Sol. Since [x 2 ] π is an integral multiple of π for 0 ≤ x ≤ 1. Therefore, sin [x 2 ] π = 0. Hence

f(x) =

Since f(x) is differentiable in [0, 2], So, it is continuous and differentiable at x = 1 also.

Now,

f(x) is continuous at x= 1

⇒ f(x) = f(x) = f(1)

⇒ ax 3 + b = 2 cos π x + tan –1 x = a + b

⇒ a + b = 2 cos π + tan –1 1

⇒ a + b = –2 + ... (i)

Since f(x) is differentiable at x = 1

∴ (LHD at x = 1) = (RHD at x = 1)

⇒ =

⇒ =

⇒ a

= [Using (i)]

⇒ a =

⇒ 3a = 2 +

⇒ 3a = 2 +

⇒ 3a = 2 × 0 + ×

⇒ 3a = ... (ii)

Solving (i) and (ii), we get

a = and b = –