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Teko Classes,

Teko Classes,Bhopal Ph.: (0755) 32 00 000

IIT JEE/AIEEE MATHS by SHUAAG SIR Bhopal, Ph. (0755)32 00 000.www.tekoclasses.com Question on 2 D Collection # 1 Page: 1 of 39

THIS FILE CONTAINS (COLLECTION # 1)

Very Important Guessing Questions For IIT JEE 2010 With Detail Solution

Junior Students Can Keep It Safe For Future IIT JEEs

�� Two Dimensional Geometry (2D)

�� The Point

�� Straight Lines

�� Circles

�� Parabola

�� Ellipse

�� Hyperbola

Index

For Collection # 1 Question (Page 2 to 39)

�� Single Correct Answer Type Question

�� Comprehension Type Quetions

�� Assertion Reason Type Question

�� More Correct Answers Type Questions

�� Subjective (Up to 4 Digits)

�� Detiail Solution By Genuine Method (But In) Classroom I Will Give

Short Tricks )

For Collection # 2 Question (Page 41 to 56)

�� Same As Above

Here Solutions Given Are Long (Genuine)

Method But In “15 Days Class ” I Will

Give SHORT TRICKS, Which Are Must

For IIT / AIEEE Just To Save Your Time

Teko Classes,



Single Correct Type

Que. 1. The shortest distance from the line 3x 4y 25+ = to the circle 2 2x y 6x 8y+ = − is equal to

(a) 7/5 (b) 9/5 (c) 11/5 (d) 32/5 (code-V2T1PAQ7)

Que. 2. ( )The graph of y x against (y + x) is as shown

Which one of the following shows the graph of y against x ?

−

(y+x)

(y x)−

(a) x

y

(b) x

y

O

(c) x

y

O

(d) x

y

O (code-V2T3PAQ3)

Que. 3. If H represent the harmonic mean between the abscissae, and K that between the ordinates of the

points, in which a circle 2 2 2x y c+ = is cut by a chord x my ,+ = δl where l and m are the direction

consines of the unit vector in the xy plane, then H mK+l has value equal to

(a) 2c

2δ −δ

(b) 2c

2δ −

δ(c)

22cδ −

δ(d)

2c2

2δ −

δ (code-V2T12PAQ6)

Que. 4. Area of the triangle formed by the line x + y = 3 and the angle bisectors of the line pair

2 2x y 4y 4 0− + − = is (code-V2T13PAQ15)

(a) 1/2 (b) 1 (c) 3/2 (d)2

Que. 5. The shaded area enclosed by 2f (x) 12 ax x= + − coordinate axes

and the ordinate at x 3= is 45 square units. If m and n are the x-axis

intercepts of the graph of y f (x)= then the value of (m n a)+ +

equals (a) 0 (b) 4 (c) 6 (d) 8 (code-V2T13PAQ18)

Que. 6. The vertices of a triangle ABC are ( ) ( ) ( )2 2 2A p , p , B q ,q ,C r , r .− − The area of the triangle ABC is

(a) ( )( )( )1

p q q r r p2

+ + + (b) ( )( )( )1

p q q r r p2

− + + (code-V2T14PAQ2)

(c) ( )( )( )1

p q q r r p2

+ − − (d) ( )( )( )1

p q q r p r2

+ + −

Que. 7. The least integral value of k for which ( ) ( ) ( )2 1 1k 2 x 8x k 4 sin sin12 cos cos12

− −− + + + > + for all

x R,∈ is (code-V2T14PAQ3)

(a) – 7 (b) – 5 (c) – 3 (d) 5

Que. 8. The equation 3x t 9= + and 33t

y 64

= + represents a straight line where t is a parameter. The y-

intercept of the line is (code-V2T14PAQ9)

(a) 3/ 4− (b) 9 (c) 6 (d) 1

Que. 9. If 2

4x

1

λ=

+ λ and

2

2

2 2y

1

− λ=

+ λ where λ is real parameter then 2 2x xy y− + lies between [a,b] then

(a+b) is (code-V2T14PAQ10)

(a) 8 (b) 10 (c) 13 (d) 25

Teko Classes,



Que. 10. Let C1 and C

2 are circles difined by 2 2x y 20x 64 0+ − + = and 2 2x y 30x 144 0.+ + + = The length of

the shortest line segment PQ this is tangent to C1 at P and to C

2 at Q is (code-V2T14PAQ14)

(a) 15 (b) 18 (c) 20 (d) 24

Que. 11. A variable line moves in such way that the product of the perpendiculars form (a,0) and (0,0) is

equal to k2. The locus of the feet of the perpendicular from (0,0) upon the variable line is a circle, the

square of whose radius is ( )Given : | a | 2 | k |< (code-V2T14PAQ18)

(a) 2

2ak

4+ (b)

2 2a k

4

+(c)

22 k

a4

+ (d) 2 2a k

2

+

Que. 12. The tangent and normal at the extremities of the latus rectum of a parabola 2y 4x= form a quadri-

lateral whose area is (code-V2T14PAQ23)

