Story Transcript
fnYyh v/khuLFk lsok p;u vk;ksx izf'kf{kr Lukrd f'k{kd (efgyk oxZ) ijh{kk] 2018 gy iz'u&i= 1. lehdj.k x3 – 13x2 + 15x + 189 = 0 dks
gy
2. (C) fdUgha
nks ifjes; la[;kvksa ds chp ,d vkSj ifjes; la[;k gksrh gS] bls ?kuRo xq.k dgrs gSaA
djas] ;fn ,d ewy nwljs ls 2 vf/kd gSA (A) (3, –7, –9)
(B) (–3, 7, 9)
(C) (3, 7, 9)
(D) (–3, –7, –9)
1. (B) leh- x3 – 13x2 + 15x + 189 = 0
esa ewy a, a + 2, b gSA
— x 2 dk x.qkkad a x 3 dk xq.kkd 2a + 2 + b = 13
a + a + 2 + b =
b = 11 – 2a
...(i)
a(a + 2) + (a + 2) b + ab
x dk x.qkkad = 3 a x dk xq.kkd a2 + 2a + ab + 2b + ab = 15 a2 + 2a + 2b + 2ab = 15
0 1/8 1/4 3/8 1/2 3. fuEufyf[kr esa ls dkSu&ls osDVj (lfn'kksa) dk lewg R3 esa jSf[kd :i ls Lora= gSA
(1) (2) (3) (4)
(B) 2 vkSj 3
(C) 1 vkSj 4
(D) 3 vkSj 4
3. (B) lHkh osDVj dk ;fn Determinant (lkj&
f.kd) ;fn 0 gks] rc osDVj jSf[kd :i ls Lora= gksaxsA
a + 2a + 2(11 – 2a) + 2a (11 – 2a) 2
a2 + 2a + 22 – 4a + 22a – 4a2
= 15
3a2 – 20a – 7 = 0
[(1,0,0), (0,1,0), (1,1,0)] [(1,0,0), (0,1,0), (0,0,1)] [(0,1,0), (1,0,1), (1,1,0)] [(0,0,1), (0,1,0), (0,1,1)]
(A) 1 vkSj 2
= 15
tSlsµ 0 vkSj 14 ds chp 18 la[;k gSA
;fn] 1 0 0
lewg (1) = 0 1 0 1 1 0
3a – 21a + a – 7 = 0
= 1 (0) – 0 – 0
3a (a – 7) + 1(a – 7) = 0
=0
2
(a – 7) (3a + 1) = 0
a–7=0
;k 3a + 1 = 0
a=7
;k
tc a = 7 rc leh- (i) ls
1 a = – 3
b = 11 – 2(7)
;g jSf[kd :i ls Lora= ugha gSA
= 1(1 – 0) – 0 + 0 =1
= 11 – 14
β=–3
\ a + 2 = 7 + 2 = 9
leh- ds ewy 7, 9, –3 gSaA
2. fdUgha
nks ifjes;] la[;kvksa ds chp ,d vkSj ifjesa;] la[;k gksrh gS] bls dgk tkrk gSA
(A) vkfdZfeMh;u (B) vlekurk (C) ?kuRo
xq.k
(fo"kerk)
xq.k
(D) e/;&eku
xq.k
lewg
1 0 0 (2) = 0 1 0 0 0 1
0 1 0
lewg (3) ÷ 1 0 1 1 1 0
= 0 – 1(0 – 1) + 0 =1 0 0 1
lewg (4) ÷ 0 1 0 0 1 1
= 0 – 0 + 1 (0) =0
vr% fodYi (2) rFkk (3) lR; gaS
ijh{kk frfFk % 22-11-2018 (f}rh; ikyh) 4. 143 vkSj 481 dk m, n cjkcj gSµ (A) – 9, 3
