IIT JEE Maths Paper 2

Transcription

1 IIT JEE Maths Paper A. Question paper format: 1. The question paper consists of 4 sections.. Section I contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only one is correct. 3. Section II contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which one or more is/are correct. 4. Section III contains questions. Each question has four statements (A, B, C and D) given in column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 5. Section IV contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. The answer will have to be appropriately bubbled in the ORS as per the instructions given at the beginning of the section. B. Marking scheme: 6. For each question in Section I you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles is darkened. In case of bubbling of incorrect answer, minus one (-1) mark will be awarded. 7. For each question in Section II, you will be awarded 4 marks if you darken the bubble(s) corresponding to the correct choice(s) for the answer, and zero mark if no bubble is darkened. In all other cases, Minus one (-1) mark will be awarded. 8. For each question in Section III, you will be awarded marks for each row in which you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marking for incorrect answer(s) for this section. 9. For each question in Section IV, you will be awarded 4 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (-1) mark will be awarded.

2 SECTION I 0. A line with positive direction cosines passes through the point P(, 1, ) and makes equal angles with the coordinate axes. The line meets the plane x + y + z = 9 at point Q. The length of the line segment PQ equals (A) 1 (B) (C) 3 (D) (C) ( + a, 1 + a, + a) on x + y + z = 9 for a = 1 l = 3 a = The normal at a point P on the ellipse x + y = 16 meets the x axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points (A) (C) ± 3 5, ± 7 (B) ± 3 5, ± ± 3, ± (D) 7 ± 3, ± 7 (C) P(4 cos θ, sin θ), m PQ = tan θ, Q(3 cos θ, 0) M(cos θ /, sin θ) (x / 7) + y = 1 where x = ± 3, y = ± 1 / 7.. If the sum of first n terms of an A.P. is cn, then the sum of squares of these n terms is (A) n ( 4 n 1 ) c 6 (B) n ( 4 n + 1 ) c 3 (C) T n = c (n (n 1) ) = c (n 1) c (C) n ( 4 n 1 ) c 3 (D) 4 n ( n + 1 )( n + 1 ) 4 n ( n + 1 ) nc + n = ( 4 n 1 ). 6 3 n ( 4 n + 1 ) c 6 3. The locus of the orthocentre of the triangle formed by the lines (1 + p) x py + p (1 + p) = 0, (1 + q) x qy + q (1 + q) = 0, and y = 0, where p q, is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line (D) Vertices (p, 0), (q, 0). Altitudes: px + (1 + p)y = pq and qx (1 + q) y = pq Orthocentre: ( pq, pq) Locus is y + x = 0.

3 SECTION II 1 4. For the function f ( x ) = x cos, x 1, x (A) for at least one x in the interval [1, ), f(x + ) f(x) < (B) Lim f ( x ) = 1 x (C) for all x in the interval [1, ), f(x + ) f(x) > (D) f ' (x) is strictly decreasing in the interval [1, ) (BCD) f ( x ) = cos + sin, f ( x ) = cos x x x 3 x x f '' (x) < 0 f ' (x) is strictly decreasing in the interval [1, ) For x, f ' (x) cos sin 0 = 1 f ' (x) > 1 x f ( x + ) f ( x ) By LMVT, = f ( c ) > 1 ( x + ) x f(x + ) f() > x. 5. The tangent PT and the normal PN to the parabola y = 4ax at the point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose (A) vertex is a, 0 3 a (C) latus rectum is 3 (B) directrix is x = 0 (D) focus is (a, 0) (AD) P(at, at), T( at, 0), N(a + at, 0) 3x = a + at and 3y = at 3x a = a (3y / a) or, 9x = 1ax 8a y = 4 (a / 3) (x (a / 3)) Vertex: (a / 3, 0), directrix: x = a / 3, latus rectum = 4a / 3 and focus is (a, 0).

4 π sin nx 6. If I n = dx, n = 0, 1,,..., then x ( 1 + π ) sin x π (A) I n = I n + (B) I 10 m + 1 = m = 1 10 π (C) 10 m = 0 m = 1 I (D) I n = I n + 1 π π 1 (ABC) Replacing x with x and adding, sin nx = = sin nx I n dx dx sin x sin x π 0 π sin nx sin( n ) x π I n I n = dx = cos ( n 1 ) dx = 0 sin x 0 0 Also, I 1 = π and I 0 = 0 I odd = π and I even = For 0 < θ < π /, the solution(s) of 6 m = 1 ( m 1 cosec θ + 4 ) π m π cosec θ + = 4 4 (are) (A) π / 4 (B) π / 6 (C) π / 1 (D) 5π / 1 (CD) Using cosec α cosec β = (cot β cot α) / sin (α β) 6 ( m 1 ) π m π π cot θ + cot θ + = 4 sin = m = 1 is 6 π cot θ cot θ + = 4 cot θ + tan θ = 4 cot θ = ± An ellipse intersects the hyperbola x y = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that the hyperbola. If the axes of the ellipse are along the coordinate axes, then (A) equation of ellipse is x + y = (B) the foci of ellipse are (± 1, 0) (C) equation of ellipse is x + y = 4 (D) the foci of ellipse are (±, 0) (AB) For orthogonal intersection, foci coincide. For ellipse, e = 1 / and foci (± 1, 0) a =, b = 1 Ellipse is x + y =.

