JEE-ADVANCED MATHEMATICS. Paper-1. SECTION 1: (One or More Options Correct Type)

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1 JEE-ADVANCED MATHEMATICS Paper- SECTION : (One or More Options Correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE are correct.. Let f ( x) (4a )( x log5) ( a 7)sin x. Then f ( x) has (A) critical points, if 4/ a (B) no critical points if a 4/ (C) critical points if a (D) If a 4/, then cos x f ( x) (4a )( x log5) ( a 7)sin( x) f ( x) (4a ) ( a 7)cos( x)( ) For f ( x) 0 (Critical points) (4a ) ( 7 a)cos( x) 0 ( 4 a) cos( x) ( a 7) cos( x) ( 4 a) ( a 7) a 7 ( 4 a) 7 a 4 a 4 For a, cos( x) Hence, the correct options are (A), (B) and (C).. Let x f ( x) a a x a x a x a x a x where x f ( x) lim e, then x 4 i 0 a 6 is equal to (A) 5 (B) (C) (D) 6 a0, a,... a are real and a if

2 For finite limits, Therefore, x f ( x) a a x a x a x a x a x f x 0 6 ( ) x a a x a x a x a 0, a 0, a 0, a x f x a4x a5 x a6 x ( ) / x x [ a x a x a x ] f ( x) lim ( a 4x a5 x a6 x ) x a 4 e a 4 x 4 5 a6x a x a x 4 5 a6x a x a x x Now Hence, the correct answer is (B). 4 i0 a a a a a a i A line l is passing though the point (, 6). If the product of the intercepts of l on the axes is, then the equation of l is (A) x y 5 0 (B) x y 4 0 (C) 4x y 0 (D) 9x y 0 ab = Equation of l is

3 x y a b or bx ay ab or bx ay () Since () passes through (, 6) b() a( 6) or 6a b 0 () or or 6a a 0 a 6a a a, 6 b, Possible lines are x y y First x 9x y or 9x y 0 D is correct x y Second y or x or 4x y or 4x y 0 Hence, the correct options are (C) and (D). 4. Let cos A cos B sin A sin B x sin A sin B cos A cos B Then (A) x = 0 if n is an odd positive integer n (B) x tan ( A B) / if n is an even positive integer n (C) x cot ( A B) / if n is an even positive integer (D) x = 0 if n is an even positive integer n cos( A B ) cos( AB ) X sin ( A B ) sin ( B A ) n n n

4 For n = odd positive For n = even positive cot Hence, the correct options are (A) and (C). AB A B ( ) cot n n n ( ) n = X = 0 (A) n A B x cot d Let D 0 d 0 which of the following are true? 0 0 d (A) D is a symmetric matrix (B) If ddd 0, 0 0 d then D 0 0 d 0 0 d (C) Trace of D d d d (D) D commutes with every order matrix (A) Here D is symmetric matrix about diagonal. (B) D d d 0 0 d (C) D d d d Hence, the correct options are (A), (B) and (C). 6. Area of the region in which point p x, y, 0 tan y x is x lies; such that y 6 x and

5 6 8 (A) (B) 8 (C) 4 (D) none of these Required area is the area of shaded region APOQ =area of OAQ area of sector OAP (0, 0) O P Q (4, 0) A x = 4 Hence, the correct option is (B). 7. If S n... n, N S S S is equal to... n n, then (a) ns n n (b) ns n (c) n S n n (d) ns n n S S S... n = +... n Adding vertically: = n n n n n... n = n... [...] n =

6 ns n n nsn n Hence, the correct options are (A) and (D). 8. If k is a constant such that x xy k e satisfies the xdy differential equation x x y x, then dx k may be (a) (b) 0 (c) (d) Differentiating the given equation of curve dy x x e. x y (i) dx From given differential equation dy x x x y x (ii) dx For Validity of (i) and (ii) Hence k Hence, the correct option is (A). e x xy SECTION : (Paragraph Type) This section contains 4 Paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D). Paragraph for Questions 9 and 0: The greatest or least values of a function f ( x ) on an interval [a, b] may be attained either at a critical point of f or at the end points of the interval. Answer the following three questions. 9. The greatest value of f ( x) x x x on the interval [, 5/] is (A) 0 (B) 5 (C) (D) 8 f ( x) ( x ) x 6x 6x 6( x x ) 6( x )( x )

