63487 [Q. Booklet Number]

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1 WBJEE - 0 (Answers & Hints) 687 [Q. Booklet Number] Regd. Office : Aakash Tower, Plot No., Sector-, Dwarka, New Delhi-0075 Ph. : Fa : ANSWERS & HINTS for WBJEE - 0 by & Aakash IIT-JEE MULTIPLE CHOICE QUESTIONS SUB : MATHEMATICS. The eccentricity of the hyperbola 9y = 6 is (B) 5 y 9 a, b a b e a 9. The length of the latus rectum of the ellipse 6 + 5y = 00 is 5/6 unit (B) /5 unit 6/5 unit 5/ unit b 6 Length of latus rectum = a y 00 y ; 5 6 a 5; b 6. The verte of the parabola y + 6 y + = 0 is y 6 y 6 6 Verte, (, ) (B) (, ), 7, - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-75 Ph.: Fa : ()

2 WBJEE - 0 (Answers & Hints). The coordinates of a moving point p are (t +, t + 6). Then its locus will be a circle (B) straight line parabola ellipse t, y t 6, y 6 y t 6 t y 6 y 6 8 y 6 5. The equation 8 + y + y = 0 represents an ellipse (B) a hyperbola a parabola a circle a by hy g fy c 0 represents ellipse if h ab 0 y y 0 h 0, a, b h ab 0 6. If the straight line y = m lies outside of the circle + y 0y + 90 = 0, then the value of m will satisfy m < (B) m < m > m > m 0m 90 m 0m 90 0 D 0 00m 90 m 0 0m 60 m 9 ; m 7. The locus of the centre of a circle which passes through two variable points (a, 0), ( a, 0) is = (B) + y = a + y = a = 0 Ans : (0,h) (,0) (,0) Centre lies on y-ais locus = 0 - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

3 WBJEE - 0 (Answers & Hints) 8. The coordinates of the two points lying on + y = and at a unit distance from the straight line + y = 0 are (, ), (7, ) (B) (, ), ( 7, ) (, ), (7, ) (5, ), (, ) Let ph, h h h 0 5 h 5 h, 7 ; p,,, 7, 9. The intercept on the line y = by the circle + y = 0 is AB. Equation of the circle with AB as diameter is + y = (B) ( ) +y(y ) = 0 + y = ( )( )+(y )+(y )= ,; y 0, 0,0,, as diametric ends 0 y 0y 0 y y 0 0. If the coordinates of one end of a diameter of the circle +y + 8y+5=0, is (,), the coordinates of the other end is ( 6, 7) (B) (6, 7) ( 6, 7) (7, 6) y 9 8y 5 0 Centre circle (, ) (,) (h,k) (,) h h 6 k k 7 h,k 6, 7. If the three points A(,6), B(, ) and C(, y) are collinear then the equation satisfying by and y is 5 + y = 0 (B) 5 + y + 5 = 0 5 y + 5 = 0 5y +5 = 0 - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

4 WBJEE - 0 (Answers & Hints) 6 0 y y y y y 6 0 y 0 0 y 5 t. If sin and θ lies in the second quadrant, then cosθ is equal to t t t (B) t t t t t t θ in nd quad Cosθ < 0 cos t t t t t cos t. The solutions set of inequation cos < sin is [, ] (B) Ans :, [0, ], cos sin cos sin,, cos sin 0. The number of solutions of sin + cos = is (B) infinite No solution Ans : 5 No solution 5. Let a tan a and tan then α + β is a (B) π - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

5 WBJEE - 0 (Answers & Hints) a tan, tan a a a a a a a a a a a a tan a a a a a a a a a a 6. If, then ( tan )( tan ) is equal to (B) 5/ / tan tan tan tan tan 7. If sinθ and cosθ are the roots of the equation a b + c = 0, then a, b and c satisfy the relation a + b + ac = 0 (B) a b + ac = 0 a + c + ab = 0 a b ac = 0 b sin cos a sin.cos c a b a c a b a ac a b ac 0 8. If A and B are two matrices such that A+B and AB are both defined, then A and B can be any matrices (B) A, B are square matrices not necessarily of the same order A, B are square matrices of the same order Number of columns of A = number of rows of B Addition is defined if order of A is equal to order of B A B nm nm is defined if m = n A, B are square matrices of same order 9. If A is a symmetric matri, then the value of is (B) - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (5)

