MATHEMATICS. 1. The line parallel to x- axis and passing through the intersection of the lines ax + 2by + 3b =0 And bx -2ay -3a = 0 where (a, b) 0.

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1 1. The line parallel to x- axis and passing through the intersection of the lines ax + 2by + 3b =0 And bx -2ay -3a = 0 where (a, b) 0. a) Below x- axis at a distance 2/3 from it b) Below x- axis at a distance 3/2 from it c) Above x- axis at a distance 2/3 from it d) Above x- axis at a distance 3/2 from it

2 Sol : The required equation is of the form Ax+2by+3b + k (bx-2ay-3a) = 0 If line is parallel to x axis, then a + bk = 0 k = -a/b (i.e coefficient of x = 0) ( 2b + 2a2/b ) y +(3b + 3a2/b) = 0 2y + 3 = 0 which is a line parallel to x axis and lies below x axis at a distance 3/2 from origin

3 2. Let P(-1,0), Q(0, 0) and R= (3, 3 3)be three points. Then the equation of the bisector is. a) 3/2 X + Y=0 b) X + 3Y=0 c) 3X+ Y=0 d) X + 3/2 Y=0 MATHEMATICS

4 Sol : PQR = RQM = =120 0 (tan _RMQ= RM/QM=3 3/3 = 3) ( RQM =60 0 ) Therefore inclination of the bisector = = The equation of bisector is Y-0= - 3 (X-0) 3X + Y = 0 The correct option is C.

5 3. The line joining A(1-3, 1) and B (1, 2) is rotated about A through an angle 15 0 in anti-clock-wise direction. The equation of AB in new position is a) x+y= 3 b) x-y= 3 c) x-y+ 3=0 d) none of these MATHEMATICS

6 Sol : Slope of AB = (2-1) / (1-1+ 3) = 1/ 3 AB makes 30 0 with positive direction of X axis In new position AB would be = 45 0 Its slope = 1 its equation is y 1 = x x-y = 0 which is (iii) So the answer is C.

7 4. Let A 0 A 1 A 2 A 3 A 4 A 5 be a regular hexagon inscribed in a circle of Unit radius. Then the Product of the lengths of the line segments A 0 A 1, A 0 A 2 and A 0 A 4 is a) ¾ b) 3 3 c) 3 d) 3 3/2 MATHEMATICS

8 Sol : Let o be centre of the circle Since _A 0 OA 1 = /6= 60 0 _A 0 OA 1 is equilateral triangle => A 0 A 1 = 1 (radius of circle = 1) Also A 0 A 2 = A 0,A 4 =2OD =2 (OA 0 ), Sin60 0 =>2(1) 3/2= 3 Now (A 0,A 1 ) (A 0,A 2 ) (A O A 4 ) = (1)( 3) ( 3) = 3 Which is (iii)

9 5. If One of the line of the pairs ax 2 +2hxy+by 2 =0 bisects the angle between the positive directions of the axes, then a, b, h satisfy the relation a) a+b = 2 h b) a+b = -2h c) a-b = 2 h d) (a-b) 2 = 4h 2 MATHEMATICS

10 Sol : One of the lines of ax 2 +2hxy+by 2 =0 Bisects the angle between positive directions of the axes, Then one of the line must be x-y=0 (ax-by) (x-y) = ax 2 +2hxy+by 2 MATHEMATICS => ax 2 - (a+b)xy+by 2 = ax 2 +2hxy+by 2 => -(a+b)= 2h or a+b=-2h Which is (b)

11 6. The pair of lines represented by 3ax 2 + 5xy + (a 2-2)y 2 =0 are perpendicular to each other for a) two values of a b) a c) for one value of a d) for no value of a

12 Sol: The lines will be perpendicular if, coeff of x 2 + coeff of y 2 =0 3a+(a 2-2) = 0 => a 2 +3a-2=0 Clearly this gives us two values of a. answer is (a)