e-l- 143m + 481n gS] rc (B) – 8, 3
(C) – 10, 3
(D) – 7, 3
4. (C) 143 rFkk 481 dk
e-l-
481 = 143 × 3 + 52 143 = 52 × 2 + 39 52 = 39 × 1 + 13 39 = 13 × 3 + 0 143 481 3 429 52 143 2 104 39 52 1 39 13 39 3 39 ×
...(1) ...(2) ...(3) ...(4)
leh- (3) ls] 13 = 52 – 39 × 1 13 = 52 – (143 – 52 × 2) × 1 [leh- (2) ls] 13 = 52 – 143 × 1 + 52 × 2 = 52 × 3 – 143 × 1 = (481 – 143 × 3) × 3 – 143 × 1 [leh- (1) ls] = 481 × 3 – 143 × 9 – 143 × 1 = 481 – 3 – 143 × 10 = 143 × (–10) + 481 × 3 ...(5)
fn;k x;k] 13 = 143m + 48n dh rqyuk leh- (5) ls djus ij] m = – 10, n = 3 2 –6 5. b 3 l
dks fdl la[;k ls foHkkftr fd;k tk, –6 fd HkkxQy b 32 l ds cjkcj gSµ
2 6 (B) b 3 l 2 12 3 12 (C) b 3 l (D) b 2 l 5. (D) ekuk] la[;k = x 3 6 (A) b 2 l
iz'ukuqlkj]
b2l 3 x
–6
3 –6 = b 2 l
b2l 3 –6 = x b3l 2 –6
isij | 1
–6 b 2 × 2 l = x 3 3
2 b 3 l
2×– 6
–2+5+6–9=0
10. 360 ds
= x
2 –12 b 3 l = x
3 12 x = b 2 l V R SSS 1 1 1 WWW 6. vkO;wg SS a b c WW dk in gSµtgk¡ a = b SS 2 2 2WW Sa b c W X T ≠ c gSµ (A) 0 (B) 1 (C) 2 (D) 3 1 1 1 6. (C) a b c a2 b2 c2
0 0 0 1 1 1 a+b b+c c+a
gS %
(A) 7 (C) 0
cjkcj
8. lehdj.kksa 2x3 + 5x2 – 6x – 9 = 0 vkSj 3x3 + 7x2 – 11x – 15 = 0 ds nks mHk;fu"B ewy gS]
3 (B) b1, 2 l –3 (C) (1, –3) (D) b –1, 2 l 8. (A) lehdj.k 2x3 + 5x2 – 6x – 9 = 0 (A) (–1, –3)
2 |
x = – 1 – 3 + 7 + 11 – 15 = 0
x = – 1 j[kus
ij]
lHkh xq.ku[kaMksa dk ;ksx = 1170. Alternative Method : 360 = 23 × 32 × 51
\
xq.ku[k.Mksa dh la[;k
= (3 + 1) (2 + 1) ( 1 + 1) = 4 × 3 × 2 = 24
xq.ku[k.Mksa dk ;ksx
=
(x + 1) (3x2 + 4x – 15) (x + 1) (3x2 + 9x – 5x – 15) (x + 1) [3x(x + 3) –5 (3x – 1)] x = – 1, – 3, 5/3 nks u ka s lehdj.kks a ds mHk;fu"B (– 1, – 3) gSA
ew y
/kukRed iw.kk±d ds fy, 5
1 1 1 7. (A) 1 2 3 ! 0 1 4 k 1(2k – 12) + 1(3 – k) + 1(4 – 2) ≠0 2k – 12 + 3 – k + 2 k – 7 ≠ 0 k ≠ 7
45, 60, 72, 90, 120, 180 vkSj 360.