6 WBJEE - 0 (Answers & Hints) 0. If A = A T or i 5i z i 5 i then i 5i 5 i 7 z is purely real (B) z is purely imaginary z z 0 z i 5i z i is purely imaginary z i 5 i 6 i 7 i 5i 5 i 5i i 5 i 5i 5i 5 i 7 = Real. The equation of the locus of the point of intersection of the straight lines sin θ + ( cos θ) y = a sin θ and sin θ ( + cos θ) y + a sin θ = 0 is y ± a (B) = ± ay y = + y = a Ans : y = a sin θ = a cos θ. y a. If sinθ + cosθ = 0 and 0 < θ < π, then θ 0 (B) Ans : sin θ + cos θ = 0 tan θ =. The value of cos 5 o sin 5 o is 0 (B) cos 5 o sin 5 o = cos60 o. The period of the function f() = cos + tan is (B) LCM, - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (6)

7 WBJEE - 0 (Answers & Hints) 5. If y = + 5, then for = and = 0. value of y is.00 (B) dy 6 d dy y.9 d 6. The approimate value of 5 correct to decimal places is.0000 (B) dy y y d 80 y = The value of cos sin d is (B) 0 Ans : cos sin d d 8. For the function f() e cos, Rolle s theorem is applicable when (B) applicable when 0 applicable when 0 applicable when f f 9. The general solution of the differential equation d y dy 8 6y 0 d d is (A + B )e 5 (B) (A + B)e (A + B )e (A + B )e d y d dy 8 6 y 0 d auilary equation m + 8m + 6 = 0 m = y a b e Solution dy 0. If + y =, then y d (B) 0 dy y 0 d - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (7)

8 WBJEE - 0 (Answers & Hints). d 8 tan + c (B) tan c + tan + c tan c dy tan. 6 sin d 0 (B) sin d 5 cos 0 0 dy dy. The degree and order of the differential equation y are respectively d d, (B),,, dy dy y d d 0, 0. f() =, The function f () is 0 increasing when 0 (B) strictly increasing when > 0 Strictly increasing at = 0 not continuous at = 0 and so it is not increasing when > 0 y 5. The function f() = a + b is strictly increasing for all real if a > 0 (B) a < 0 a = 0 a 0 f () = a f () > 0 a > 0 6. cos d cos sin + log sec + tan + C (B) sin log sec tan + c sin log sec + tan + C sin + log sec tan + C - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (8)

9 WBJEE - 0 (Answers & Hints) cos d sin log sec tan cos sin cos d sin cos sin C (B) sin C sin C sin C sin cos d cos d sin C dy 8. The general solution of the differential equation loge y is d e + e y = C (B) e + e y = C e y + e = C e + e y = C dy e.e y y d e y dy. e d e e c A d y 9. If y B, then = d y (B) y y y d y A d B y 0. If one of the cube roots of be ω, then i i ω (B) i 0 Ans : C C C C C + C C C + ωc C C C. boys and girls occupy seats in a row at random. Then the probability that the two girls occupy seats side by side is (B) 6 n(e) = 5. n(s) = 6 5. p 6 = 6 = - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (9)

10 WBJEE - 0 (Answers & Hints). A coin is tossed again and again. If tail appears on first three tosses, then the chance that head appears on fourth toss is 6 (B) 8 p =... =. The coefficient of n in the epansion of e e e 7 is ( ) n n n (B) n n n n n n n n ( ) n Ans : e e e 7 e e Co-efficient of n n n () () n ( ) = n! n! n n ( ) n!. The sum of the series... is... log e + (B) log e log e log e s = =... 5 = log 5. The number (0) 00 is divisible by 0 (B) (0) 00 = 00 C C 00² + 00 C 00³ C = 00 [ + 00 C + 00 C ] = (0 ) 6. If A and B are coefficients of n in the epansions of (+ ) n and (+) n respectively, then A/B is equal to (B) 9 6 A = n C n. B = n C n - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (0)

11 WBJEE - 0 (Answers & Hints) A B C n n n n Cn n = 7. If n > is an integer and 0, then ( + ) n n is divisible by n (B) n n ( + ) n = n C 0 + n C + n C ² + n C +... = + n + ² ( n C + n C +...) ( + ) n n = ² ( n C + n C +...) 8. If n C, n C 5 and n C 6 are in A.P., then n is 7 or (B) 7 or n C, n C 5, n C 6 are in AP. n C 5 = n C + n C 6 5(n 5) (n ) 0 by solving n = or 7 9. The number of diagonals in a polygon is 0. The number of sides of the polygon is 5 (B) n C n = 0 n = C C C5 (B) C + 5 C C 5 = 5 C = 5 5. Let a, b, c be three real numbers such that a + b + c = 0. Then the equation a + b + c = 0 has both the roots comple (B) hat its roots lying within < < 0 has one of roots equal to a b c 0 a b c 0 has its roots lying within < < 6 = ab 5. If the ratio of the roots of the equation p + q + r = 0 is a : b, then (a b) p qr Let roots are aα and bα pr (B) q q pr pq r q a b p - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

