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Examine the differentiability of f, where f is defined byf(x) = ,if,if{x2sin 1x, if x≠00,if x=0 at x = 0 - Mathematics

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Question

Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin  1/x",",  "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0

Sum
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Solution

Given that, f(x) = `{{:(x^2 sin  1/x",",  "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0

For differentiability we know that:

Lf'(c) = Rf'(c)

∴ Lf'(0) = `lim_("h" -> 0)  ("f"(0 - "h") - "f"(0))/(-"h")`

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

= `("h"^2 sin  (- 1/"h"))/(-"h")`

= `"h"* sin (1/"h")`

= `0 xx [-1 ≤ sin  (1/"h") ≤ 1]`

= 0

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

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

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

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

= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`

= 0

So, Lf'(0) = Rf'(0) = 0

Hence, f(x) is differentiable at x = 0.

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Chapter 5: Continuity And Differentiability - Exercise [Page 109]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 21 | Page 109

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