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A-PDF Watermark DEMO: Purchase from www.A-PDF.com to remove the watermarkIIT-JEE-2005-S-1 Note: FIITJEE solutions to IIT JEE, 2005 Screening Test is based on Screening Test paper created using memory retention of select FIITJEE students appeared in this test and hence may not exactly be the same as the original paper. However, every effort has been made to reproduce the original paper in the interest of the aspiring students. FIITJEE solutions to IIT JEE, 2005 Screening 1. The locus of z which lies in shaded region is best represented by (A) z : |z + 1| > 2, |arg(z + 1)| < /4 (B) z : |z - 1| > 2, |arg(z - 1)| < /4 (C) z : |z + 1| < 2, |arg(z + 1)| < /2 (D) z : |z - 1| < 2, |arg(z - 1)| < /2 Ans. A Sol. 2. P( 2-1, 2) A (-1, 0) (1, 0) m oom Q( 2-1,- 2) .c .c a a l l The points (1, 0), ( 2 1, 2 ) and ( 2 1, 2 ) are equidistant from the point (- 1, 0). ee The shaded area belongs to the region outside the m sector of circle |z + 1| = 2, lying between the line rays ddm arg (z + 1) = and arg (z + 1) = . aa 4 4 o l lo n In an equilateral triangle, 3 coins of radii n 1 unit each are kept so that they touch each other andw also the sides of the triangle. Area of the triangle is oow (A) 4 + 2 3 (B) 6 + 4 3 .d .d 7 3 7 3 w (C) 12 + (D) 3 + w 4 4 w w w w Ans. B Sol. The line joining the vertex of the triangle and the centre of the coin makes angle triangle. The length of each of the sides of the equilateral triangle is 2 + 2 cot Hence its area is 3. Ans. Sol. 4. with the sides of the 6 = 2 (1 + 6 3 ). 3 4(1 + 3)2 = 6 + 4 3 . 4 If a, b, c are integers not all equal and w is a cube root of unity (w 1), then the minimum value of |a + bw + cw2| is (A) 0 (B) 1 3 1 (D) (C) 2 2 B 2 3 1 b c 2 ( (a b)2 + (b c)2 + (c a)2 ) . a + (c b) = 4 2 2 2 This is minimum when a = b and (b c)2 = (c a)2 = 1 The minimum value is 1. |a + bw + cw2| = A rectangle with sides 2m 1 and 2n 1 is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is (A) (m + n + 1)2 (B) 4m+n-1 2 2 (C) m n (D) mn(m + 1)(n + 1) for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-2 Ans. Sol. C There are 2m vertical (numbered 1, 2, ., 2m) and 2n horizontal lines (numbered 1, 2, .2n). To form the required rectangle we must select two horizontal lines, one even numbered and one odd numbered and similarly two vertical lines. The number of rectangles is then mC1. mC1. nC1. nC1 = m2n2. Alternate solution: Number of rectangles possible is (1 + 3 + 5 + . +(2m 1)) (1 + 3 + 5 + . + (2n 1)) = m2n2. 5. A circle is given by x2 + (y 1)2 = 1, another circle C touches it externally and also the x-axis, then the locus of its centre is (A) {(x, y) : x2 = 4y} {(x, y) : y 0} (B) {(x, y) : x2 + (y 1)2 = 4} {x, y) : y 0} (C) {(x, y) : x2 = y} {(0, y) : y 0} (D) {(x, y) : x2 = 4y} {(0, y) : y 0} Ans. Sol. D Let the circle touching the x-axis be x2 + y2 2ax 2by + a2 = 0 with centre at (a, b) and radius b. m 6. Ans. Sol. 7. Ans. m 2 . Since it touches the circle x2 + (y 1)2 = 1, | b + 1|= a 2 + (b 1)o o c c . 2 2 2 b + 2b + 1 = a + b 2b + 1 . a la the circle lies on the y-axis, then y 0. 