The Maths Tailor

When a question asks for the greatest or least value, as a parameter varies, of something that is itself a derivative at a fixed point (a gradient, a normal slope, a stationary value, a distance to a stationary point), there are two nested layers. First differentiate in $x$ and substitute the fixed point to get the target as an expression in the parameter alone. Then optimise that expression over the parameter, usually by completing the square if it is quadratic, otherwise by differentiating with respect to the parameter.

Trigger: 'least/greatest value of the gradient (or normal slope, or distance) as $a$ (or $p$ , $q$ , $k$ ) varies'.

Instances:
(i) get the gradient at the fixed $x$ as a quadratic in the parameter, then complete the square to read off its minimum;
(ii) for a normal slope $- 1/ f^{'} (x_{0})$ , optimise the slope by optimising the single quadratic in the denominator;
(iii) differentiate the target with respect to the parameter and set that to zero when completing the square is awkward;
(iv) substitute $B$ for $b^{2}$ to turn a quartic-in- $b$ distance into a quadratic in $B$ before minimising.

The least possible value of the gradient of the curve

y = (2 x + a) (x - 2 a)^{2}

at the point where

x = 1

, as

a

varies, is

(A)

- \frac{49}{4}

(B)

- 8

(C)

- \frac{25}{4}

(D)

\frac{7}{4}

(E)

\frac{47}{16}

Expand:

y = (2 x + a) (x - 2 a)^{2} = (2 x + a) (x^{2} - 4 a x + 4 a^{2}) = 2 x^{3} - 8 a x^{2} + 8 a^{2} x + a x^{2} - 4 a^{2} x + 4 a^{3} = 2 x^{3} - 7 a x^{2} + 4 a^{2} x + 4 a^{3}

. Differentiate:

\frac{d y}{d x} = 6 x^{2} - 14 a x + 4 a^{2}

. At

x = 1

: gradient

= 6 - 14 a + 4 a^{2}

. Minimise over

a

by completing the square:

4 a^{2} - 14 a + 6 = 4 (a^{2} - \frac{7}{2} a) + 6 = 4 (a - \frac{7}{4})^{2} - \frac{49}{4} + 6 = 4 (a - \frac{7}{4})^{2} - \frac{25}{4}

. The minimum value is

- \frac{25}{4}

, attained at

a = \frac{7}{4}

. Hence the answer is C.

A curve

C

has equation

y = f (x)

where

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

and

p

is real. The gradient of the normal to the curve

C

at the point where

x = - 1

M

. What is the greatest possible value of

M

p

varies?

(A)

- \frac{3}{2}

(B)

- \frac{2}{3}

(C)

- \frac{1}{2}

(D)

\frac{1}{4}

(E)

\frac{2}{3}

(F)

\frac{3}{2}

Differentiate (treating

p

as a constant):

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

. At

x = - 1

f^{'} (- 1) = - 6 p^{2} - 6 p - 3 = - 3 (2 p^{2} + 2 p + 1) .

The gradient of the normal is:

M = \frac{- 1}{f ^{'} ( - 1 )} = \frac{1}{3 ( 2 p ^{2} + 2 p + 1 )} .

To maximise

M

, minimise

2 p^{2} + 2 p + 1

. Completing the square:

2 p^{2} + 2 p + 1 = 2 (p + \frac{1}{2})^{2} + \frac{1}{2} .

The minimum is

\frac{1}{2}

p = - \frac{1}{2}

, giving:

M_{m a x} = \frac{1}{3 \times \frac{1}{2}} = \frac{2}{3} .

Hence the answer is E.

The curve

C

has equation

y = x^{2} + b x + 2

, where

b \geq 0

. Find the value of

b

that minimises the distance between the origin and the stationary point of the curve

C

.

(A)

b = 0

(B)

b = 1

(C)

b = 2

(D)

b = \frac{6}{2}

(E)

b = 2

(F)

b = 6

The stationary point of

y = x^{2} + b x + 2

is where

\frac{d y}{d x} = 2 x + b = 0

, giving

x = - \frac{b}{2}

and

y = 2 - \frac{b ^{2}}{4}

. The squared distance from the origin is:

D^{2} = (- \frac{b}{2})^{2} + (2 - \frac{b ^{2}}{4})^{2} = \frac{b ^{2}}{4} + 4 - b^{2} + \frac{b ^{4}}{16} = \frac{b ^{4}}{16} - \frac{3 b ^{2}}{4} + 4.

Let

B = b^{2} \geq 0

: we minimise

h (B) = \frac{B ^{2}}{16} - \frac{3 B}{4} + 4

. Differentiating:

h^{'} (B) = \frac{B}{8} - \frac{3}{4} = 0 ⟹ B = 6

. So

b^{2} = 6

, giving

b = 6

(since

b \geq 0

). Hence the answer is F.

A curve has equation

y = (2 q - x^{2}) (2 q x + 3)

The gradient of the curve at

x = - 1

is a function of

q

. Find the value of

q

which minimises the gradient of the curve at

x = - 1

.

(A)

- 1

(B)

- \frac{3}{4}

(C)

- \frac{1}{2}

(D)

0

(E)

\frac{1}{2}

(F)

\frac{3}{4}

(G)

1

Expand:

y = (2 q - x^{2}) (2 q x + 3) = 4 q^{2} x + 6 q - 2 q x^{3} - 3 x^{2}

. Differentiating:

\frac{d y}{d x} = 4 q^{2} - 6 q x^{2} - 6 x

. At

x = - 1

: gradient

= 4 q^{2} - 6 q (- 1)^{2} - 6 (- 1) = 4 q^{2} - 6 q + 6

. To minimise over

q

, differentiate with respect to

q

\frac{d}{d q} (4 q^{2} - 6 q + 6) = 8 q - 6 = 0

, giving

q = \frac{3}{4}

. This is a minimum since the second derivative is

8 > 0

. Hence the answer is F.

A family of quadratic curves is given by

y_{k} = 2 (x - \frac{k}{2})^{2} + \frac{k ^{2}}{2} + 4 k + 3

where

k

is any real number and

y_{k}

is a function of

x

. All these curves are sketched, and the point with the lowest

y

-coordinate among all the curves

y_{k}

(a, b)

. Find the value of

a + b

.

(A)

- 1

(B)

- 3

(C)

- 5

(D)

- 7

(E)

- 9

The curve

y_{k}

is in completed-square form, so its minimum is at

x = \frac{k}{2}

with minimum value

\frac{k ^{2}}{2} + 4 k + 3

. Thus the locus of minima is

a = \frac{k}{2}

and

b = \frac{k ^{2}}{2} + 4 k + 3

. To minimise

b

over all

k

, differentiate:

\frac{d b}{d k} = k + 4 = 0

gives

k = - 4

. Then

a = \frac{- 4}{2} = - 2

and

b = \frac{16}{2} + 4 (- 4) + 3 = 8 - 16 + 3 = - 5

. So

a + b = - 2 + (- 5) = - 7

. Hence the answer is D.