Đề thi thử đại học môn Toán năm 2015 (14)

2 y CT 1 2 cos 2 x cos 2 10x 1 3 cos11x.sin 9x z 1 i w =1+ z + z 3 + z 4 4 x 9 x = 5(ln x 1)(3 x 2 x ) + + = + + + − − = + − − ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ∈ I x −x − x dx ∫ ⊥ AB = a,BC = a 3 6 3 P x y + z = + = x − y = ( nx 5 1 x 2 ) n n C n n min , , P x y z y z x 3 2 . z x y z z 2 xy yz zx 2 y CT 1 2 y ' = 2x 3 − 2m 2 x; y' = 0 ⇔ x = 0 x 2 = m 2 ⎡ ⎣ ⎢ ⎢ m = 0 ⇒ y' = 2x 3 ; y ' = 0 ⇔ x = 0 x = 0; y CT = 1 ≠ 1 2 m ≠ 0 ±m; y CT = y(±m) = 1− m 4 2 1− m 4 2 = 1 2 ⇔ m 4 = 1 ⇔ m = −1;m = 1 m = −1;m = 1 cos 2 x + cos 2 10x = 1+ 3 cos11x.sin 9x z = 1+ i w =1+ z + z 3 + z 4 1+ cos2x +1+ cos20x 2 = 1+ 3 cos11x.sin9x ⇔ cos2x + cos20x 2 = 3 cos11x.sin9x ⇔ cos11x.cos9x = 3 cos11x.sin9x ⇔ cos11x = 0 tan9x = 1 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⇔ x = π 22 + k π 11 x = π 54 + k π 9 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ,k ∈ ! x = π 22 + k π 11 ;x = π 54 + k π 9 ,k ∈ ! w =1+ (1+ i ) + (1+ i ) 3 + (1+ i) 4 = −4+ 3i ⇒ w = 5 4 x 9 x = 5(ln x 1)(3 x 2 x ) x > 0 (2 x −3 x )(2 x + 3 x ) = 5(ln x −1)(3 x − 2 x ) ⇔ (2 x −3 x )(2 x + 3 x + 5ln x −5) = 0 ⇔ 2 x = 3 x 2 x + 3 x + 5ln x −5 = 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⇔ x = 0(l ) x = 1(t / m) ⎡ ⎣ ⎢ ⎢ x = 1 (x + y)( 1 xy + 3) = 6(x 2 + y 2 ) + 4 2(x 2 + y 2 ) 4− x 2 − y 2 = 2 2xy + 2− x 2 − y 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ (x, y ∈ !) xy > 0,x 2 + y 2 ≤ 2 3(x + y − 2(x 2 + y 2 )) + 1 x + 1 y − 4 2(x 2 + y 2 ) = 0 ⇔ 3(x + y − 2(x 2 + y 2 )) + 2(x 2 + y 2 )(x + y)−4xy xy 2(x 2 + y 2 ) = 0 ( 2(x 2 + y 2 ) − x − y) −3+ 2(x 2 + y 2 )(x + y)−4xy xy 2(x 2 + y 2 )( 2(x 2 + y 2 ) − x − y) ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ = 0 ⇔ ( 2(x 2 + y 2 ) − x − y) A−3 ( ) = 0 (*) A = ( 2(x 2 + y 2 )(x + y)−4xy)( 2(x 2 + y 2 ) + x + y) xy 2(x 2 + y 2 )(x − y) 2 = ( 2(x 2 + y 2 )(x + y)−4xy)( 2(x 2 + y 2 ) + x + y) xy 2(x 2 + y 2 )(x − y) 2 = 2(x + y) + 2(x 2 + y 2 ) xy 2(x 2 + y 2 ) xy ≤ x 2 + y 2 2 ≤1 2(x + y) + 2(x 2 + y 2 ) xy 2(x 2 + y 2 ) ≥ 3(x + y) xy 2(x 2 + y 2 ) ≥ 6 2xy(x 2 + y 2 ) ≥ 6 2.1.2 = 3 (*) ⇔ x = y = 1 x = y ⎡ ⎣ ⎢ ⎢ x = y =1 x = y 4− 2x 2 = 2 2x + 2− 2x 2 ⇔ ( 2x −1)( 2x + 3)+ ( 2− 2x 2 −1) = 0 ⇔ ( 2x −1)( 2x + 3− 1+ 2x 2− 2x 2 +1 ) = 0 ⇔ ( 2x −1)( 2x 2− 2x 2 + 2+ 2 2− 2x 2 ) = 0 ⇔ 2x −1= 0 ⇔ x = 1 2 (x; y) = ( 1 2 ; 1 2 ) 2(x 2 + y 2 ) − x − y (x − y) 2 (x + y)(1+ 7 20xy ) = 10(x 2 + y 2 ) +7 5 2(x 2 + y 2 ) 2− x 2 − y 2 = 2 2xy −1 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ (x; y) = ( 1 2 ; 1 2 ) (2− x )(2− y) + x 2 + y 2 2 = 2 2− x 2 − y 2 + 2 2xy = 4− x 2 − y 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ (x; y) = ( 1 2 ; 1 2 ) HD: (2− x )(2− y) = 2− x 2 + y 2 2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 2 ⇔ (x − y) 2 2 2(x 2 + y 2 ) + x + y − 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = 0 ⇔ x = y 2 2(x 2 + y 2 ) + x + y − 1 2 ≥ 2 2+ 2 − 1 2 = 0 I = (x +1) −x 2 −2x + 3 dx 0 1 ∫ t = 3− x 2 − 2x ⇒ t 2 = 3− x 2 − 2x ⇒ 2tdt = (−2x − 2)dx = −2(x +1)dx I = t.