Chuyen de van dung cao tich phan mon toan lop 12 ixavi (1)

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+ Một số trường hợp đặt t  ( )x khi đó ta giải được x  ( )t Thế ngược lại vào phương trình ta được f t( ) g( ( ))  t , khi đó ta có hàm số f x( ) g( ( ))  x

+ Trong một số trường hợp cần sử dụng các phép biến đổi thích hợp để đưa về dạng f( ( ))  x h( ( ))  x Khi đó hàm số cần tìm sẽ có hạng f x( ) h x( )

Dạng 2: Giả thiết cho a x f u x( ) ( ( )) b x f v x( ) ( ( ))  w( )x Phương pháp giải:

Đặt ẩn phụ u x( ) v t( ) để thu thêm một phương trình nữa đó là

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hàm hằng, tức là f(x)=1 khi đó giá trị tích phân là

2

a

Đáp án B

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3 2

3 2

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2 (1) 2(2) ( )

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1 1

Bài 8 Cho hàm số f(x) có f’(x) liên tục trên [0;1], và thỏa mãn

f x  f x   x x  Tính

1 0

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Giải thích chỗ này A+B tức là f(4)  f(3)  f( 2)   f( 3)   f(4)  f(3) Đáp án A

Bài 10 Cho hàm số f(x) có đạo hàm liên tục trên [0;1] thỏa mãn

1 (1) 0, ( '( )) 7, ( ) , ( )

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1

7 '( ) 7

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f x   I  dx Đáp án B

Bài 13 Cho hàm số y  f x( ) có đạo hàm và liên tục trên khoảng

(0;  ); ( )f x  0với mọi x thuộc khoảng (0;) và thỏa mãn

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nhất M và giá trị nhỏ nhất m của hàm số y  f x( ) trên đoạn [1;3] Biết rằng giá trị của biểu thức P  2M m có dạng a 11 b 3 c

(a, b,c ).Tính giá trị của biểu thức S a b c.

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1122

1201 45

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3 (1) 2

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2017 0

2017 0

2017 cos ln

2017 0

2017 sin ln

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Bài 25 Gọi là hình phẳng giới hạn bởi đồ thị của hàm số

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 

1 0

I   f x dx

A 3.

1

3

1 5

V V

5 27

5 19

5 24

5 32

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Bài 28 (Lam Sơn – Thanh Hóa) Cho hàm số f x có đạo hàm dương, liên  

tục trên đoạn   0;1 thỏa mãn f  0  1 và

1

53 50

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    1   1

1 1

0 1

3

I 

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0

9

4 9

4 3

2 3

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