Bac Blanc n°1 ES page
4 pages
English

Bac Blanc n°1 ES page

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4 pages
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Niveau: Secondaire, Lycée
Bac Blanc n°1 ES 2011 page 1/4 Exercise 1. (5 pts) Baccalauréat ES Asia june 2011 1. ( )f 1' = slope of the tangent at D rise 6 6 run 1 = = = (between G and D) ( )f 2 0' = since the tangent at E is horizontal 2. The equation of the tangent to the graph fC at D is y 6x 4= ? (since it's slope is 4 and it's y- intercept -4). 3. On the interval [ ]1 0;? : f is continuous f is strictly decreasing ( )f 1 6 0? = > and ( )f 0 4 0= ? < Then, according to the intermediate value theorem, the equation ( )f x 0= has one unique solution . 4. 1 2 3x 0 5 x 0 7 x 2 9. ; . ; .≈ ? ≈ ≈ 5. f' has to be negative on [ ]1 0;? and [ [2;+∞ (where f is decreasing) , so C1 is the graph of f' f being negative on [ ]1 2x x; and [ [3x ;+∞ , F has to be decreasing on these intervals (since F'=f), so C2 is the graph of F (according to the values of 1 2 3x x x, , of question 4) CORRIGE BB1 x 1? 1x 2x 3x

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CORRIGEBB1 Exercise 1. (5 pts)Baccalauréat ES Asia june 2011 rise 6 1.f'1slope of the tangent at D1 16 run 1 (between G and D) f'2 0since the tangent at E is horizontal 2.The equation of the tangent to the graphCat D isy 6x%4(since it’s slope is 4 and it’s y-intercept -4). 3.On the interval%1;0]: f is continuous f is strictly decreasing f 11620andf 0%400Then, according to the intermediate value theorem, the equationf x0has one unique solution . 4.1xxx¥1 23 Sign off  00 0 x» %0.5;x»0.7;x»2.9 1 23 5.f’ has to be negative on%1;0]and2;¥(where fis decreasing) , soC1is the graph of f’ f be, F has to be decreasing on these intervals (since ing negativeonx;x]andx3;¥ 1 2 F’=f), soC2xF (according to the values ofis the graph of,x,x ofquestion 4) 1 2 3
Bac Blanc n°1ES 2011
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