INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com AIEEE- 2003 Mathematics 1. A function f front he set of natural numbers to integers defined by 1, when is odd 2,when is even 2 n n f(n) n n − = − (1) One – one but not onto (2) Onto but not one -one (3) One – one and onto both (4) Neither one – one nor onto Sol. (3) : f N I → (1) 0 , (2) 1, (3) 1, (4) 2, (5) 2 f f f f f = =− = − = − = and (6) 3 f = − so on. In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is one-one and onto function. 2. Let 1 z and 2 z be two roots of the equation. 2 0, z az b z + + = being complex. Further, assume that the origin, 1 z and 2 z form an equilateral triangle. Then (1) 2 a b = (2) 2 a=2b (3) 2 3 a b = (4) 2 4 a b = Sol. (3) 2 0 z az b + + = 1 2 1 2 & z z a zz b + =− = 1 2 0, , z z form an equilateral ∆ ∴ 2 2 2 1 2 1 1 2 2 0 0. . .0 z z z z z z + + = + + (for eq. ∆; 2 2 2 1 2 3 12 2 3 31 )z z z zz zz zz + + = + + 2 2 1 2 12 z z zz + = 2 1 2 1 2 ( ) 3 z z zz + = ∴ 2 3 a b = . 3. If z and ω are two non – zero complex numbers such that | | 1, zω = and Arg( ) , 2π ω = then z ωis equal to (1) 1 (2) –1 (3) i (4) –I Sol. (4) | || | 1 z ω = …… (1) As Arg 2 z z i π ω ω = ∴ = ∴ 1 zω = (2) from (1) and (2) |z|=| ω|=1 and 0; z z ω ω + = 0 z z ω ω + = 1 2 3 4 5 0 – 1 1 2 – 2 3 6 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com . . z z z ω ω ωω ω − = − = 2 | | z i i ω ω = − = − . 4. If 1 1, 1 x ii + = − then (1) 4 , x n = where n is any positive integer (2) 2 , x n = where n is any positive integer (3) 4 1 x n = + , where n is any positive integer (4) 2 1, x n = + where n is any positive integer Sol. (1) 2 2 1 (1 ) 1 1 1 1 x x i i i i + + = ⇒ = − − 2 1 2 1 () 1 1 1 x x i i i + + = ⇒ = + ∴ 4 ; x n n I+ = ∈ . 5. If 2 2 2 2 2 2 11 0 1 a a a b b b c c c ++ = + and vectors ( 2 2 1, , ), (1, , ) a a b b and 2 (1, , ) c c are non – coplanar, then the product a b c equals. (1) 2 (2) – 1 (3) 1 (4) 0 Sol. (2) 2 3 2 3 2 3 11 0 1 a a a b b b c c c ++ = + ⇒ 2 2 3 2 2 3 2 2 3 11 0 1 a a a a a b b b b b c c c c c + = ( ) ( ) ( ) ( )( )( ) 0 a b b c c a abc a b b c c a − − − + − − − = ( 1)[( )( )( )] 0 abc a b b c c a + − − − = As 222 11 0 1 a a b b c c ≠ (given condition) ∴ 1 abc = − . 6. If the system of linear equations 2 0 x ay az + + = 3 0 x by bz + + = 4 0 x cy cz + + = has a non – zero solution, then a,b,c (1) Are in A.P. (2) Are in G.P. (3) Are in H.P. (4) Satisfy a +2b+3c =0 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Sol. (3) 1 2 1 3 0 1 4a a b b c c = 2 2 3 2 c c c → − 3 3 2 1 0 11 2 a b b R R R c c ⇒ → − , 2 2 1 R R R → − ⇒ 1 0 0 0 0 2 a b b a c b c b − = − − ( ) ( )(2 ) 0 b c b b a c b − − − − = On simplification 2 1 1 b a c = + ∴ a, b, c are in Harmonic Progression. 7. If the sum of the roots of the quadratic equation 2 0 ax bx c + + = is equal to the sum of the squares of their reciprocals, then , a b c a and cb are in (1) Arithmetic Progression (2) Geometric Progression (3) Harmonic Progression (4) Arithmetic – Geometric Progression Sol. (3) 2 0 ax bx c + + = ;b c a a α β αβ − + = = As for given condition 2 2 1 1 α β α β + = + 2 2 2 2 α β α β α β + + = 22 22 2 b c b a a c a a− − = On simplification 2 2 2 2a c ab bc = + ⇒ 2a c b b a c = + ∴ , & a b c c a b are in H.P. 8. The number of real solutions of the equation 2 3 | | 2 0 x x − + = is (1) 2 (2) 4 (3) 1 (4) 3 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Sol. (2) 2 3 | | 2 0 x x − + = (| | 2)(| | 1) 0 x x − − = | | 1; 2 x = 1; 2 x = ± ± ∴ No. of Solution = 4. 9. The value of ‘a’ for which one root of the quadratic equation 2 2 ( 5 3) (3 1) 2 0 a a x a x − + + − + = is twice as large as the other, is (1) 23 (2) 23 − (3) 13 (4) 13 − Sol. (1) 2 2 2 1 3 2 3 &2 5 3 5 3 a a a a a α α − = = − + − + 2 2 2 2 1 (1 3 ) 2 2 9 ( 5 3) 5 3 a a a a a − = − + − + 2 2 (1 3 ) 9 ( 5 3) a a a − = − + 2 2 9 6 1 9 45 27 a a a a − + = − + 39 26 a = a 23 = . 