Question

Let ℝ → ℝ be function defined as
f(x) = αsin($(\frac{π[x]}{2})$+[2-x],α∈R
where [t] is the greatest integer less than or equal to t.
If lim x→1 f(x) exists, then the value of$∫_0^4$ f(x)dx
is equal to

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (B)

-2

Answer 2

The correct answer is (B):
f(x) = αsin($\frac{-2π}{3}$)+[2-x]α∈R
Now,
∵ limx→-1 f(x) exists
∴ limx→-1 f(x) = limx→-1+f(x)
⇒ αsin($\frac{-2π}{3}$)+3 = αsin($\frac{-2π}{3}$)+2
⇒ -α=1
⇒ α = -1
Now, ∫40f(x)dx = ∫40(-sin($\frac{-2π}{3}$)+[2-x])dx
= ∫10 1dx + ∫21-1dx+∫32-1dx+∫43(1-2)dx
= 1-1-1-1
= -2