(−t dt ) 3 0 ∫ = t 2 dt 0 3 ∫ = t 3 3 3 0 = 3 ⊥ AB = a,BC = a 3 6 3 OC = AC 2 = a 2 + 3a 2 2 = a SO = SC 2 −OC 2 = 6a 2 − a 2 = a 5 V S .ABCD = 1 3 SO.S ABCD = 1 3 .a 5.a.a 3 = a 3 15 3 IH SO = CH CO = CI CS = 1 3 ,IH = 1 3 SO = a 5 3 d (SB;AI ) = d(SB;(AIM )) = d (B;(AIM )) = 2d (C ;(AIM )) = 2. CA HA d (H ;(AIM )) = 12 5 d (H ;(AIM )) HK ⊥ (AIM ) ⇒ HK = d (H ;(AIM )) S AMC = 1 2 S ABC = 1 2 . 1 2 .a.a 3 = a 2 3 4 ,AM = a 2 + 4a 2 3 = a 21 3 d (C ;AM ) = 2S AMC AM = 2. a 2 3 4 a 21 3 = 3a 7 14 HE = 5 6 d (C ; AM ) = 5 6 . 3a 7 14 = 5a 7 28 1 HK 2 = 1 IH 2 + 1 HE 2 = 9 5a 2 + 112 25a 2 = 157 25a 2 ⇒ HK = 5a 157 d (SB;AI ) = 12 5 HK = 12a 157 (P ) : x − 2y + 2z −1= 0 d (I ;(P )) = 1− 2.2+ 2.(−1)−1 1 2 + (−2) 2 + 2 2 = 2 (S ) : (x −1) 2 + (y− 2) 2 + (z +1) 2 = 4 H ( 1 2 ;1; m −1 2 ) 1 2 − 2.1+ 2. m −1 2 −1 = 0 ⇔ m − 7 2 = 0 ↔ m = 7 2 ⇒ M (0;0;− 7 2 ) !"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=% ">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%% 03A%?AW?6%X<?3@A8IOYZ8! 6! 0[R%\%]^_C%SA`=aY%Trong!mặt!phẳng!với!trục!toạ!độ!Oxy!cho!hình!chữ!nhật!ABCD!có!tâm!I.!Gọi! M!là!trung!điểm!cạnh!CD,!H!là!hình!chiếu!vuông!góc!của!D!trên!AM!và!N!là!trung!điểm!AH.! Tìm!toạ!độ!các!đỉnh!B,D!biết!phương!trình!đường!tròn!ngoại!tiếp!tam!giác!IMN!là! (C ) : (x − 5 2 ) 2 + ( y− 9 2 ) 2 = 5 2 !và!đỉnh!D!có!hoành!độ!nguyên!nằm!trên!đường!thẳng! x − 2y = 0 .! ! Đường!tròn!(C)!có!tâm J ( 5 2 ; 9 2 ) ,!bán!kính! R = 10 2 .! !!+)!Ta!có!tứ!giác!IMDN!nội!tiếp!đường!tròn:! Chứng)minh.) Gọi!E!là!trung!điểm!đoạn!HD,!ta!có! NE = IM = 1 2 AD NE / /IM ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ !nên! tứ!giác!IMEN!là!hình!bình!hành.! Do!đó!E!là!trực!tâm!tam!giác!MND,!và! EM ⊥ ND ⇒ IN ⊥ ND .!!! Hay!góc! IND ! = IMD ! = 90 0 ,!do!vậy!IMDN!nội!tiếp!đường!tròn.!! +)!Do!D!thuộc!(C)!nên!toạ!độ!D!thoả!mãn!hệ: (x − 5 2 ) 2 + (y− 9 2 ) 2 = 5 2 y − 2x = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ x = 3, y = 6 x = 8 5 , y = 16 5 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⇒ D(3;6) .! +)!Do!J!là!trung!điểm!của!ID!nên!I(2;3),!và!I!là!trung!điểm!của!BD!nên!B(1;0).! !W?%@RP86!Vậy! B(1;0),D(3;6) .!! ! ! 03b%JY!Nếu!cho!thêm!giả!thiết!điểm! H ( 63 17 ; 82 17 ) !ta!tìm!được: A(−1;2),B(1;0),C (5;4),D(3;6) .!! 0[R%D%]C_c%SA`=aY!Tìm!số!hạng!chứa! x 10 !trong!khai!triển! ( nx 5 − 1 x 2 ) n !biết!n!là!số!tự!nhiên!thoả! mãn! n 2 +C n 2 = 145 (n ≥ 2) .!! +)!Ta!có:! ! n 2 + n(n −1) 2 = 145 ⇔ 3n 2 −n −290 = 0 ⇔ n =10(t / m) n = − 29 3 (l ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ .! +)!Vì!vậy! ( nx 5 − 1 x 2 ) n = (2x − 1 x 2 ) 10 = C 10 k (2x ) 10−k (− 1 x 2 ) k k=0 10 ∑ = C 10 k .(−1) k .2 10−k .x 10−3k k=0 10 ∑ .! Chọn! 10−3k = 10 ⇔ k = 0 suy!ra!số!hạng!cần!tìm!là! C 10 0 .2 10 = 1024 .!!!!! ! !"