10. If A= a b b a and 2 , A α β β α = then (1) 2 2, a b ab α β = + = (2) 2 2, 2 a b ab α β = + = (3) 2 2 2 2 , a b a b α β = + = − (4) 2 2 2 , ab a b α β = = + Sol. (2) 2 A α β β α = = a b a b b a b a 2 2; a b α= + 2ab β = . 11. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 form the first five question. The number of choices available to him is (1) 140 (2) 196 (3) 280 (4) 346 Sol. (2) As for given question two cases are possible. (i) Selecting 4 out of first five question & 6 out of remaining 8 Question = 8 4 6 5 140 C C × = choices. (ii) Selecting 5 out of first five Question & 5 out of remaining 8 Qs. = 5 8 5 5 56 C C × = choices. ∴ Total No. of choices = 140 + 56 = 196. 12. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by (1) 6 ! 5! × (2) 30 (3) 5!×4! (4) 7!×5! Sol. (1) No. of ways in which 6 mm can be arranged at a round table = (6–1)! Now women can be arranged in 6! Ways. × × × × × × M M M M M M W W W W W W INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Total Number of ways = 6! × 5! 13. If 1, 2 , ω ω are the cube roots of unity, then 2 2 2 1 1 1n n n n n n ω ω ω ω ω ω ∆ = is equal to (1) 0 (2) 1 (3) ω (4) 2 ω Sol. (1) Applying 1 1 2 3 R R R R → + + As, 2 1 0 n n ω ω + + = (if you don’t remember this, put n=1 and you will get the answer) ∴ 0 ∆ = . 14. If n r C denotes the number of combinations of n things taken r at a time, then the expression 1 1 2 n n n r r r C C C + − + +× equals (1) 2 n r C + (2) 2 1 n r C + + (3) 1 n r C + (4) 1 1 n r C + + Sol. (2) Using 1 1 n n n r r r C C C + + − = 1 1 n n n n r r r C C C Cr + − + + + 1 1 n n n r r r C C C + ++ + 1 1 1 n n r r C C + + + + 2 1 n r C + + ⇒ . 15. The number of integral terms in the expansion of ( )256 8 3 5 + is (1) 32 (2) 33 (3) 34 (4) 35 Sol. (2) 256 8 ( 3 5) + 256 256 8 1 ( 3) ( 5) r r r r T C − + = 256 256 /8 2 1 (3) (5) r r r r T C − + = Terms would be integral if 256 & 2 8 r r − both are +ve Integers. As 0 256 r ≤ ≤ ∴ 0, 8, 16, 24,.......256 r = For above values of 256 , 2 r r − is also an integer. ∴ Total no. of Values of r = 33. 16. If x is positive, the first negative term in the expansion of ( )27/5 1 x + is (1) 7th term (2) 5th term (3) 8th term (4) 6th term INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Sol. (3) 27 5 (1 ) x + 1 ( 1)( 2).....( 1)( ) ! r r n n n n r T x r + − − − + = For first negative term 1 0 n r − + < 32 7 5 r r > ∴ = ∴ First negative term is 8 T . 17. The sum of the series 1 1 1 .......... 1.2 2.3 3.4 − + − upto∞ is equal to (1) 2 log 2 e (2) log 2 1 e − (3) log 2 e (4) 4 loge e Sol. (4) 1 1 1.......... 1.2 2.3 3.4 − + ∞ Let 1 1 1 ( 1) 1 n T n n n n = = − + + 1 2 3 4 5............. S T T T T T = − + − + ∞ 1 1 1 1 1 1 1 1 ............ 1 2 2 3 3 4 4 5 − − − + − − − ⇒ 1 1 1 1 1 2 .......... 2 3 4 5 − − + − ∞ [ ] 1 2 log(1 1) 1 − − + + 4 2log 2 1 log e − = . 18. Let f (x) be a polynomial function of second degree. If f(1) = f(–1) and a, b, c are in A.P. then f’(1), f’(2) and f’(3) are in (1) A.P. (2) G.P. (3) G.P. (4) Arithmetic – geometric Progression Sol. (1) 2 ( ) f x ax bx c = + + (1) ( 1) f f a b c a b c = − ⇒ + + = − + 0 b = ∴ 2 ( ) f x ax c = + '( ) 2 f x ax = Now '( ); '( )& '( ) f a f b f c are 2 ( ); 2 ( );2 ( ) a a a b a c If a, b, c are in A.P. then '( ), '( )& '( ) f a f b f c are also in A.P. 19. If 1 2 3 , , , x x x and 1 2 3 , , y y y are both in G.P. with the same common ratio, then the pints 1 1 ( , ) x y 2 2 ( , x y ) and 3 3 ( , ) x y (1) Lie on a straight line (2) Lie on an ellipse (3) Lie on a circle (4) Are vertices of a triangle INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Sol. (1) Taking co-ordinates as , ; (, ) & ( , ) x y x y xr yr r r Above co-ordinates satisfy the relation y mx = ∴ Lies on the straight line. 20. The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is (1) cot a nπ (2) cot 2 2 a n π (3) cot 2 a n π (4) cot 4 2 a n π Sol. (2) tan 2a n r π = sin 2a n R π = cot cosec 2a r R n n π π + = + .cot 2 2 a r R n π + = . 