12 WBJEE - 0 (Answers & Hints) ab p r ab r p. (a b) p q ab rp (a b) q 5. If α and β are the roots of the equation + + = 0, then the equation whose roots are α 9 and β 7 is = 0 (B) + = 0 + = = 0 Ans : α and β are the roots of + + = 0 α = ω β = ω α 9 = ω β 7 = ω (α 9 + β 7 ) + α 9 β 7 = 0 Thou, (ω + ω ) + ω. ω = = 0 5. For the real parameter t, the locus of the comple number z = ( t²) + i t in the comple plane is an ellipse (B) a parabola a circle a hyperbola Given z = ( t²) + i t Let z = + iy = t y = + t Thus, + y = y = y = ( ) Thus parabola 55. If cos, then for any integer n, n n cos nθ (B) sin nθ i cos nθ i sin nθ cos Let = cos θ + sin θ cos sin Thus n cos n n 56. If ω is a cube root of unity, then the sum of the series S = + ω + ω ² nω n is n (B) n(ω ) n 0 - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

13 WBJEE - 0 (Answers & Hints) s = + ω + ω² n ω n sω = ω + ω (n )ω n + nω n s( ω) = + ω + ω² ω n nω h = 0 n n s = n 57. If log + log y = + log and log ( + y) =, then =,y = 8 (B) = 8, y = =, y = 6 = 9, y = log + log y = + log.y = 8 log ( + y) = + y = 9 we will get = and y = If log 7 = λ, then the value of log 9 (8) is (λ + ) (B) (λ + ) log 9 8 = log 7 7 ( ) (λ + ) log 7 log 7 7 a a 59. The sequence log a, log, log,... is b b a G.P. (B) an A.P. a H.P. both a G.P. and a H.P log a. (log a log b)(log a log b) = T T = log a log b = T T = log a log b 60. If in a triangle ABC, sin A, sin B, sin C are in A.P., then the altitudes are in A.P. (B) the altitudes are in H.P. the angles are in A.P. the angles are in H.P. ap = bp = cp = a = H.P. b c p p p 6. a b b c c a b c c a a b c a a b b c 0 (B) c c + c + c - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

14 WBJEE - 0 (Answers & Hints) 6. The area enclosed between y = and y = is Ans : sq. units (B) units units 6 units = = A d 0 / ( ) 0 o = [ 0] [ 0] 6 6 O (OC) A (, ) 6. Let f() = e, > 0. Then the maimum value of f() is e (B) e 7e 9 f() =.e = f () = e + e ( ) = e [ ] = 0, = Maimum at = f() = e 6. The area bounded by y = and = y is A = 0 sq. unit (B) 6 sq. unit sq. unit 0 = 0 d. )( ). / 0 0 cos sq. unit A () =. / 0 [ 0] 5/ = 65. The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is 8m/sec. The time taken by the particle to move the second metre is sec (B) sec ( ) = 0 m m sec sec - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

15 WBJEE - 0 (Answers & Hints) S ut at.8.t t t S ut at 8.T T T Time = 66. The solution of = dy y y tan is d = c sin(y/) (B) = c sin(y) y = c sin(y/) y = c sin (/y) dy y tan y d Put y, y = θ dy d d d d. tan, d d dy tan cot d d log sinθ = log + logc sinθ =.c., sin y.c = c.sin y 67. Integrating Factor (I.F.) of the defferential equation dy y sin () d is Ans : e (B) log( + ) + d If pd e e = e log( ) log( ) e = ( + ) = - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (5)

16 WBJEE - 0 (Answers & Hints) 68. The differential equation of y = ae b (a & b are parameters) is yy y y = a.e b... (i) y = abe b y = by... (ii) y = by... (iii) (B) yy y yy y yy y Dividing (ii) & (iii) 69. The value of y y y y yy y n r lim is r n n r log (/) (B) e log (/) e log e log e Lt. n r r n n n n =. d log( ) 0 0 = (log log) log The value of sin cos d is 0 0 (B) π/ π/ 50 9 I sin.cos d f () f (a ) I sin ( cos ()) sin.cos = I = I I = (f () f ()log )d is f () + C (B) f() + C (log )f() + C (log ) f() + C = f() I f ()d f () log d - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (6)

17 WBJEE - 0 (Answers & Hints) 7. Let f() = tan. Then f () + f () is = 0, when is equal to 0 (B) + i i f() = tan f () f (). ( ), + =, ( ) = 0 = 7. If y = tan, then y () = / (B) / / / y tan Put = tanθ sec = tan cos tan tan sin sin tan tan tan = sin cos =.tan, y ( ) y (). n... n 7. The value of lim is n (B) n n(n ) n(n ) Lt ( ) ( ) ( )...( ) = n = n(n ) 75. lim 0 sin( sin ) π (B) π π π Ans : - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (7)

