Tuyển chọn các bài toán về hàm số

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(KHÓA LUYỆN THI 2015 – 2016)

Sách hay, chỉ TẶNG chứ không BÁN!

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CHỦ ĐỀ 1 TIẾP TUYẾN CỦA ĐỒ THỊ HÀM SỐ

DẠNG 1 TIẾP TUYẾN TẠI MỘT ĐIỂM THUỘC ĐỒ THỊ

Câu 1: [ĐVH] Cho hàm số 2

2

x y x

=

− , có đồ thị ( )C Vi ế t ph ươ ng trình ti ế p tuy ế n c ủ a ( )C t ạ i các giao đ i ể m c ủ a ( )C v ớ i đườ ng th ẳ ng y= 3x− 3

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x

x Tìm M thuộc ( )C sao cho ti ế p tuy ế n qua M c ắ t Ox và

Oy l ầ n l ượ t t ạ i A và B sao cho OA= 4OB.

m

m m

+

+ +

Ph ươ ng trình giao đ i ể m v ới Ox:

( ) 2( )

0

0 1 1

y

m

x m

m m

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1

y x

= +

a) Ph ươ ng trình tr ụ c hoành là: y= 0 Do đ ó y0 = 0 ⇒x0 = 2 Khi đ ó: ( )

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ph ươ ng trình ti ế p tuy ế n c ủ a ( )C qua M∈ ( )C bi ế t 5

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−

= + , có đồ thị ( )C Vi ế t ph ươ ng trình ti ế p tuy ến d củ a ( )C

bi ế t ti ế p tuy ế n song song v ớ i đườ ng th ẳ ng x− 28y+ = 4 0

( ) ( )

0

0 2

x

 = + =

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Ta có y' = 2x2− − 8x 1 Theo giả thiết, tiếp tuyến song song với đường thẳng 7x+ − =y 1 0 ⇒ y x' ( ) 0 = − 7

−

= + , có đồ thị ( )C Vi ế t ph ươ ng trình ti ế p tuy ế n c ủ a ( )C bi ế t

ti ế p tuy ế n vuông góc v ớ i đườ ng th ẳ ng 5x+ − =y 12 0

1

y x

= +

Theo gi ả thi ế t, ti ế p tuy ế n vuông góc v ớ i đườ ng th ẳ ng 5x+ − =y 12 0 ' ( )

0

1 5

4 4

x m

−

= + , có đồ thị ( )C m Tìm m đề ti ế p tuy ế n c ủ a ( )C m t ạ i giao

đ i ể m c ủ a đồ th ị hàm s ố v ớ i tr ụ c tung song song v ớ i đườ ng th ẳ ng 5x− +y 17 = 0

0

0 2

x

x

x m y

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0 4

2 0

3 0 8

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0 0

0 2

2 1

x x

x x

a)Vi ế t ph ươ ng trình ti ế p tuy ến d củ a ( )C t ạ i đ i ể m có hoành độ x0 = − 3

b) V ớ i đườ ng th ẳng d ở câu a hãy viế t ph ươ ng trình ti ế p tuy ế n ∆ c ủ a ( )C bi ế t ti ế p tuy ế n song song v ớ i đườ ng th ẳng d.

Do v ậ y ph ươ ng trình ti ế p tuy ế n là: y= 16 (x− + 2 ) 1 hay y= 16x− 31

DẠNG 3 TIẾP TUYẾN ĐI QUA MỘT ĐIỂM

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; 1

3

1 1

x

x x

3

1 1

x x x x

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• V ớ i x0 = 2 ph ươ ng trình ti ế p tuy ế n là: y= 20 (x− + 2 ) 4 hay y= 20x− 36

• V ớ i x0 = − 2 ph ươ ng trình ti ế p tuy ế n là: y= − 20 (x+ + 2 ) 4 hay y= − 20x− 36

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2

2 2

x x

2 2

x

x x

−

−

0 0

• V ớ i x0 = 1 ph ươ ng trình ti ế p tuy ế n là: y= 0 (x− + 1 ) 3 hay y= 3

• V ớ i x0 = + 1 3 ph ươ ng trình ti ế p tuy ế n là: y= − ( 9 6 3 )(x− − 1 3 ) − + 6 3 3 hay

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a)Vi ế t l ạ i đườ ng th ằng d: y= − + 2x 3

Xét ph ươ ng trình hoành độ giao đ i ể m ta có:

1 3

y= − −x

Câu 10: [ĐVH] Cho hàm s ố 2 1

, ( ) 1

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0 3

2 1

x

x x

0 2

1

2

x x

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x= > nên để ( )C giao d tạ i 3 đ i ể m phân bi ệ t, trong đ ó có đ úng hai đ i ể m có hoành độ

d ươ ng thì ph ươ ng trình g x( ) = 0 có 2 nghi ệ m trái d ấ u khác 3

2 1

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Để đồ th ị ( )C c ắ t tr ục Ox tạ i 2 đ i ể m phân bi ệ t ⇔ ( ) 1 có 2 nghi ệ m phân bi ệ t.

m −

b) Đểđồ th ị ( )C c ắ t tr ục Ox tạ i 3 đ i ể m phân bi ệ t ⇔ ( ) 1 có 3 nghi ệ m phân bi ệ t

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0.

3 2

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Với x= 2 ⇒y= 0 ⇒ A( ) 2; 0 ứng với đề bài đã cho.

Khi đó d và ( )C cắt nhau tại A( ) 2; 0 , , B C phân biệt ⇔ ( ) 1 có hai nghiệm phân biệt khác 2

th ẳ ng d: y = 3x+ 2 Tìm m để ( )C giao d tạ i 3 đ i ể m phân bi ệ t A( − − 1; 1 , , ) B C sao cho ,B C đố i

x ứ ng nhau qua đườ ng th ẳ ng x+ 3y− = 2 0

m

m y

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3

m m

th ẳ ng d: y = −x 7 Tìm m để ( )C giao d tạ i 3 đ i ể m phân bi ệ t A( − − 4; 11 , , ) B C sao cho di ệ n tích

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m x m x m x x m m y