Đề kiểm tra học kì 1, môn toán, lớp 11, THPT chuyên doc

§Ò kiÓm tra häc kú 1 M«n: To¸n 11 (chuyªn) A. §¹i sè (6 ®iÓm): Bµi 1 (2 ®iÓm): ): T×m c¸c giíi h¹n sau: a, lim x →−1 x3 + 3 x 2 + 4 x + 2 sin( x + 1) 3 b, lim 8 − x − 2 x + 1 x→0 x Bµi 2 (2 ®iÓm):Gi¶i c¸c ph¬ng tr×nh sau: a, 32+ x − x − 3x − x + 8 = 0 2 2 b, log 3 ( x − 1) = 8 x − 3x 2 − 3 2 2x +1 Bµi 3 (2 ®iÓm): Cho d·y sè (un) tháa m·n: u1 = 5, u2 = 12, u3 = 32 un+3 = 4un+ 2 - un+1 - 6un + 4 T×m un = ? B. H×nh häc §¸p ¸n §Ò kiÓm tra häc kú 1 M«n: To¸n 11 (chuyªn) Bµi 1(2 ®iÓm): C©u a,(1 ®iÓm) Ta cã: lim x →−1 x3 + 3 x 2 + 4 x + 2 ( x + 1)( x 2 + 2 x + 2) = lim x →−1 sin( x + 1) sin( x + 1) lim ( x + 2 x + 2) 2 = x →−1 lim x +1→ 0 sin( x + 1) x +1 1 1 = =1 C©u b,(1 ®iÓm) Ta cã: 8 − x − 2 x +1 = ( 3 8 − x − 2) + 2(1 − x + 1) lim lim x→0 x→0 x x 3 = lim 8 − x − 2 + 2 lim 1 − x + 1 x→0 x →0 x x 3 0.25 0.5 0.25 0.25 0.25 = lim x→0 8 − x − 23 + 2xlim →0 1 − ( x + 1) x(1 + x + 1) x ( 3 (8 − x) 2 + 2 3 8 − x + 4) 1 1 = - lim - 2xlim 2 x→0 3 3 → 0 (8 − x) + 2 8 − x + 4 1+ x +1 1 1 13 =- -2 =12 2 12 Bµi 2(2 ®iÓm) C©u a(1 ®iÓm) §Æt t = 3x − x , t > 0, ph¬ng tr×nh trë thµnh 0.25 0.25 2 9t - 1 + 8 = 0 ⇔ 9t2 + 8t - 1 = 0 t  t = −1 < 0 ⇔ 1 t =  9 1 ph¬ng tr×nh ⇔ 3x2 − x = 3−2 ⇔ x2 - x = -2 9 2 ⇔ x − x + 2 = 0 ph¬ng tr×nh v« nghiÖm Víi t = x ≠ 1  C©u b(1 ®iÓm) §iÒu kiÖn:  1  x > − 2 2 ph¬ng tr×nh ⇔ log3 ( x − 1) + 1 = 8 x − 3x 2 − 2 2x + 1 3( x − 1) 2 = 2 x + 1 − 3( x − 1) 2 2x +1 ⇔ log 3 [3( x − 1) 2 ] + 3( x − 1) 2 = log 3 (2 x + 1) + 2 x + 1 (*) ⇔ log 3 §Æt f(t) = log3t + t, ta cã f(t) lµ hµm ®ång biÕn vµ (*) ⇔ f [3( x − 1)2 ] = f (2 x + 1) ⇔ 3( x − 1) 2 = 2 x + 1  4 + 10 x = 3 ⇔ 3x 2 − 8 x + 2 = 0 ⇔  , tháa ®iÒu kiÖn  4 − 10 x = 3  C©u 3(2 ®iÓm) Theo bµi ra ta cã Un+3 - 1 - 4(Un+2 - 1) + Un+1 - 1 + 6(Un - 1) = 0 §Æt Vn = Un - 1, ta cã: V1 = 4, V2 = 11, V3 = 31 vµ Vn+3 - 4Vn+2 + Vn+1 + 6Vn = 0 Gièng ®Ò 1, cã Vn = 2n −1 + 3n ⇒ U n = 2n−1 + 3n + 1 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25  ... Un+3 - - 4(Un+2 - 1) + Un+1 - + 6(Un - 1) = §Æt Vn = Un - 1, ta cã: V1 = 4, V2 = 11, V3 = 31 vµ Vn+3 - 4Vn+2 + Vn+1 + 6Vn = Gièng ®Ò 1, cã Vn = 2n −1 + 3n ⇒ U n = 2n−1 + 3n + 0.25 0.25 0.25 0.25