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KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2016 Date : 09-11-2016
Time: 3 Hours
Max. Marks : 160.
STREAM - SB/SX INSTRUCTIONS 1. 2. 3.
Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited. The Test Booklet consists of 120 questions. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. MARKING SCHEME :
PART-I : MATHEMATICS Question No. 1 to 20 consist of ONE (1) mark for each correct response. PHYSICS Question No. 21 to 40 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 41 to 60 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 61 to 80 consist of ONE (1) mark for each correct response. PART-II :
4.
5. 6. 7. 8.
MATHEMATICS Question No. 81 to 90 consist of TWO (2) marks for each correct response. PHYSICS Question No. 91 to 100 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 101 to 110 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 111 to 120 consist of TWO (2) marks for each correct response. Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. for Part-I 0.25 marks will be deducted for indicating incorrect response of each question and for Part-II 0.50 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the Answer sheet. No candidate is allowed to carry any textual material, printed or written, bits of papers, paper, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page. On completion of the test, the candiate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. Do not fold or make any stray marks on the Answer Sheet.
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PART-I : (One Marks Questions)
MATHEMATICS 1.
Sol.
2.
Sol.
The number of triples (x, y, z) of real numbers satisfying the equation x 4 + y4 + z4 + 1 = 4xyz is (A) 0 (B) 4 (C) 8 (D) more than 8 OkkLrfod la[;kvksa esa cuh dqy fdruh (x, y, z) frdfM;k¡ (triples) laHko gS] tks lehdj.k x4 + y4 + z4 + 1 = 4xyz dks larq"B djrh gS : (A) 0 (B) 4 (C) 8 (D) 8 ls T;knk x4 + y4 + z4 + 1 = 4xyz applying AM GM in L.H.S x 4 y 4 z 4 1 xyz 4 4 4 4 x + y + z + 1 4xyz Now equality holds iff x 4 = y4 = z4 = 1 x = ±1, y = ± 1, z = ±1 and xyz = 1 (x, y, z) = (1, 1, 1) or (1, – 1, –1) or (–1, 1, –1) or (–1, –1, 1) hence 4 triples one possible x option (B) Let P(x) be a polynomial with real coefficients such that P(sin2x) = P(cos2x), for all x [0, /2]. Consider the following statements : (I) P(x) is an even function (II) P(x) can be expressed as a polynomial in (2x – 1)2 (III) P(x) is a polynomial of even degree. then (A) all are false (B) only I and II are true (C) only II and III are true (D) all are true eku yhft, fd P(x) okLrfod xq.kkadksa ls cuk ,d cgqin (polynomial) gS] tks lHkh x [0, /2] ds fy, P(sin2x) = P(cos2x) dks larq"B djrk gS] fuEu okD;ksa dks i x 0 < x < 2 1 and f(x) < x < x 1 2 Also f(x) is increasing 1 1 further f(x) 0x 2 2 1 1 f(x) x ,1 2 2
(II) vuar x [0, 1] laHko gS ;fn limn fn(x) =
Sol.
(D) I, II, III ,oa IV
1 So x 0, 2 f(f(x)) > f(x) (f in increasing) f(f(f(x))) > f(f(x)) > f(x) Hence f n(x) increases as n increasing for a given x and it has an appear 1 1 1 based , so lim fn(x) = x 0, n 2 2 2 1 Similarly for x ,1 2 1 lim f n(1) = n 2 x
12.
The limit limx x2 e t
2
–x2
dt equals
0
x
limx x2 e t
2
–x2
dt dk eku D;k gksxk ?
0
(A)
1 3
(B) 2 x
Sol.
lim x
x
2
e
t3 – x3
(C)
(D)
2 3
dt
0
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PAGE - 7
x
x 2. e t
3
form e Applying L–H Rule, 0
= lim
x3
x
x
3
3
x 2 .e x 2x e t dt
0
lim
2
x
3x e
x3
x
e
1 2 + . lim x 3 3x
t3
dt
0
ex
3
1 3 Hence option (A) is correct
=
x
2
Alter : lim x . x
x
e = lim
x
t3
e
t3 – x3
dt form
0
x3
dt
0
2
e /x Apply L–H Rule 3
lim
x
13.
