JEE MAIN 01 Mathematics 1. The circle passing through (1, ) and touching the axis of x at (, 0) also passes through the point (1) (, 5) () (5, ) () (, 5) (4) ( 5, ) The equation of the circle due to point (, 0) touching the axis of x is given by ( x ) ( y 0) y 0 It is given that the circle passes through point (1, ). Therefore, (1 ) ( ) ( ) 4 4 0 4 Therefore, the equation of the circle is: ( x ) y 4y 0 from which it is clear that (5, ) satisfies. Hence, the correct option is ().. ABCD is a trapezium such that AB and CD are parallel and BC CD. If ADB, BC pand CD q, then AB is equal to p q cos (1) pcos qsin p q () p cos q sin ( p q )sin () ( pcos qsin ) ( p q )sin (4) pcos qsin Using sine rule in the triangle ABD, as shown in the following figure, we get AB BD p q p q sin AB sin sin( ) sin( ) sin cos cos sin p q sin ( p q )sin. [(sin g) / ( p q )] [(cos p) / p q ] ( pcos qsin ) Copyright Wiley India Page 1
Hence, the correct option is (4).. Given: A circle, x + y = 5 and a parabola, y 4 5 x. Statement-I: An equation of a common tangent to these curves is y x 5. 5 Statement II: If the line, y mx ( m 0), is their common tangent, then m satisfies m m 4 m + = 0. (1) Statement I is True; Statement II is true; Statement II is not a correct explanation for Statement I. () Statement I is True; Statement II is False. () Statement I is False; Statement II is True. (4) Statement I is True; Statement II is True; Statement II is a correct explanation for Statement I. Let us consider that the common tangent to the parabola be y mx 5 m ( m 0). Its distance from the centre of the circle, (0, 0) must be equal to the radius of the circle, 5. Therefore, 4 m m m m m 0 m Hence, ( m 1)( m ) 0 m 1. Therefore, the common tangents are obtained as y x 5 and y x 5. Both statements are correct as the condition m 1 satisfies Statement-II. 4. A ray light along x y gets reflected upon reaching x-axis, the equation of the reflected ray is (1) yx () y x () y x 1 (4) y x Let us consider a point (0, 1) on the line, that is, on the ray of light. So this point also lies on the image of this line. So the equation of the reflected ray is Copyright Wiley India Page
1 ( y 0) tan 0 ( x ) y x y x x y Therefore, the equation of reflected ray is yx. 5. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given? (1) Median () Mode () Variance (4) Mean Before the grace marks were given, the variance of marks of the students is expressed as ( xi x) 1 (1) N After the grace marks were given, the variance of marks of the students is expressed as i [( xi 10) ( x 10)] ( x x) () N N From (1) and (), we get 1. Hence, variance will not change even after the grace marks were given; however, mean, median and mode will increase by 10. Hence, the correct option is (). 6. If x, y, z are in AP and tan 1 x, tan 1 y and tan 1 z are also in AP, then (1) x = y = 6z () 6x = y = z () 6x = 4y = z (4)x = y = z If x, y, z are in AP, we have, y = x + z (1) If tan 1 x, tan 1 y and tan 1 z are also in AP, we have, Therefore, tan y x z 1 y 1 xz 1 1 tan 1 1 1 tan y tan x tan z () Copyright Wiley India Page
Using Eq. (1) in Eq. (), we get, 1 x z 1 x z tan tan y xz or x z 0 x y z 1 y 1 xz Hence, the correct option is (4). 5 f ( x) dx ( x) 7. If ( ), then x f x dx is equal to 1 (1) ( ) ( ) x x x x dx C 1 () ( ) ( ) x x x x dx C 1 () ( ) ( ) x x x x dx C 1 (4) ( ) ( ) x x x x dx C We have, f ( x) dx ( x) Let x = t and x dx = dt/. Therefore, 5 1 1 1 x f ( x ) dx tf ( t) dt t ( t) dt 1 f ( t) dtdt x ( x ) x ( x ) dx C. Hence, the correct option is (). 8. The equation of the circle passing through the foci of the ellipse centre at (0, ) is (1) x y 6y 7 0 () x y 6y 5 0 () x y 6y 5 0 (4) x y 6y 7 0 Foci of the ellipse is given by ( ae, 0). We have Radius of the circle as, r ( ae) b (1) x y 16 9 1, and having where a = 4; b = ; e 1 (9 /16) 7 / 4 ae 7 Therefore, from Eq. (1), we get, r ( ae) b 7 9 4 Therefore, equation of circle with centre 0, and radius 4 is, ( x 0) ( y ) 4 That is, x y 6y 7 0 Hence, the correct option is (4). 9. The x-coordinate of the incentre of the triangle that has the coordinates of midpoints of its sided as (0, 1) (1, 1) and (1, 0) is (1) ()1 () 1 (4) From the following figure of the given triangle, the x-coordinate of the incentre is obtained as follows: Copyright Wiley India Page 4