18 WBJEE - 0 (Answers & Hints) sin sin sin Lt 0 0 = π 76. If the function (A ) A for f () for is continuous at =, then A = 0 (B) A = A = A = (A ) A A Put A = [] [ ], when f () when If f() is continuous at =, the value of λ will be (B) 0 LHL Lt [ h] [ ( h)] = h 0 h0 Lt ( h) RML = Lt [ h] ( ( h)) h 0 = + ( h) = = λ = 78. The even function of the following is a f () a a a f (). a a (B) f () a a f () log f ( ) ( ) a a a = ( n) a = n (a ) (a ) = f() 79. If f( + y, y) = y, then f(, y) is equal to y (B) ( y ) 8 ( y ) ( + y ) - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (8)

19 WBJEE - 0 (Answers & Hints) y a y b a b y a b a b a b a b f (a,b) The locus of the middle points of all chords of the parabola y = a passing through the verte is a straight line (B) an ellipse a parabola a circle h =, k = y y = a k = ah y = a - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (9)

20 WBJEE - 0 (Answers & Hints). The harmonic mean of two numbers is. Their arithemetic mean A and the geometric mean G satisfy the relation A + G = 7. find the numbers. Ans. (, 6) Sol : Let the number be a, b DESCRIPTIVE TYPE QUESTIONS SUB : MATHEMATICS A.H G H G = A A + G = 7 A + A = 7 7 A 6 G = 8 G = 8 a.b = 8 a + b = 9 a 6 or a b b 6. If the area of a rectangle is 6 sq. unit, find the minimum value possible for its perimeter. Ans. Sol. Let the dimesions be a, b Area = ab Paimeter = (a + b) We have ab = 6 b = 6 a Perimeter as function of a 6 P (a) = a a for maima or minimum 6 Pa 0 a a = ± 8 = P a 0 a 8 P (8) is minimum Minimum P (8) = ( 8 + 8) = - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : (0)

21 WBJEE - 0 (Answers & Hints). Find the image of the point ( 8, ) with respect to the line + 7y + = 0 Ans. ( 6, ) A (-8, ) Sol. P (h - 8/, k + /) ( + 7y + = 0) A (h, k) Image h 8 k 7 0 h 6 + 7k + + = 0 h + 7k + 78 = 0 h 7k 78...(i) nd equation, we can get Slope of AA = 7 k 7 n 8 k 8 = k 7h = 0...(ii) Solving (i) & (ii) Equation (i) 7 + Equation (ii) 8h 9k 56 8h 6 k 6 k h 6 65 k 0 A ( 6, ) is the image of ( 8, ). How many triangles can be formed by joining 6 points lying on a circle? Ans. 0 Sol. Number of triangle 6 C = If r = + y + z, then prove that yz z y tan tan tan r ry rz A. S S tan S - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

22 WBJEE - 0 (Answers & Hints) z y S = r r r = 0 6. Determine the sum of imaginary roots of the equation Ans. ( + ) ( + ) = 6 Sol. Put + = y ( ) ( ) = 6, on solving + + = 0 7. If cos A + cos B + cos C = 0, prove that cos A + cos B + cos C = cos A cos B cos C A. L.H.S = cos A cos A cos A = cos A. cos B. cos C cos A 8. Let IR be the set of real numbers and f : IR IR be such that for all, y IR, f() f(y) y. Prove that f is a constant function. A. y f y f y y = f () 0 f () = 0 f () = constant 9. Find the general solution of ( + logy) dy + y d = 0 Ans. y + y lny y = 0 Sol. dy + y d + log y dy = 0 d y log y dy 0 y + y lny y = 0 0. Prove that I / 0 sec d cosec sec A. I sec cos ec d cos ec sec cos ec sec 0 0 I d I 0 - Regd. Office: Aakash Tower, Plot No.-, Sector-, Dwarka, New Delhi-0075 Ph.: Fa : ()

WBJEEM Answer Keys by Aakash Institute, Kolkata Centre MATHEMATICS

WBJEEM Answer Keys by Aakash Institute, Kolkata Centre MATHEMATICS WBJEEM - 05 Answer Keys by, Kolkata Centre MATHEMATICS Q.No. μ β γ δ 0 B A A D 0 B A C A 0 B C A * 04 C B B C 05 D D B A 06 A A B C 07 A * C A 08 D C D A 09 C C A * 0 C B D D B C A A D A A B A C A B 4