21. If in a triangle A B C 2 2 3 cos cos , 2 2 2 C A b a c + = then the sides a, b, and c (1) Are in A. P. (2) Are in G.P. (3) Are in H.P. (4) Satisfy a + b = c Sol. (1) If 2 2 3 cos cos 2 2 2 c A b a c + = [ ] [ ] cos 1 cos 1 3 a c c A b + + + = ( ) ( cos cos ) 3 a c a c c b b + + + = 3 a c b b + + = 2 a c b + = , , a b c are in A.P. 22. Ina triangle ABC, medians AD and BE are draw. If AD =4 6 DAB π ∠ = and , 3 ABE π ∠ = then the area of the ABC ∆ is (1) 83 (2) 16 3 (3) 32 3 (4) 64 3 Sol. () No given option is correct o 8 /3 tan 60 x = 8 3 3 x = Area of 1 8 16 4 2 3 3 3 3 ABD ∆ = × × = 60o 90o 8/3 4/3 P E A B x D C 30o INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com ∴ Area of 16 32 2 3 3 3 3 ABC ∆ = × = 23. The trigonometric equation 1 1 sin 2 sin , x a − − = has a solution for (1) 1 1 | | 2 2 a < < (2) all real values of a (3) 1 | | 2 a < (4) 1 | | 2 a ≥ Sol. () 1 1 sin 2sin x a − − = ∵ 1 sin 2 2 x π π − − ≤ ≤ ∴ 1 2sin 2 2 a π π − − ≤ ≤ 1 sin 4 4 a π π − − ≤ ≤ ∴ 1 1 2 2 a − ≤ ≤ ∴ 1 | | 2 a ≤ (As 1 12 2 > ) Out of given four options no one is absolutely correct but 3. Could be taken into consideration. 1 | | 2 a → ≤ is correct, if 1 | | 2 a < is taken as correct then it domain satisfy for 13 a = but equation is satisfied. 1 1 1.2 2 3 > > 24. The upper 34 th portion of a vertical pole subtends an angle 1 3 tan 5 − at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is (1) 20m (2) 40m (3) 60m (4) 80m Sol. (2) θ = α + β β = θ – α tan tan tan 1 tan tan θ α β θ α − = + 3 40 160 5 1 . 40 160 h h h h − = + h2 – 200h + 6400 = 0 h = 40 or 160 meter ∴ possible height. = 40 metre 25. The real number x when added to its inverse gives the minimum value of the sum at x equal to (1) 2 (2) 1 (3) – 1 (4) –2 Sol. (2) 1 y x x = + α β θ h − 43 h/4 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 2 1 1 dy dx x = − For max. or min. 2 1 1 0 1 x x − = ⇒ =± 2 2 2 3 2 2 2 2 ( minima) x d y d y ve dx x dx = = ⇒ = + ∴ x = 1 26. If f : R→R satisfies f (x+y) =f (x) + f (y), for all x, y ∈R and f (1) = 7, then 1 ( ) n r f r − ∑ is (1) 72n (2) 7( 1) 2 n + (3) 7 ( 1) n n+ (4) 7 ( 1) 2 n n+ Sol. (4) ( ) () ( ) f x y f x f y + = + Let ( ) f m α α = ( ) 7; 7 f m ⊥ = ∴ = ( ) 7 f x x = 1 1 ( ) 7 n n r f r r = = ∑ ∑ = 7 ( 1) 2 n x+. 27. If f (x) = n x , then the value of f(1) – '(1) ''(1) '''(1) ( 1) (1) .... 1! 2! 3! ! n n f f f fn − + − + + is (1) 2n (2) 1 2n− (3) 0 (4) 1 Sol. (3) ( ) (1) 1 n f x x f = ⇒ = 1 '( ) '(1) n f x nx f n − = ⇒ = 2 ''( ) ( 1) ''(1) ( 1) n f x n n x f n n − = − ⇒ = − …… ……. ( ) ! (1) ! n n f x n f n = ⇒ = ( 1) ( 1) ( 2) ! 1 ....... ( 1) 1 ! 2 ! 3 ! ! n n nn nn n nn − − − = − + − + + − 0 1 2 3 ...... ( 1) n n n n n n n C C C C C = − + − + +− = 0 Alternatively, put n=1 and n=2 and in both the cases you will get the answer as 0. 28. Domain of definition of the function 3 10 2 3 ( ) log ( ), 4 f x x x x = + − − is INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com (1) (1,2) (2) (–1,0) ∪ (1,2) (3) (1,2) (2, ) ∪ ∞ (4) (–1,0) (1, 2) (2, ) ∪ ∪ ∞ Sol. (4) 3 10 2 3 ( ) log ( ) 4 f x x x x = + − − 2 4 0 x − ≠ 3 0 x x − > 4 x ≠ ± ( 1,0) (1, ) { 4} D ∴ = − ∪ ∞− ( 1, 0) ( , 2) (2, ) D= − ∪ ⊥ ∪ ∞ . 29. [ ] [ ]3 2 1 tan 1 sin 2 lim 1 tan –2 2 x x x x x π π → − − + is (1) 18 (2) 0 (3) 1 32 (4) ∞ Sol. (3) 2 tan .(1 sin ) 4 2 lim ( 2 )3 x x x x π π π → − − − Let ; 0 2 x y y π = + → 0 lim y→ 3 tan .(1 cos ) 2( 2 ) y y y − − − − 0 lim y→ 2 3 tan .2sin 2 2 ( 8). .8 8 y y y − − 0 lim y→ 2 tan 1 sin/2 2 . 32 /2 2 y y y y ⇒ 1 32 . 30. If ( ) ( ) 0 log 3 – log 3 – lim , x x x k x → + = the value of k is (1) 0 (2) 13 − (3) 23 (4) 23 − Sol. (3) 0 log(3 ) log(3 ) lim x x x K x → + − − = (By L Hospital rule) – + – + 1 0 – 1 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 0 1 1 3 3 lim 1 x x x K → − − + − = 23 K ∴ = . 