ax1 bx cx 0 0 4. abc 4 10. The intercepts on x-axis made by tangents to the curve, parallel to the line y = x, are equal to (1) ± () ± () ±4 (4) ±1 Slope of the tangent to the curve will be. dy So we can equate the slope as, x x dx For x =, we have, For x =, we have, y y t dt 0 0 t dt x y t dt, x R, which are Therefore, one tangent passes through the point (, ) and has slope : y ( x ) y x The other tangent passes through the point (, ) and has slope : y ( x ) y x Substituting y = 0, we get x-intercepts as, x = 1 and 1. Hence, the correct option is (4). 11. The sum of first 0 terms of the sequence 0.7, 0.77, 0.777,, is 7 0 (1) (99 10 7 ) 0 () (179 10 ) 9 81 7 0 () (99 10 7 ) 0 (4) (179 10 ) 9 81 have t r = 0.777,, r, which is expressed as r 1 1 10 1 10 1 r 10 10 1 1 10 r 7 r 7(10 10 10 10 ) 7 7 (1 10 ) 1 110 9 9 0 Copyright Wiley India Page 5
0 0 7 r 7 1 0 7 0 Therefore, S0 tr 0 10 0 (1 10 ) (179 10 ). r1 9 r1 9 9 81 Hence, the correct option is (). 1. Consider Statement-I: ( p q) ( p q) is a fallacy. Statement-II: ( p q) ( q p) is a tautology. (1) Statement -I is True; Statement II is true; Statement-II is not a correct explanation for Statement-I () Statement-I is True; Statement-II is False. () Statement-I is False; Statement-II is True (4) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I. Table 1 p q ~p ~q p q p q ( p q) ( p q) T T F F F F F T F F T T F F F T T F F T F F F T T F F F Table p q ~p ~q p q q p ( p q) ( q p) T T F F T T T T F F T F F T F T T F T T T F F T T T T T Tautology From Table 1, it is obvious that ( p q) ( p q) is fallacy. From Table, it is obvious that ( p q) ( q p) is tautology. Although both statements are true, Statement-II is not a correct explanation for Statement-I. 1. The area (in square units) bounded by the curves y x, y x 0, x-axis, and lying in the first quadrant is (1) 6 () 18 () 7 (4) 9 4 First solving the equations, We have, x x (1) Squaring on both side of Eq. (1), we get, 4x x 6x 9 x 10x 9 x 9, x 1. Copyright Wiley India Page 6
Since x 1 intersects the parabola below the x-axis, this point is extraneous. So for x 9 we have, y. Therefore, the required area under the curve (see the following figure) is y (y ) y y y 9 9 9 9. 0 0 Hence, the correct option is (4). tan A cot A 14. The expression can be written as 1cot A 1tan A (1) seca coseca + 1 () tana + cota () seca + coseca (4) sina cosa + 1 We have tan A cot A 1 cot Acot A 1 cot A 1cot A 1 cot A 1 tan A cot A(1 cot A) cot A (1 tan A) cot A(1 cot A) (1 cot A) cot A(1 cot A) cosec A cot A 1 sec Acosec A. cot A 15. The real number k for which the equation x + x + k = 0 has two distinct real roots in [0, 1] (1) lies between and () lies between 1 and 0 () does not exist (4) lies between 1 and When the given equation, x + x + k = 0 has two distinct real roots in [0, 1], then f (x) will change sign, but f (x) = 6x + > 0, for all values of x R. Therefore, no value of k exists. Hence, the correct option is (). (1 cos x)( cos x) lim 16. x0 xtan 4x (1) 1 () 1 is equal to () (4) 1 4 Copyright Wiley India Page 7
(1 cos x)( cos x) sin x cos x 1 We have, lim lim x0 0 xtan 4x x x tan 4x 4x 1 4 4x Hence, the correct option is (). 17. Let T n be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn1 Tn 10, then the value of n is (1) 5 () 10 () 8 (4) 7 If T 1 T 10, then the value of n is obtained as follows: n n 1 1 1 n1 n n n n n n n C C 10 10 nn 1 n 1 n 60 6 6 nn 1 0 nn 1 5 4 n 5 18. At present, a firm is manufacturing 000 items. It is estimated that the rate of change of dp production P w.r.t. additional number of workers x is given by 100 1 dx x. If the firm employs 5 more workers, then the new level of production of items is (1) 000 () 500 () 4500 (4) 500 Given that, dp 100 1 dx x dp 100 1 x dx Therefore, the new level of production of items is P 5 1 / dp (100 1 x) dx ( P 000) 5100 (5) P 500. 000 0 Hence, the correct option is (). 19. Statement-I: The value of the integral b / Statement-II: f ( x) dx f ( a b x) dx. b a a dx /6 1 tan is equal to. x 6 (1) Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I. () Statement-I is True; Statement-II is False. () Statement-I is False; Statement-II is True. (4) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I. We have Copyright Wiley India Page 8
