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CLASS XII (STREAM SX) CAREER POINT

Kota : CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 1 KVPY EXAMINATION 2017 CAREER POINT CLASS XII (STREAM SX) KVPY QUESTION PAPER-2017 (STREAM SX) Date : 05 /11/2017 Part A-Mathematics 1. Let BC be a fixed line segment in the plane. The locus of a POINT A such that the triangle ABC is isosceles, is (with finitely many possible exceptional points) [2017] (A) a line (B) a circle (C) the union of a circle and a line (D) the union of two circles and a line Sol.

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Transcription of CLASS XII (STREAM SX) CAREER POINT

1 Kota : CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 1 KVPY EXAMINATION 2017 CAREER POINT CLASS XII (STREAM SX) KVPY QUESTION PAPER-2017 (STREAM SX) Date : 05 /11/2017 Part A-Mathematics 1. Let BC be a fixed line segment in the plane. The locus of a POINT A such that the triangle ABC is isosceles, is (with finitely many possible exceptional points) [2017] (A) a line (B) a circle (C) the union of a circle and a line (D) the union of two circles and a line Sol.

2 [D] Case (i) : AB C If B = C locus of A is bisector of BC So it is straight line Case (ii) : AB C If A = C BC fixed B(a, 0), C(0, a) BC = AB So, (x a)2 + y2 = 2a2 Circle Case (iii) : A = B AC = BC 22)ak(h = 2a2 x2 + (y a)2 = 2a2 also a circle So union of two circle and a line. 2. The number of solution pairs (x, y) of the simultaneous equations log1/3 (x + y) + log3 (x y) = 2 and 2y2 = 512x+1 is [2017] (A) 0 (B) 1 (C) 2 (D) 3 Sol. [B] )yx(log13 + log3(x y) = 2 log3(x + y) + log3(x y) = 2 log3 yxyx = 2 yxyx = 9 1x9y)2(22 )1x(9y222 y2 = 9(x + 1) Solve eliminate y 16x2 225x 225 = 0 x = 15, 1615 At x = 15, y = 12 x = 1615 , y = 43 (not possible) only sol.

3 X = 15, y = 12 only one sol. 3. The value of the limit xlim x2x x42 is [2017] (A) (B) 41 (C) 0 (D) 41 CAREER POINT Kota : CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 2 KVPY EXAMINATION 2017 CAREER POINT CLASS XII (STREAM SX) Sol. [D] Rationalise x2xx4x2xx4)x2xx4(lim222x x2x14|x|xlimx at x |x| = x x2x14xxlimx = 221 = 41 4. Let R be a relation on the set of all natural numbers given by a R b a divides b2. Which of the following properties does R satisfy ?

4 I. Reflexivity II. Symmetry III. Transitivity [2017] (A) I only (B) III only (C) I and III only (D) I and II only Sol. [A] (I) This relation is reflexive relation because every natural no. divides square of itself a R a a divides a2 (II) not symmetric eg. 5 R 10 5 Divide 100 But 10 R 5 10 Divide 25 (III) Not transitivity for example if 8 R 4 & 4 R 2 8 R 2 only (I) Option 5. The fractional part of a real number x is x [x], where [x] is the greatest integer less than or equal to x. Let F1 and F2 be the fractional parts of (44 2017)2017 and (44 + 2017)2017 respectively.

5 Then F1 + F2 lies between the numbers [2017] (A) 0 and (B) and (C) and (D) and Sol. [C] I + F2 = 2017442017 F2 = 2017442017 ; 0 < F2 < 1 I + F2 F2 = 2 ..)44(2017C201612017 F2 = F2 F2 = ( )2017 Now, F1 = 2017)201744( = 2017) ( Fractional part can not ve. So, F1 = 1 ( )2017 So, F1 + F2 = 1 1 lie Between & 6. The number of real solutions of the equation 2sin 3x + sin 7x 3 = 0 which lie in the interval [ 2 , 2 ] is [2017] (A) 1 (B) 2 (C) 3 (D) 4 Sol. [B] only possible when sin 3x = 1 & sin 7x = 1 sin 3x = 1 sin 3x = sin (4n + 1)2 , n I 3x = (4n + 1)2 x = (4n + 1)6 sin 7x = sin(4m + 1)2 , m I x = (4m + 1)14 for common solution (4n + 1)6 = (4m + 1)14 Solving these 1 = 3m 7n First solution is m = 5, n = 2 Second solution is m = 12, n = 5 So two solutions are possible 7.