7 Hence, the correct option is (D). f ( x) 0 at x, f ( x) x 6 f ( ) 0 f ( ) ( ) () ( ) 8 0. The least value of the function sin xsin x on is (A) (B) 4 (C) (D) 0 y sin( x)sin( x) when y x x x 0, sin [cos( ) cos ( )] Hence, the correct option is (B). y cos( x)sin x cos( x)sin( x) sin( x) 0 or cos ( x) 0 x 0, cos( x), sin x y 4 Paragraph for Questions and : Consider the parabola y 6x.. Let P(, 4), Q, and R(4,8) be the three points on a parabola. Then the area of the triangle 4 formed by the tangents to the parabola at points P, Q and R is (A) (B) 4 (C) (D) y 6x or y 4 4x a 4

8 Area of triangle formed by tangents (Area of triangle PQR ) (4 )(8 4)( 8) Hence, the correct option is (B).. The locus of the mid-points of chords of the parabola which subtend right angle at the vertex is (A) y (C) y 4( x 6) 8( x 6) (B) (D) y 8( x 4 a) y 8( x 8) ( y y )( y )( ) y y y 8a at 0 t Slope m of OA at 0 t t

9 at 0 Slope m of OB at 0 t Now OA OB tt 4 t t Now or By using t t = 4, () and () or or Hence, the correct option is (C). ( ) h a t t, k at at 4{ t t } t t h () 4( t t ) k () y k h ( 4) 4 8( x 6) k h 8 6 { k 8} h 6 8 8h k 8 or Paragraph for Questions and 4: Let X, Y and Z be the three sets of complex numbers defined as follows: X = {z: Im(z) } Y = {z: z i = } Z = { z : Re[ z( - i)] = }. The number of elements in the set X ÇY Ç Z is (A) 0 (B) (C) (D) Infinite y : ( x ) ( y ) 9 Hence, x Ç y Ç z = Hence, the correct option is (B). z = x + y = 4. Let z be any point in X ÇY Ç Z. Then z + i + z 5 i lies between (A) 5 and 9 (B) 0 and 4 (C) 5 and 9 (D) 40 and 44

10 This lies b/w 5 and 9. Hence, the correct option is (C). z i z 5 i ( x ) i( y ) ( x 5 ) i( y ) ( x ) ( y ) ( x 5 ) ( y ) Paragraph for Questions 5 and 6: A box B contains white ball, red balls and black balls. Another box B contains white balls, red balls and 4 black balls. A third box B contains white balls, 4 red balls and 5 black balls. 5. If ball is drawn from each of the boxes B, B and B, the probability that all drawn balls are of the same colour is (A) (C) 558 (B) (D) Following table shows the probability of draining balls from the boxes: B B B W W W R R 4 R B 4 B 5 B P( WWW RRR BBB) Hence, the correct option is (A). 6. If balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these balls are drawn from box B is (A) 6 8 (C) 65 8 (B) 6 8 (D) 55 8

11 WR P P( B ) B B P WR WR WR WR P P( B ) P P( B ) P P( B ) B B B C C 9 C 4 C C C C C C 9 9 C C C Hence, the correct option is (D). SECTION - : (Matching List Type) Each question contains two columns, Column I and Column II. Column I has four entries (A), (B), (C) and (D). Column II has four entries (P), (Q), (R), (S) and (T). Match the entries in Column I with the entries in Column II. Only one entry in Column I may match with one or more entries in Column II. 7. Match the items in Column I with those in Column II. (A) If (B) If Column I sin x cos x, then x = 6 tan x tan tan 8, then x = Column II (P) 6 (Q) 5 (C) If tan(sec x) sin(tan ) and x > 0, then x is equal to (R) 5 (D) If tan (4/ x) tan (9 / x) /, then x = (S) (T)

12 (A) Let y x (B) or, (C) or, (D) sin ( x) sin ( x) 6 7 sin ( x) sin x x 6 x x x ( x ) 6 4 x x 6 y y 0 6 y x 8 tan ( x) tan 4 x 8 x8 4x x x 5 tan (sec x) tan cos x tan tan sin sin x 5 x x 5 4 tan 49/ x x 6 x 5 x 6 0