(a) 4 2 (b) 8 (c) 8 2 (d) 16

Que. 13. If the lines

( )

( )

( )

x sin y cos 0

x cos y sin 0

x sin y cos 0

λ + α + α =

+ α + α =

− α + α =

pass thorugh the same point where Rα∈ then λ lies in the

interval (code-V2T17PAQ2)

(a) [ ]1,1− (b) 2, 2 − (c) [ ]2,2− (d) [ ],−∞ ∞

Que. 14. The range of values of m for which the line y mx= and the curve 2

xy

x 1=

+ enclose a region, is

(a) (–1,1) (b) (0,1) (c) [0,1] (d) (1, ∞ ) (code-V2T17PAQ3)

Que. 15. A(1,0) and B(0,1) and two fixed points on the circle 2 2x y 1.+ = C is a varible point on this circle.

As C moves, the locus of the orthocentre of the triangle ABC is (code-V2T17PAQ5)

(a) 2 2x y 2x 2y 1 0+ − − + = (b) 2 2x y x y 0+ − − =

(c) 2 2x y 4+ = (d) 2 2x y 2x 2y 1 0+ + − + =

Que. 16. Mr. Shuag Kariya lives at origin on the cartesian plain and has his office at (4,5). His friend Mr.

Vivek Jain lives at (2,3) on the same plane. Mr. Shuag Kariya can go to his office travelling one block

at a time either in the +y or +x direction. If all possible paths are equally likely then the probability

that Mr. Shuag Kariya passed his friends house is

(a) 1/2 (b) 10/21 (c) 1/4 (d) 11/21 (code-V2T20PAQ5)

Que. 17.If the left hand side of the equation 2 2x y x 3y sec 0− + − + θ = can be factorised into two linear

factors then the value θ is (code-V1T4PAQ3)

(a) 3

π(b)

7

6

π(c)

4

3

π(d)

5

6

π

Que. 18. Let x, y, z, t be real numbers 2 2 2 2x y 9; z t 4 and xt yz 6+ = + = − = then the greatest value of

P xz,= is (code-V1T5PAQ5)

(a) 2 (b) 3 (c) 4 (d) 6

Que. 19. If the two vertices of a trianlge are (7,2) and (1,6) and its centroid is (4,6) then the corrdinate

of the third vertex are (a,b). The value of (a + b), is (code-V1T7PAQ2)

(a) 13 (b) 14 (c) 15 (d) 16

Teko Classes,



Que. 20. Number of values of ‘a’ for which the lines 2x y 1 0

ax 3y 3 0

3x 2y 2 0

+ − =

+ − =

+ − =

are concurrent (code-V1T7PAQ4)

(a) 0 (b) 1 (c) 2 (d) infinite

Que. 21. Number of straight lines parallel to the line 3x 6y 7 0+ + = and have intercept of length 20

between the coordinate axes (code-V1T7PAQ11)

(a) 1 (b) 2 (c) 4 (d) Infinite

Que. 22. A circle has radius of ( )2

10log a and a circumference of ( )4

10log b . The value of a

log b is equal

to (code-V1T10PAQ3)

(a) 1

4π(b)

1

π(c) π (d) 2π

Que. 23. The points (x,y) lies on the line 2x 3y 6.+ = The smallest value of the quantity 2 2x y+ , is

(a) 6 13

13(b) 6 (c)

113

2(d) 13 (code-V1T12PAQ6)

Que. 24. If ( 2,7)− is the highest point on the graph of 2y 2x 4ax k,= − − + then k equals

(a) 31 (b) 11 (c) –1 (d) – 1/3 (code-V1T13PAQ2)

Que. 25. If the point ( )P u, v= is on the graph of 2y ax bx c,a 0,= + + ≠ which of the following is

also on the graph ? (code-V1T13PAQ9)

(a) b

u, va

−

(b)

bu, v

a

− −

(c)b

u, va

− +

(d) b

u, va

+

Que. 26. Locus of all point P(x,y) satisfying 2 2x y 3xy 1+ + = consists of union of (code-V2T13PAQ19)

(a) a line and an isolated point (b) a line pair and an isolated point

(c) a line and a circle (d) a circle and an isolated point.