13 7. If the pair of lines ax 2 +2(a+b) xy + by 2 = 0 lies along the diameters of a circle and divide the circle into four sectors such that the area of the one of the sector is thrice the area of the another sector then. a) 3a ab +3b 2 =0 b) 3a 2 +2ab+3b 2 =0 c) 3a 2-10ab+3b 2 =0 d) 3a 2-2ab+3b 2 =0 MATHEMATICS

14 Sol : Clearly the angle between pair of lines is /4 Tan 1 = (a+b) 2 = 4[(a+b) 2 -ab] 3a 2 +2ab+3b 2 =0 Therefore the correct answer is (B)

15 8. The area enclosed by pair of lines xy=0, the lines x-4=0 and y=-5 is a) 10 sq units b) 20 sq units c) 0 sq units d) 5/4 sq units MATHEMATICS

16 Sol : xy=0 => x=0, y=0 the required area is ABCO where A= (0, -5), B= (4, -5) C= (4, 0) Area OABC = OA. OC = 4.5 =20 sq units Which is (b)

17 9. If Sinѳ 1 + Sin ѳ 2 + Sin ѳ 3 =3 then CoSѳ 1 + CoSѳ 2 +CoSѳ 3 is equal to a) 3 b) c) 2 1 d) 0

18 Sol : Sin ѳ 1 + Sin ѳ 2 + Sinѳ 3 =3 (max. value of sinѳ =1) Sin ѳ 1 = Sin ѳ 2 = Sin ѳ 3 =1 ѳ 1 =ѳ 2 = ѳ 3 = /2 Cosѳ 1 + Cosѳ 2 +Cosѳ 3 =0 Correct option is (d) MATHEMATICS

19 10. If tanѳ = -4/3 then SinѳIs a) -4/3 but not 4/5 b) -4/5 or 4/5 c) 4/5 but not -4/5 d) none of these MATHEMATICS Sol : tanѳ = - 4/3 <0 => ѳ II or IV quadrant Sinѳ>0 or Sinѳ< 0 Now tanѳ = -4/3 => Sinѳ= -+4/5 Correct answer is (b)

20 11. The maximum value of sin (x+ /6) Cos (x+ /6 ) in the interval (0, /2 ) is attained at a) /12 b) /6c) /3 d) /2 Sol : Y= sin(x+ /6) Cos (x+ /6 )= 1/2. sin (2x+ /3) Max value of y=1/2 This is attained at 2x+ /3 = /2 => x= /12 Correct answer is (a)

21 12. If K = sin sin 5 sin 7 then the numerical value of K is a) 1/2 b) 1/3 c) 1/8 d) 1/6 Sol: Consider K= ½[sin (2sin5 sin7 )] = 1/2[sin π/18 (cos π/9 - cos 2 π/3)] = 1/2[sin (cos +1/2 )] = 1/4 [2cos sin +2(1/2)sin ] =1/4 [sin -sin + sin ] = 1/8 So the answer is C.

22 13. If A +B + C =,then sin 2 A + sin 2 B + sin 2 C = a) 2+2 cosa cos B cos C b) 2+2 sin A sin B sin C c) 2-2 cos A cos B cos C d) 2 + cos A cos B cos C

23 Sol : Sin 2 A+Sin 2 B+Sin 2 C =1/2 [3-(cos2 A+cos2B+cos2C)] = 1/2[3-(-1-4cosAcosBcosC)] = 2+2 cosa cosb cosc The correct answer is A. Using 1-Cos 2A =2Sin 2 A

24 14. Cos α.sin (β γ) + Cos β. sin (γ α) + Cosγ. sin (α β) = MATHEMATICS a)0 b) ½ c) 1 d) 4 cos α cos β cos Sol : G.E.= Σcosα.sin (β γ) = Σcosα.(Sinβ. cosγ) - cos β.sinγ =0 Which is (a)