Sol.
14.
Sol.
ex x4 3
3
x2 (3x 2 .e x ) – e x (2x)
= lim
x
1 x4 = 4 3 3x – 2x
The polynomial equation x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 has (A) exactly one real root for any real a (B) three real root for any real a (C) three real roots for any a 0, and exactly one real root for any a < 0 (D) three real roots for any a 0, and exactly one real root for any a > 0 cgqin lehdj.k x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 dk (A) okLrfod a ds fy, dsoy ,d okLrfod e3wy laHko gS (B) okLrfod a ds fy, rhu okLrfod ewy laHko gS (C) a 0 ds fy, rhu okLrfod ewy ,oa a < 0 ds fy, dsoy ,d okLrfod ewy laHko gS (D) a 0 ds fy, rhu okLrfod ewy ,oa a > 0 ds fy, dsoy ,d okLrfod ewy laHko gS x3 – 3ax2 + (27a2 + 9)x + 2016 = 0 Let, 3 2 2 y = x – 3ax + (27a + 9)x + 2016 dy >0xR dx f(x) is monotonically increasing function Hence exactly one real root exist for all values of 'a' Hence option (A) is correct The area of the region bounded by the curve y = |x3 – 4x2 + 3x| and the x-axis, 0 x 3, is oØ y = |x3 – 4x2 + 3x| ,oa x-v{k, 0 x 3 ds chp dk {ks=kQy D;k gksxk ? 37 9 37 (A) (B) (C) (D) 0 6 4 12 3 2 y = |x – 4x + 3x| = |x(x – 1)(x – 3)|
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PAGE - 8
1
0
Area =
3
1
3
3 2 ( x – 4x 3x ) dx +
(x
0
=
3
– 4 x 2 3 x ) dx
1
1 4 3 80 104 24 – + – 4 3 2 4 3 2
5 8 37 + = 12 3 12 Hence option (C) is correct
=
1
15.
2
The number of continuous functions f : [0, 1] [0, 1] such that f(x) < x for all x and
f ( x) dx = 0
(A) 0
(B) 1
(C) 2
1 is 3
(D) infinite 1
lrr Qyu f : [0, 1] [0, 1] dh dqy la[;k,¡ D;k gksxh tks x ds lHkh eku ds fy, f(x) < x2 vkSj f ( x) dx = 0
1 dks 3
larq"V djrh gS ? Sol.
(A) 0 2 f(x) < x 1
(C) 2
(D) infinite
1
f ( x) dx <
0
f ( x) dx < 0
x
2
dx
0
1
(B) 1
1 3
Hence number such function exists Hence option (A) is correct 16.
On the real line R, we define two functions f and g as follows : f(x) = min {x – [x], 1 – x + [x]}, g(x) = max {x – [x], 1 – x + [x]}, Where [x] denotes the largest integer not exceeding x. The positive integer n for which n
(g( x ) – f ( x )) dx = 100 is 0
,d okLrfod js[kk R ij nks Qyu f ,oa g bl izdkj ls ifjHkkf"kr gS : f(x) = min {x – [x], 1 – x + [x]}, g(x) = max {x – [x], 1 – x + [x]}, n
tgk¡ [x] vf/kdre iw.kkZad gS tks x ls vf/kd ugha gS (g( x ) – f ( x )) dx = 100 esa n dk D;k eku gksxk 0
Sol.
(A) 100 (B) 198 f(x) = min. { x},1 – { x} g(x) = max {{x}, 1 – {x}} g(x) – f(x) will be periodic with period = 1
(C) 200
(D) 202
1
0
1/2 1
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PAGE - 9
1
shaded area is =
(g( x ) – f ( x )) dx 0
1 = 2 n
(g( x ) – f ( x )) dx = 0
n 2
n = 200 Hence option (C) is correct
17.
