Question -
Answer -
We know that the maximum value of A cos α + B sin α +C is C + √(A2 +B2),
And the minimum value is C – √(a2 +B2).
(i) 12 sin x – 5 cos x
Given: f(x) = 12 sin x – 5 cos x
Here, A = -5, B = 12 and C = 0
–√((-5)2 + 122) ≤ 12 sin x – 5cos 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
Hence, the maximum and minimum values of f(x) are 13and -13 respectively.
(ii) 12 cos x + 5 sin x + 4
Given: 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
Hence, the maximum and minimum values of f(x) are -9and 17 respectively.
(iii) 5 cos x + 3 sin (π/6 – x) + 4
Given: f(x) = 5 cos x + 3 sin (π/6 – x) + 4
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
So, 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
Hence, the maximum and minimum values of f(x) are -3and 11 respectively.
(iv) sin x – cos x + 1
Given: f(x) = sin x – cos x + 1
So, 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
Hence, the maximum and minimum values of f(x) are 1– √2and 1 + √2respectively.