(x + 1) (2x2 + 3x – 9) (x + 1) (2x2 + 6x – 3x – 9) (x + 1) (x + 3) (2x – 3) 3 x = –1, –3, 2 leh- 3x3 + 7x2 – 11x – 15 = 0
9. n izR;sd
(B) 6 (D) 5
Kkr dhft,A
(D) 12, 810
2 3 + 1 - 1 3 2 + 1 - 1 51 + 1 - 1 2 -1 × 3 -1 × 5 -1 15 26 24 = 1 × 2 × 4 = 15 × 13 × 6 = 1170
ewyksa ds
?kuksa dk ;ksxQy Kkr dhft,A
0 0 0 = a–b b–c c–a ]a – bg]a + bg ]b – cg]b + cg ]c – ag]c + ag
vr% vkO;wg dk in 2 gSA 7. fuEufyf[kr ;qxir~ lehdj.k gS % x + y + z = 3 x + 2y + 3z = 4 x + 4y + kz = 6 k ds fy, ,d vf}rh; gy ugha gksxkA
(B) 24, 1080
(C) 24, 1170
11. lehdj.k x3 – 6x2 + 11x – 6 = 0 ds
0 0 0 a–b b–c c–a a2 – b2 b2 – c2 c2 – a2
= (a – b)(b – c)(c – a)
(A) 12, 540
10. (C) 360 ds xq.ku[kaM = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40,
C 1 → C 1 – C 2, C 2 → C 2 – C 3, C 3 → C3 – C1 =
xq.ku[k.Mksa dh la[;k vkSj mudk ;ksx
gS %
3
n n7 + n5 + 23n – 105 gS % (A) fo"ke iw.kk±d (B) iw.kk±d (C) ½.kkRed okLrfod la[;k (D) ifjes; la[;k
n n 7 n 5 2n 3 9. (B) 7 + 5 + 3 – 105 tgk¡ n /kukRed iw.kk±d gSA
15n 7 + 21n5 + 70n3 – n 105 tc n = 1
...(i)
15 + 21 + 70 – 1 105 105 = 105 =1
tks ,d iw.kk±d gSA blh izdkj] tc n = 2 leh- (i) esa j[kus ij]
15 × 128 + 21 × 32 + 70 × 8 – 2 105 1920 + 672 + 560 – 2 = 105 3150 = 105 = 30 tks Hkh ,d iw.kk±d gSA
=
(A) 31 (B) 32 (C) 35 (D) 36 11. (D) fn;k x;k] leh- x3 – 6x2 + 11x – 6 = 0 ...(i) leh- (i) esa x = 1 j[kus ij ⇒ 1 – 6 + 11 – 6 = 0 vr% (x – 1), x3 – 6x2 + 11x – 6 = 0 dk ,d xq.ku[k.M gSA rc]
(x – 1) (x2 – 5x + 6) (x – 1) (x2 – 3x – 2x + 6) (x – 1) (x – 2) (x – 3) leh- ds rhuksa ewy 1, 2 rFkk 3 gaSA
iz'ukuqlkj] leh- ds rhuksa ewyksa ds ?kuksa dk ;ksx =
13 + 23 + 33 = 1 + 8 + 27 = 36 12. x esa f}?kkr cgqin dks tc (x – 1), (x – 2), (x – 3) ls foHkkftr fd;k tkrk gSA rc Øe'k% 4, 4, 0 'ks"kQy cprk gSA f}?kkr cgqin Kkr
dhft;sA
(A) –2x2 + 6x + 5 (B) –2x2 + 6x (C) –2x2 + 6x – 5 (D) –2x2 + 6x + 3
12. (B) ekuk] f (x) = ax2 + bx + c dksbZ
f}?kkr
tc cgqin f (x) dks x – 1 ls Hkkx nsrs gaSA
rc]
a2 = ^3 + 2 2 h = 9 + 8 + 12 2 2 a = 17 + 12 2
f (1) = 4
a + b + c = 4
...(1)
f (2) 4a + 2b + c = 4
...(2)
f (3) 9a + 3b + c = 0
...(3)
leh- (3) dks leh- (2) ls ?kVkus ij]
...(4) blh] izdkj leh- (2) dks leh- (1) esa ls ?kVkus ij] – 3a – b = 0 leh- (4) vkSj (5) ls]
...(5)
a dk
eku leh- (5) esa j[kus ij 6 – b = 0
b = 6
eku leh- (i) esa j[kus ij c = 0 \ f}?kkr cgqin] ax2 + bx + c a rFkk b dk