25 15. In a triangle ABC, Σ (b+c) tan A/2. Tan((B-C)/2 )= a) a b) b c) c d) 0 Sol: By Napier s tgt rules, we have Tan ((B-C)/2 ) =(b-c)/(b+c).cot(a/2) (b+c).tan (A/2).tan ((B-C)/2 )=b-c Now Σ(b+c).tan(A/2). tan ((B-C)/2 )=Σ (b-c ) = (b-c) + (c-a) + (a-b)=0 Correct answer is (d)

26 16. In a ABC, if acos 2 (C/2) + cos 2 (A/2) = 3b/2, then the sides a,b, c are a) A, P b) G, P c) H, P d) none of these

27 Sol : G.E. = acos 2 C/2+ccos 2 =3b/2 (a+c)+ (acosc+c cosa)=3b a+b+c =3b => a+c =2b a,b and c are in A.P. Therefore the correct answer is A.

28 17. The angle of a triangle are in the ratio 4:1:1, then the ratio of the largest side to its perimeter is a) 3: 2+ 3 b) 1:6 c)1:2+ 3 d)2: 3

29 Sol : by data A:B:C=4:1:1 A= 120 0, B =30 0, C =30 0 If AB=x, then AC=x And BC= 2xcos30 0 Required ratio = = = Correct answer is (a)

30 18. If in a triangle ABC, S=2b then cot A/2 cot C/2 = a)2 b) 1 c)3 d) 2 Sol: By data, S=2b cot A/2 cot C/2 = = s/(s-b) = = 2 Therefore the correct answer is A.

31 19. If two towers of heights h 1 and h 2 subtend angles 60 0 and 30 0 respectively at the midpoint of line joining their feet then h 1 : h 2 = a) 1:2 b) 1:3 c) 2:1 d) 3:1

32 Sol : h 1 = tan 60 0 = x 3 and h 2 =xtan30 0 = x(1/ 3) = x/ 3 Now h 1 :h 2 = 3x :x(1/ 3) h1 : h2=3:1 which is (d) MATHEMATICS

33 20. An Areoplane flying with Uniform speed horizontally one kilometer above the ground is observed at an elevation of After 10 seconds if the elevation is observed 30 0, then the speed of the plane (in km/hr) is a) 240/ 3 b) c) d) 120/ 3 MATHEMATICS

34 Sol : tan 60 0 =1/AB=> AB = 1/ 3 tan 30 0 = 1/AD=> AD = 3 Now BD=AD-AB 3 1/ 3 = 2/ 3 km MATHEMATICS 10 seconds it has travelled, 2/ 3 km In one hour, it will travel 2/ 3 km X 3600/10 = km/hour Which is (c)

35 21. From a top of a building 60 meters high the angle of elevation of top of tower is found to be equal to the angle of depression of the foot of the tower, the height of tower in meters is a) 100 b) 90 c) 120 d) 105 MATHEMATICS

36 Sol : Clearly CE=ED CD=2CE=2AB CD=2(60)=120 metres MATHEMATICS Correct answer is (c)

37 22. At 2.15 O clock, the hour hand and minute hand of a clock form an angle a) 5 0 b)22 1/2 0 c)28 0 d) 30 0 Sol : In two hours, hour hand rotates = 60 0 In 15 minutes, its describes 7 0 In one Minute, minute hand traces -6 0 In 15 minutes, minute hand traces Required angle ½ 0 = 22 1/2 0 Therefore the correct answer is B.

38 23. If Sin (ѳ + α)= cos (ѳ + α) then the value of is MATHEMATICS a) - tan ѳ c) Zero b) tan ѳ d) Cosѳ Sol : Given Sin (ѳ + α)= Cos (ѳ + α) tan (ѳ + α)=1 => (ѳ + α) = /4 => α = /4 -ѳ Now = = = tan ѳ Correct answer is (b)

39 24. If a non zero numbers a, b, c are in H P then the point of concurrency of the straight lines x/a + y/b + 1/c = 0 is a) (1, 2) b) (1, -2) c) (-1, 2) d) (-1, -2)

40 Sol: By data 1/a, 1/b, 1/c are in A.P => 2/b =1/a + 1/c Compare with y/b = x/a + 1/c we get x = 1, y = -2 => the point of concurrency = (1, -2) The correct answer is (b)