Que. 27. The length of a line segament AB is 10 units. If the coordinates of one extremity are (2, –3)

and the abscissa of the other extremity is 10 then the sum of all possible values of the ordinate of the

other extremity is (code-V1T15PAQ2)

(a) 3 (b) – 4 (c) 12 (d) – 6

Que. 28. The value of k for which the points ( ) ( ) ( )A k 1,2 k ;B 1 k, k and C 2 k,3 k+ − − − + − are collinear is

(a) 0 (b) 1

2(c) 1 (d) 2 (code-V1T15PAQ3)

Que. 29. A point P(x,y) moves so that the sum of the distances from P to the coordinate axes is equal to

the distance from P to the point A(1,1). The equation of the locus of P in the first quadrant is

(a) (x 1)(y 1) 1+ + = (b) (x 1)(y 1) 2+ + = (c) (x 1)(y 1) 1− − = (d) (x 1)(y 1) 2− − = (code-V1T15PAQ7)

Que. 30. If x, y R∈ satisty the equation 2 2x y 4x 2y 5 0,+ − − + = then the value of the expression

( )2

x y 4 xy

x xy

− +

+ is (code-V1T15PAQ9)

(a) 2 1+ (b) 2 1

2

+(c)

2 1

2

−(d)

2 1

2

+

Teko Classes,



Que. 31. The ordinate of a point P on the line 6x y 9,+ = which is closest to the point (–3, 1) can be

expresed in the form a/b. Where a,b N∈ and are in lowest form, the value (a+b) equals

(a) 86 (b) 44 (c) 65 (d) 100 (code-V1T17PAQ4)

Que. 32. Consider a circle 2 2x y ax by c 0+ + + + = lying completely in first quadrant. If m

1 and m

2 are

the maximum and minimum values of y/x for all ordered pairs (x,y) on the circumference of the

circle then the value of ( )1 2m m+ is (code-V1T17PAQ5)

(a) 2

2

a 4c

b 4c

−

−(b) 2

2ab

b 4c−(c) 2

2ab

4c b−(d) 2

2ab

b 4ac−

Que. 33. Let A(a,0) and ( )B b,0 be fixed distinct points on the x -axis, none of which coincides with

the origin O(0,0), and let C be a point on the y-axis. Let g be a line through the origin O(0,0) and

perpendicular to the line AC. The locus of the point of intersection of the lines g and BC if C varies

along the y-axis, is (Provided 2c ab 0+ ≠ ) (code-V1T17PAQ6)

(a) 2 2x y

xa b

+ = (b) 2 2x y

ya b

+ = (c) 2 2x y

xb a

+ = (d) 2 2x y

yb a

+ =

Que. 34. A rectangular billiard table has vertices at P(0,0), Q(0,7), R(10,7) and S (10,0). A small bil-

liard ball starts at M(3,4) and moves in a straight line to the top of the table, bounces to the right side

of the table, then comes to rest at N(7,1). The y-cordinate of the pont where it hits the right side, is

(a) 3.7 (b) 3.8 (c) 3.9 (d) 4 (code-V1T17PAQ7)

Que. 35. A triangle formed by 3 lines denoted by equation 3 2 2 35x 11x y 6xy y 0− + − = will always be

(a) acute angled (b) abtuse angled (c) right angled (d) none (code-V1T19PAQ1)

Que. 36. A point is selected at random inside an equilateral triangle. From this point perpendiculars are

dropped to each side. The sum of these perpendiculars is (code-V1T19PAQ2)

(a) half the sum of the sides of the triangle

(b) equal to the altitude of the triangle.

(c) least when the point is the centroid to the triangle

(d) maximum when the point is centroid of the triangle

Que. 37. One diagonal of a square is the portion of the variable line ( ) 2x 1 y ; 0; 1λ + λ − = λ − λ λ > λ ≠

which is intercepted between the axes. If the area of the square is 17

4 then the number ofvertices of

the square whose both thecoordinates are integers, is (code-V1T19PAQ5)

(a) one (b) two (c) four (d) none

Que. 38. Let nC be a circle of radius

n2

n

centeredat the origin for n 0,1,2,.........= and AAn be the area of

the region that is inside the circle C2n

and out side the circle 2n 1C + for n = 0, 1, 2........ The value of the

sum n

n 0

A∞

=

∑ equals. (code-V1T19PAQ6)

(a) 3

5

π(b)

9

13

π(c)

7

15

π(d)

10

13

π

Que. 39. A circle of radius 2 has its centre at (2,0) and another cricle of radius 1 has its centre at (5,0). A line

is tangent to the two circles at points in the first quadrant. The y-intercept of the tangent line is

(a) 2 (b) 2 2 (c) 3 2 (d) 4 2 (code-V1T20PAQ1)

Teko Classes,



Que. 40. The smallest distance between the circle ( ) ( )2 2

5 x y 3 1− + + = and the line 5x 12y 4 0,+ − = is

(a) 1

13(b)

2

13(c)

3

15(d)

4

15 (code-V1T20PAQ2)

Que. 41. A convex quadrilateral is drawn such that each of its vertices (x,y) satisty the equaation 2 2x y 73+ =

and xy 24.= The are of the quadrilateral is (code-V1T20PAQ4)

(a) 64 (b) 55 (c) 55 2 (d) 110

Que. 42. If the vertices of a ABC∆ are ( ) ( )A 5,0 ,B 3,4 and ( )C 5,2 5 then the coordinates of the

orthocentre is (code-V1T20PAQ6)

