Question

If f(x) `{:(= sin x - cos x, "if"  x ≤ pi/2),(= 2x - pi + 1, "if"  x > pi /2):}` Test the continuity and differentiability of f at x = `π/2`

Answer 1

f(x) `{:(= sin x - cos x,","  x ≤ pi/2),(= 2x - pi + 1,","  x > pi /2):}`

Continuity at x = `pi/2`:

`lim_(x -> pi^-/2) "f"(x) = lim_(x -> pi^-/2) (sinx - cosx)`

= `sin  pi/2 - cos  pi/2`

= 1 – 0

= 1

`lim_(x -> pi^+/2) "f"(x) = lim_(x -> pi^+/2) (2x - pi + 1)`

= `2(pi/2) - pi + 1`

= 1

`"f"(pi/2) = sin  pi/2 - cos  pi/2 = 1 - 0` = 1

∴ `lim_(x -> pi^-/2) "f"(x) = lim_(x -> pi^+/2) "f"(x) = "f"(pi/2)`

∴ f(x) is continuous at x = `pi/2`

Differentiability at x = `pi/2`:

`"Lf'"(pi/2) = lim_("h" -> 0^-)  ("f"(pi/2 + "h") - "f"(pi/2))/"h"`

= `lim_("h" -> 0^-) (sin(pi/2 + "h") - cos (pi/2 + "h" ) - 1)/"h"`

= `lim_("h" -> 0^-) (cos "h" + sin "h" - 1)/"h"`

= `lim_("h" -> 0^-) (sin"h"/"h" - (1 - cos "h")/"h")`

= `lim_("h" -> 0^-) (sin"h"/"h" - (2sin^2  "h"/2)/"h")`

= `1 - lim_("h" -> 0^-) ((sin^2("h"/2))/("h"/2))`

= 1 – 0

= 1

`"Rf'"(pi/2) = lim_("h" -> 0^+) ("f"(pi/2 + "h") - "f"(pi/2))/"h"`

= `lim_("h" -> 0^+) ([2(pi/2 + "h") - pi + 1] - 1)/"h"`

= `lim_("h" -> 0^+) ((2"h")/"h")`

= `lim_("h" -> 0^+) 2`  ...[∵ h → 0, h ≠ 0]

= 2

∴ `"Lf'"(pi/2) ≠ "Rf'"(pi/2)`

∴ f(x) is not differentiable at x = `pi/2`.