Let v be a vector in the plane such that | v – i | = | v – 2 i | = | v – j |. Then | v | lies in the interval
;fn ,d ry (plane) esa v ,d ,slk lfn'k gS tks | v – i | = | v – 2 i | = | v – j | dks larq"V djrk gS] rks | v | dk varjky (interval) D;k gS ? Sol.
(A) (0, 1] (B) (1, 2] |v – i | = |v – 2 i | = |v– i | Let v = a ˆi + b ˆj
(C) (2, 3]
(D) (3, 4]
(a – 1)2 + b2 = (a – 2)2 + b2 = a2 = a2 + (b – 1)2 on solving we set, 3 3 a= and b = 2 2 9 9 9 Hence | v | = = 4 4 2 | v | (2, 3) Hence option (C) is correct 18.
A box contains b blue balls and r red balls. A ball is drawn randomly from the box and is returned to the box with another ball of the same colour. The probability that the second ball drawn from the box is blue is ,d cDls esa b uhys jax ds vkSj r yky jax ds xsans gSA ;kn`PN :i ls cDls esa ,d xsan fudkyh tkrh gS vkSj mlh jax
dh nwljh xsan ds lkFk okil cDls esa Mky nh tkrh gS bldh izkf;drk D;k gksxh fd nwljh xsan tc cDls ls fudkyh tk, rks og uhys jax dh gks ? (A) Sol.
19.
Sol.
b r b
(B)
b2 (r b ) 2
(C)
b 1 r b 1
(D)
b(b 1) (r b)(r b 1)
blue balls b Re d balls r Case-I : Blue, Blue Case-II : Red, Blue b (b 1) r b b 2 b rb b Prob. = × + × = = (b r ) (b r 1) (b r ) (b r 1) (b r )(b r 1) b r Hence option (A) is correct
The number of noncongruent integer-sided triangles whose sides belong to the set {10, 11, 12, ...... 22} is iw.kkZad Hkqtkvksa okys vlokZaxle (noncongruent) f=kHkqtksa dh dqy la[;k,¡ fdruh gksxh] ;fn mudh Hkqtk,¡ leqPp; {10, 11, 12, ...... 22} ds vo;o gks ? (A) 283 (B) 446 (C) 448 (D) 449 13 number of scalene = C3 – 3 = 283 13 number of isosceles = ( C2× 2) – 4 = 152 number of equilateral = 13C1 = 13 total no. of D = 448 Hence option (C) is correct ®
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20.
Suppose we have to cover the xy-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes : equilateral triangle, square, regular pentagon, regular hexagon. Then the tiling can be done with tiles of (A) all four shapes (B) exactly three of the four shapes (C) exactly two of the four shapes (D) exactly one of the four shapes eku yhft;s fd xy-ry dks ,dleku VkbZyks (tiles) ls bl izdkj Yyh ds fo"k; esa vlR; gS\ (A) ;g izksVhu vkSj fyfiM ls fey dj cuh gksrh gSA (B) ifj/kh; izksVhu bl f>Yyh ls f'kfFkyrk ls tqM+s gksrs gSaA (C) vfuok;Z izksVhu fyfiM dh f}Lkrg ds vkj ikj tkrs gSaA (D) fyfiM vkSj f>Yyh ds izksVhu] lajpukRed vkSj fØ;kRed vlefefr ugha iznku djrs gSaA Ans.
(D)
65.
The percentage of sunlight captured by plants is (A) 2-10% (B) 10-20% (C) 60-80%
(D) 100%
ikS/ks lw;Z ds izdk'k dk fdruk izfr'kr xzg.k djrs gS\a Ans. 66.
Ans.
(A) 2-10% (A)
(B) 10-20%
(C) 60-80%
(D) 100%
The hard outer layer of pollens, named exine, is made of (A) cellulose (B) Tapetum (C) Sporopollenin
(D) pectin
ijkxd.kksa dh dBksj cká lrg] ,Dt+kbu] fuEu esa ls fdldh cuh gksrh gS\ (A) lSY;wykst+ (B) VSihVe (C) Liksjksysfuu
(D) isfDVu
(C)
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PAGE - 25
67.