(a) ( )8,4 2 5+ (b) ( )8 5, 5− (c) ( )8 5,4 2 5+ + (d) ( )8 5, 2 5−

Que. 43. A particle P moves from the point A(0,4) to the point ( )B 10, 4 .− The particle P can travel the upper

half plane ( ){ }x, y | y 0≥ at the speed of 2 m/s and travel the lower half plane ( ){ }x, y | y 0≤ at the

speed of 2 m/s. The coordinates of a point on the x-axis, if the sum of the squares of the travel times

of the upper and lower half planes is minimum, is (code-V1T20PAQ8)

(a) (1,0) (b) (2,0) (c) (4,0) (d) (5,0)

Comprehesion Type# 1 Paragraph for Q. 1 to Q. 3 (code-V2T2PAQ1,2,3)

Consider a variable line L which passes through the point of intersection ‘P’ of the lines

3x 4y 12 0+ − = and x 2y 5 0+ − = , meeting the coordinate axes at the points A and B.

Que. 1. Locus of the middle point of the segment AB has the equation

(a) 3x 4y 4xy+ = (b) 3x 4y 3xy+ = (c) 4x 3y 4xy+ = (d) 4x 3y 3xy+ =

Que. 2. Locus of the feet of the perpendicular from the origin on the variable line ‘L’ has the equation

(a) ( )2 22 x y 3x 4y 0+ − − = (b) ( )2 2

2 x y 4y 3x 0+ − − =

(c) 2 2x y 2x y 0+ − − = (d) 2 2x y x 2y 0+ − − =

Que. 3. Locus of the centroid of the varible triangle OAB has the equation (where ‘O’ is the origin)

(a) 3x 4y 6xy 0+ + = (b) 4x 3y 6xy 0+ − = (c) 3x 4y 6xy 0+ − = (d) 4x 3y 6xy 0+ + =

# 2 Paragraph for Q. 4 to Q. 6 (code-V2T4PAQ7,8,9)

Let C be a circle of radius r with centre at O. Let P be a point outside C and D be a point on C.A. line

through P intersects C at Q and R,S is the midpoint Of QR.

Que. 4. For different choices of line through P, the curve on which S lies, is

(a) a straight line (b) an arc of circle with P as centre

(c) an arc of circle with PS as diameter (d) an arc of circle with OP as diameter

Que. 5. Let P is situated at a distance ‘d’ form centre O, then which of the folloiwng does not equal the

product (PQ)(PR)?

(a) 2 2d r− (b) 2PT , where T is a point on C and PT is tangent to C

(c) ( ) ( )( )2

PS QS RS− (d) ( )2

PS

Que. 6. Let XYZ be an equilateral triangle inscribed in C. If , ,α β γ denote thedistances of D from vertices

X, Y, Z respectively, the value of prodict ( )( )( ) ,β + γ − α γ + α −β α + β − γ is

(a) 0 (b) 8

αβγ(c)

3 3 33

6

α + β + γ − αβγ(d) None of these

Teko Classes,




The base of an isoceles triangle is equal to 4, the base angle is equal to o45 . A straight line cuts the

externsion of the base at point M at the angle θ and bisects the lateral side of the triangle which is

nearest to M.

Que. 7. The area ‘A’ of the quadrilateral which the straight line cuts off from given triangle is

(a) 3 tan

1 tan

+ θ

+ θ(b)

3 2 tan

1 tan

+ θ

+ θ(c)

3 tan

1 tan

+ θ

− θ(d)

3 5 tan

1 tan

+ θ

+ θ

Que. 8. The range of values of ‘A’ for differnt values of ,θ lie in the interval,

(a) 5 7

,2 2

(b) ( )4,5 (c) 9

4,2

(d) ( )3,4

Que. 9. The length of portion straight line inside the triangle may lie in the range :

(a) ( )2,4 (b) 3

, 32

(c) ( )2,2 (d) ( )2, 3


Let C be curve difined by 2a bxy e .+= The curve C passing through the point P(1,1) and the slope of the

tangent at P is (–2). Also C1 and C

2 are the circles 2 2(x a) (y b) 3− + − = 2 2(x 6) (y 11) 27− + − = re-

spectively.

Que. 10. The value of 2 2a b+ is equal to

(a) 2 (b) 8 (c) 18 (d) 32

Que. 11. The length of the shortest line segment AB which is tangent to C1 at A and to C

2 at B is

(a) 9 3 (b) 10 3 (c) 11 (d) 12

Que. 12. If f is a real valued derivable function satisfying x f (x)

fy f (y)

=

with f '(1) 2.= Then the value of the

integral ( )a

b

f (x)d n x∫ l is equal to

(a) 0 (b) 2 2e e

2

−−(c)

2 2e e

2

− −(d) 2


Given the continuous function

2

2

2

x 10x 8, x 2

y f (x) ax bx c 2 x 0,a 0

x 2x x 0

+ + ≤ −

= = + + − < < ≠

+ ≥

If a line L touches the graph of y f (x)= at three points then

Que. 13. The gradient of the line ‘L’ is equal to

(a) 1 (b) 2 (c) 4 (d) 6

Que. 14. The value of ( )a b c+ + is equal

(a) 5 2 (b) 5 (c) 6 (d) 7

Que. 15. If y f (x)= is differentiable at x 0= then the value of b

(a) is –1 (b) is 2 (c) is 4 (d) ican not be determined

Teko Classes,




Let ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD

respectively so that AE = AF. Let P be a point inside the square ABCD.