13 x6 Hence, the correct matching is (A) (S); (B) (R); (C) (Q); (D) (P) 8. PQ is a double ordinate of the parabola y 4x. If the normal at P meets the line passing through Q and parallel to axis at G. Then, the locus of G is a parabola. For this parabola, match the items of Column I with those of Column II. Column I (A) Length of the latus rectum of the locus of G (P) 5 (B) Abscissa of the vertex (Q) (C) Abscissa of the focus (R) 4 (D) The directrix is x a where a is equal to (S) 6 (T) Column II Parabola is y = 4()x a = Equation of normal at P(t, t) is () y tx t t Equation of line through θ (the other end of a double ordinate) is y t () Now eliminating t from () and () y yx or y 0 8 or y 4yx 6y 0 or y( y 4x 6) 0 Now y 4x 6 0 Now shifting origin to (4, 0) yx y y y or yx y y y 8 y = 0 or y 4x + 6 = 0 y 4( x 4)

14 x X 4 y Y 0 Equation of locus in new co-ordinates is Y = 4X Latus rectum = 4 Vertex (0, 0) x 0 4 y = In old system vertex is (4, 0) Abscissa of the vertex is 4. New System Old System Directrix X = x 4 = or x = a = Focus (, 0) x 4 (5,0) Abscissa of vertex is 5. y 0 Hence, the correct matching is (A) (R); (B) (R); (C) (P); (D) (Q) 9. Match the items of Column I with those of Column II. Column I (A) If a b c 0, a, b 5 and c 7 then the angle between a and b is Column II (P) (B) If a and b are unit vectors such that a b is also unit vector, then the angle (Q) 4 between a and b is (C) In a regular tetrahedron (i.e., all edges are equal) the angle between opposite pair of edges is (R) 6 (D) If a and b are two non-collinear vectors such that a b a b, then the angle (S) between a and b is (A) Let a x, i y, j, b xi y j, c xi y j Now, x x x i j( y y y ) 0

15 a( x y ), b ( x y ) (B) x x ( x )and ( y y ) ( y ) x x y, y x ( x x ) y ( y y ) x x y y 5 Angle between a and b = 5 5 cos, Now, x y and x y, x y y y cos, x y x y (C) The angle b/w opposite edge = (D) x y y y x x y y x x y y 0 Angle b/w a and b Hence, the correct matching is (A) (S); (B) (S); (C) (P); (D) (P) 4 0. Consider the lines : y z, : y z L L and the planes P :7x y z, P x 5y 6z 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L and L, and perpendicular to planes P and P. Match Column I with Column II Column I Column II (A) a = (P) (B) b = (Q) (C) c = (R) (D) d = (S) We have x y z L t; x 4 y z L t. For finding point of intersection, we have

16 t 4 t () and t t () Solving, we get t =, t =. The point of intersection is (5,, ). The equation of plane P will be as follows: x 5 y z ( 6)( x 5) 48( y ) ( z ) 0 ( x 5) ( y ) ( z ) 0 x y z Therefore, a =, b =, c =, d =. Hence, the correct matching is (A) (R); (B) (Q); (C) (S); (D) (P) Paper SECTION : (Single Option Correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE is correct.. Tan (/ ) Tan (/ ) is equal to (A) (B) (C) 4 (D) 6 tan tan 5 tan tan tan () 5 4 Hence, the correct option is (B).. Let A (,, 5), B (,, ) and C (x, 5, y) be the vertices of ABC. If the median AD is equally inclined to the coordinate axes, then (A) x = 0, y = 7 (B) x = 7, y = 0 (C) x = 7, y = 0 (D) x = 0, y = 7

17 Position vector of D = x y,4, 5 x 8 y AD,, As AD is equally included to the -axis so, 5 x 8 y,, (,, ) 5 x, x 7, 8 y, y 0. Hence, the correct option is (C).. Let complex numbers and lie on circles (x x0) + (y y 0) = r and (x x 0) + (y 0 y 0) = 4r, respectively. If z 0 = x 0 + iy 0 satisfies the equation z r, then (A) (C) 7 As α satisfies z z0 r, we have Therefore, Also, since satisfies z z0 r, we have (B) (D) z r ( z )( z ) r 0 0 z z z r () z 0 r z0 z0 4r ( z )( z ) 4r z z z 4 r ()

