Tamil Nadu Board of Secondary EducationHSC Commerce Class 11th

# Verify the continuity and differentiability of f(x) = {1-xif x<1(1-x)(2-x)if1≤x≤23-xifx>2 at x = 1 and x = 2. - Business Mathematics and Statistics

Sum

Verify the continuity and differentiability of f(x) = {(1 - x if  x < 1),((1 - x)(2 - x) if 1 <= x <= 2),(3 - x if x > 2):} at x = 1 and x = 2.

#### Solution

"L"[f(x)]_(x=1) = lim_(x->1) f(x) = lim_(h->0) "f"(1 - "h")

= lim_(h->0) 1 - (1 - "h")  [∵ f(x) = 1 - x if x < 1]

= lim_(h->0) h = 0     ...(1)

"R"[f(x)]_(x=1) = lim_(x->1) f(x) = lim_(h->0) "f"(1 + "h")

= lim_(h->0) [1 - (1 + h)][2 - (1 + h)]   [∵ f(x) = (1 - x)(2 - x) if 1 ≤ x ≤ 2]

= lim_(h->0) (cancel(1) - cancel(1) - "h")(2 - 1 - "h")

= lim_(h->0)(-h)(1 - h)

= lim_(h->0) - h + h2 = 0 + 0 = 0   ....(2)

Also f(1) = (1 - 1)(2 - 1) ...[∵ f(x) = (1 - x)(2 - x) when x = 1]

= 0   ...(3)

From (1), (2) and (3),

"L"[f(x)]_(x=1) = "R"[f(x)]_(x=1) = f(1) = 0

∴ f(x) is continuous at x = 1

"L"["f"'(1)] = lim_(x->1^-) ("f"(x) - "f"(1))/(x - 1)

= lim_(h->0) ([1 - (1 - "h")] - [(1 - 1)(2 - 1)])/(1 - "h" - 1)

= lim_(h->0) ([1 - (1 - "h")] - [1 - (1 - "h")(2 - (1 - "h"))])/(-"h")    ...[∵ f(x) = (1 - x)(2 - x)]

= lim_(h->0) (1 - 1 + "h")/(-"h") = - 1   ....(4)

"R"["f"'(1)] = lim_(x->1^+) ("f"(x) - "f"(1))/(x - 1)

= lim_(h->0) ("f"(1 + "h") - "f"(1))/(1 + "h" - 1)

= lim_(h->0) (1 - (1 + "h")(2 - (1 + "h")) - 0)/"h"

= lim_(h->0) ((-"h")(1 - "h"))/"h" = - 1   ...(5)

From (4) and (5), "L"["f"'(1)] = "R"["f"'(1)]

∴ f(x) differentiable at x = 1

Now

"L"[f(x)_(x=2)] = lim_(x->2^-) "f"(x) = lim_("h" -> 0)f(2 - h)

= lim_("h" -> 0) [1 - (2 - "h")][2 - (2 - "h")] ...[f(x = (1 - x)(2 - x)) when x < 2]

= lim_("h" -> 0)(- 1 + h)(h)

= lim_("h" -> 0) = - h + h2 = 0 + 0 = 0   ....(6)

"R"[f(x)_(x=2)] = lim_(x->2^+) "f"(x) = lim_("h" -> 0)f(2 + h)

= lim_(h->0) 3 - (2 + h)   ...[∵ f(x) = 3 - x when x > 2]

= lim_(h->0) 1 - h = 1 - 0 = 1

and f(2) = (1 - 2)(2 - 2) = 0    ....(7)

From (6) and (7), f(x) is not continuous at x = 2.

"L"["f"'(2)] = lim_(x->2^-) ("f"(x) - "f"(2))/(x - 2)

= lim_(h->0) ("f"(2 - "h") - "f"(2))/(2 - "h" - 2)

= lim_(h->0) ([1 - (2 - "h")][2 - (2 - "h")] - [(1 - 2)(2 - 2)])/(-"h")   ....[∵ f(x = (1 - x)(2 - x)) when x < 2]

= lim_(h->0) ((- 1 + "h")(+"h") - 0)/(- "h")

= lim_(h->0) (- 1 + "h")/(-1)

= 1    ...(8)

"R"["f"'(2)] = lim_(x->2^+) ("f"(x) - "f"(2))/(x - 2)

= lim_(h->0) ("f"(2 + "h") - "f"(2))/(cancel(2) + "h" - cancel(2))

= lim_(h->0) ([3 - (2 + "h")] - [(1 - 2)(2 - 2)])/"h"   ...[∵ f(x) = 3 - x when x > 2]

= lim_(h->0) ((1 - "h") - 0)/"h"

= lim_(h->0) (1 - 0)/0 = ∞    ...(9)

therefore "L"["f"'(2)] ne "R"["f"'(2)]`

⇒ f(x) is not differentiable at x = 2.

Concept: Limits and Derivatives
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