= –2x2 + 6x 13. ;fn a ,d iw.kZ la[;k vkSj p ,d vHkkT; la[;k gS] rks Fermat dh izes; ds vuqlkj% (A) 2p – 1 – 2, p ls foHkkT; gSA (B) 2p – 1, p ls foHkkT; gSA (C) 2p – 2, p ls foHkkT; ugha gSA (D) 2p – 2, p ls foHkkT; gSA 13. (D) Fermat dh
izes; vuqlkj]
ap ≡ a(mod p)
2p ≡ 2(mod p)
rc fodYi (D) lR; gksxkA
1 1 14. ;fn a = , b= 3–2 2 3+ 2 2 a2 + b2 dk eku gSµ (A) 36 (B) 37 (C) 34 (D) 35 14. (C) fn;k
gS] rks
x;k gS]
1 a = 3–2 2
rFkk
1 b= 3+ 2 2
3+ 2 2 1 × 3– 2 2 3+ 2 2 3+ 2 2 = 9–8 7]a + bg]a – bg = a2 – b2A a =
(C) iz R ;s d
a = 3 + 2 2
2
cgqin gSA
...(1)
blh izdkj] b=
3–2 2 1 × 3+ 2 2 3– 2 2
3–2 2 9–8 b = 3 – 2 2 b2 = 9 + 8 – 12 2 b2 = 17 – 12 2 ...(2) leh- (1) o (2) dks tksM+us ij] a2 + b2 = 17 + 12 2 + 17 – 12 2 = 34 15. ;fn ,d lfn'k lef"V V3 ls okLrfod la[;k R3 ds lHkh f=Hkqtksa dk leqPp; (x1, x2, x3) gS] rc ,d mèokZ/kj ry y = x ls fu:fir b =
mi&lef"V leqPp;ksa ds jSf[kd la;kstu }kjk izkIr dh tk ldrh gSµ (A) (1, 1, 0) vkSj (0, 0, 1) (B) (1, 0, 1) vkSj (0, 0, 1) (C) (1, 0, 0) vkSj (0, 1, 0) (D) (1, 1, 0) vkSj (1, 0, 0) 15. (A) iz'u esa fn;k x;k gS] fd R3 ,d lfn'k lef"V gS rFkk milef"V {(x1, x2, x3) ∈
R3 : y = x}
rc] lHkh f=xq.kksa esa ls fodYi (A) bldks lUrq"V djrk gSA 16. ;fn µ y ;wyj dk VksfV;aV Qyu gS] rc y f (92) gS%
(A) 44 (B) 46 (C) 48 (D) 42 16. (A) j(92) 92 = 23 × 2 × 2 j(23). j(4) ;wyj dks VksfV;aV Qyu ds vuqlkj] j(P) = P – 1 (tgk¡ P ,d vHkkT; la[;k gSA) = 22 × 2 = 44 17. 39312 dks fdrus rjhdkas ls nks xq.ku[kaMksa esa
foHkkftr fd;k tk ldrk gS] tks ,d&nwljs ls vHkkT; gSµ (A) 6 (B) 10 (C) 8 (D) 4 17. (C) 39312 = 24 × 33 × 7 × 13 ;s pkjksa i`Fkd la[;k gS] nh xbZ
la[;k ds xq.ku[kMa gSA rjhdksa dh la[;k = 23 = 8. 18. fuEufyf[kr esa ls dkSu&lk vlR; gSµ (A) 2 ,d ifjes; la[;k gSA (B) izR;sd vifjes; la[;k ,d okLrfod la[;k gSA
iw . kk± d ,d okLrfod la [ ;k gSA (D) izR;sd ifjes; la[;k ,d okLrfod la[;k gSA 18. (D) okLrfod l[;k,¡ ifjes; la[;k,¡ ugha gksrh] mUgsa vifjes; la[;k dgrs gSa] tSlsµ 2 , 8 , p vkfnA 19. pkbuht 'ks"kQy izes; }kjk X = 3(mod 5), x = 5(mod 7) dks gy djsaA mHk;fu"B gy gSµ (A) X = 33 (mod 35) (B) X = 31 (mod 5)
(C) X = 27 (mod 35) (D) X = 28 (mod 35) 19. (A) X = 3(mod 5) X = 5 (mod 7) M = m1 m 2 = 5 × 7 = 35 M 35 Z1 = m1 = 5 Z1 = 7
blh izdkj]
Z2 = 5 yi = (Zi)–1 (mod Mi) 1 y1 = 7 (mod 5) 7y1 = 1 (mod 5)