31. Let f(1) = g(1) = k and their nth derivatives fn(1), gn(1) exist and are not equal for some n. Further if ( ) ( ) – ( ) – ( ) ( ) ( ) lim 4 ( ) – ( ) x a f a g x f a g a f x g a g x f x → + = (1) 4 (2) 2 (3) 1 (4) 0 Sol. (1) (9( ) ( ) lim 4 9( ) ( ) x ak x kf x k f x → − = − (By L Hospital’ rule) 9'( ) '( ) lim 4 9'( ) '( ) x a x f x k x f x − − = − ∴ 4 k = . 32. The function f(x) = log (x+ 2 1), rx + is (1) An even function (2) An odd function (3) A periodic function (4) Neither an even nor odd function Sol. (2) 2 ( ) log( 1) f x x x = + + 2 ( ) log( 1) f x x x − =− + + ( ) ( ) f x f x − =−− f(x) is odd function. 33. If f(x) = 1 1 | | , 0 then f(x) is 0 , 0 x x xe x x − + ≠ = (1) Continuous as well as differentiable for all x (2) Continuous for all x but not differentiable at x =0 (3) Neither differentiable nor continuous at x = 0 (4) Discontinuous every where Sol. (2) f(0) = 0 1 1 | | ( ) x x f x xe − + = R.H.L. 2/2/0 0 lim(0 ) lim 0 h h h h h h e e − → → + = = L.H.L. 1 1 0 lim(0 ) 0 h h h h e − − → − = ∴ ( ) f x is continuous. INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com R.H.D. 1 1 1 1 0 (0 ) lim 0 h h h h h h e h e h − + − + → + − = L.H.D. 1 1 1 1 0 (0 ) lim 1 h h h h h h e h e h − − − + → − − = − ∴ L.H.D. ≠ R.H.D. ( ) f x is not differentiable at x = 0. 34. If the function f(x)= 3 2 2 2 –9 12 1, x ax ax + + where a>0, attains its maximum and minimum at p and q respectively such that 2 , p q = then a equals (1) 3 (2) 1 (3) 2 (4) 12 Sol. (3) 3 2 2 ( ) 2 9 12 1 f x x ax a x = − + + 2 2 '( ) 6 18 12 f x x ax a = − + ''( ) 12 18 f x x a = − for max. or min. 2 2 2 2 6 18 12 0 3 2 0 x ax a x ax a − + = ⇒ − + = x = a or x = 2a, at x = a max. and at x = 2a min ∵ p2 = q a2 = 2a ⇒ a = 2 or a =0 but a > 0, therefore a = 2. 35. If f (y) = , y e g(y)=y; y = 0 and 0 ( ) ( – ) ( ) , t f t f t y g y dy = ∫ then (1) ( ) 1– (1 ) t F t e t − = + (2) ( ) –(1 ) t F t e t = + (3) f(t) = t t e (4) ( ) t F t te− = Sol. (2) 0 ( ) ( ) ( ) t F t f t y g y dy = − ∫ 0 0 . t t t y t y II I e y dy e e y dy − − = = ∫ ∫ 0 0 t y y t t y y t e ye e e ye e − − − − = − − =− + 0 1 t t t e te e − − − + −− 1 (1 ) t t t t t e e e t e + − = = − + 36. If ( – ) (), f a b x f x + = then ( ) ba x f x dx ∫ is equal to (1) ( ) 2 ba a b f b x dx + − ∫ (2) ( ) 2 ba a b f x dx + ∫ (3) ( ) 2 ba b a f x dx + ∫ (4) INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Sol. (2 & 4) ( ) b a I x fx dx = ∫ ( )( ) b a I a b x f a b x dx = + − + − ∫ ( ) ( ) ( ) b b a a a b f a b x dx x f a b x dx = + + − − + − ∫ ∫ ( ) ( ) () b b a a a b f a b x dx x f x dx = + + − − ∫ ∫ 2 ( ) ( ) b a I a b f x dx = + ∫ ( ) ( ) 2 b a a b I fx dx + = ∫ ( ) 2 b a a b I fa b x dx + = +− ∫ 37. The value of 2 2 0 0 sec sin x x t dt lim x x → ∫ is (1) 3 (2) 2 (3) 1 (4) 0 Sol. (3) 2 2 2 2 0 0 0 sec sec .2 lim limsin cos ( sin ) x x x d t dt x x dx d x x x x x dx → →= + ∫ (by L hospital’s rule) 2 2 0 2sec 2 1 lim 1 sin 1 1 cos x x x x x → × = = + + 38. The value of the integral 10 (1 )n I x x dx = − ∫ is (1) 1 1 n + (2) 1 2 n + (3) 1 1 1 2 n n − + + (4) 1 1 1 2 n n + + + Sol. (3) 1 0 (1 )n I x x dx = − ∫ 1 1 0 0 (1 ) (1 1)(1 ) n n I x x dx x xdx − = − − = − − − ∫ ∫ 1 1 1 0 0 (1 ) (1 ) n n x dx x dx + = − − − ∫ ∫ 1 1 2 1 0 0 (1 ) (1 ) 1 1 ( 2) ( 1) 2 1 n n x x n n n n + + − − = − = − − + − + + + 1 1 1 2 I n n = − + + 39. 4 3 4 3 3 3 5 5 1 2 3 ..... 1 2 3 ......... lim lim n n n n n n →∞ →∞ + + + + + + + + − INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com (1) 1 30 (2) Zero (3) 14 (4) 15 Sol. (4) 4 4 4 4 3 3 4 4 4 1 1 2 3 1 1 2 lim ..... lim ..... n n n n n n n n n n n n n →∞ →∞ + + + − + + 1 5 1 4 0 0 1 ( ) 0 5 5 x x dx− = = ∫ 40. Let sin ( ) , 0. x d e F x x dx x = > If 3 4 sin 1 3 ( )– (1), x e dx Fk F x = ∫ then one of the possible values of k, is (1) 15 (2) 16 (3) 63 (4) 64 Sol. (4) sin ( ) x d e F x dx x = 3 3 2 4 4 sin sin 3 1 1 3 3 x x x e dx e dx x x = ∫ ∫ Put x3 = t, 3x2 dx = dt When x = 1, t=1 & x = 4, t = 64 sin 64 64 1 1 ( ) ( ) t e F t