4b = a2 so that locus of (a, b) is x2 = 4y. If the centre of l ee cos ( - ) = 1 and cos ( + ) = 1/e, where , [- , m ]. Pairs of , which satisfy both the equations is/ are ddm (A) 0 (B) 1 aa (D) 4 (C) 2 o lo D nnl For cos ( - ) = 1, = so that w cos ( + ) = 1/e + = cos-1 1/e w 1 oo , can be satisfied by 4 sets of values. 2 = cos 1 [ 2 , 2 ]. e .d .d In ABC, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is w w given by w C A A B w B C w (A) (b c) sin (B) (b c) cos = a sin w = a cos 2 2 2 2 A B+C (C) (b + c) sin = a cos 2 2 B b c sin B sin C = = Here a sin A Sol. A B+C (D) (b c) cos = 2a sin 2 2 B C B + C sin B C cos 2 . 2 2 = A A A 2sin cos cos 2 2 2 2sin 30 30 30 30 30 30 30 30 The value of + ...... + is, where 0 10 1 11 2 12 20 30 30 30 (B) (A) 10 15 8. 60 (C) 30 n n = Cr . r 31 (D) 10 Ans. Sol. A The given expression is the coefficient of x20 in the product (1 + x)30 (1 x)30 = (1 x2)30 the given expression = 30C10. 9. A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the centroid D 1 1 1 (x, y, z) of triangle ABC satisfies the relation 2 + 2 + 2 = k, then the value of k is x y z (A) 3 (B) 1 (C) 1/3 (D) 9 for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-3 Ans. D Sol. x y z Let + + = 1 be the variable plane so that a b c 10. If 1 1 1 1 =1. + 2+ 2 2 a b c The plane meets the coordinate axes at A (a, 0, 0), B (0, b, 0), C (0, 0, c). The centroid D of the triangle ABC is a b c , , 3 3 3 a b c 1 1 1 1 1 1 x = , y = , z = and 2 + 2 + 2 = 1 2 + 2 + 2 = 9. 3 3 3 a b c x y z 1 t ( f ( t ) ) dt = (1 sin x ) , then f 2 sin x 1 is 3 (A) 1/3 (C) 3 Ans. Sol. 11. Ans. Sol. 12. (B) 1/ 3 (D) 3 m oom .c C .c a Differentiating both sides with respect to x, we get a ll 1 ee 2 - sin x f (sin x). cos x = - cos x f (sin x) = 2 m sin x m d 1 1 aad f (x) = 2 f =3. x 3 lo lo n n In the quadratic equation ax2 + bx + c = 0, if = b2 4ac and + , 2 + 2, 3 + 3 are in G.P. where , are w 2 the roots of ax + bx + c = 0, then w oo (A) 0 (B) b = 0 d . .d (C) c = 0 (D) = 0 C w w 2 We have ( 2 + 2) = ( + ) ( 3 + 3) ( - )2 = 0 w w c = 0. w w A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is (A) 5/11 (B) 5/6 (C) 6/11 (D) 1/6 Ans. A Sol. The required probability = 3 = 1 5 1 5 + + ...... 6 6 6 6 5 1 5 5 2 1 + + ... = . 6 6 6 11 13. If f(x) is a twice differentiable function and given that f(1) = 1, f(2) = 4, f(3) = 9, then (A) f (x) = 2, for x (1, 3) (B) f (x) = f (x) = 5 for some x (2, 3) (C) f (x) = 3, x (2, 3) (D) f (x) = 2, for some x (1, 3) Ans. Sol. D Let g(x) = f(x) x2. We have g(1) = 0, g(2) = 0, g(3) = 0. Hence by Rolle s theorem g (x) = 0 for some c (1, 2) and g (x) = 0 for some d (2, 3). Again, by Rolle s theorem g (x) = 0 at some x (c, d) f (x) = 2 for some values x (1, 3). 0 14. (x 3 ) + 3x 2 + 3x + 3 + (x + 1) cos ( x + 1) dx is equal to 2 (A) 4 (C) 4 (B) 0 (D) 6 for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-4 Ans. C 0 Here I = [x 3 + 3x 2 + 3x + 3 + ( x + 1) cos(x + 1)]dx Sol. 2 Put x + 1 = t 1 = t ( 1 1 = 3 ) + t cos t + 2 dt 2dt = 4 . 