(a) Transformations:

Solution:

1. Translation: Replace \( x \) with \( x+1 \) → \( y = \arctan(x+1) \) (shift left by 1 unit)

2. Horizontal stretch: Replace \( x \) with \( 2x \) → \( y = \arctan(2x+1) \) (scale factor 1/2)

3. Vertical translation: Add \( \frac{\pi}{4} \) → \( y = \arctan(2x+1) + \frac{\pi}{4} \) (shift up by π/4)

[3 marks]

(b) Proof of arctangent addition formula:

Solution:

Let \( \theta = \arctan p \) and \( \phi = \arctan q \), so \( \tan \theta = p \) and \( \tan \phi = q \).

Using tangent addition formula:

\[ \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 – \tan \theta \tan \phi} = \frac{p+q}{1-pq} \]

Therefore:

\[ \theta + \phi = \arctan\left(\frac{p+q}{1-pq}\right) \]

Which proves:

\[ \arctan p + \arctan q = \arctan\left(\frac{p+q}{1-pq}\right) \]

[4 marks]

(c) Verification:

Solution:

Using part (b) with \( p = \frac{x}{x+1} \) and \( q = 1 \):

\[ \arctan\left(\frac{x}{x+1}\right) + \arctan 1 = \arctan\left(\frac{\frac{x}{x+1} + 1}{1 – \frac{x}{x+1}}\right) = \arctan(2x+1) \]

Since \( \arctan 1 = \frac{\pi}{4} \), we have:

\[ \arctan(2x+1) = \arctan\left(\frac{x}{x+1}\right) + \frac{\pi}{4} \]

[3 marks]

(d) Proof by induction:

Solution:

Base case (n=1):

\[ \text{LHS} = \arctan\left(\frac{1}{2}\right) = \text{RHS} \]

Inductive step:

Assume true for \( n = k \):

\[ \sum_{r=1}^k \arctan\left(\frac{1}{2r^2}\right) = \arctan\left(\frac{k}{k+1}\right) \]

For \( n = k+1 \):

\[ \text{LHS} = \arctan\left(\frac{k}{k+1}\right) + \arctan\left(\frac{1}{2(k+1)^2}\right) \]

Using part (b):

\[ = \arctan\left(\frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1 – \frac{k}{2(k+1)^3}}\right) = \arctan\left(\frac{k+1}{k+2}\right) = \text{RHS} \]

Therefore, by induction, the statement holds for all \( n \in \mathbb{Z}^+ \).

[6 marks]