dt F t dt t = = ∫ ∫ = F (64) – F (1) K = 64 41. The area of the region bounded by the curves y=| x – 1| and y =3 –|x| is (1) 2 sq. units (2) 3 sq. units (3) 4 sq. units (4) 6 sq. units Sol. (3) { } { } { } 0 1 2 1 0 1 (3 ) ( 1) (3 ) ( 1) (3 ) ( 1) A x x dx x x dx x x dx − =+ − − + + − − − + + − − − ∫ ∫ ∫ 0 1 2 1 0 1 (2 2 ) 2 (4 2 ) x dx dx x dx − = + + + − ∫ ∫ ∫ [ ] 0 2 1 2 2 0 1 1 2 2 4 x x x x x − = − + + − 0 ( 2 1) (2 0) (8 4) (4 1) = − − + + − + − − − = 1 + 2 + 4 – 3 = 4 sq. units 42. Let f(x) be a function satisfying f’(x) = f(x) with f(0) =1 and f(x) be a function that satisfies f(x) + f(x) = x2. Then the value of the integral 1 0 ( ) ( ) , f x g x dx ∫ is (1) 2 5 – 2 2 e e − (2) 2 3 – 2 2 e e + (3) 2 3 – – 2 2 e e (4) 2 5 2 2 e e+ + Sol. (3) Let f (x) = ex 1 1 2 0 0 ( ) ( ) ( ) x x f x g x dx e x e dx ∴ = − ∫ ∫ 1 1 2 2 0 0 x x x e dx e dx = − ∫ ∫ 1 1 1 2 2 0 0 0 1 2 2 x x x x x e xe e e = − − − (0, 3) (1, 0) (–1, 2) y = x – 1 y = – x + 1 y = 3 + x y = 3 – x (2, 1) INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com [ ] 2 1 2 1 2 2 e e ee = − − − − + 2 3 2 2 e e = − − 43. The degree and order of the differential equation of the family of all parabolas whose axis is x – axis, are respectively (1) 2,1 (2) 1, 2 (3) 3, 2 (4) 2, 3 Sol. (2) 2 4 ( ) y ax h = − 2 1 1 1 2 2 4 2 0 yy a yy a y yy = ⇒ = ⇒ + = Degree =1, order = 2 44. The solution of the differential equation ( )( )1 2 tan – 0 y dy a y x e dx − + + = , is (1) 1 tan ( –2) y x ke− = (2) 1 1 tan 2 tan 2 y x e e y k − − = + (3) 1 tan 1 tan y s e y k − − = + (4) Sol. (2) 1 2 tan (1 ) ( ) 0 y dy y x e dx − + + − = 1 2 tan (1 ) y dx y x e dy − + + = 1 tan 2 2 (1 ) (1 ) y dx x e dy y y− + = + + I.F. 1 2 1 tan 1 dy y y e e− + ∫ = = 1 1 1 tan tan tan ( ) 1 y y y e x e e dy y − − − = + ∫ 1 1 2tan tan ( ) 2 y y e x e C − − = + ∴ 1 1 tan 2tan 2 y y xe e k − − = + 45. If the equation of the locus of a point equidistant form the points 1 1 ( , ) a b and 2 2 ( , ) a b is 1 2 1 2 ( – ) ( – ) 0, a a x b b y c + + = then the value of ‘c’ is (1) 2 2 2 2 2 2 1 1 1( – – ) 2 a b a b + (2) 2 2 2 2 1 2 1 2 – – a a b b + (3) 2 2 2 2 1 2 1 2 1( ) 2 a a b b + + + (4) 2 2 2 2 1 1 2 2 – – a b a a + Sol. (1) 2 2 2 2 1 1 2 2 ( ) ( ) ( ) ( ) h a k b h a k b − + − = − + − 2 2 2 2 1 2 1 2 2 2 1 1 1 ( ) ( ) ( ) 0 2 a a x b b y a b a b − + − + + − − = 2 2 2 2 2 2 1 1 1( ) 2 C a b a b = + − − 46. Locus of centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, – b cos t) and (1,0) where t is a parameter ,is (1) 2 2 2 2 (3 –1) (3 ) – x y a b + = (2) 2 2 2 2 (3 –1) (3 ) x y a b + = + INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com (3) 2 2 2 2 (3 1) (3 ) x y a b + + = + (4) (3x +1)2+ (3y)2 2 2 a b + Sol. (2) cos sin 1 cos sin 3 1 3 t b t x a t b t x + + = ⇒ + = − sin cos sin sin 3 3 a t b t y a t b t y − = ⇒ − = Squaring & adding (3x – 1)2 + (3y)2 = a2 + b2 47. If the pair of straight line 2 2 – 2 – 0 x pxy x = and 2 2 – 2 – 0 x pxy y = be such that each pair, then (1) p = q (2) p = – q (3) pq =1 (4) pq = – 1 Sol. (4) Equation of bisectors of both pair of st. lines 2 2 2 0 px xy py + − = …… (1) 2 2 2 0 qx xy qy + − = …… (2) From (1) & (2) 2 1 2 1 q q p − = = − − ⇒ pq = – 1 48. A square of side a lies above the x – axis and has one vertex at the origin. The side passing through the origin makes an angle (0 ) 4π α α < with the positive direction of x – axis. The equation of its diagonal not passing through the origin is (1) (cos sin ) — (sin cos ) y x a α α α α − − = (2) (cos sin ) (sin cos ) y x a α α α α + + − = (3) (cos sin ) (sin cos ) y x a α α α α + + + = (4) (cos sin ) (sin – cos ) y x a α α α α + + = Sol. (4) co-ordinates of A = (a cos α, a sin α) Equation of OB tan 4 y x π α = + ∵ CA ⊥ r to OB ∴ slope of cot 2 4 CA π = − + Equation of CA sin cot 2 ( cos ) 4 y a x a π α α − =− + − (sin cos ) (cos sin ) y x s a α α α α + + − = 49. If the two circles 2 2 2 ( –1) ( –3) x y r + = and 2 2–8 2 8 0 x y x y + + +=intersect in tow distinct points, then (1) 2< r< 8 (2) r < 2 (3) r = 2 (4) r > 2 Sol. (1) 1 2 1 2 | | r r CC − < for intersection 3 5 8 r r ⇒ − < ⇒ < …… (1) and r1 + r2 > C1C2 B A C O X Y α π/4 INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com r + 3 > 5 ⇒ r = 2 ……. (2) From (1) & (2) 2 < r < 8 50. The lines 2x – 3y =5 and 3x – 4y =7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is (1) 2 2 2 2 62 x y x y + + − = (2) 2 2 2 2 47 x y x y + + − = (3) 2 2 2 2 47 x y x y + − + = (4) 2 2 2 2 62 x y x y + − + = Sol. (3) 2 154 7 r r π = ⇒ = For centre on solving equation 2 3 5 & 3 4 7 x y x y − = − = x = 1, y = 1 centre = (1, – 1) Equation of circle 2 2 2 ( 1) ( 1) 7 x y − + + = 2 2 2 2 47 x y x y + − + = 51. The normal at the point 21 1 ( , 2 ) bt bt on a parabola meets the parabola again in the point 22 2 ( , 2 ), bt b then (1) 2 1 1 2 t t t = − − (2) 2 1 1 2 t t t = − + (3) 2 1 1 2 t t t = − (4) 2 1 1 2 t t t = + Sol. (1) Fundamental theorem (fact) 2 1 1 2 t t t = − − 52. The foci of the ellipse 2 22 1 16x yb + = an the hyperbola 2 2 1 144 81 25 x y − = coincide. Then the value of the b2 is (1) 1 (2) 5 (3) 7 (4) 9 Sol. (3) 2 2 1 144 81 25 x y − = 1 144 81 81 15 5 , , 1 25 25 144 12 4 a b e = = = + = = Foci = (3, 0), focus of ellipse = (3, 0) ⇒ 34 e = 2 9 16 1 7 16 b = − = 53. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C (– 1, 1, 2). Then the angle between the faces OAB and ABC will be (1) 1 19 cos 35 − (2) 1 17 cos 31 − (3) 30o (4) 90o Sol. (1) Vector perpendicular to the face OAB 1 2 1 5 3 2 1 3 i j k OA OB i j k = × = = − − Vector perpendicular to the face ABC INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 1 1 2 5 3 2 1 1 i j k AB AC i j k = × = − = − − − − Angle between the faces = angle between their normals 5 5 9 19 cos 35 35 35 θ + + = = 1 19 cos 35 θ − = 54. The radius of the circle in which the sphere 2 2 2 2 2 4 19 0 x y z x y z + + + − − − = is cut by the plane 2 2 7 0 x y z + + + = is (1) 1 (2) 2 (3) 3 (4) 4 Sol. (3) centre of sphere = (– 1, 1, 2) Radius of sphere 1 1 4 19 5 = + + + = Perpendicular distance from centre to the plane 1 2 4 7 12 4 3 1 4 4 OC d − + + + = = = = + + 2 2 2 2 2 5 4 9 AC AO OC = − = − = AC = 3 55. The line 2 3 4 1 1 x y zk − − − = =− and 1 4 5 2 1 x y z k− − − = = are coplanar if (1) k = 0 or – 1 (2) k = 0 or – 1 (3) k = 0 or – 3 (4) k = 3 or – 3 Sol. (3) 2 2 2 1 2 1 1 1 1 2 2 2 0 x x y y z z l m n l m n − − − = 1 1 1 0 0 1 1 1 0 2 1 0 2 1 2 1 1 k k k k k − − − − = ⇒ + − = + 2 2 3 0 ( 3) 0 k k kk + = ⇒ + = k = 0 or – 3 56. The two lines , x ay b z cy d = + = + and ' ', ' ' x a y b z c y d = + = + will be perpendicular, if and only if (1) ' ' ' 1 0 aa bb cc + + + = (2) ' ' ' 0 aa bb cc + + = (3) ( ')( ') ( ') 0 a a b b c c + + + + = (4) ' ' 1 0 aa cc + + = Sol. (4) 3 1 x b y d a c − − = = ' 3 ' ' 1 ' x b y d a c − − = = For perpendicular ' 1 ' 0 aa cc + + = O C A INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 57. The shortest distance from the plane 12 4 3 327 x y z + + = to the sphere 2 2 2 4 2 6 155 x y z x y z + + + − − = is (1) 26 (2) 4 1113 (3) 13 (4) 39 Sol. (3) ∵ shortest distance = perpendicular distance 2 12 4 1 3 3 327 26 144 9 16 − × + × + × − = = + + ∴ shortest distance [ ] 26 4 1 15 9 26 r = − + + + − ∵ = 26 – 13 = 13 58. Two systems of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and ', ', ' a b c from the origin, then (1) 2 2 2 2 2 2 1 1 1 1 1 1 0 ' ' ' a b c a b c + + + + + = (2) 2 2 2 2 2 2 1 1 1 1 1 1 0 ' ' ' a b c a b c + − + + − = (3) 2 2 2 2 2 2 1 1 1 1 1 1 0 ' ' ' a b c a b c − − + − − = (4) 2 2 2 2 2 2 1 1 1 1 1 1 0 ' ' ' a b c a b c + + − − − = Sol. (4) Equation of planes be 1 x y z a b c + + = and 1 1 1 1 x y z a b c + + = (Perpendicular distance on plane from origin is same) ∴ 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 a b c a b c − − = + + + + ∴ 2 21 1 1 0 a a Σ −Σ = 59. , , a b c are 3 vectors, such that 0, a b c + + = | | 1, | | 2, | | 3, a b c = = = then . . . a b b c c a + + is equal to (1) 0 (2) – 7 (3) 7 (4) 1 Sol. (2) 0 a b c + + = ( ) ( ) . 