Question

If the function f(x) = `(log (1 + "ax") - log (1 - "bx))/x, x ≠ 0` is continuous at x = 0 then, f(0) = _____.

विकल्प

• log a - log b

• a + b

• log a + log b

• a - b

Answer 1

If the function f(x) = `(log (1 + "ax") - log (1 - "bx))/x, x ≠ 0` is continuous at x = 0 then, f(0) = a + b.

Explanation:

Given function

f(x) = `(log (1 + "a"x) - log(1 - "b"x))/x, x ne 0`

is continuous at x = 0

`therefore lim_(x->0) "f"(x) = "f"(0)`    ...(i)

Here,

`lim_(x->0) "f"(x) = lim_(x->0) (log (1 + "a"x) - log(1 - "b"x))/x   ...(0/0 "form")`

`= lim_(x->0) ("a"/(1 + "ax") + ("b")/(1 - "bx"))/1`   (Using L' Hospital's Rule)

`= lim_(x->0) ("a"/(1 + "bx") + "b"/(1 - "bx"))` = a + b

From equation (i), we get

f(0) = a + b