Transcript
GENERAL INSTRUCTIONS ã The Test Booklet consists of 120 questions. ã There are Two parts in the question paper. The distribution of marks subjectwise in eachpart 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 : 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. Date : 27-10-2013 Duration : 3 Hours Max. Marks : 160 KISHORE VAIGYANIK PROTSAHAN YOJANA - 2013 STREAM - SB/SX KVPY QUESTION PAPER - STREAM (SB / SX) PART-IOne Mark QuestionsMATHEMATICS 1. The sum of non-real roots of the polynomial equation x 3 + 3x 2 + 3x + 3 = 0.(A) equals 0(B) lies between 0 and 1(C) lies between –1 and 0 (D) has absolute value bigger than 1 Sol. f(x) = x 3 + 3x 2 + 3x + 3 = 0f ’(x) = 3x 2 + 6x + 3 = 3(x + 1) 2 ! 0f(x) is Increasing functionNowf( – 3) = – 6 < 0 f( – 2) = 1 > 0 real root # lies between – 3 and – 2 $ + % + # = – 3;– 3 < # < – 2 $ + % – 3 < $ + % + # < $ + % – 2 $ + % – 3 < – 3 < $ + % – 2 – 1 < $ + % < 0 Ans.(C)2. Let n be a positive integer such thatlog 2 log 2 log 2 log 2 log 2 (n) < 0 < log 2 log 2 log 2 log 2 (n).Let ! be the number of digits in the binary expansion of n. Then the minimum and the maximum possiblevalues of ! are(A) 5 and 16(B) 5 and 17(C) 4 and 16(D) 4 and 17 Sol. log 2 log 2 log 2 log 2 log 2 n < 0 < log 2 log 2 log 2 log 2 n16 < n < 2 16 no. of digits in the binary expansion of 16 is 5no. of digits in the binary expansion of 2 16 is 17so no. of digits in the binary expansion of n lies in 5 to 16 Ans. (A)3. Let ' be a cube root of unity not equal to 1. Then the maximum possible value of | a + bw + cw 2 | where a,b, c ( {+1, –1} is (A) 0(B) 2(C) 3 (D) 1+ 3 Sol. |a + bw + cw 2 ||a – c + (b – c)w| , for maximum valuetakinga = 1 , c = –1, b = 1 |a + bw + cw 2 | = |2 + 2w| = 2|w 2 | = 2 Ans. (B) KVPY QUESTION PAPER - STREAM (SB / SX) 4. If a, b are positive real numbers such that the lines ax + 9y = 5 and 4x + by = 3 are parallel, then the leastpossible value of a + b is(A) 13(B) 12(C) 8(D) 6 Sol. )* b94a ab = 36using AM ! G.M )!+ ab2ba a + b ! 12 Ans. (B)5. Two line segments AB and CD are constrained to move along the x and y axes, respectively, in such a waythat the points A, B, C, D are concyclic. If AB = a and CD = b, then the locus of the centre of the circlepassing through A, B, C, D in polar coordinates is(A) 4bar 222 +* (B) 4ba2cosr 222 ,*- (C) r 2 = 4(a 2 + b 2 )(D) r 2 cos2 - = 4(a 2 – b 2 ) Sol. 2cg 2 , = a2 cf 2 , = bPolar coordinates of centre of circle be (rcos - , rsin - )g = – r cos - and g 2 – f 2 = 4ba 22 , f = – r sin - r 2 cos 2 - = 4ba 22 , Ans. (B)6. Consider a triangle ABC in the xy -plane with vertices A = (0,0), B = (1,1) and C = (9, 1). If the line x = adivides the triangle into two parts of equal area, then a equals(A) 3(B) 3.5(C) 4(D) 4.5 Sol. C(19, 1)B(1, 1)A(0, 1)x = aQP P(a, 1)Q . / 012 3 9a,a area of PQC = 21 area of ABC )18( 21219a1)a9( 21 44*. / 012 3 ,, (9 - a) 2 = 36 ) a = 3 Ans. (A) KVPY QUESTION PAPER - STREAM (SB / SX) 7. Let ABC be an acute-angled triangle and let D be the midpoint of BC. If AB = AD, then tan(B)/tan(C) equals(A) 2 (B) 3 (C) 2(D) 3 Sol. Using M - N Rule(1 + 1) cot ( 5 - B) = 1.cot B - cot C3cotB = cot C CtanBtan = 3 Ans. (D)8. The angles $ , % , # of a triangle satisfy the equations 2sin $ + 3cos % = 23 and 3sin % + 2cos $ = 1. Thenangle # equals(A) 150 ° (B) 120 ° (C) 60 ° (D) 30 ° Sol. 2sin $ + 3cos % = 3 2 ....(1)3sin % + 2cos $ = 1....(2)sum of squares of equation (1) and (2)4 + 9 + 12 sin ( $ + % ) = 19sin ( $ + % ) = 21 ) $ + % = 150 ° or 30° If $ + % = 30 ° ) % = 30 - $ put in equation (1) and (2)we get 7 sin $ + 3 3 cos $ = 6 2 7 cos $ - 3 3 sin $ = 2cos $ = 37697 + = .8 < 23cos $ < cos 30 ° $ > 30 ° $ + % 6 30 ° Ans. (D)9. Let f : R 7 R be a function such that 87 x lim f(x) = M > 0. Then which of the following is false? (A) 87 x lim xsin(1/x) f(x) = M(B) 87 x lim sin(f(x)) = sin M(C) 87 x lim xsin(e –x ) f(x) = M(D) 0)x(f. xxsinlim x * 87 Sol. )x(f exe)esin(xlim)x(f)esin(x xlim xxxx ,,, 87*87 = 1 × (0) × M = 0 Ans. (C)
