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प्रश्न
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`
उत्तर
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`.
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