Que. 16. The maximum possible area of quadrilateral CDFE is

(a) 1

8(b)

1

4(c)

5

8(d)

3

8

Que. 17. The value of ( ) ( ) ( ) ( )2 2 2 2

PA PB PC PD− + − is equal to

(a) 3 (b) 2 (c) 1 (d) 0

Que. 18. Let a line passing through point A divides the square ABCD in to two parts so that area of one

portion is double the other, then the length of portion of line inside the square is

(a) 10

3(b)

13

3(c)

11

3(d)

2

3


Consider a triangle PQR coordinates of its vertices ( ) ( ) ( )P 8,5 ;Q 15, 19 and R 1, 7 .− − − − The bisec-

tor of the interior angle of P has the equation which can be written in the form ax 2y c 0+ + = .

19. The distance between the orthocentre and the circumcentre of the triangle PQR is

(a) 1

122

(b) 1

152

(c) 1

104

(d)3

114

20. Radius of the incircle of the triangle PQR is

(a) 4 (b) 5 (c) 6 (d) 8

21. The sum of the coefficients (a + c)

(a) 129 (b) 78 (c) 89 (d) 99


Consider the family of lines passing through the intersection of the lines

1U ; 3x 4y 7 0+ + = and 2U : 4x 3y 1 0− + =

22. A member of the family which bisects the angle between them and is closer to the origin, is

(a) x 7y 6 0− − = (b) 7x y 8 0+ + = (c) 7x y 6 0− + = (d) y 7x 4 0+ + =

23. A member of this family with gradient minus 2 has y-intercept equal to

(a) 2 (b) –3 (c) 1 (d) – 2

24. A member of this family whose slope is not difined is

(a) y 1 0+ = (b) x 1= (c) 3x 4= (d) x 1 0+ =


Consider 3 non collinear point A (9,3); B(7,–1) and C(1,–1). Let P(a,b) be the centre and ‘R’ is the

radius of the circle ‘S’ pasing through A,B,C. Also ( )H x, y are the coordinates of the orthocentre of

the triangle ABC whose are be denoted by ∆ .

25. If D, E and F area the middle points of BC, CA and AB respectively then the area of the triangle DEF

is

(a) 12 (b) 6 (c) 4 (d) 3

26. The value ( )a b R+ + equals

(a) 3 (b) 12 (c) 13 (d) None

27. The ordered pair ( )x, y is

(a) (9,5) (b) (–9,6) (c) (9,–6) (d) (9,–5)

Teko Classes,




Consider a circle 2 2x y 4+ = and a point P(4,2). θ denotes the angle enclosed by the tangents from

P on the circle and A,B, are the points of contact of the tangents from P on the circle.

Que. 28. The value of θ lies in the interval

(a) ( )o0,15 (b) ( )o o15 ,30 (c) ( )o o30 ,45 (d) ( )o o45 ,60

Que. 29. The intercept made by a tangent on the x - axis is

(a) 9/4 (b) 10/4 (c) 11/4 (d) 12/4

Que. 30. Locus of the middle points of the portion of the tangent of the circle terminated by the coordinate

axes is

(a) 2 2 2x y 1− − −+ = (b) 2 2 2x y 2− − −+ = (c) 2 2 2x y 3− − −+ = (d) 2 2 2cx y 4− − −= =


Consider a family of lines ( ) ( ) ( )4a 3 x a 1 y 2a 1 0+ − + − + = where a R∈

Que. 31. The locus of the foot of the perpendicular from the origin on each member of this family, is

(a) ( ) ( )2 2

2x 1 4 y 1 5− + + = (b) ( ) ( )2 2

2x 1 y 1 5− + + =

(c) ( ) ( )2 2

2x 1 4 y 1 5+ + − = (d) ( ) ( )2 2

2x 1 4 y 1 5− + − =

Que. 32. A member of this family with positive gradient making an angle of / 4π with the line 3x 4y 2,− = is

