Question

Let f: R→R be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈ R. If I₁ = `int_0^8f(x)dx` and I₂ = `int_(-1)^3f(x)dx`, then the value of I₁ + 2I₂ is equal to ______.

Options

• 15

• 16

• 17

• 18

Answer 1

Let f: R→R be a continuous function such that f(x) + f(x + 1) = 2, for all x ∈ R. If I₁ = `int_0^8f(x)dx` and I₂ = `int_(-1)^3f(x)dx`, then the value of I₁ + 2I₂ is equal to 16.

Explanation:

Given that f: R→R is continuous if

f(x) + f(x + 1) = 2  ...(i)

Replace x by x + 1

f(x + 1) + f(x + 2) = 2  ...(ii)

Now, from equation (ii) – (i), we have

f(x + 1) + f(x + 2) – f(x) – f(x + 1) = 2 – 2

⇒ f(x + 2) – f(x) = 0

⇒ f(x) = f(x + 2)  ...(iii)

If f(x) = f(x + T) then f(x) will be a periodic function with period T.

Using the concept we can say that given function is periodic with period 2.

Now, I₁ = `int_0^8f(x)dx`

⇒ I₁ = `int_0^(2 xx 4) f(x)dx`

⇒ I₁ = `4int_0^2 f(x)dx`  ...(iv)

Also I₂ = `int_(-1)^3 f(x)dx`

⇒ I₂ = `int_0^4 f(x + 1)dx`

Using Equation (i)

I₂ = `int_0^4 (2 - f(x))dx`

⇒ I₂ = `int_0^4 2dx - int_0^4 f(x)dx`

⇒ I₂ = `2int_0^4 dx - int_0^4 f(x)dx`

⇒ I₂ = `2(x)_0^4 - int_0^(2 xx 2) f(x)dx`

⇒ I₂ = `2(4 - 0) - 2int_0^2f(x)dx`

Now Using Equation (iv)

⇒ I₂ = `8 - 2(I_1/4)`

⇒ I₂ = `8 - I_1/2`

⇒ 2I₂ = 16 – I₁

⇒ I₂ + 2I₂ = 16