41 25. The Orthocentre of the triangle with vertices (0, 0), (3, 4), and (4, 0) is a) (-3, 3/4 ) b) (3, 3/4 ) c) (3, - 3/4) d) (3, - 4/3 )

42 Sol : Orthocenter of le formed by origin, (x 1, y 1 ) and (x 2, y 2 ) is [(y1-y2)(x1x2+y1y2)/x1y2-x2y1,(x1-x2)(x1x2+y1y2)/x1y2-x2y1] Now, (x1x2+y1y2)/x1y2-x2y1 = 12/-16 =-3/4, X1-X2=-1 and Y1-Y2=4 So, Ortho centre = (4 (-3/4), (-1) (-3/4)]= (-3, 3/4) Correct answer is (a)

43 26. The equation of a straight lines passing through the point (-5, 4) and which cuts off an intercept of + 2 Units between the lines x+y+1=0 and x+y-1=0 is a) x-2y+13=0 b) 2x-y+14=0 c) x-y+9=0 d) x-y+10=0 MATHEMATICS

44 Sol : MATHEMATICS Clearly the given two lines are parallel lines and distance between them is 2 Also the point (-5, 4) lies on x+y+1=0 Thus required line is the line through (-5, 4) and perpendicular to the line x+y+1=0 This is given by y-4 = 1(x+5) x-y+9=0 which is (c).

45 27. The three lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 are concurrent only when a) a+b+c=1 b) a 2 +b 2 +c 2 =ab+bc+ca c) a 3 +b 3 +c 3 =3abc d) a 3 +b 3 +c 3 = abc MATHEMATICS

46 Sol : The given lines are concurrent if =0 a (bc-a 2 ) b (b 2 -ac)+c(ab-c 2 )=0 abc-a 3 -b 3 + abc+abc-c 3 =0 a 3 +b 3 +c 3 = 3 abc Which is (c)

47 28. The incentre of the triangle with the vertices (1, 3), (0, 0) and (2, 0) is a) (1, 3/2) b) (2/3, 1/ 3) c) (2/3, 3/2) d) (1, 1/ 3)

48 Sol : Let A(1, 3), O= (0,0) and B (2, 0) AO= 1+3 =2 ), OB=2, AB= 1+3 =2 ) The triangle is equilateral Therefore, Incentre coincide with centroid I= ((1+0+2)/3, ( 3+0+0)/3 ) = (1, 1/ 3) Therefore, Ans is (d)

49 29. The equation of the pair of lines through (1, -1) and perpendicular to the pair of lines x 2 -xy-2y 2 =0 is a) 2x 2 -xy + y 2 +5x +y +2 =0 b) 2x 2 -xy - y 2-5x y + 2 =0 c) x 2 xy +2y 2-5x y +2 =0 d) 2x 2 xy -y 2 +5x +y -2 =0 MATHEMATICS

50 Sol : Equation of pair of lines through origin and perpendicular to the pair x 2 -xy-2y 2 =0 is -2x 2 +xy+y 2 =0 (1) The pair of lines through (1,-1) and parallel to the pair -2x 2 +xy+y 2 =0 (hence Perpendicular to the given pair) is given by, -2(x-1) 2 + (x-1)(y+1)+(y+1) 2 =0 2x 2 -xy-y 2-5x-y+2=0 Correct answer is (b) MATHEMATICS

51 30. The centroid of the triangle formed by the pair of lines 12x 2-20xy+7y 2 +0 and line 2x-3y+4=0 is MATHEMATICS a) (-7/3, 7/3) b) (-8/3, 8/3) c) (8/3, 8/3) d) (4/3, 4/3)

52 Sol : The centroid of the triangle formed by the pair of lines represented by the equation ax 2 +2hxy+by 2 =0 and the line lx+my=n is given by, = = MATHEMATICS here a =12, b=7, h=-10, l=2, m=-3, n=-4 we have X/-16=Y/-16=-1/6=> (x,y)=(8/3,8/3) Correct answer is (c) (OR)