Insectivorous plants such as Venus fly trap catch and digest insects in order to supplement the deficiency of (A) sulphur (B) nitrogen (C) potassium (D) Phosphorus
dhV Hk{kh ikS/ks] tSls dh ohul ¶ykbZ VªSi] fuEufyf[kr esa ls fdldh deh ls iwjk djus ds fy, dhVksa dks idM+ dj ikpu djrs gSa\ (A) lYQj (B) ukbVªkstu (C) iksVSf'k;e (D) QkWLQksjl Ans.
(B)
68.
Which of the following statements about nucleosome is TRUE ? (A) It consist of only DNA (B) It is a nucleus-like structure found in prokaryotes. (C) It consists of DNA and proteins (D) It consists of only histone proteins
U;wfDy;kslkse ds lanHkZ esa dkSulk dFku lR; gS \ (A) blesa dsoy DNA ik;k tkrk gSa (B) ;g iwoZdsanzdh; dksf'kdkvksa esa ik;h tkus okyh dsUnzd tSlh lajpuk gS (C) blesa DNA vkSj izksVhUl nksuksa gh ik, tkrs gSa (D) blesa dsoy fgLVksUl ik;s tkrs gSa Ans.
(C)
69.
Epithelial cells in animals are held by specialized junctions, one of them being "Gap junction". Function of a "Gap junction" is to (A) facilitate cell-cell communication by rapid transfer of small molecules (B) cement the neighbouring cells (C) stop substances from leaking (D) provide gaps in the tissue to facilitate uptake of small molecules across tissues tUrqvksa esa midyk dksf'kdk,¡ vkil esa ,d nwljs ls fo'ks"k izdkj ds tksM+ksa (junctions) tSls dh xSi (gap) tksM+ ls
tqM+s gksrs gSa] xSi tksM+ dk dke D;k gS\ (A) NksVs v.kqvksa ds Rofjr LFkkukUrj.k }kjk dksf'kdkvksa ds e/; lapkj lqyHk djkukA (B) iM+kslh dksf'kdkvksa dks ,d LFkk ck¡/ks j[krh gSA (C) inkFkksZa ds fjlko dks jksdrh gSA (D) Årdksa esa fjfDr (gap) iznku dj muds vkj ikj NksVs v.kqvksa dk mn~xzg.k (uptake) lqyHk djkukA Ans.
(A)
70.
Which of the following statements is TRUE about glandular epithelium in salivary gland? (A) It consists of isolated single cells. (B) It consists of multicellular cluster of cells. (C) Its secretions are endocrine. (D) It consists of squamous epithelial cells.
xzafFk;ksa dh xzaFkh; midyk ds fo"k; esa dkSulk dFku lR; gS \ (A) blesa foykfxr ,dy dksf'kdk,¡ gksrh gSA (B) buesa cgqdksf'kdh;] dksf'kdkvksa ds xqPNs gksrs gSaA (C) buds òko] var%òkoh gksrs gSa] (D) buesa 'kYdh; midyk dksf'kdk,¡ gksrh gSA Ans.
(B)
71.
Which one of the following ion pairs is involved in nerve impulses? + + + + + 2+ (A) Na , K (B) Na , Cl (C) K , Cl (D) K , Ca fuEufyf[kr esa ls dkSulk vk;u ;qXe] raf=kdk vkosxksa (impulses) esa dk;Zjr gS \ (A) Na+, K+ (B) Na+, Cl(C) K+, Cl(D) K+, Ca2+ (A)
Ans. 72.
Which of the following hormones that controls blood pressure is secreted by human heart? (A) Erythropoetin (B) Atrial natriuretic factor (C) ACTH (D) Glucocorticoid
fuEufyf[kr esa ls fdl gkWeksZUl] tks fd jDr pki dks fu;af=kr djrk gS] dk òko.k ekuo ân; ls gksrk gS \ (A) bfjFkzksiksbZfVu (B) ,Vªh;y uSVªh;qjsfVd QSDVj (C) ACTH (D) XyqdksdksVhZDok;M Ans.