(a) 7x y 5 0− − = (b) 4x 3y 2 0− + = (c) x 7y 15+ = (d)5x 3y 4 0− − =

Que. 33. Minimum area of the triangle which a member of this family with negative gradient can make with

the positive semi axes, is

(a) 8 (b) 6 (c) 4 (d) 2


Consider 3 circles

2 2

1

2 2

2

2 2

3

S : x y 2x 3 0

S : x y 1 0

S : x y 2y 3 0

+ + − =

+ − =

+ + − =

34. The radius of the circle which bisect the circumferences of the circles 1 2 3S 0; S 0; S 0= = = is

(a) 2 (b) 2 2 (c) 3 (d ) 10

35. If the circle S = 0 is orthogonal to 1 2 3S 0; S 0 and S 0= = = and has its centre at (a,b) and radius

equals to ‘r’ then the value of ( )a b r+ + equals

(a) 0 (b) 1 (c) 2 (d) 3

36. The radius of the circle touching 1 2S 0 and S 0= = at (1,0) and passing through (3,2) is

(a) 1 (b) 12 (c) 2 (d) 2 2


An altutude BD and a bisector BE are drawn in the trianlge ABC from the vertex B. It is known that

the length of side AC = 1, and the magnitudes of the angles BEC, ABD, ABE, BAC form an arithmetic

progression.

37. The area of circle circumscribing ABC∆ is

(a) 8

π(b)

4

π(c)

2

π(d) π

Teko Classes,



38. Let ‘O’ be the circumcentre ABC,∆ the radius of circle inscribed in BOC∆ is

(a) 1

8 3(b)

1

4 3(c)

1

2 3(d)

1

2

39. Let B' be the image of point B with respect to side AC of ABC∆ , then the length BB ' is equal to

(a) 3

4(b)

2

4(c)

1

2(d)

3

2

Assertion & Reason Type

In this section each que. contains STATEMENT-1 (Assertion) & STATEMENT-2(Reason).Each

question has 4 choices (A), (B), (C) and (D), out of which only one is correct.

Bubble (A) STATEMENT-1 is true, STATEMENT-2 is True; STATEMENT-2 is a correct

explanation for STATEMENT-1.

Bubble (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct

explanation for STATEMENT-1.

Bubble (C) STATEMENT-1 is True, STATEMENT-2 is False.

Bubble (D) STATEMENT-1 is False, STATEMENT-2 is True.

Que. 1. Consider the following statements (code-V2T3PAQ11)

Statement 1: The equation 2 2x 2y 2 3x 4y 5 0+ − − + = represents two real lines on the cartesian plane.

because

Statement 2: A general equation of degree two 2 2ax 2hxy by 2gx 2fy c 0+ + + + + = denotes a line pair

if 2 2 2abc 2fgh af bg ch 0+ − − − =

Que. 2. Consider the folloiwing statements (code-V2T3PAQ13)

Statement 1: The area of the triangle formed by the points A(20,22);(B(21,24) and C(22, 23) is the

same as the area of the triangle formed by the point P(0,0);Q(1,2) and R(2,1).

because

Statement 2: The area of the triangle is invariant w.r.t. the translation of the coordinate axes.

Que. 3. Statement 1: The circle 2 2

1C : x y 6x 4y 9 0+ − − + = bisects the circumference of the circle

2 2

2C : x y 8x 6y 23 0.+ − − + =

because

Statement 2: Centre of the circle C1 lies on the circumference of C

2. (code-V2T6PAQ4)

Que. 4. Passing through a point A(6,8) a variable secant line L is drawn to the circle

2 2S: x y 6x 8y 5 0.+ − − + = Form the point of intersection of L with S, a pair of tangent lines are drawn

which intersect at P. (code-V2T6PAQ6)

Statement 1 : Locus of the point P has the equation 3x 4y 40 0.+ − =

because

Statement 2 : Point A lies outside the circle.

Que. 5. Statement 1 : The equation ( ) ( )2 2

x 2 y 3 1+ + − = − does not represent a circle. (code-V2T15PAQ5)

bacause

Statement 2 : ( ) ( )2 2

x 2 y 3 1+ + − = − represents no real locus.

Teko Classes,



Que. 6. Consider the curves 2

2 2 3

1 2

yC : x a and C : xy c

3− = = (code-V2T16PAQ12)

Statement 1: C1 and C

2 are orthogonal curves.

because

Statement 2: C1 and C

2 intersect at right angles everywhere wherever they intersect.

Que. 7. Consider a general expression of degree two variables as 2 2f (x, y) 5x 2y 2xy 6x 6y 9= + − − − +

Statement 1: f (x, y) can be resolved into two linear factor over real coefficients.

because (code-V1T6PAQ4)

Statement 2: Discriminant of f (x,y) i.e. 2 2abc 2fgh af ch 0+ − − = .

Que. 8. Let triangle ABC be an acute triangle and ‘O’be its circumcentre. D, E and F are the foot of the

perpendiculars dropped from ‘O’ to BC, CA and AB respectively. (code-V1T6PAQ5)

Statement 1:Area of ABC∆ is four times the area of DEF∆

because

Statement 2: Ratio of the areas of two similiar triangle is the ratio of proportional sides.

Que. 9. Statement 1: If the diagonals of the quadrilateral formed by the lines px qy r 0,+ + =

p 'x q 'y r 0,px qy r ' 0,p ' x q ' y r ' 0+ + = + + = + + = are at right angles, then 2 2 2 2p q p ' q ' .+ = +


Statement 2: Diagonals of a rhombus are bisected and prependicular to each other.