y1 = 3 1 y2 = 5 ]mod 7g 5y2 = 1 (mod 7)
y2 = 3
X = (a1 y1 Z1 + a2 y2 Z2) (mod M) = (3 × 3 × 7 + 5 × 3 × 5)
(mod 35)
= 63 + 75 (mod 35) = 138 (mod 35) X = 33 mod (35) 20. fuEufyf[kr
esa ls dkSu&lk leqPp; milef"V
W = )=
x y G, x + 2y + t = 0, y + t = 0 3 0 t
dk vk/kkj gSµ –1 1 G3 (A) )= 2 –1 1 –1 G3 (B) )= 0 1 1 0 0 1 0 0 (C) )= G, = G, = G3 0 0 0 0 0 1 2 1 1 –1 G, = G3 (D) )= 0 –1 0 1
isij | 3
20. (B) W
,d nks vk;keh milef"V gSA ge tkurs gSa] fd nks jSf[kd Lora= osDVj dsoy vkèkkj ij gSA vr% fodYi 1 ∉ W fodYi 3 ∉ W fodYi esa 2 ls T;knk vk;keh milfe"V gSA vr% fodYi (B) lR; gSA 21. lekUrj
Js.kh ds izFke 20 inksa dk ;ksxQy gS] ftldk izFke in 5 vksj lkoZvUrj 4 gSµ
(A) 830
(B) 850
(C) 820
(D) 860
(
dk xq.kkad ) x dk xq.kkd a
— x2
(a + b)2 = a2 + b2 + 2ab (12)2 = a2 + b2 + 2 × 32 a2 + b2 = 144 – 64 a2 + b2 = 80
α2 + β2 80 rc] α + β = 12 20 = 3 23. ;fn a1x + b1y + c1 = 0 vkSj a2x + b2y + c2 = 0 esa vuUr gy gS] rks fuEufyf[kr esa dkSu&lk c ! c12 c ! c12 c = c1 2 c1 =c 2
1 :0, 1 D (C) :– 2 , 1D (D) 2 24. (C) a2 + b2 + c2 = 1 (fn;k gS)
leh- (1) rFkk (2) dk xq.kk djus ij
⇒ 19 (a + b + c) b 1a + 1b + 1c l >
1 1 1 1/ 3 (abc)1/3 b a . b . c l
a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca > 0 1 – (ab ≠ bc + ca) > 0
ab + bc + ca < 1
1 ab + bc + ca :– 2 , 1D
ds vUrjky esa
lehdj.k ax + bx + c = 0 ds nks ewy ]a – bg ]b – cg (a – b) vkSj (b – c) gS] rc c–a dk eku gSµ 2
b c (A) b (B) c bc ab (C) a (D) c 25. (A) leh- ax2 + bx + c ds nks ewy (a – b) rFkk (b – c) gSaA (a – b) + (b – c) = – ba b α + β = –b l a –b a – c = a ...(1) c (a – b) (b – c) = a b αβ = c l a ]a – bg]b – cg c a c – a = a × b c = b
1 1 1 26. ]a + b + cg b a + b + c l (A) 9 ls de vkSj cjkcj
gSµ
(B) 3 ls
1 (abc)1/3 ]abcg1/3 1 + 1 + 1l ^a + b + c h b a b c $9
⇒
;k 9 ls vf/kd vkSj cjkcj
27. ;fn Øfer ;qXe lehdj.k 2x – 3y = 18 vkSj 4x – y = 16 dks lUrq"V djrk gS] lehdj.k 5x – py – 23 = 0 dks Hkh lUrq"V djrk gSA rc p dk eku Kkr dhft,µ (A) –1
(B) 1
(C) 2
(D) –2
27. (C) fn;s
fLFkr gSA
25. ;fn
⇒ 19 (a + b + c) b 1a + 1b + 1c l >
2
x;s leh-
2x – 3y = 18
...(1)
4x – y = 16
...(2)
leh- (1) eas 2 dk xq.kk djds leh- (2) ls ?kVkus
–20 5 ,y=–4 y dk eku leh- (2) esa j[kus ij 4x – (– 4) = 16 4x + 4 = 16 4x = 16 – 4 4x = 12 x = 3 x rFkk y dk eku leh- 5x – py = 23 esa j[kus ij] 5 × 3 – p × (4) = 23 28. ;fn