(B) ®
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73.
Oxytocin and vasopressin are synthesized in (A) hypothalamus (B) adrenal gland (C) pituitary gland (D) ovary
vkWDlhVkslhu vkSj oSlksizsflu dk fuekZ.k dgk¡ gksrk gS \ (A) gkbiksFkSysel (B) vf/ko`Dd xzafFk (C) ih;w"k xzafFk (D) v.Mk'k; Ans.
(C)
74.
If you exhale multiple times into a conical flask containing lime water through a single inlet fixed through a stop cock, lime water will. (A) become cooler (B) turn milky (C) remain unchanged (D) turn yellow ,d 'kaDokdkj ¶ykLd] ftlesa pwuk ikuh (lime water) gS] ds eq[k ij jks/kuh yxkdj ,dy fuxZe (single outlet) }kjk dbZ ckj mPNo'ku (exhale) djrs gSa bl izfØ;k ls pwuk ikuh esa dSlk cnyko vk;sxk \ (A) B.Mk gks tk,xk (B) nwf/k;k gks tk,xk (C) dksbZ cnyko ugha gksxk (D) ihyk gks tk,xk (B)
Ans. 75.
The path of passage of stimulus when you accidentally touch a hotplate is (A) receptor brain muscles (B) muscles spinal cord receptor (C) muscles brain receptor (D) receptor spinal cord muscles
vpkud xeZ ros dks Nw ysus ij mn~nhiu ds izokg dk iFk fuEu esa ls dSlk gksxk \ (A) xzkgh efLr"d ek¡lisf'k;k¡ (B) ek¡lisf'k;k¡ es: jTtw xzkgh (C) ek¡lisf'k;k¡ efLr"d xzkgh (D) xzkgh es: jTtw ek¡lisf'k;k¡ Ans.
(D)
76.
In the presence of glucose and lactose. Escherichia coli utilizes glucose. However, lactose also enters the cells because (A) low level of permease is always present in the cell (B) it uses the same transporter as glucose (C) it diffuses through the bacterial cell membrane (D) it is transported through porins Xywdkst+ vkSj ySDVkst+ dh mifLFkfr esa Escherichia coli , Xywdkst+ dk mi;ksx djrk gS] fQj Hkh ySDVkst+ dksf'kdk ds
vUnj izos'k djrk gS] bldk D;k dkj.k gS \ (A) dksf'kdk ds vUnj ijfe;st+ dh lw{e ek=kk igys ls gh ekStwn gksrh gSA (B) ySDVkst] Xywdkst ds okgd dk gh mi;ksx djrk gSA (C) ySDVkst] thok.kq dh dksf'kdk f>Yyh ls vklkuh ls folfjr gks tkrk gSA (D) ySDVkst dk ogu iksjhu izksVhu ds ek/;e ls gksrk gSA Ans.
(A)
77.
Passive immunization is achieved by administering (A) heat killed vaccines (B) toxoids (C) live attenuated vaccines (D) antibodies
fuEu esa ls fdlds lekos'ku ls fuf"Ø; izfrj{kk izkIr dh tk ldrh gS \ (A) Å"ek&e`r Vhdk (B) fo"k (toxoids) (C) thfor {kh.k Vhdk (D) izfrj{kh Ans.
(D)
78.
Which of the following anions neutralize the acidic pH of the chyme that enters into the duodenum from the stomach? (A) H2PO 4 (B) HSO4 (C) HCO3 (D) CH3 COO fuEu esa ls dkSulk _.k&vkos'k (anions) vkek'k; ls xzg.kh esa izos'k djus okys vEyh; pH okys dkbZe (chyme) ds
mnklhuhdj.k ds fy, mÙkjnk;h gS \ Ans.
(A) H2PO 4 (C)
(B) HSO4 ®
(C) HCO3
(D) CH3 COO
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79.