Que. 10. Statement 1: The joint equation of lines 2 2 2 2y x and y x is y x i.e., x y 0= = − = − + =


Statement 2: The joint equation of lines ax by 0+ = and cx dy 0+ = is (ax by)(cx by) 0+ + = where

a,b,c,d are constant.

Que. 11. Given a ABC∆ whose vertices are ( ) ( ) ( )1 1 2 2 3 3A x , y ;B x , y ;C x , y . Let there exists a point

P(a,b) such that 1 2 3 1 2 36a 2x x 3x ; 6b 2y y 3y= + + = + + (code-V1T16PAQ11)

Statement 1: Area of triangle PBC must be less than the area of ABC∆

because

Statement 2: P lies inside the triangle ABC

Que.12. Let points A, B, C are represented by ( )i ia cos ,a sin i 1,2,3θ θ = and

( ) ( ) ( )1 2 2 3 3 1

3cos cos cos .

2θ − θ + θ − θ + θ − θ = − . (code-V1T18PAQ10)

Statement 1: Orthocentre of ABC∆ is at origin

because

Statement 2 : ABC∆ is equilateral triangle.

Que. 13. Let C1 denotesa family of circles with centre on x-axis and touching the y-axis at the origin.

and C2 denotes a family of circles with centre on y-axis and touching the x-axis at the origin.

Statement 1: Every member of C1 intersects any member of C

2 at right anglesat the point

other than origin.


Stastement 2: If two circles interesect at 90o at one point of their intersection, then they must

intersectat 90o on the other point of intersection also.

Teko Classes,



Que. 14. Consider the lines 1 2 3

x 2y x 3y: 1 0; : 1 0; : 5x 3y 1 0

3 3 2 4− + = − + = − + =l l l

Statement 1: The lines 1 2 3, andl l l are concurrent. (code-V1T19PAQ8)

because

Statement 2: The area of the trianlge formed by the points 1 2 1 3

, , ,3 3 2 4

− −

and (5,–3) vanishes.

Que. 15. Consider the lines 1 2L :3x 4y 2 and L :5x 12y 7+ = − = (code-V1T10PAQ11)

Statement 1: Every point on the line 64 8y 61− = is equidistant from 1L and 2L .

because

Statement 2: Is the bisector of the angle between 1L and 2L .which contains the origin in its region.

Que. 16. Consider the line L : 3x y 4 0= + + = and the points A( 5,6)− and ( )B 3,2 (code-V1T10PAQ13)

Statement 1: There is exactly one point on the line L which is equidistant form the point A and B.

because

Statement 2: The point A and B are on different sides of the line.

More than One Correct Type

Que. 1. Three distinct lines are drawn in a plane. Suppose there exist exactly n circles in the plane tangent

to all the three lines, then the possible values of n is/are

(a) 0 (b) 1 (c) 2 (d) 4 (code-V2T1PAQ13)

Que. 2. Consider the points ( ) ( )O, 0,0 , A 0,1 and ( )B 1,1 in the x-y plane. Suppose that point C(x,1) and

D(1,y) are chosen such that 0 < x < 1 and such that O,C and D are collinear. Let sum of the area of

triangles OAC and BCD be denoted by ‘S’ then which of the following is/are correct ?

(a) Minimum value of S is irrational lying in (1/3, 1/2)

(b) Minimum value of S irrational in (2/3, 1)

(c) The vlaue of x for minimum value of S lies in (2/3, 1)

(d) The value of x for minimum value of S lies in (1/3, 1/2) (code-V2T1PAQ14)

Que. 3. If θ is the angle between the pair of tangents drawn from (c, 0) to the corcle 2 2x y 1+ = then which

of the following conclusion(s) is/are true ?

(a) If ( )5, c 1, 6 2

6

π θ∈ π ⇒ ∈ +

(b) If ( )3

, c 1, 5 15

π θ∈ π ⇒ ∈ −

(c) If ( ), c 1, 22

π θ∈ π ⇒ ∈

(d) If ( ), c 1, 2

3

π θ∈ π ⇒ ∈

(code-V2T2PAQ13)

Que. 4. If θ is eliminated from the equation a sec x tan yθ − θ = and bsec0 y tan x+ θ = (a and b are

constant) then the eliminant denotes the equation of (code-V2T15PAQ12)

(a) The director circle of the hyperbola 2 2

2 2

x y1

a b− = (b) anuxiliary circle of the ellipse

2 2

2 2

x y1

a b+ =

(c) Director circle of the ellipse 2 2

2 2

x y1

a b+ = (d) Director circle of the circle

2 22 2 a b

x y .2

++ =

Teko Classes,



Que. 5. If the vertices P,Q,R of a triangle PQR are rational points, which of the following points of the

triangle PQR is/are always rational point(s) (code-V2T17PAQ15)