/ / / / dx dx dx dx I /61 tan x /6 /6 /61 cot x 1 tan x 1 tan x 6 / / dx tan x I dx I /6 1 /61 tan x 1 tan x Hence, the correct option is (). / /6 1 tan x 1 tan x 1 dx I I. 6 1 1 0. If p 1 is the adjoint of a matrix A and A 4, then α is equal to 4 4 (1) 11 () 5 () 0 (4) 4 1 We have, p 1 4 4 Therefore, Adj A A Adj A 16 1(1 1) (4 6) (4 6) 16 6 16 11. 1. The number of values of k, for which the system of equations ( k 1) x 8y 4k kx ( k ) y k 1 has no solution, is (1) 1 () () (4) infinite For no solution the lines must be parallel for which we have, k k k k 1 1 8 4 k k k k k k 4 8 4 0 1, k 1 is rejected since it gives coincident lines. Therefore number of such values of k is just one.. If 1 y sec(tan x) (1) 1, then dy dx () 1 at x = 1 is equal to Copyright Wiley India Page 9
1 () (4) 1 We have, y sec(tan x) dy 1 1 1 dy 1 1 Therefore, sec(tan x) tan(tan x). 1. dx 1 x dx x1 Hence, the correct option is (4).. 4 1 4 5 If the lines y z and y z are coplanar, then k can have 1 1 k k 1 (1) exactly one value () exactly two values () exactly three values (4) any value For lines to be coplanar, scalar triple product of vectors joining the two given points of the lines and the parallel vectors to the line must be 0. We have, 1 1 1 1 1 k 0 1(1 k) 1(1 k ) 1( k) 0 k 1 k 1 k 0 k 1 k k 0 ( k)( k ) 0 Therefore, there are two values of k. Hence, the correct option is (). 4. Let A and B be two sets containing elements and 4 elements, respectively. The number of subsets of A B having or more elements is (1) 0 () 19 () 11 (4) 56 We know that A B will have eight elements. Out of these 8 elements, the total number of subsets containing or more elements is, 8 8 8 8 8 8 8 8 8 8 C C4 C5 C6 C7 C8 C0 C1 C 56 1 8 8 19. Hence, the correct option is (). 5. If the vectors AB iˆ 4kˆ and AC 5iˆ ˆj 4kˆ are the sides of a triangle ABC, then the length of the median through A is (1) 7 () () 45 (4) 18 AB AC From the following figure, we see that, AM AM 4iˆ ˆj 4kˆ Therefore, AM 16 16 1. Copyright Wiley India Page 10
Hence, the correct option is (). 6. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is 1 11 (1) 5 () 5 10 17 () 5 (4) 5 We have p = Correct answer = 1/; q = Incorrect answer = /. Therefore probability of either 4 or 5 correct answers is, 51 1 50 4 1 5 5 5 5 5 C4 C5 C4 C5 5 5 5 1 1 1 1 5 1 11. () () Hence, the correct option is (). 7. If z is a complex number of unit modulus and arguments θ, then 1 z arg 1 z equals (1) () () (4) We have, z 1 zz 1. Therefore, 1 z 1 z z. Therefore arg z 1z 1 (1/ z) Hence, the correct option is (). 8. If the equations x + x + = 0 and ax + bx + c = 0, a, b, cr, have a common root, then a : b : c is (1) : : 1 () 1 : : () : 1 : (4) 1 : : Discriminant = 4 1 < 0 and 1,, R Therefore the equation has complex conjugate roots, which means both roots are common. Copyright Wiley India Page 11
a b c Hence all coefficients must be proportional,. 1 Therefore, a:b:c = 1::. Hence, the correct option is (4). 9. Distance between two parallel planes x + y + z = 8 and 4x + y + 4z + 5 = 0 is (1) 5 () 7 () 9 (4) The two parallel planes can be written as, 4x y 4z 16;4x y 4z 5 Let a point on the first plane is 0, 0, 4 Therefore, its distance from the other plane is obtained as, Hence, the correct option is (). x1 x1 0. The term independent of x in expansion of x x 1 x x (1) 10 () 10 () 10 (4) 4 We have x1 x1 / 1/ 1/ x x 1 x x / 1/ x x 1 10r Therefore, 10 1/ / 1/ ( x 1) ( x x 1) / 1/ 1/ 0 0 16 5 1 1 7. 4 4 6 6 10 is 1 ( x 1) ( x1). x ( x 1) x x x x x x 1 1/ x 1 1/ 1 1/ 1/ 10 1 1 1 ( ) 1 r 10 1 1 r 10 (05 r)/6 r1 r ( 1) r T C x x C x For T r 1 to be independent of x, 0 5r 0 r 4 Therefore, 4 10 T T 1 C 10. 5 41 4 Hence, the correct option is (). 10 Copyright Wiley India Page 1