Mathematics Tuition for GCSE & A-Level

01 · Learn the shortcut

1. Take the number before the final 5

   For 75, that number is 7.

2. Multiply it by the next whole number
   $$7\times 8=56$$
3. Append 25 to the result

   Write 56 followed by 25: 5625.

   Appending means writing the digits together, not adding 25.

02 · Why the shortcut works

Any non-negative whole number ending in 5 can be written as $10n+5$.

Multiplying $n(n+1)$ by 100 leaves two places for the final digits 25.

01 · Find the right pair

This method starts with a quadratic whose $x^2$ coefficient is 1.

1. Look for two numbers that multiply to 12

   Candidate pairs: 1 and 12, 2 and 6, 3 and 4.

2. Choose the pair that adds to 7
   $$3\times 4=12$$
   $$3+4=7$$
3. Write one number in each bracket
   $$x^2+7x+12$$
   $$=(x+3)(x+4)$$

02 · Check, then handle signs

Expand to confirm every term:

A negative constant

We need a product of −12 and a sum of −1. The pair is −4 and 3.

01 · Update the bag after each draw

A bag contains 3 red and 2 blue counters. Draw two at random, without replacing the first.

R = red; B = blue. Read the branches from left to right.

02 · Multiply paths, add alternatives

1. Two red counters: follow one path

   After a red is drawn, 2 red counters remain out of 4.

   $$\mathrm{P}\left(\text{RR}\right)=\frac{3}{5}\times \frac{2}{4}$$
   $$=\frac{3}{10}$$
2. One of each: use both possible orders

   Red then blue:

   $$\mathrm{P}\left(\text{RB}\right)=\frac{3}{5}\times \frac{2}{4}$$
   $$=\frac{3}{10}$$

   Blue then red:

   $$\mathrm{P}\left(\text{BR}\right)=\frac{2}{5}\times \frac{3}{4}$$
   $$=\frac{3}{10}$$
3. Add the two mutually exclusive paths
   $$\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$$

01 · Find the maximum height

A ball’s height h metres, t seconds after launch, is modelled by:

1. Complete the square
   $$h=-2(t^2-6t)+5$$

   $$=-2{(t-3)}^2+23$$
2. Interpret the turning point
   Since ${(t-3)}^2\ge 0$, the maximum height is 23 m, reached at t = 3 s.
3. Find when it reaches the ground

   Set h = 0 and solve:

   $${(t-3)}^2=\frac{23}{2}$$

   $$t=3\pm \sqrt{\frac{23}{2}}$$

   The positive time is approximately 6.39 s.

02 · Interpret the model

01 · Your method toolkit

Power

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^n\right)=n{x}^{n-1}$$

Product

$${\left(uv\right)}^{\prime }=u^{\prime }v+u{v}^{\prime }$$

Quotient

$${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$$

For v ≠ 0.

Chain

$$\frac{\mathrm{d}}{\mathrm{d}x}f\left(g\left(x\right)\right)$$

$$=f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)$$

02 · Apply the chain rule

1. Identify the inner function
   $$u=3x^2+1$$

   $$\frac{\mathrm{d}u}{\mathrm{d}x}=6x$$
2. Differentiate the outer function
   $$f=u^5$$

   $$\frac{\mathrm{d}f}{\mathrm{d}u}=5u^4$$
3. Multiply the derivatives
   $$f^{\prime }\left(x\right)=5{(3x^2+1)}^4\times 6x$$

   $$=30x{(3x^2+1)}^4$$

01 · Integrate each term

1. Raise each power by one

   Add 1 to the exponent. The power rule gives:

   $$\int a{x}^n\mathrm{d}x=\frac{a{x}^{n+1}}{n+1}+C$$

   For n ≠ −1.

2. Divide by the new power
   $$\frac{6x^3}{3}-\frac{4x^2}{2}+3x$$
3. Include the constant of integration
   $F\left(x\right)=$$2x^3-2x^2+3x+C$

02 · Check and extend

Differentiating returns the original integrand, confirming the result.

A definite integral

01 · Work through the method

1. Differentiate and set the gradient to zero
   $$y^{\prime }=3x^2-12x+9$$

   $$=3(x-1)(x-3)$$

   Set this equal to zero: x = 1 or x = 3.

2. Find the coordinates
   $$y\left(1\right)=8$$

   $$y\left(3\right)=4$$
3. Classify using the second derivative
   $$y^{″}=6x-12$$

   $$y^{″}\left(1\right)=-6$$

   Maximum at (1, 8).

   $$y^{″}\left(3\right)=6$$

   Minimum at (3, 4).

02 · Read the curve

01 · Calculate modulus and argument

1. Distance from the origin
   $$|z|=\sqrt{3^2+4^2}=5$$
2. Angle from the positive real axis
   $$\theta =\arctan \left(\frac{4}{3}\right)$$

   $$\approx 0.927\text{rad}$$
3. Write in polar form
   $$z=5(\cos \theta +i\sin \theta )$$

02 · Interpret the geometry

01 · Build the equations

A target moves with position (in metres):

An interceptor starts at the origin and travels at 5 m/s in a fixed direction θ.

1. Equate the two coordinates
   $$5t\cos \theta =4+3t$$

   $$5t\sin \theta =2+t$$
2. Eliminate the angle
   $${(4+3t)}^2+{(2+t)}^2$$

   $$=25t^2$$

   $$15t^2-28t-20=0$$
3. Choose the positive time
   $$t=\frac{14+4\sqrt{31}}{15}$$

   $$\approx 2.42\text{s}$$

02 · Recover the direction

At interception, both objects are at approximately (11.25, 4.42) metres. The interceptor travels in a straight line to this point.

Extend the model

For an unspecified speed v:

For every finite t > 0, $v>\sqrt{10}$. The speed approaches $\sqrt{10}$ m/s as t increases, but this bound is never attained at a finite interception time.