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Continuity And Differentiability

Question
CBSEENMA12035498
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Verify Rolle's Theorem for the function f(x) = sin2 x in [0, straight pi].

Solution

Here f(x) = sin2 x
(i) We know that sin x is continuous in [0, straight pi]
Now sin2x, being the product of two continuous function sin x and sin x, is continuous in [0, straight pi].
(ii) f'(x) = 2 sin x cos x, which exists in (0, straight pi)
∴ f(x) is derivable in (0, π).
(iii) f(0) = sin2 0 = 0, f(straight pi) = sin2 π = 0
∴ f(0) = f(straight pi)
∴ f{x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that
f'(c) = 0 where 0 < c < straight pi.
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 2 space sin space straight c space cos space straight c equals 0 therefore space cos space straight c equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space sin space straight c not equal to 0 space in space left parenthesis 0 comma space straight pi right parenthesis right square bracket therefore space straight c equals straight pi over 2 element of left parenthesis 0 comma space straight pi right parenthesis therefore space Rolle apostrophe straight s space Theorem space is space verified.

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