Question

If the function f defined as f(x) = `1/x - (k - 1)/(e^(2x) - 1)` x ≠ 0, is continuous at x = 0, then the ordered pair (k, f(0)) is equal to ______.

Options

• (3, 1)

• (3, 2)

• `(1/3, 2)`

• (2, 1)

Answer 1

If the function f defined as f(x) = `1/x - (k - 1)/(e^(2x) - 1)` x ≠ 0, is continuous at x = 0, then the ordered pair (k, f(0)) is equal to (3, 1).

Explanation:

If the function is continuous at x = 0, then

`lim_(x rightarrow 0)` f(x) will exist and f(0) = `lim_(x rightarrow 0)` f(x)

Now, `lim_(x rightarrow 0) f(x) = lim_(x rightarrow 0) (1/x - (k - 1)/(e^(2x) - 1))` 

= `lim_(x rightarrow 0)((e^(2x) - 1 - kx + x)/((x)(e^(2x) - 1)))`

= `lim_(x rightarrow 0)[((1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + ...) - 1 - kx + x)/((x)((1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + ...) - 1))]`

= `lim_(x rightarrow 0)[((3 - k)x + (4x^2)/(2!) + (8x^3)/(3!) + ...)/((2x^2 + (4x^3)/(2!) + (8x^3)/(3!) + ...))]`

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator.

Therefore, coefficient of x in the numerator is equal to zero.

`\implies` 3 – k = 0

`\implies` k = 3

So the limit reduces to

`lim_(x rightarrow 0) ((x^2)(4/(2!) + (8x)/(3!) + ...))/((x^2)(2 + (4x)/(2!) + (8x^2)/(3!) + ...)`

= `lim_(x rightarrow 0) (4/(2!) + (8x)/(3!) + ...)/(2 + (4x)/(2!) + (8x^2)/(3!) + ...)` = 1

Hence, f(0) = 1