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Question: A continuous function f: [0, 1) → [0, ∞) satisfies $f(\frac{x+1}{2})=f(x)+1$ and $f(1-x)=\frac{1}{f(...

A continuous function f: [0, 1) → [0, ∞) satisfies f(x+12)=f(x)+1f(\frac{x+1}{2})=f(x)+1 and f(1−x)=1f(x)∀x∈(0,1)f(1-x)=\frac{1}{f(x)} \forall x \in (0,1).

Then TRUE Statement is/are

A

∫01f(x)dx=32\int_0^1 f(x) dx = \frac{3}{2}

B

∫01f(x)dx=23\int_0^1 f(x) dx = \frac{2}{3}

C

∫01f(x)dx=43\int_0^1 f(x) dx = \frac{4}{3}

D

∫01f(x)dx=34\int_0^1 f(x) dx = \frac{3}{4}

Answer

∫01f(x)dx=32\int_0^1 f(x) dx = \frac{3}{2}

Explanation

Solution

Let I=∫01f(x)dxI = \int_0^1 f(x) dx. Using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx, we have I=∫01f(1−x)dxI = \int_0^1 f(1-x) dx. Using the given condition f(1−x)=1f(x)f(1-x) = \frac{1}{f(x)}, we get I=∫011f(x)dxI = \int_0^1 \frac{1}{f(x)} dx.

Split the integral II into two parts: I=∫01/2f(x)dx+∫1/21f(x)dxI = \int_0^{1/2} f(x) dx + \int_{1/2}^1 f(x) dx.

Consider the second integral ∫1/21f(x)dx\int_{1/2}^1 f(x) dx. Let x=y+12x = \frac{y+1}{2}, so dx=12dydx = \frac{1}{2} dy. When x=1/2x=1/2, y=0y=0. When x=1x=1, y=1y=1. ∫1/21f(x)dx=∫01f(y+12)12dy\int_{1/2}^1 f(x) dx = \int_0^1 f(\frac{y+1}{2}) \frac{1}{2} dy. Using the condition f(y+12)=f(y)+1f(\frac{y+1}{2}) = f(y)+1: ∫1/21f(x)dx=∫01(f(y)+1)12dy=12∫01f(y)dy+12∫011dy=12I+12\int_{1/2}^1 f(x) dx = \int_0^1 (f(y)+1) \frac{1}{2} dy = \frac{1}{2} \int_0^1 f(y) dy + \frac{1}{2} \int_0^1 1 dy = \frac{1}{2} I + \frac{1}{2}.

Substitute this back into the expression for II: I=∫01/2f(x)dx+(12I+12)I = \int_0^{1/2} f(x) dx + (\frac{1}{2} I + \frac{1}{2}). This implies ∫01/2f(x)dx=12I−12\int_0^{1/2} f(x) dx = \frac{1}{2} I - \frac{1}{2}.

Now consider the integral ∫01/2f(x)dx\int_0^{1/2} f(x) dx. Let x=1−ux = 1-u, so dx=−dudx = -du. When x=0x=0, u=1u=1. When x=1/2x=1/2, u=1/2u=1/2. ∫01/2f(x)dx=∫11/2f(1−u)(−du)=∫1/21f(1−u)du\int_0^{1/2} f(x) dx = \int_1^{1/2} f(1-u) (-du) = \int_{1/2}^1 f(1-u) du. Using f(1−u)=1/f(u)f(1-u) = 1/f(u): ∫01/2f(x)dx=∫1/211f(u)du\int_0^{1/2} f(x) dx = \int_{1/2}^1 \frac{1}{f(u)} du.

So, 12I−12=∫1/211f(x)dx\frac{1}{2} I - \frac{1}{2} = \int_{1/2}^1 \frac{1}{f(x)} dx.

We also have I=∫011f(x)dx=∫01/21f(x)dx+∫1/211f(x)dxI = \int_0^1 \frac{1}{f(x)} dx = \int_0^{1/2} \frac{1}{f(x)} dx + \int_{1/2}^1 \frac{1}{f(x)} dx. Substitute the expression for ∫1/211f(x)dx\int_{1/2}^1 \frac{1}{f(x)} dx: I=∫01/21f(x)dx+(12I−12)I = \int_0^{1/2} \frac{1}{f(x)} dx + (\frac{1}{2} I - \frac{1}{2}). This implies ∫01/21f(x)dx=12I+12\int_0^{1/2} \frac{1}{f(x)} dx = \frac{1}{2} I + \frac{1}{2}.

We have the relations:

  1. ∫01/2f(x)dx=I2−12\int_0^{1/2} f(x) dx = \frac{I}{2} - \frac{1}{2}
  2. ∫1/21f(x)dx=I2+12\int_{1/2}^1 f(x) dx = \frac{I}{2} + \frac{1}{2}
  3. ∫01/21f(x)dx=I2+12\int_0^{1/2} \frac{1}{f(x)} dx = \frac{I}{2} + \frac{1}{2}
  4. ∫1/211f(x)dx=I2−12\int_{1/2}^1 \frac{1}{f(x)} dx = \frac{I}{2} - \frac{1}{2}

Consider the integral ∫01/2f(x)dx\int_0^{1/2} f(x) dx. Let x=2ux=2u, so dx=2dudx=2du. ∫01/2f(x)dx=∫01f(2u)12du\int_0^{1/2} f(x) dx = \int_0^1 f(2u) \frac{1}{2} du. So, 12∫01f(2u)du=I2−12\frac{1}{2} \int_0^1 f(2u) du = \frac{I}{2} - \frac{1}{2}. ∫01f(2u)du=I−1\int_0^1 f(2u) du = I - 1.

If we assume I=3/2I = 3/2, then ∫01f(2u)du=3/2−1=1/2\int_0^1 f(2u) du = 3/2 - 1 = 1/2. This is consistent with ∫01/2f(x)dx=3/22−12=34−12=14\int_0^{1/2} f(x) dx = \frac{3/2}{2} - \frac{1}{2} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}, and ∫1/21f(x)dx=3/22+12=34+12=54\int_{1/2}^1 f(x) dx = \frac{3/2}{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}. And ∫01/2f(x)dx+∫1/21f(x)dx=14+54=64=32=I\int_0^{1/2} f(x) dx + \int_{1/2}^1 f(x) dx = \frac{1}{4} + \frac{5}{4} = \frac{6}{4} = \frac{3}{2} = I.

The value I=3/2I=3/2 is consistent with the derived relations.