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F(x) = ,if,if{1-cos2xx2,if x≠05,if x=0 at x = 0 - Mathematics

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Question

f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0

Sum
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Solution

We have, `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0

At x = 0

L.H.L. = `lim_(x -> 0^-) (1 - cos 2x)/x^2`

= `lim_("h" -> 0) (1 - cos2(0 - "h"))/(0 - "h")^2`

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

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

 R.H.L. = `lim_("h" -> 0^+) (1 - cos 2x)/x^2`

= `lim("h" -> 0) (1 - cos2(0 + "h"))/(0 + "h")^2`

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

Also f(0) = 5 ... (Given)

Since, L.H.L. = R.H.L. ≠ f(0)

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

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Chapter 5: Continuity And Differentiability - Exercise [Page 107]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 3 | Page 107

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