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प्रश्न
Verify Rolle’s theorem for the following function:
`f(x) = e^(-x) sinx " on" [0, pi]`
उत्तर
`f(x) = e^(-x). sin x`
1) f (x) is continuous on `[0,pi]` because `e^(-x)` and sin x are continuous function on its domain
2) `e^(-x)` and sin x is differentiable on `(0, pi)`
3) `f(0) = e^(-0). sin 0 = 0`
`f(pi) = e^(-pi). sin pi = 0`
4) Let c be number such that f'(x) = 0
`:. f'(x) = e^(-x). cos x _+ sin x. e^(-x) (-1)`
`:. f'(c) = e^(-c) (cos c - sin c)`
`:. f'(c) = 0`
`e^(-c) (cos c - sin c) = 0`
`:.e^(-c) = 0`
`:. cos C - sin C = 0
tan C =1
`:. C = pi/4, (3pi)/4 , (5pi)/4,.....`
`:. pi/4 in [0, pi]`
∴ Rolle's theorem verify
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