علشان ما نعملش مواضيع كتير عن امتحان التفاضل تعالوا ندردش مع بعض هنا+ اجابة الامتحان

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اجابة امتحان التفاضل والتكامل


للصف الثالث الثاوى2009 – لغات

1- (a) * tan x – 3 sin 2x + c


* ( 3x – 1 )5 + c

(b) F/(x) = 0 x = ± 1 C.point. ------- +++ -----


L.max. value( 5) , L.min. value ( -3)


F//(x) = 0 then x = 0 Inflection . point ( 0 , 1)


++++ -------


[ F(X) is incr. at ] 0 , ∞ [ , decr. At ] - ∞ , 0

2- (a) ∫ 3 +4x dx = ∫ 6 y dy then 3x +2x2 + c = 3y2 C =2


Equation of curve : 3x +2x2 + 2 = 3y2


(b) y / = 1-2x /2-2y at y = 0 x2 – x -2 = 0 then x = -1 , x = 2


At (-1 , 0) at ( 2 , 0 )


m = 3/2 m = -3/2


equation of normal:


3 y + 2 x + 2 = 0 3 y – 2 x + 4 = 0

3- (a) A = (4x/5 )(3x/5 )


dA/dt = 24/25 (x) dx/dt then -60 = 24/25 (x) (-2.5)


then x =25 cm dimensions are : 20 , 15 cm


Area of rectangle = 20 * 15 = 300 cm2


x2 – 4 x


(b)(i) F(x) =


4 x - x2


F/(x)+ = lim h2 -4h – 0 = - 4

h 0 h


F/(x)- = lim -h2 +4h – 0 = 4 f(x) not diff.

h 0 h


(ii) F/(x) = 0 then x = 2 C.P.


F( 1 ) = -3


F( 4 ) = 0


F( 1 ) = -4 absolute max. value ( 0 ) , absolute min. value ( - 4 )

4- (a) y (2x+y) = 3 by diff. W. R .to x


y/ (2x+y) + ( 2+ y/ ) y = 0 by diff. W. R .to


y// (2x+y) + ( 2+ y/ ) y/ + y// y + y/ ( 2 +y/ ) = 0


2y//y +2 y2/ + 4 y/ + 2 x y// = 0


2 y / + y// ( x + y ) + y2/ = 0


2 dy/dx + d2y/dx2 ( x+ y ) + (dy/dx )2 = 0

(b) let Ө angle between two sides a , b


A = 0.5 a b sin Ө by diff. W. R .to Ө

dA/d Ө = 0.5 a b cos Ө

cos Ө = 0 then Ө = 900


Then: the third side is a diameter of circle passing its vertices.

5- (a)* at x € ] 1 , 2 [ : it is cont.


at x € ] 2 , 5 [ : it is cont .


* discuss cont . at x =1 : F(1)+ = F(1) = 1 + 1 = 2 it is cont.

* discuss cont . at x =1: F(1)- = F(2) = 2 +1 = 3


F(1)+ = (2)2 + 1 = 5 it is not cont


F(x) is cont . at [ 1 , 5 [ - 2

(b) x y = a y/ = - a x -2


By diff. w.r.to y


dx/dy . y + x = 0


dx/dy = - x/y

d2x/dy2 . y + dx/dy + dx/dy = 0


d2x/dy2 = -2 dx/dy


y


x y = a x/ = - a y -2


By diff. w.r.to x


dy/dx . x + y = 0


dy/dx = - y/x

d2y/dx2 . x + dy/dx + dy/dx = 0


d2y/dx2 = -2 dy/dx


x

- 2 dy/dx . -2dx/dy > - y . –x


X y x y


4 dy/dx . dx/dy > x y 4 (- a x -2 ) (- a y -2 ) > x y

4 a2 > x y a2 > x3y3


x2y2 4

a > x3y3

4

مع دوام النجاج الباهر


أ / أحمد البدرى