If 14CO2 is added to a suspension of photosynthesizing chloroplasts, which of the following will be the first compound to be radioactive? (A) ATP (B) NADPH (C) NADH (D) 3-phospho glycerate ;fn gfjr yod ds ,d izyacu (suspension) esa] tks izdk'k la'ys"k.k dj jgk gS] 14CO2 feyk;k tkrk gS rks
fuEufyf[kr esa ls dkSu lk izFke jsfM;k/kehZ ;kSfxd gksxk\ Ans. 80.
Ans.
(A) ATP (D)
(B) NADPH
(C) NADH
(D) 3-QkWLQks fXyljsV
Which of the following species makes the largest true flower in the world? (A) Amorphophallus titanium (B) Rafflesia arnoldii (C) Nelumbo nucifera (D) Helianthus annuus fuEu esa ls dkSu lh iztkfr fo'o ds lcls fo'kky lR; iq"i (true flower) dk fuekZ.k djrh gS\ (A) Amorphophallus titanium (B) Rafflesia arnoldii (C) Nelumbo nucifera (D) Helianthus annuus (B)
PART-II : (Two Marks Questions)
MATHEMATICS 81.
Sol.
The remainder when the polynomial 1 + x2 + x4 + x6 + ..... + x22 is divided by 1 + x + x2 + x3 + .... + x11 is ;fn cgqin 1 + x2 + x4 + x6 + ....... + x22 dks 1 + x + x2 + x3 + .... + x11 ls Hkkx fn;k tk, rks 'ks"k D;k gksxk (A) 0 (B) 2 (C) 1 + x2 + x4 + ....... + x10 (D) 2(1 + x2 + x4 + ...... + x10) 2 4 6 22 1 x x x ...... x 1 x x 2 x 3 ...... x 11 (1 – x 24 )(1 – x ) = [G.P. formula] (1 – x 2 )(1 – x12 ) 1 x12 1 x remainder on division we get = 2 hence option (B) is corrects
=
82.
1 1 The range of the polynomial p(x) = 4x3 – 3x as x varies over the interval – , is 2 2 1 1 ;fn cgqin p(x) = 4x3 – 3x, esa x dk eku – , vUrjky esa gks rks cgqin dk ijkl (range) fuEu esa ls dkSu lk gS ? 2 2
(A) [–1, 1]
(B) (–1, 1]
Sol.
(C) (–1, 1)
1 1 (D) – , 2 2
+1 1/2 –1/2 –1
f(x) = 4x3 – 3x 2 = x(4x – 3) f'(x) = 0 1 1 x= ,– 2 2 1 1 if x – , 2 2 f(x) (–1, 1) Hence option (C) is correct ®
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83.
Ten ants are on the real line. At time t = 0, the k-th ant starts at the point K2 and travelling at uniform speed, reaches the point (11– k)2 at time t = 1. The number of distinct times at which at least two ants are at the same location is nl phfV;k¡ ,d okLrfod js[kk ij gS t = 0 le; ij k-phaVh fcUnq K2 ls 'kq: djds leku pky ls pyrs gq, t = 1 le; esa fcUnq (11– k)2 ij igq¡prh gS fdrus fHkUu le;ksa ij nks phfV;k¡ ,d gh fcUnq ij ikbZ tk,axh ? (A) 45 (B) 11 (C) 17 (D) 9
Sol.
x
O
Initial positions of ants were (1, 0), (4, 0), (9, 0), ........, (64, 0), (81, 0), (100, 0) at t = 1 there respective position are (100, 0), (88, 0), ..........., (9, 0), (4, 0), (1, 0) 84.
A wall is inclined to the floor at an angle of 135°. A ladder of length is resting on the wall. As the ladder slides down. its mid-point traces an are of an ellipse. Then the area of the ellipse is fp=k esa n'kkZ, vuqlkj ,d nhokj (wall) Q'kZ ls 135° dks.k ij >qdh gS yEckbZ dh ,d lh 0 ds fy, f'(x) 2f(x), rks [0, ) esa (A) f(x) ges'kk fLFkj Qyu gS (B) f(x) dsoy c