0 a b c a b c + + + + = 2 2 2 | | | | | | 2( . . . ) 0 a b c ab bc ca + + + + + = 1 4 9 . . . 7 2 a b b c c a − − − + + = =− 60. If , u v and w are three non-coplanar vectors, then ( ).( )( ) u v w u v v w + − − × − equals (1) 0 (2) . u v w × (3) . u w v × (4) 3 . u v w × Sol. (b) ( ).( ) u v w u v u w v v v w + − × − × − × + × = ( ).( ) u v w u v u w v w + − × − × + × INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com = . ( ) .( ) . ( ) .( ) .( ) .( .( ) .( ) . ( ) 0 0 0 0 0 0 u u v u u w v u v v v w w u w wu u v w v u w w u v × × × × × × − + × + − × + − × + − = .( ) . ( ) .( ) u v w v u w w u v × − × − × = [ ][ ][ ] u v w v w u w u v + − = . ( ) u v w× 61. Consider point A, B, C and D with position vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 7 4 7, 6 10 , 3 4 i j k i j k i j k − + − + − − + and ˆ ˆ ˆ 5 5 i j k − + respectively. Then ABCD is a (1) Square (2) Rhombus (3) Rectangle (4) Parallelogram but not a rhombus Sol. () No option satisfied wrong. A = (7, – 4, 7), B = (1, – 6, 10), C (– 1, – 3, 4) and D (5, – 1, 5) 2 2 2 (7 1) ( 4 6) (7 10) 36 4 9 7 AB= − +− + + − = + + = 2 2 2 (1 1) ( 6 3) (10 4) 4 9 36 7 BC= + +− + + − = + + = 2 2 2 ( 1 5) ( 3 1) (4 5) 36 4 1 41 CD= − − + − + + − = + + = 2 2 2 (5 7) ( 1 4) (5 7) 4 9 4 17 DA= − +−+ + − = + + = 62. The vectors ˆ ˆ 3 4 , AB i k = + and ˆ ˆ ˆ 5 2 4 AC i j k = − + are the sides of a triangle ABC. The length of the median through A is (1) 18 (2) 72 (3) 33 (4) 288 Sol. (3) Position Vector of (3 5) (0 2) (4 4) 2 i j k AD + + − + + = = 4i – j + 4k | | 16 16 1 33 AD= + + = 63. A particle acted on by constant forces ˆ ˆ ˆ 4 3 i j k + − and ˆ ˆ ˆ 3i j k + − is displaced from the point ˆ ˆ ˆ 2 3 i j k + + to the point ˆ ˆ ˆ 5 4 i j k + + . The total work done by the force is (1) 20 units (2) 30 units (3) 40 units (4) 50 units Sol. (3) 1 2 7 2 4 F F F i j k + + = + − . . d PV = of . . B PV − of A = 4 2 2 i j k + − . W F d = = 28 + 4 + 8 = 40 unit 64. Let ˆ ˆ, u i j = + ˆ ˆ v i j = − and ˆ ˆ ˆ 2 3 w i j k = + + . If ˆ n is a unit vector such that ˆ . 0 u n= and ˆ . 0 v n= , then ˆ | . | w n is equal to (1) 0 (2) 1 (3) 2 (4) 3 Sol. (4) n ∵ is perpendicular u and v n u v = × A B D C 5i – 2j + 4k 3i + 4j INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 1 1 0 ˆ 1 1 0 2 ˆ ˆ 2 2 2 i j k k n k − − = = = − × ˆ | . | |( 2 3 ).( )| | 3| 3 w n i j k k = + + − = − = 65. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new set (1) Is increased by 2 (2) Is decreased by 2 (3) Is two times the original median (4) Remains the same as that of the original set Sol. (4) n = 9 then median term 9 1 2 th + = = 5th term last four observations are increased by 2 ∵ the median is 5th observation which is remaining unchanged. ∴ There will be no change in median. 66. In an experiment with 15 observations on x, the following results were available 2 2830, 170 x x Σ = Σ = One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is (1) 78.00 (2) 188.66 (3) 177.33 (4) 8.33 Sol. (1) 2 170, 2830 x x Σ = Σ = increase in 10, x Σ = then ' 170 10 180 x Σ = + = Increase in 2 900 400 500 x Σ = − = then 2 ' 2830 500 3330 x Σ = + = Variance 2 2 1 1 ' ' x x n n = Σ − Σ 2 1 1 3330 180 15 15 = × − × = 222 – 144 = 78 67. Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is (1) 45 (2) 35 (3) 15 (4) 25 Sol. (4) 5 2 ( ) n S C = 2 2 1 1 ( ) n E C C = + 2 2 1 1 5 2 ( ) 2 ( ) ( ) 5 C C n E p E n S C + = = = 68. Events A, B, C are mutually exclusive events such that 3 1 1 ( ) , ( ) 3 4 x x P A P B + − = = and 1 2 ( ) . 