18 Subtracting Eq. () from Eq. (), we have That is, Also, we have That is, Dividing Eq. () by Eq. (4), we get Hence, the correct option is (C). z0 r ( ) ( ) ( 4 ) z0 r ( )( ) ( 4 ) () 0 r 0 r z ( z ) (4) The equation of the line through the point of intersection of the lines x y + = 0 and x + 5y 9 = 0, and whose distance from the origin is 5 distance is (A) x y 5 0 (B) x y 5 0 (C) x y 5 0 (D) x y 5 0 Equation of the line through the point of intersection of the given lines is x y (x 5y 9) 0 or x( ) y(5 ) ( 9 ) 0 () Given that this line is at a distance 5 from origin, so 0( ) 0(5 ) ( 9 ) 5 ( ) (5 ) 9 or or or or 7 (8 7) 0 8 The required equation is x y or x y or x y 55 0

19 or x y 5 0 Hence, the correct option is (C). x 5. If the lines y z and x y 5 z 6 are at right angles to each 5 other, then is equal to (A) (B) (C) Since lines are to each other, so sum of products of drs = 0 ( 5) 0 or 9 0 or Hence, the correct option is (C). (D) If each of a, b and c takes values from the set {,,, 4, 5, 6}, then the probability that the equation (A) 5 6 ax bx c 0 has real roots is equal to (B) 4 6 Hence, the correct option is (B). 7. The area enclosed by the curves (C) 6 Sample space ns 666 For real roots, b 4ac This is true for b,, 4,5,6 and a, c,,, 4,5,6, ne 4 4 pe y sin x cos x and y cos x sin x over the interval 0, is (A) 4( ) (B) ( ) (D) 6 6 (C) ( ) (D) ( ) From the following figure that depicts the area enclosed by the given curves, we have

20 (sin x cos x) dx (cos x sin x) dx (sin x cos x) dx / /4 / 0 0 /4 cos x sin x sin x cos x cos x sin x / / /4 /4 / / /4 /4 (0 ) ( 0) 0 4 ( ) Hence, the correct option is (B). 8. A line drawn through the point (, 4) meets the positive coordinate axes. Then, the minimum value of the sum of the intercepts on the axes is (A) 9 (B) 8 (C) 0 (D) Let x and y intercept are a and b then equation of line x y a b as this passes from (, 4) 4 a b Now, Let p = sum of intercept = a + b b p b b 4 dp ( b 4)(b ) ( b b) db ( b 4) dp when 0, b and 6 db when b = 6, a = so, a + b = 9 Hence, the correct option is (A).

21 9. In ABC, if cos A cos B, then tan A B (A) a b a b (B) a b a b (C) a b a b (D) a b a b Since, cos A cos B cos B cos A cos B cos A cos B cos A sin[( A B) / ]sin[( A B) / ] cos[(a B) / ]cos[(a B) / ] Hence, the correct option is (B). A B A B tan cot A B tan[( A B) / ] tan tan[( A B) / ] sin A sin B sin A sin B a b a b 0. An ellipse of eccentricity / is inscribed in a circle. A point inside the circle is selected at random. The probability that the point lies outside the ellipse is (A) (B) (C) (D) 4 Area of ellipse Here, b a A 4 a x dx a 0 4b a sin a x = asin dx = a cos x x a x ab sin a a a a a o

22 Here ( a / ) Probability that point lies inside ellipse a So, probability that point, outside ellipse Hence, the correct option is (A). A ab a b a 9 a b a a A a SECTION : (One or More Options Correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE are correct.. Let Then (A) a = /e (B) b = /e (C) a = e (D) b = e cosecx (cos x sin x) if x0 f ( x) a if x 0 / x / x / x e e e if 0 x / x / x ae be We have We have lim f ( x) lim(cos h sin h) x00 h0 h0 h0 h0 cosech lim( cos h sin h ) lim[ (cos h sin h )] lim e e (cos hsin h)/sin h e cosech (cos hsin h)cosech (cos hsin h)

