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Verify Rolle’s theorem for the function, f(x) = sin 2x in [0,π2]. - Mathematics

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प्रश्न

Verify Rolle’s theorem for the function, f(x) = sin 2x in `[0, pi/2]`.

योग

उत्तर

Consider f(x) = sin 2x in `[0, pi/2]`

Note that:

(i) The function f is continuous in `[0, pi/2]`, as f is a sine function, which is always continuous.

(ii) f′(x) = 2cos 2x, exists in `(0, pi/2)`, hence f is derivable in `(0, pi/2)`.

(iii) f(0) = sin0 = 0 and `"f"(pi/2)` = sinπ = 0

⇒ f(0) = `"f"(pi/2)`.

Conditions of Rolle’s theorem are satisfied. Hence there exists at least one c ∈ `(0, pi/2)` such that f′(c) = 0.

Thus 2 cos 2c = 0

⇒ 2c = `pi/2`

⇒ c = `pi/4`.

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अध्याय 5: Continuity And Differentiability - Solved Examples [पृष्ठ ९७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Solved Examples | Q 17 | पृष्ठ ९७

वीडियो ट्यूटोरियलVIEW ALL [3]

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