Question

lf f(x) = `(20^x + 3^x - 6^x - 10^x)/(1 - cos8x)`, for x ≠ 0

= `(k/16)log(10/3).log2`, for x = 0

is continuous at x = 0, then the value of k is ______ 

पर्याय

• sin²30°

• `3^{log_3(1/2)}`

• `root(3)(1/4)`

• `(log_2 2)/3`

Answer 1

lf f(x) = `(20^x + 3^x - 6^x - 10^x)/(1 - cos8x)`, for x ≠ 0

= `(k/16)log(10/3).log2`, for x = 0

is continuous at x = 0, then the value of k is `underline(3^{log_3(1/2)})`.

Explanation:

Since, f(x) is continuous at x = 0.

∴ f(0) = `lim_{x→0}f(x)`

⇒ `(k/16)log(10/3) . log2`

= `lim_{x → 0} (20^x + 3^x - 6^x - 10^x)/(1 - cos8x)`

= `lim_{x → 0} ((10^x - 3^x)(2^x - 1))/(2sin^2 4x)`

= `lim_{x → 0} (((10^x - 1)/x - (3^x - 1)/x) . ((2^x - 1)/x))/(2 xx (sin^2 4x)/(16x^2) xx 16`

= `((log10 - log3)(log2))/32`

∴ `(k/16)log(10/3) . log2 = 1/32log(10/3)log2`

⇒ `k/16 = 1/32`

⇒ `k = 1/2`

⇒ k = `3^{log_3(1/2)}` ........`[∵ a^{log_a^x} = x]`