=3x^{2}-5$ – correct limiting argument with conclusion; no errors seen (reject $h=0$ substituted with no limit stated)A1*
Answer: $\displaystyle f'(x)=\lim_{h\to0}\left(3x^{2}+3xh+h^{2}-5\right)=3x^{2}-5$
Examiner tip: The final mark needs the limit made explicit – write '$\lim_{h\to0}$' or 'as $h\to0$'; just setting $h=0$ or crossing out the $h$ terms isn't enough.

3StandardCalculator6 marks

The curve $C$ has equation $y=x^{3}-4x^{2}+5$. The point $P(3,-4)$ lies on $C$.

(a) Find an equation of the tangent to $C$ at $P$, giving your answer in the form $y=mx+c$. [4]

(b) Find an equation of the normal to $C$ at $P$, giving your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers. [2]

(a) for differentiating, $x^{n}\to x^{n-1}$M1
(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-8x$A1
(a) for substituting $x=3$ into their $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ (gradient $3$) and using it with $(3,-4)$ in a line equation, e.g. $y+4=3(x-3)$ or $y=3x+c$M1
(a) $y=3x-13$A1
(b) for using gradient $-\dfrac{1}{\text{their }m}$ (i.e. $-\tfrac13$) with $(3,-4)$, e.g. $y+4=-\tfrac13(x-3)$M1
(b) $x+3y+9=0$ (accept any integer multiple, e.g. $-x-3y-9=0$)A1
Answer: (a) $y=3x-13$ (b) $x+3y+9=0$
Examiner tip: Substitute $x=3$ into $\frac{\mathrm{d}y}{\mathrm{d}x}$, not into $y$, to get the gradient.

4StandardCalculator7 marks

The curve $C$ has equation $$y=2x^{3}-9x^{2}+12x+1.$$

(a) Find the coordinates of the stationary points of $C$. [4]

(b) Determine the nature of each stationary point, justifying your answer. [2]

(a) for differentiating: $6x^{2}-18x+12$M1
(a) for setting $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ and solving a 3TQ: $6(x-1)(x-2)=0$M1
(a) $x=1$ and $x=2$A1
(a) $(1,6)$ and $(2,5)$A1
(b) for finding $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=12x-18$ and evaluating at one of their $x$ values (or another valid test)M1
(b) $x=1$: $-6<0$ so maximum at $(1,6)$; $x=2$: $6>0$ so minimum at $(2,5)$A1
(c) $1B1ft
Answer: (a) $(1,6),\ (2,5)$ (b) $(1,6)$ maximum, $(2,5)$ minimum (c) $1
Examiner tip: Justify the nature: quote the sign of $\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$ ($-6<0$) and then name it – a bare 'max' gets no credit.

5HardCalculator9 marks

An open-topped box (it has no lid) is in the shape of a cuboid with a square base of side $x$ cm and height $h$ cm, as shown. The volume of the box is $32\,000\text{ cm}^{3}$.

(a) Show that the external surface area, $A\text{ cm}^{2}$, of the box is given by $$A=x^{2}+\frac{128\,000}{x}.$$ [3]

(b) Use calculus to find the value of $x$ for which $A$ is stationary, and find the corresponding value of $A$. [4]

(a) $x^{2}h=32000$B1
(a) for $A=x^{2}+4xh$ (base plus four sides, no lid) and substituting $h=\dfrac{32000}{x^{2}}$M1
(a) $A=x^{2}+\dfrac{128000}{x}$ with no errorsA1*
(b) for differentiating: $\dfrac{\mathrm{d}A}{\mathrm{d}x}=2x-128000x^{-2}$M1
(b) for setting their derivative $=0$ and solving to $x^{3}=k$M1
(b) $x=40$A1
(b) $A=4800\text{ cm}^{2}$A1
(c) for finding $\dfrac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}}=2+256000x^{-3}$ and substituting their $x$M1
(c) $=6>0$, so $A$ is a minimumA1
Answer: (b) $x=40$, $A=4800\text{ cm}^{2}$ (c) $\frac{\mathrm{d}^{2}A}{\mathrm{d}x^{2}}=6>0$ so minimum
Examiner tip: An open box has only one square face – writing $2x^{2}$ loses the method mark in (a).

6StandardCalculator5 marks

The curve $C$ has equation $y=ax^{2}+bx$, where $a$ and $b$ are constants.

The point $(2,2)$ lies on $C$, and the gradient of $C$ at this point is $5$.

Find the value of $a$ and the value of $b$. [5]

for substituting $(2,2)$: $4a+2b=2$M1
for differentiating to $2ax+b$ and setting $=5$ at $x=2$: $4a+b=5$M1
for solving their two equations simultaneouslyM1
$a=2$A1
$b=-3$A1
Answer: $a=2,\ b=-3$
Examiner tip: Two unknowns need two equations: one from the point on the curve, one from the gradient.

7HardCalculator6 marks

The curve $C$ has equation $y=x^{3}-2x^{2}+x$. The point $P$ on $C$ has $x$-coordinate 2.

The tangent to $C$ at $P$ meets $C$ again at the point $Q$.

Using algebra, find the coordinates of $Q$. [6]

for differentiating (at least one term $x^{n}\to x^{n-1}$), e.g. $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-4x+1$M1
for substituting $x=2$ into their $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ (gradient $5$) and into $y$ to find $P(2,2)$M1
tangent $y=5x-8$A1
for equating: $x^{3}-2x^{2}+x=5x-8\Rightarrow x^{3}-2x^{2}-4x+8=0$M1
for using $(x-2)$ (a repeated factor) to factorise: $(x-2)^{2}(x+2)=0$M1
$Q(-2,-18)$A1
Answer: $Q(-2,-18)$
Examiner tip: You already know $x=2$ is a root (a double root, since the line is a tangent) – use that to factorise the cubic quickly.

8StandardCalculator4 marks

$$f(x)=x^{3}+3x^{2}+6x-2$$

Prove that $f(x)$ is an increasing function for all real values of $x$. [4]

Differentiates: at least one term $x^{n}\to x^{n-1}$ correct (e.g. $3x^{2}$ or $6x$ seen)M1
$f'(x)=3x^{2}+6x+6$ (constant $-2$ correctly removed)A1
Attempts to show their 3-term quadratic $f'(x)$ has no real roots / a positive minimum: completing the square of the form $3(x\pm1)^{2}\pm k$ or $3[(x+1)^{2}\pm k]$, or evaluates $b^{2}-4ac$ for $f'(x)$ (or for $x^{2}+2x+2$), or solves $f''(x)=6x+6=0$ and finds $f'(-1)$M1
Correct work and full conclusion, e.g. $f'(x)=3(x+1)^{2}+3$ and $(x+1)^{2}\geqslant0\Rightarrow f'(x)\geqslant3>0$ for all $x$, so $f$ is increasing. Accept: $b^{2}-4ac=36-72=-36<0$ (or $4-8=-4<0$) and $f'(x)$ is a positive quadratic (e.g. $f'(0)=6>0$ or leading coefficient $3>0$) so $f'(x)>0$ for all $x$, hence increasing; or minimum of $f'$ is $f'(-1)=3>0$. Reject a discriminant argument with no reference to $f'$ being positive (only showing 'no roots').A1
Answer: $f'(x)=3(x+1)^{2}+3>0$ for all $x$
Examiner tip: End with the logic chain: $(x+1)^{2}\geqslant0\Rightarrow f'(x)\geqslant3>0\Rightarrow$ increasing.

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