1 15. If P (x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that P (1) = 1, P (0) = 0 and P (x) > 0 x [0, 1], then (A) S = (B) S = {(1 a) x2 + ax 0 < a < 2 2 (D) S = {(1 m a)m x2 + ax 0 < a < 1 (C) (1- a) x + ax a (0, ) Ans. Sol. 16. Ans. oo c . .c B a Let the polynomial be P (x) = ax2 + bx + c a l P (0) = 0 c = 0 and P (1) = 1 a + b = 1 so that eel P (x) = 2(1 b)x + b > 0 x m m b (0, 2). d d 2 aa S = {(1 a)x + ax, a (0, 2)} o l lo 2 2 n nthe tangent to the ellipse x2 + y2 = 1 and coordinate axes is The minimum area of triangle formed by a b w oow 2 2 a +b sq. units (A) ab sq. units (B) d .d 2 . w (a + b)2 a 2 + ab + b 2 (C) sq. units w (D) sq. units w 3 2 w w w A A tangent of the given ellipse is y = mx + a 2 m 2 + b 2 . 2 2 2 It meets the axes at a m + b , 0 and 0, a 2 m 2 + b 2 . m 2 2 2 2 1 1 Hence the area of the triangle is a m + b = a 2 m + b ab . 2 2 m m Alternate: x cos y sin The equation of tangent at (acos , bsin ) is + = 1. a b It meets the coordinate axes at A (0, b cosec ), B (a sec , 0). ab ab Area of triangle = = ab . 2 sin cos sin 2 Sol. ( ) 17. If the functions f(x) and g(x) are defined on R R such that x rational x irrational 0, 0, , g (x) = , then (f g)(x) is f(x) = x irrational x rational x, x, (A) one-one and onto (B) neither one-one nor onto (C) one-one but not onto (D) onto but not one-one Ans. A Sol. x irrational x ; Let h(x) = f(x) g(x) = x rational x ; the function h(x) is one-one and onto. for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-5 18. The area bounded by the parabolas y = (x + 1)2 and y = (x 1)2 and the line y = 1/4 is (A) 4 sq. units (B) 1/6 sq. units (C) 4/3 sq. units (D) 1/3 sq. units Ans. D 3 1 1 3 The parabolas meet at (0, 1) and intersect the line y = 1/4 at x = , , and . 2 2 2 2 1/ 2 1/ 2 1 2 1 1 = Hence the required area = 2 ( x 1)2 dx = ( x 1)3 4 4 3 3 0 0 Sol. 19. The function given by y = ||x| - 1| is differentiable for all real numbers except the points (A) {0, 1, 1} (B) 1 (C) 1 (D) 1 Ans. Sol. A From the graph, the function is not differentiable at x = 1, 0, 1. 20. Ans. Sol. m oom .c .c a a l (0, l e e1) m ddm O (1, 0) aa ( 1, 0) o l o nnl If y = y(x) and it follows the relation x cos y + y cos x = , then y (0) w (A) 1 (B) 1 w o o (C) (D) - d .d . C w w x cos y + y cos x = , y(0) = . w w dy dy - x sin y w = 0 y (0) = 1. + cos y y sin x + cos x dx w dx Again differentiating and using y (0) = 1 and y(0) = , we get y (0) = . The solution of primitive integral equation (x2 + y2) dy = xy dx, is y = y (x). If y(1) = 1 and y(x0) = e, then x0 is 21. (A) 2(e2 1) 2(e2 + 1) (C) 3e Ans. (B) (D) e2 + 1 2 C dy xy . = dx x 2 + y 2 Solving the homogenous differential equation by writing y = vx , we get Sol. We have x2 2y 2 1 + ln y = . 2 For y = e , 2 x0 2e 2 + ln e = 1 0 0 A = 0 1 1 , I = 0 2 4 (A) - 6, -11 (C) - 6, 11 22. Ans. 1 x02 = 3e2 x0 = 2 3 e. 1 0 0 0 1 0 and A-1 = 1 2 , then the value of c and d are ( A + cA + dI ) 6 0 0 1 (B) 6, 11 (D) 6, - 11 C for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-6 We evaluate A2 and A3 and write the given equation as AA-1 = I = Sol. 1 3 [ A + cA 2 + dA ] . 6 Comparing the corresponding elements on both the sides we get c = - 6, d = 11. Alternatively, we may use Cayley Hamilton Theorem. 3 2 If P = 1 2 23. 