2 x P C − = The set of possible values of x are in the interval (1) 1 1 , 3 2 (2) 1 2 , 3 3 (3) 1 13 , 3 3 (4) [ ] 0, 1 Sol. (1) 3 1 1 1 2 ( ) , ( ) , ( ) 3 4 2 x x x P A P B P C + − − = = = ∵ These are mutually exclusive INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 3 1 1 0 1, 0 1 3 4 x x + − ≤ ≤ ≤ ≤ and 1 2 0 1 2 x − ≤ ≤ 1 3 2 , 3 1 x x − ≤ ≤ − ≤ ≤ and 1 2 1 x − ≤ ≤ 1 2 1 1 3 1,and 3 3 2 2 x x x − ≤ ≤ ≤− ≤ ≤ − ≤ ≤ Also 1 3 1 1 2 0 1 3 4 2 x x x + − − ≤ + + ≤ 0 13 3 12 x ≤ − ≤ ⇒ 1 3 13 x ≤ ≤ ⇒ 1 13 3 3 x ≤ ≤ max. 1 1 1 2 113 , 3, , min , 1, , 3 2 3 3 2 3 x − − − ≤ ≤ 1 1 11, 3 2 32 x x ≤ ≤ ⇒ ∈ 69. The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then ( 1) P X = is (1) 1 32 (2) 1 16 (3) 18 (4) 14 Sol. (1) 4 1 1 , , 8 2 2 2 np q p n npq= ⇒ = = = = 7 8 1 1 1 ( 1) 2 2 p X C = = 8 5 1 1 1 8.2 2 32 = = = 70. The resultant of force P and Q is . R If Q is double then R is doubled. If the direction of Q is reversed, then R is again doubled. Then 2 2 2 : : P Q R is (1) 3 : 1 : 1 (2) 2 : 3 : 2 (3) 1 : 2 : 3 (4) 2 : 3 : 1 Sol. (2) 2 2 2 2 cos R P Q PQ θ = + + ……. (1) 2 2 2 4 4 4 cos R P Q PQ θ = + + ……. (2) 2 2 2 4 2 cos R P Q PQ θ = + − ……. (3) On (i) + (ii), 2 2 2 5 2 2 R P Q = + ……. (4) On (iii) × 2 + (ii), 2 2 2 12 3 6 R P Q = + ……. (5) 2 2 2 2 2 5 P Q R + − ……. (6) 2 2 2 3 6 12 P Q R + − ……. (7) by cross multiplication 2 2 2 24 30 24 15 12 6 p Q R = = − + − − INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com 2 2 2 6 9 6 P Q R = = 2 2 2 : : 2:3:2 P Q R = 71. Let R1 and R2 respectively be the maximum ranges up and down an inclined plane and R be the maximum range on the horizontal plane. Then R1, R, R2 are in (1) Arithmetic-Geometric Progression (A.G.P.) (2) A.P. (3) G.P. (4) H.P. Sol. (4) Let β be the inclination o the plane to the horizontal and u be the velocity of projection of the projectile 2 1 (1 sin ) u R g β = + and 2 2 (1 sin ) u R g β = − 2 1 2 1 1 2g R R u + = 1 2 1 1 2 R R R + = 2 u R g = ∵ ∴ R1, R, R2 are in H.P. 72. A couple is of moment G and the force forming the couple is P . If P is turned through a right angle, the moment of the couple thus formed is . H If instead, the force P are turned through an angle α, then the moment of couple becomes (1) sin cos G H α α − (2) cos sin H G α α + (3) cos sin G H α α + (4) sin cos H G α α − Sol. (3) a r p = × | | sin a rp θ = | | cos H rp θ = [ sin(90 ) cos ] v θ + = ∵ G = rp sin θ …… (1) H = rp cos θ …… (2) x = rp sin ( θ + α) …… (3) from (1), (2) and (3) cos sin x a H α α = + 73. The particles start simultaneously from the same point and move along two straight lines, one with uniform velocity u and the other from rest with uniform acceleration . f Let α be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time (1) sin u f α (2) cos f u α (3) u sin α (4) cos u f α Sol. (4) After t; velocity = f × t ( ) BA V f t u = +− 2 2 2 2 cos BA V ft u fut α = + − For max. and min. 2 2 ( ) 2 2 cos 0 BA dV ft fu dt α = − = Y P X θ O r A B O u A α fINDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com cos u t f α = 74. Two stones are projected from the top of a cliff h metres high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then tan θ equals (1) 2u gh (2) 2 u g h (3) 2 u h g (4) 2 u gh Sol. (4) 2 ( cos ) h R u u t g θ = = × 1 2 cos h t g θ = …… (1) Now 2 1 ( sin ) 2 h u t gt θ = − + ‘t’ from (1) 2 sin 2 1 2 cos 2 cos u h h h g g g θθ θ = − + 2 2 tan sec h h u h g θ θ = − + 2 2 tan tan h h u h h g θ θ = − + + 2 2 tan tan 0 u hg θ θ − = ∴ 2 tan u hg = 75. A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by (1) 1 1 2s f r + (2) 2 1 1 s f r + (3) 2 ( ) s f r + (4) 1 1 2s f r + Sol. (4) Portion OA, OB corresponds to motion with acceleration ‘f’ and retardation ‘r’ respectively. Area of ∆OAB = S and OB = t. Let OL = t1, LB = t2, and AL = v 1 . 2 S OBAL = 1 . 2t v = 2S v t = h O R X Y u θ O t1 t2 L B t V A v INDIANET @IIT-JEE, Where technology meets education! http://www.123iitjee.com OR http://www.indianetgroup.com Also 1 v f t = , 1 2 v s t f tf = = and 2 2 2 , r v v s r t t r t = = = 1 2 2 2 S S t t t tf tr = + = + 1 1 2S t f r t = + ⇒ 1 1 2 t s f r = +