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Discuss the continuity of the cosine, cosecant, secant and cotangent functions. - Mathematics

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प्रश्न

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

योग
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उत्तर

(i) f(x) = cos (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cos c) = (cos c) = (cos c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(ii) f(x) = cosec (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x ->^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cosec c) = (cosec c) = (cosec c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(iii) f(x) = sec (x)

Let c be any real number.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (sec c) = (sec c) = (sec c)

This statement is true; that is, f(x) is continuous at every point on the real number line.

(iv) f(x) = cot (x)

Let c be any real number such that (n − 1)π < x < nπ, where n represents an integer point.

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ (cot c) = (cot c) = (cot c)

This statement is true; that is, f(x) is continuous at every point on the real number line between (n − 1)π and nπ.

Now if we consider c such that c = nπ, where n represents an integer point, then:

If f(x) is continuous at x = c, this implies:

f(c) = `lim_(x -> c^+)` f(x) = `lim_(x -> c^-)` f(x)

⇒ ±∞ = ±∞ = ±∞

That is, f(x) is continuous at every point on the real number line except at the nπ-type points.

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अध्याय 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १६०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.1 | Q 22 | पृष्ठ १६०

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