1 2 ,A= 3 2 1 1 T T 2005 P, then x is equal to 0 1 and Q = PAP and x = P Q 1 2005 (A) 1 0 (C) Ans. Sol. 24. Ans. Sol. 1 1 2 + 3 4 2 3 1 .c .c a A a l PTP = I eel T Q = PAP so that m x = PTQ2005P = PT(PAPT)2005P m d T T T 2004 d = P PAP (PAP ) P aa 1 2005 o 2005 l o =A = 1 0 nnl w 6 at w a point P (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Tangent to the curve y = x2 +o o Then the coordinates of Q are .d (A) ( 6, 11) (B) ( 9, 13) .d w (C) ( 10, 15) (D) ( 6, 7) w w w D 2x-y+5=0 w Equation of tangent w to the parabola at (1, 7) is Q (y + 7) + 6 = 0 2x y + 5 = 0. 2 Centre (- 8, - 6) Equation of CQ = x + 2y + k = 0 8 12 + k = 0 k = 20 PQ 4x 2y + 10 = 0 CQ x + 2y + 20 = 0 = 5x + 30 = 0 x = - 6 6 + 2y + 20 = 0 y = 7 Hence the point of contact is ( 6, 7). x P(1,7 C(-8,-6) 1 If f(x) is a continuous and differentiable function and f = 0 n 1 and n I, then n (A) f (x) = 0, x (0, 1] (B) f (0) = 0, f (0) = 0 (C) f (0) = 0 = f (0), x (0, 1] (D) f(0) = 0 and f (0) need not to be zero 25. Ans. Sol. 4 + 2005 3 6015 (B) 4 2005 3 2005 1 2005 2 3 m (D) m 4 3 2005 2 +o o B 1 Given f = 0 n 1 and n I. n This indicates that f(x) has a wavy behaviour. Amplitude of the wave either (a) is constant (b) increases or (c) decreases. In case of (a) and (b), function will not be differentiable at 0. Amplitude has to decrease such that f (0) = 0. for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942 IIT-JEE-2005-S-7 G G G G G G b a G b a G G G G G G If a, b, c are three non-zero, non-coplanar vectors and b1 = b G 2 a, b 2 = b + G 2 a, |a| |a| G G G G G G G G G G G G G G G G G G G G G c a G b c G G c a G b1 c G G c a G b c G G c a G b c G c1 = c G 2 a + G 2 b1 , c2 = c G 2 a G 2 b1 , c3 = c G 2 a + G 2 b1 , c4 = c G 2 a G 2 b1 , then |a| |c| |a| | b1 | |c| |c| |c| |b| the set of orthogonal vectors is G G G G G G (B) (a, b1 , c2 ) (A) (a, b1 , c3 ) G G G G G G (C) (a, b1 , c1 ) (D) (a, b 2 , c2 ) 26. Ans. Sol. 27. Ans. Sol. B G G G G b a G G G G Obviously a b1 = b a G 2 a a = 0 a G G G G & a c2 = 0 and b1 c2 = 0 . G G G a, b1 , c2 are orthogonal vectors. m m o o ( ) .c .c a a l R, y > 0, y = y(x), y(1) = 1, then y(-3) is x l For the primitive integral equation ydx + y2 dy = x dy;e e (A) 3 (B) 2 m (C) 1 (D) 5 m dd aa A o l lo dx xdy y = dy n 2 n y w x oow = y + c y .d .d y(1) = 1 c = 2 w w y2 2y + x = 0 w y(-3): w y2 2y 3 = 0 (y 3)(y + 1) = 0 w w y = 3, -1 28. X and Y are two sets and f : X Y. If {f(c) = y; c X, y Y} and {f-1(d) = x; d Y, x X}, then the true statement is (B) f-1 (f(a)) = a (A) f(f-1(b)) = b -1 (C) f(f (b)) = b, b y (D) f-1(f(a)) = a, a x Ans. Sol. D The given data is shown in the figure Since f 1(d) = x f(x) = d Now, if a x, f(a) d f 1(f(a)) = a. X Y y c d x a f 1 f(a) Analyse your performance in Screening Test for evaluation of your preparation for Mains. A comprehensive analysis of your preparation on different topics would be couriered to you. Fill this sheet as per answers you have made in the IIT-JEE Screening Examination as per the sequencing provided in the solution booklet and send to nearest FIITJEE s office immediately. for more papers please visit www.downloadmela.com FIITJEE Ltd. ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 2686 5182, 26965626, 2685 4102, 26515949 Fax : 26513942

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