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=% ">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%% 03A%?AW?6%X<?3@A8IOYZ8! 7! 0[R%^C%]^_C%SA`=aY!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn! z = min x, y, z { } .!Tìm!giá!trị!nhỏ! nhất!của!biểu!thức! P = x y + z + y z + x + 3 2 . z x + y − z z 2 + xy + yz + zx .! Đặt! a = x z ,b = y z (a,b ≥1) !khi!đó:! ! P = a b +1 + b a +1 + 3 2(a + b) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ − 1 (a +1)(b +1) .! +)!Ta!có:! (a +1)(b +1) ≥ 2,∀a,b ≥1 ⇒− 1 (a +1)(b +1) ≥− 1 2 ,!và!! ! a b +1 + b a +1 + 3 2(a + b) = (a + b +1)( 1 b +1 + 1 a +1 ) + 3 2(a + b) − 2 ≥ 4(a + b +1) a + b + 2 + 3 2(a + b) − 2 .! Đặt! t = a + b ≥ 2 ,!ta!có:!!! a b +1 + b a +1 + 3 2(a + b) ≥ f (t) = 4(t +1) t + 2 + 3 2t − 2 ≥ f (2) = 7 4 .! Do!đó! P ≥ f (t) − 1 2 ≥ 5 4 .!!Dấu!bằng!đạt!tại! x = y = z .!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.!!! 0dN3%G6!Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!! z z 2 + xy + yz + zx = z (z + x)(z + y) ≤ 1 2 z z + x + z z + y ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ .! Suy!ra!! x y + z + y z + x − z z 2 + xy + yz + zx ≥ x y + z + y z + x − 1 2 z z + x + z z + y ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = 1 2 2 y − z z + x + 2x − z z + y ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ .! Sử!dụng!bất!đẳng!thức!Cauchy!–!Shcwarz!ta!có:! ! 2y − z z + x + 2x − z z + y = (2y − z) 2 (2y − z)(z + x ) + (2x − z ) 2 (2x − z )(z + y) ≥ (2y − z + 2x − z) 2 (2y − z)(z + x )+ (2x − z )(z + y) = 4(x + y − z ) 2 4xy + z(x + y) −2z 2 ≥ 4(x + y − z ) 2 (x + y) 2 + z(x + y)− 2z 2 = 4(x + y − z ) x + y + 2z .! Vì!vậy! P ≥ 2(x + y − z) x + y + 2z + 3 2 . z x + y .!Đặt! t = x + y z ≥ 2 ta!có! P ≥ f (t ) = 2(t −1) t + 2 + 3 2t .! Xét!hàm!số!f(t)!ta!có: f '(t) = 6 (t + 2) 2 − 3 2t 2 = 9t 2 −12t −12 t 2 (t + 2) 2 = 3(t − 2)(3t + 2) t 2 (t + 2) 2 ≥ 0,∀t ≥ 2 .!! ! ! !"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=% ">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%% 03A%?AW?6%X<?3@A8IOYZ8! 8! ! Vì!vậy! P ≥ f (t) ≥ f (2) = 5 4 .!Dấu!bằng!đạt!tại! x = y = z .!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.! e:A%?PU%?Qf89%?g%h%% e:A%Oi%C^6%Cho!x,y,z!là!các!số!thực!dương!thoả!mãn! x, y, z ≤1; x + y ≥ z +1 .!Tìm!giá!trị!nhỏ!nhất! của!biểu!thức! P = x y + z + y z + x − z z 2 + xy + yz + zx .!Đ/s:! P min = 1 2 .!! e:A%Oi%CG6!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn! z = min x, y, z { } .!Tìm!giá!trị!nhỏ!nhất!của! biểu!thức! P = x y + z + y z + x + 5 4 . z x + y − z z 2 + xy + yz + zx .!Đ/s:! P min = 9 8 .!!!!! ! ! ! . D(3;6) .! +)!Do!J!là!trung!điểm!của!ID!nên!I(2;3),!và!I!là!trung!điểm!của!BD!nên!B(1;0).! !W?%@RP86!Vậy! B(1;0),D(3;6) .!! ! ! 03b%JY!Nếu!cho!thêm!giả !thi t!điểm! H ( 63 17 ; 82 17 ) !ta!tìm!được: A(−1;2),B(1;0),C (5;4),D(3;6) .!! 0[R%D%]C_c%SA`=aY!Tìm!số!hạng!chứa!