23 Since f (x) is continuous, we have x00 h0 Hence, the correct options are (A) and (D). h0 / h / h / h e e e lim f ( x) lim e lim b ae h h be (/ h) (/ h) ae e (/ h) b a a, b e e b e. Circle(s) touching x-axis at a distance from the origin and having an intercept of length 7 on y-axis is (are) (A) x + y 6x + 8y + 9 = 0 (B) x + y 6x + 7y + 9 = 0 (C) x + y 6x 8y + 9 = 0 (D) x + y 6x 7y + 9 = 0 From the following figure, it is clear that the center is (, α) and radius is. Therefore, The radius is Therefore, x y 6x y c 0. 9 c c = 9 The intercept on y-axis is a c 7 Therefore, the equation is x y 6x 8y 9 0. Hence, the correct options are (A) and (C).. The graph of the function y cosx cos(x + ) cos (x + ) is (A) a straight line passing through (0, sin )

24 (B) a straight line passing through, sin (C) a straight line parallel to x-axis (D) a straight line parallel to y-axis or or or y cos x cos( x ) cos ( x ) cos ( x ) y cos x cos( x ) { cos( x )} y cos xcos( x ) cos x cos( x ) cos ( x ) y cos( x x ) cos( x x ) cos(x ) y cos(x ) y = sin (Equation of line parallel to x-axis) A is true line y = sin passes through (0, sin ) cos( ) cos(x ) { cos } sin B is also true. C is also true. Hence, the correct options are (A), (B) and (C). 4. In ABC, if ( a b) ab c and sin A cos A / then (A) C 0 (B) A 5

25 (C) B 45 (D) a : b : c ( ) : : 6 a b c abab ab a b c cos( C ) ab C 0 sin A cos A sin A cos A sin A sin 4 A 5 4 B= a : b : c sin A : sin B : sin C sin5 : sin 45 : sin0 ( ):: 6 Hence, the correct options are (A), (B), (C) and (D). 5. Let a > 0 and z ( / z) a( z 0) is a complex number. Then the maximum and minimum values of z are (A) a + a + 4 a a (C) Now, again, Þ (B) (D) a + a + 4 a a z - z + z + z z z Þ z - a z + z z z + a z ³ z - a z + ³ 0 æ a - a 4 ù éa + a 4 ö z Î,, ç - È ú ê è û ë ø z - a z ()

26 Þ Þ - a z - a z - a z z - a z z + a z - ³ 0 and z - a z - 0 Þ æ - a - a 4 ù é- a + a 4 ö Þ z Î,, ç - È ú ê è û ë ø And æ a - a + 4 a + a + 4 ö z Î, ç è ø æ - a + a - 4 a + a + 4 ö Þ z Î, ç è ø Common from Eqs. () and () æ a a a a ö z Î, ç è ø Hence, the correct options are (A) and (C). () SECTION : (Integer Value Correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). x y 6. The number of points on the ellipse from which a pair of perpendicular 50 0 x y tangents are drawn to the ellipse is 6 9 x y For perpendicular tangents to, the points must lie on x + y = 6+9 = i.e x + y = 5 () Now since 5 5 there are four points of director circle, which are common to ellipse x y. 50 0

27 There are four points, A, B, C, D from which perpendicular tangents can be drawn to ellipse x y Let x and xn xn for n. Then lim xn exists and is equal to x We have lim x x n Therefore, x x 5 and x 5. That is, Therefore, That is, Therefore, That is, y =. Hence, the correct answer is. 8. If a y x n... y = y = y = 0 y y y = 0 (y ) (y ) = 0 and a b 0, then a [ a { a ( a b )}] is nb where Here, a b = 0 a b 4 a 9a a b a b Now, = () 4 b, N = 6, n = 4. n is. 9. A number from the set {,,,..., 5} is selected at random. The probability that it is a proper divisor of 0 is a/b (in lowest terms). Then a + b is equal to. Divisor of 0 in,,...5,,,5, 6,0,5 7c 7 a P( E) 5c 5 b a b The integers from to 000 are written in order around a circle. Starting at, every fifteenth number is marked (that is, 6, etc.). This process is continued until a number is reached A which has already been marked. If the number of marked numbers is A, then find. 50

28 In the first round all integers, which leave remainder when divided by 5, will be marked, last number of this category is 99. Next number will be = 6. That means in second round all integers, which leave remainder 6 when divided by 5 will be marked. In short numbers of the form 5k + will be marked. A Therefore, A = 00 and 4 50.

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola)

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola) QUESTION BANK ON CONIC SECTION (Parabola, Ellipse & Hyperbola) Question bank on Parabola, Ellipse & Hyperbola Select the correct alternative : (Only one is correct) Q. Two mutually perpendicular tangents