As we know that the maximum value of A cos α + B sin α + C is C + √(A² + B²),

And here the minimum value is C – √(a² + B²).

(i) Given as f(x) = 12 sin x – 5 cos x

Here, A = -5, B = 12 and C = 0

–√((-5)² + 12²) ≤ 12 sin x – 5 cos x ≤ √((-5)² + 12²)

–√(25 + 144) ≤ 12 sin x – 5 cos x ≤ √(25 + 144)

–√169 ≤ 12 sin x – 5 cos x ≤ √169

-13 ≤ 12 sin x – 5 cos x ≤ 13

Thus, the maximum and minimum values of f(x) are 13 and -13 respectively.

(ii) 12 cos x + 5 sin x + 4

Given as f(x) = 12 cos x + 5 sin x + 4

Here, A = 12, B = 5 and C = 4

4 – √(12² + 5²) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(12² + 5²)

4 – √(144 + 25) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(144 + 25)

4 – √169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √169

-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17

Thus, the maximum and minimum values of f(x) are -9 and 17 respectively.

(iii) 5 cos x + 3 sin (π/6 – x) + 4 

Given as f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

As we know that, sin (A – B) = sin A cos B – cos A sin B

f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

= 5 cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4

= 5 cos x + 3/2 cos x – 3√3/2 sin x + 4

= 13/2 cos x – 3√3/2 sin x + 4

Therefore, here A = 13/2, B = – 3√3/2, C = 4

4 – √[(13/2)² + (-3√3/2)²] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(13/2)² + (-3√3/2)²]

4 – √[(169/4) + (27/4)] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(169/4) + (27/4)]

4 – 7 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + 7

-3 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 11

Thus, the maximum and minimum values of f(x) are -3 and 11 respectively.

(iv) sin x – cos x + 1

Given as f(x) = sin x – cos x + 1

Therefore, here A = -1, B = 1 And c = 1

1 – √[(-1)² + 1²] ≤ sin x – cos x + 1 ≤ 1 + √[(-1)² + 1²]

1 – √(1 + 1) ≤ sin x – cos x + 1 ≤ 1 + √(1 + 1)

1 – √2 ≤ sin x – cos x + 1 ≤ 1 + √2

Thus, the maximum and minimum values of f(x) are 1 – √2 and 1 + √2 respectively.