5 SECTION III 9. Match the statements/expressions given in Column I with the values in Column II Column I (A) Root(s) of the equation (p) π / 6 sin θ + sin θ = (B) Points of discontinuity of the equation (q) π / 4 6 x 3 x f ( x ) = cos π π, where [y] denotes the largest integer less than or equal to y (C) Volume of the parallelopiped with its edges (r) π / 3 represented by the vectors i ˆ + ˆ j, i ˆ + ˆ j and i ˆ + ˆ j + π k ˆ r r r r (D) Angle between vectors a and b where a, b (s) π / r and c are unit vectors satisfying r r r r a + b + 3 c = 0 (t) Column II (A) (qs); (B) (prst); (C) (t); (D) (r) (A) sin θ = cos θ cos θ (1 sin θ) = 0 cos θ = 0 or cos θ = 0 π / 4, π /. (B) At x = kπ / 6, [6x / π] is discontinuous and cos [3x / π] is non zero. π / 6, π / 3, π /, π. (C) ( i ˆ + ˆ j ) ( i ˆ + ˆ j ) = k STP = k. ( i ˆ + ˆ j + π k ˆ ) = π. (D) a r r + b r r has magnitude 3 only if angle between a and b is π / 3. π

6 30. Match the statements/ expressions in Column I with the open intervals in Column II Column I (A) The number of solutions of the equation (p) 1 xe sin x cos x = 0 in the interval 0, π (B) Value(s) of k for which the planes (q) kx + 4y + z = 0, 4x + ky + z = 0 and x + y + z = 0 intersect in a straight line (C) Value(s) of k for which (r) 3 x 1 + x + x x + = 4k has integer solution(s) (s) 4 (D) If y' = y + 1 and y(0) = 1, then the value(s) (t) 5 of y (ln ) (A) (p); (B) (qs); (C) (qrst); (D) (r) (A) xe sin x π cos x is increasing function in 0,. Since increases from 1 to πe /, exactly 1 solution. Column II k 4 1 (B) 4 k = 0 1 k + 8 6k = 0 k = or 4. (C) Expr. has minimum value = 6 for 1 x 1 4k 6. For k, solution of x = k is valid. (D) ln (y + 1) = x + c = x + ln y = e x 1 y(ln ) = 3.

7 SECTION IV 31. The centres of two circles C 1 and C each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C 1 and C and C be a circle touching circles C 1 and C externally. If a common tangent to C 1 and C passing through P is also a common tangent to C and C, then the radius of the circle C is 8 If centres be ( 3,0) & (3, 0), P be (0, 0), common tangent be x 8 y = 0, for C; if centre be (0, a) and radius r, then 8a / 3 = r and 9 + a = (r + 1) Let ABC and ABC' be two non congruent triangles with sides AB = 4, AC = AC' = and angle B = 30. The absolute value of the difference between the areas of these triangles is 4 C' A M C 30 0 B AM = 4 sin 30 = CAM = C'AM = 45 C'AC is right angled with area = 4. x 33. Let f : R R be a continuous function which satisfies f ( x ) = f ( t ) dt. 0 Then the value of f (ln 5) is 0 f ' (x) = f(x) and f (0) = 0 f(x) = ae x. Since f(0) = 0, f(x) = 0 x f(ln 5) = Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations: 3x y z = 0 3x + z = 0 3x + y + z = 0. Then the number of such points for which x + y + z 100 is 7 Solving y = 0 and z = 3x 10 x 100 x 3. x = ± 3, ±, ± 1, 0 i.e. 7 points.

8 35. If the function f(x) = x 3 + e (x / ) and g(x) = f 1 (x), then the value of g' (1) is g(f(x)) = x g' (f(x)) f ' (x) = 1 Puting x = 0, g' (1). ( 1 / ) = 1 g'(1) =. 36. The smallest value of k, for which both the roots of the equation x 8kx + 16(k k + 1) = 0 are real, distinct and have values at least 4, is D > 0, f(4) 0, b / a > 4. k 1 > 0, (k 1) (k ) 0, 4k > 4 k min =. 37. The maximum value of the function f(x) = x 3 15x + 36x 48 on the set A = {x x + 0 9x} is 7 x 9x x 5 f ' (x) = 6 (x ) (x 3) > 0 for 4 x 5. Max. Value = f(5) = 7. p ( x ) 38. Let p(x) be a polynomial of degree 4 having extremum at x = 1, and Lim 1 + =. x 0 x Then the value of p() is 0 P(x) = x + ax 3 + bx 4 P' (x) = x + 3ax + 4bx 3 Since extremum at x = 1 and x =, + 3a + 4b = 0 and + 6a + 16b = 0 b = 1 / 4, a = 1 P(x) = x x 3 + x 4 / 4 P() = = 0.