Given f’(x) = a sin x + b cos x and f’(0) = 4

On substituting x = 0 in f’(x), we get

f’(0) = asin0 + bcos0

⇒ 4 = a × 0 + b × 1

⇒ 4 = 0 + b

∴ b = 4

Hence, f’(x) = a sin x + 4 cos x

On integrating this equation, we have

We know

Recall and

On substituting x = 0 in f(x), we get

f(0) = –acos0 + 4sin0 + c

⇒ 3 = –a × 1 + 4 × 0 + c

⇒ 3 = –a + c

⇒ c – a = 3 -------------- (1)

On substituting in f(x), we get

⇒ 5 = –a × 0 + 4 × 1 + c

⇒ 5 = 0 + 4 + c

⇒ 5 = 4 + c

∴ c = 1

On substituting c = 1 in equation (1), we get

1 – a = 3

⇒ a = 1 – 3

∴ a = –2

On substituting the values of c and a in f(x), we get

f(x) = –(–2)cos x + 4sinx + 1

∴ f(x) = 2cosx + 4sinx + 1

Thus, f(x) = 2cosx + 4sinx + 1