Question

Let f: (0,2)→R be defined as f(x) = `log_2(1 + tan((πx)/4))`. Then, `lim_(n→∞) 2/n(f(1/n) + f(2/n) + ... + f(1))` is equal to ______.

Options

• 0

• 1

• 2

• 3

Answer 1

Let f: (0,2)→R be defined as f(x) = `log_2(1 + tan((πx)/4))`. Then, `lim_(n→∞) 2/n(f(1/n) + f(2/n) + ... + f(1))` is equal to 1.

Explanation:

f: (0,2)→R

f(x) = `log_2(1 + tan((πx)/4))`  ...(i)

We have to find the value of `lim_(n→∞) 2/n(f(1/n) + f(2/n) + ... + f(1))`

Let, L = `lim_(n→∞)(2(sum_(r = 1)^n(1/n)f(r/n)))`

By replacing `1/n→dx, r/n→x, lim_(n→∞)sum→int`

Lower limit = `lim_(n→∞)(1/n)` = 0

Upper limit = `lim_(n→∞)(1/n)` = 1

We get, 

L = `2int_0^1f(x)dx`  ...(ii)

Now using equation (i) and (ii)

L = `2int_0^1log_2(1 + tan((πx)/4))dx`

Using, log_ax = `(In  x)/(In  a)` we can write

L = `2int_0^1((In(1 + tan  ((πx)/4)))/(In2))dx`

⇒ L = `2/(In  2)int_0^1 In(1 + tan((πx)/4))dx`  ...(iii)

Now using the property

`int_a^bf(x)dx= int_a^bf(a + b - x)dx`,

We can write

⇒ L = `2/(In2)int_0^1 In(1 + tan((π(1 - x))/4))dx`

⇒ L = `2/(In2)int_0^1 In(1 + tan(π/4 - (πx)/4))dx`

⇒ L = `2/(In2)int_0^1 In(1 + (tan  π/4 - tan  (πx)/4)/(1 + tan  π/4.tan  (πx)/4))dx`

⇒ L = `2/(In2)int_0^1 In(1 + (1 - tan  (πx)/4)/(1 + tan  (πx)/4))dx`

⇒ L = `2/(In2) int_0^1 In(2/(1 + tan  (πx)/4))dx`  ...(iv)

After adding equation (iii) and (iv), we get

⇒ 2L = `2/(In2)int_0^1{In(1 + tan  (πx)/4) + In(2/(1 + tan  (πx)/4))}dx`

⇒ L = `1/(In  2)int_0^1 In 2dx`

⇒ L = `int_0^1dx`

⇒ L = `(x)_0^1`

⇒ L = 1 – 0

⇒ L = 1