Question

Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.

पर्याय

• 2

• 1

• 0

• Infinitely many

Answer 1

Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is 1.

Explanation:

f(x) = `(tanxlog(x - 2))/((x^2 - 4x + 3)); x∈(2, 4) - {3, π}`

`lim_(x→π)f(x) = lim_(x→π)(tanxlog(x - 2))/(x^2 - 4x + 3)`

= `(tanπ log(π - 2))/(π^2 - 4π + 3)` = 0

and f(π) = `1/6tanπ` = 0

∴ f(x) is continuous at x = π

f(3) = `1/6tan3` and `lim_(x→3) (tanxlog(x - 2))/(x^2 - 4x + 3)`

= `lim_(x→3)(tanx log(x - 2))/((x - 1)(x - 3))`

i.e., `lim_(x→3)f(x) = lim_(x→3) (tanxlog(1 + x - 3))/((x - 1)(x - 3))` 

= `lim_(x→3) tanx/(x - 1) xx lim_(x→3) (log(1 + x - 3))/(x - 3)`

∴ `lim_(x→3)f(x) = tan3/2 xx 1 = 1/2tan3`

⇒ `lim_(x→3) f(x) ≠ f(3)`

Hence, f(x) is discontinuous at x = 3.

The number of values of x is 1.