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F(x) = ,if,if{2x2-3x-2x-2,if x≠25,if x=2 at x = 2 - Mathematics

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प्रश्न

f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2

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उत्तर

We have, f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2.

At x = 2

L.H.L. = `lim_(x -> 2^-) (2x^2 - 3x - 2)/(x - 2)`

= `lim_("h" -> 0) (2(2 - "h")^2 - 3(2 - "h") - 2)/((2 - "h") - 2)`

= `lim_("h" -> 0) (8 + 2"h"^2 - 8"h" - 6 + 3"h" - 2)/(-"h")`

= `lim_("h" -> 0) (2"h"^2 - 5"h")/(-"h")`

= `lim_("h" -> 0) ("h"(2"h" - 5))/(-"h")` = 5

R.H.L. = `lim_(x -> 2^+) (2x^2 - 3x - 2)/(x - 2)`

= `lim_("h" -> 0) (2(2 + "h")^2 - 3(2 + "h") - 2)/((2 + "h") - 2)`

= `lim_("h" -> 0) (8 + 2"h"^2 + 8"h" - 6 - 3"h" - 2)/"h"`

= `lim_("h" -> 0) (2"h"^2 + 5"h")/"h"`

= `lim_("h" -> 0) ("h"(2"h" + 5))/"h"` = 5

Also f(2) = 5  ....(Given)

∴ L.H.L. = R.H.L. = f(2)

So, f(x) is continuous at x = 2.

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अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ १०७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 4 | पृष्ठ १०७

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