26. (C) ge
(A) 2 1 (C) 2
tkurs gaS] lekUrj ek/; > xq.kksÙkj ek/; a + b + c > (abc)1/3 ...(1) vkSj
3
y =
de vkSj cjkcj (C) 9 ls vf/kd vkSj cjkcj (D) 3 ls vf/kd vkSj cjkcj
4 |
2
2(a2 + b2 + c2) – 2(ab + bc + ca) > 0
ab = 32
b ! b1 2 b1 =b 2 b1 !b 2 b1 =b 2
1 :–1, 1 D (A) : 2 , 1D (B) 2
nks
esa fLFkr gSµ
1 + 2 (ab + bc + ca) > 0 1 ab + bc + ca > – 2 (a – b)2 + (b – c)2 + (c – a)2 > 0
a + b = 12
a (A) a1 2 a1 (B) a 2 a1 (C) a 2 a1 (D) a 2
a1 b1 c1 a2 = b2 = c2 24. ;fn a2 + b2 + c2 = 1 rc ab + bc + ca vUrjky
ds nks ewy a rFkk b gSA
lR; gSµ
vuUr gy ds fy;s
(a + b + c ) + 2 (ab + bc + ca) > 0
= 860
a + b =
1 b 1 + 1 + 1 l b 1 1 1 l1/3 3 a b c > a b c ...(2)
a2x + b2y + c2 = 0
2
= 10 × 86
a1x + b1y + c1 = 0
(a + b + c)2 > 0
21. (D) lekUrj Js.kh ds n inksa dk ;ksxQy Sn = 2n 62a + ]n – 1g d @ izFke in a = 5 rFkk lkoZvUrj d = 4 20 = 2 62 × 5 + ]20 – 1g × 4@ = 10[10 + 19 × 4]
22. ;fn a vkSj b leh- x2 – 12x + 32 = 0 ds α2 + β2 ewy gS] rc α + β dk eku gSµ 8 8 (A) 3 (B) –3 20 20 (C) – 3 (D) 3 22. (D) fn;k x;k leh- x2 – 12x + 32 = 0
23. (D)
15 + 4p = 23 4p = 23 – 15 82 p = 4
p=2
lekÙkj Js.kh ds 16 inksa dk ;ksx 1624 gS] vkSj izFke in lkokZUrj dk 500 xquk gS] rc lkokZUrj Kkr dhft,A 1 (B) 5 (D) 5
28. (B) lkoZUrj = d
31. (D) fn;k
iz'ukuqlkj]
gSµ
Sn = 4n2 + 3n
Sn – 1 = 4(n – 1)2 + 3(n – 1)
203 = 1015d 203 d = 1015 1 d = 5 29. ;fn lehdj.k x2 + px + 12 = 0 dk ,d ewy 4 gS] tcfd lehdj.k x2 + px + q = 0 ds nksuksa ewy cjkcj gS] rc q dk eku gSµ (A) 4 (B) 12 49 (C) 3 (D) 4 29. (B) leh- x2 + px + 12 = 0
dk ,d ewy a = 4
(4)2 + p × 4 + 12 = 0
16 + 4p + 12 = 0
4p = – 28
p = – 7
nksuksa ewy cjkcj gSaA
x2 + px + q = 0 ds
rc] lkekU; in an = Sn – Sn – 1
c2 + b2 = 2a2
= 4n2 + 3n – [4(n2 + 1 – 2n) + 3n – 3]
35. eqds'k
= 4n + 3n – 4n – 4 + 8n – 3n + 3 2
2
= 8n – 1 32. ;fn
ifjes; xq.kkadkas okys f}?kkr lehdj.k dk ,d ewy 7 – 4 gS] rks f}?kkr lehdj.k gSµ (A) x2 – 2 7 x – 9 = 0 (B) x2 – 8x + 9 = 0 (C) x2 + 8x + 9 = 0 (D) x2 – 2 7 x + 9 = 0 32. (A) igyk
ewy] a = 7 – 4
nwljk
ewy] b = 7 + 4
a + b =
ab = ^ 7 – 4 h^ 7 + 4 h
= ^ 7 h – ]4 g 2
2
= 7 – 16
x2 – 2 7 x – 9 = 0
(A) x – 7 = 0 3
(B) x3 – 5x2 + 7x – 3 = 0 (C) x3 + 7x2 – 3 = 0 (D) x3 + 7x2 + 3 = 0 30. (B) fn;k
x;k gS]
a + b + g = 5
ab + bg + ga = 7
abg = 3
⇒ x3 – (a + b + g)x2 + (ab + bg + ga) x – abg = 0
x3 – 5x2 + 7x – 3 = 0
31. lekUrj
Js