KVPY QUESTION PAPER-2017 (STREAM SA) - …

1 KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 quadrilateral has distinct integer side lengths. If the second-largest side has length 10, then the maximumpossible length of the largest side is[2017](A) 25(B) 26(C) 27(D) 28 Sol.[B]Let b = 10, c = 9, d = 8 Sum of three sides > IVth sideb + c + d > aa < largest power of 2 that divides !100!200 is[2017](A) 98(B) 99(C) 100(D) 101 Sol.[C]Exponent of 2 in 200 ! = 197 Exponent of 2 in 100 !

2 = a1, a2, a3, a4 be real numbers such that 24232221aaaa = 1. Then the smallest possible value of theexpression (a1 a2)2 + (a2 a3)2 + (a3 a4)2 + (a4 a1)2 lies in the interval[2017](A) (0, )(B) ( , )(C) ( , 3)(D) (3, )Sol.[Bonus] 1aaaa24232221 Smallest possible value of (a1 a2)2 + (a2 a3)2 + (a3 a4)2 + (a4 a1)2 = 0if a1 = a2 = a3 = a4 = 21 (It should be bonus) S be the set of all ordered pairs (x, y) of positive integers satisfying the condition x2 y2 = 12345678. Then[2017](A) S is an infinite set(B) S is the empty set(C) S has exactly one element(D) S is a finite set and has at least two [B]x2 y2 = 12345678 (x, y I+) is even, so x, y should be odd integerbut difference of square of two odd integers ismultiple of 8 but is not multiple of 8 CAREER POINT KVPY QUESTION PAPER-2017 (STREAM SA) Date : 05 /11/2017 KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd.

3 , CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 a nine-sided regular polygon with side length 2 units. The difference between the lengthsof the diagonals A1A5 and A2A4 equals[2017](A) 122 (B) 2 12(C) 6(D) 2 Sol.[D] A5 A6 A7 A8 A9 A1 A2 A3 A4 2 2 2 2 x x 92 O 4092 OAA21222x24xx92cos x2 cos 40 ) = x2 2x2 (1 cos 40 ) = 2 (i)Now 225122x2)AA(xx98cos (in A1OA5)(A1A5)2 = 2x2 98cos1= 2x2 (1 cos160 )= 4x2 sin2 80 A1A5 = 2x sin 80 ..(ii)Similarly in A2OA4A2A4 = 2x sin 40 ..(iii)(ii) (iii)A1A5 A2A4 = 2x (sin80 sin40 ) = 2 (using (i)) a1, a2.

4 An be n nonzero real numbers, of which p are positive and remaining are negative. The number ofordered pairs (j, k), j < k, for which ajak is positive, is 55. Similarly, the number of ordered pairs (j, k), j < k, forwhich ajak is negative is 50. Then the value of p2 + (n p)2 is[2017](A) 629(B) 325(C) 125(D) 221 Sol.[C]pC2 + n pC2 = 55552)1pn)(pn(2)1p(p ..(i)Also, p(n p) = 50 ..(ii) KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.)

5 , Ph: 0744-5151200 in (i)1101p50p50)1p(p 110p50p50pp22 110p50p50pp22 110p50p100p50p2 t2 t 210 = 0t = 15 or 14 (not true) 15p50p To find p2 + (n p)2 = 22p50p = 100p50p2 = 125 (using (iii)) a, b, c, d are four distinct numbers chosen from the set {1,2,3 ..,9}, then the minimum value of dcba is[2017](A) 83(B) 31(C) 3613(D) 7225 Sol.[D]72258192 72x. 48y = 6xy, where x and y are nonzero rational numbers, then x + y equals[2017](A) 3(B) 310(C) 3(D) 310 Sol.[D]xyyx64872 xyxyy4x3yx23223 KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd.

6 , CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 +y = (i) & 3x + 4y = (ii)From (i) & (ii) x = 3yput in (i) 5y = 3y2 35y so 310y2yx AB be a line segment of length 2. Construct a semicircle S with AB as diameter. Let C be the midpoint ofthe arc AB. Construct another semicircle T external to the triangle ABC with chord AC as diameter. The areaof the region inside the semicircle T but outside S is[2017](A) 2 (B) 21(C) 2 (D) 21 Sol.[B] C A B 2 1 1 1 O ACB = 90 AC = 2 Required area = Area of semicircle having ACas diameter Area under arc OAC but outsidetriangle AOC= 21422212= 212144 r(x) be the remainder when the polynomial x135 + x125 x115 + x5 + 1 is divided by x3 x.

7 Then[2017](A) r(x) is the zero polynomial(B) r(x) is a nonzero constant(C) degree of r(x) is one(D) degree of r(x) is twoSol.[C]x135 + x125 x115 + x5 + 1 = k(x3 x) + Ax2 + Bx + Cput x = 0 1C Put x = 13 = A + B + 1 KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 + B = (i)put x = 1 1 = A B + 1A B = (ii)(i) + (ii)A = 0B = is given that the number 43361 can be written as a product of two distinct prime numbers p1 , p2.

8 Further,assume that there are 42900 numbers which are less than 43361 and are co-prime to it. Then, p1 + p2 is[2017](A) 462(B) 464(C) 400(D) 402 Sol.[A]43361 = 131 ABC be a triangle with C = 90 . Draw CD perpendicular to AB. Choose points M and N on sides AC andBC respectively such that DM is parallel to BC and DN is parallel to AC. If DM = 5, DN = 4, then AC and BCare respectively equal to[2017](A) 541,441(B) 539,439(C) 538,438(D) 537,437 Sol.[A] A 5 t D B C y N 5 4 M x 4 z 41CD DNB~AMD tzx45y 20xy ..(i)In BCD(y + 5)2 = 41 + (ii)In ADC(x + 4)2 = 41 + (iii) 222tz41)4x(41)5y( 222x441)4x(415x20 (use (i)) KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd.

9 , CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 1641)4x(x41)x520(222 1641x816xx41x200x25400222 16x2 + 200x + 400 = 16x2 + 128x 400 32x2 72x 800 = 0 4x2 9x 100 = 0425x so 51625420y A, G and H be the arithmetic mean, geometric mean and harmonic mean, respetively of two distinctpositive real numbers. If is the smallest of the two roots of the equationA(G H)x2 + G (H A) x + H(A G) = 0, then[2017](A) 2 < < 1(B) 0 < < 1(C) 1 < < 0(D) 1 < < 2 Sol.[B]x = 1 is a root as sum of coefficient = 0 Now )HG(A)GA(H Put = 1 AHAGHGGAHAGHGHA2 =)HG(A)HG(G = 1AG [as > ] the figure, ABCD is a unit square.

10 A circle is drawn with centre O on the extended line CD and passingthrough A. If the diagonal AC is tangent to the circle, then the area of the shaded region is[2017] O D X C B A (A) 6 9 (B) 6 8 (C) 4 7 (D) 4 6 Sol.[D] KVPY EXAMINATION 2017 CAREER POINT CLASS XI (STREAM SA) CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 O D X C B A 45 45 1 1 1 1 OAC = 90 as AC is tangent and OA is radiusas CAD = 45 So OAD = 45 = AOD OA =2 Area of shaded region= 1 (area of sector OAX area of OAD)= 2142211= 4232141 sum of all non-integer roots of the equation x5 6x4 + 11x3 5x2 3x + 2 = 0 is[2017](A) 6(B) 11(C) 5(D) 3 Sol.