54 31.In a triangle ABC, Cos((B+2C+3A)/2) + cos((a-b)/2 ) a) -1 b) 0 c) 1 d) 2

55 Sol: Cos( ) + cos( ) (Since, B+C =Π-A) => Cos( ) + Cos( ) Cos ( ) + Cos( ) - Cos( ) + Cos ( ) = 0 Which is (b)

56 32. If angle of a triangle are 30 0 and 45 0 and the included side is ( 3+ 1) the area of triangle is, a)2( 3 + 1) b) ( 3 + 1) c) 1/2 ( 3 +1) d) None of these MATHEMATICS

57 Sol: Let C =30 0 B =45 0 a= 3+ 1 Also A =105 0 Now b = = =2 Now area =1/2 ab sin C =1/2 ( 3+1)2 sin 30 0 = ½(2)( 3+1)(1/2)= ( 3+1)/2 Therefore the correct option is C.

58 33. If for a triangle ABC, =0 then the value of sin 2 A + sin 2 B + sin 2 C = a) 9/4 b) 4/49 c) (3 3)/2 d) 1

59 Sol : =0 applying R 1 2 = R 2 -R 1 and R 1 3 = R 3 -R 1 = 0 [(c-a)(c-b)+(a-b) 2 ]=0 a 2 +b 2 +c 2 = ab+bc+ca=> a=b=c ABC is an equilateral triangle A= B = C = 60 0 = Π/3 Sin 2 A+ Sin 2 B+ Sin 2 C=3/4+ ¾+ ¾ =9/4 Which is (a)

60 34. In a triangle ABC, + = 0 then B=? a)π/2 b) Π/4 c) 2Π/3 d) Π/3

61 Sol : G.E. = ab 2 -a 3 +cb 2 -c 3 =0 ab 2 +cb 2 -(a 3 +c 3 )=0 b 2 (a+c)(a 3 +c 3 )=0 b 2 (a+c)(a+c)(a 2 -ac+c 2 )=0 (a+c) (b 2 -a 2 +ac-c 2 )=0 (b 2 -a 2 +ac-c 2 )=0 =1/2 => B=60 0 or Π/3 Which is (d)

62 35. If 1 + sin x + sin 2 x + to = , 0 < x < Π then x = a) Π/6 b) Π/4 c) 3Π/4 d)π/3 or 2Π/3

63 Sol : Sum to of GP = a/(1-r) if r <1 Therefore 1/(1-sinx) = sinx= (4-2 3)/(16-12) = 1-3/2 sinx=1-1+ 3/2 = 3/2 Therefore x is Π/3 or 2Π/3 Correct answer is (d)

64 36. If for all real values of x, cosѳ = x+1/x then, a) ѳ is an actual angle b) ѳ is right angle c) ѳ is an obtuse angle d) no value of ѳ is possible

65 Sol : Cosѳ = x+1/x => x 2 (Cosѳ)x -1 = 0 If x is real then b 2 4ac 0 cos 2 ѳ 4 0 => cos 2 ѳ 4 Which is not possible as cos 2 ѳ 1 No value of ѳ is possible. Therefore the correct option is D.

66 37. Cot = a) b) c) d)

67 Sol: MATHEMATICS Cot = = ) ( ) = = = Which is C.

68 38. From the top of a light house, the angle of depression of two stations on opposite sides of the light house at a distance a apart are α and β. The height of the light house is, a) a(tanα +tanβ) b) a/(cotα+ cotβ) c) a/(tanα +tanβ) d) a (cotα+ cotβ) MATHEMATICS

69 Tanβ = h/x => h= tanβ (1) tan α = h/(a-x) => h=(a-x)tanα (2) => x.tanβ= a.tanα x.tanα x= atanα /(tanα+ tanβ) h=(a tanα.tanβ) /(tanα+tanβ) h=a/(cotα+ cotβ) MATHEMATICS which is (b)

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0 Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y