6 Suppose p, q, r are real numbers such that q = p (4 p), r = q (4 q), p = r (4 r). The maximum possible value of p + q + r is [2017] (A) 0 (B) 3 (C) 9 (D) 27 Sol. [C] Add all these p + q + r = 3rqp222 for maximum value p = 3, q = 3, r = 3 Answer is 9. 8. The parabola y2 = 4x + 1 divides the disc x2 + y2 1 into two regions with areas A1 and A2. Then | A1 A2 | equals [2017] (A) 31 (B) 32 (C) 4 (D) 3 Kota : CAREER POINT Ltd., CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 3 KVPY EXAMINATION 2017 CAREER POINT CLASS XII (STREAM SX) Sol.

7 [B] A1 = dx1x4204/1 + dxx12102 A1 Solve A1 = 31 + 2 A2 = (1)2 A1 = 2 31 |A1 A2| = 32 9. A shooter can hit a given target with probability 41. She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval [2017] (A) ( , ) (B) ( , ) (C) ( , ) (D) ( , ) Sol. [D] 3rd time target will hit in sixth time So, In first 5 attempt these will be 3L, 2W and at 6th attempt shot will be hit So, 5C3343 241 41 = 4096270 = 10.

8 Consider the following events : E1 : Six fair dice are rolled and at least one die shows six. E2 : Twelve fair dice are rolled and at least two dice show six. Let p1 be the probability of E1 and p2 be the probability of E2. Which of the following is true ? [2017] (A) p1 > p2 (B) p1 = p2 = (C) p1 < p2 (D) p1 = p2 = Sol. [A] p1 = 1 (no die show six) 1 665 = p2 = 1 (no die shown two + one die shown two) p2 = 1 111112126165C65 = p1 > p2 11. For how many different values of a does the following system have at least two distinct solutions ?

9 Ax + y = 0 x + (a + 10) y = 0 [2017] (A) 0 (B) 1 (C) 2 (D) Infinitely many Sol. [C] 1a = )10a(1 a2 + 10a 1 = 0 two value of a 12. Let R be the set of real numbers and f : R R be defined by f(x) = 2]x[1}x{ , where [x] is the greatest integer less than or equal to x, and {x} = x [x]. Which of the following statements are true ? I. The range of f is a closed interval II. f is continuous on R. III. f is one-one on R. [2017] (A) I only (B) II only (C) III only (D) None of I, II and III Sol. [D] f(x) = 2]x[1}x{ f(x) = onSo3x2;52x2x1;21x1x0;x0x1;21x Now check accordingly Kota : CAREER POINT Ltd.

10 , CP Tower, IPIA, Road , Kota (Raj.), Ph: 0744-5151200 4 KVPY EXAMINATION 2017 CAREER POINT CLASS XII (STREAM SX) 13. Let xn = (2n + 3n)1/2n for all natural numbers n. Then [2017] (A) nlimxn = (B) nlimxn =3 (C) nlimxn =3 + 2 (D) nlimxn =5 Sol. [B] n2/1nn2/1nn132)3(lim Put 3limn 14. One of the solutions of the equation 8 sin3 7 sin +3cos = 0 lies in the interval [2017] (A) (0, 10 ] (B) (10 , 20 ] (C) (20 , 30 ] (D) (30 , 40 ] Sol. [B] 6 sin 2 sin 3 7 sin + 3cos = 0 3cos sin = 2 sin 3 It can be written as 2 (sin (60 )) = 2 sin 3 sin (60 ) = sin 3 60 = 4 = 15 is one of the value 15.))))


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