(a) centriod (b) incentre (d) circumcentre (d) orthocentre

Que. 6. The origin, the intersection of the lines 2 22x 5xy 3y 3x 5y 2 0+ − + − − = and the points in which

these lines are cut by the line 3x 5y 2,− = are the vertices of a (code-V2T18PAQ11)

(a) parallelogram (b) rectangle (d) rhombus (d) square

Que. 7. The equations to the lines through the point of intersection of 3x y 20 and x 2y 5 0+ = − − = which

are at a distance 5 from the origin, is/are (code-V1T7PAQ12)

(a) 4x 3y 25+ = (b) 3x 4y 25− = (c) 4x 3y 25− = (d) 4x 3y 25+ =

Que. 8. A circle centred at ‘O’ has radius 1 and contains the points the A. Segment AB is tangent to the

circle at A and AOB .∠ = θ If point C lies on OA and BC bisects the angle ABO then OC equals

(a) ( )sec sec tanθ θ − θ (b) cos

1 sin

2 θ

− θ(c)

1

1 sin+ θ (d) 2

1 sin

cos

− θ

θ (code-V1T12PAQ12)

Que. 9. Let a, b,c Q+∈ satisfying a b c.> > Which of the following statement(s) hold true for the quadratic

polynomial ( ) ( ) ( )2f (x) a b 2c x b c 2a x c a 2b ?= + − + + − + + − (code-V1T14PAQ6)

(a) The mouth of the parabola y = f(x) opens upwards.

(b) Both roots of the equation f(x) = 0 are rational.

(c) x-coordinate of vertex of the graph is positive.

(d) Product of the roots is always negative.

Que. 10. If ( ) 2sin x 2x b 2,α − + ≥ for all real values of x 1≤ and ( ) ( )0, 2 2, ,α ∈ π ∪ π π then possible

real values of ‘b’ is/are (code-V1T14PAQ10)

(a) 2 (b) 3 (c) 4 (d) 5

Que. 11. If ( )2

7 ay log 2x 2x a 3−= + + + is difined x R,∀ ∈ then possible integral value(s) of a is/are

(a) –3 (b) –2 (c) 4 (d) 5 (code-V1T14PAQ11)

Que. 12. If the equation 2 2ax 2cxy by d+ + = represents two real and distinct straight lines then the

necessary and suffcient conditions can be (code-V1T17PAQ12)

(a) d is zero and 2c ab> (b) 2c ab and d R {0}= ∈ −

(c) 2c 4ab and d R= ∈ (d) 2d 0 and c ab= =

Que. 13. If 2 2 24a c b 4ac+ = − then the variable line ax + by + c = 0 always passes through one or the

other of the two fixed ponts. The coordinates of the fixed point can be (code-V1T17PAQ13)

(a) ( 2, 1)− − (b) (2, 1)− (c) ( 2,1)− (d) (2,1)

Que. 14. If the lines (code-V1T17PAQ14)

2

1

2

2

2

3

u : x y 1 0

u : x y 1 0

u : x y 0

λ − − =

− λ + =

+ − λ =

Passes through the same point then the value(s) of λ equals (code-V1T17PAQ15)

(a) 1 (b) 2 (c) 2 (d) 2−

Teko Classes,



Que. 15. Let 1 2L :3x 4y 1& L :5x 12y 2 0+ = − + = be two given lines. Let image of every point on L1

with reespect to a line L lies on L2, then possible equations of L can be (code-V1T17PAQ16)

(a) 14x 112y 23 0+ − = (b) 64x 8y 3 0− − =

(c) 11x 4y 0− = (d) 52y 45x 7− =

Que. 16. Let A(1,1) and B(3,3) be two fixed points and P be a variable point such that area of PAB∆remains constant equal to 1 for all positions of P, then locus of P is given by (code-V1T17PAQ17)

(a) 2y 2x 1= + (b) 2y 2x 1= − (c) y x 1= + (d) y x 1= −

Match Matrix TypeQue. 1. Column - I Column - II (code-V2T19PBQ2)

A. The lines y = 1; x – 6y + 4 = 0 and x+6y–9=0 P. a cyclic quadraliteral

constitute a figure which is

B. The points ( ) ( ) ( )A a,0 ,B 0,b ,C c,0 and ( )D 0,d Q. a rhombus

are such that ac bd= and a,b,c,d are all non-zero.

The points A,B,C and D always constitute

C. The figure formed by the four lines R. a square

( )ax by c 0 a b ,± ± = ≠ is

D. The line pairs 2x 8x 12 0− + = and 2y 14y 45 0− + = S. a trapezium

constitute a figure which is

Que. 2. Column - I Column - II (code-V1T8PBQ2)

A. Four lines x 3y 10 0, x 3y 20 0+ − = + − = P. a quadrilateral which is neither

3x y 5 0 and 3x y 5 0− + = − − = form a a parallelogram nor a trapezium nor

figure which is a kite

B. The point A(1,2), B(2,–3), C(–1,–5) and Q. a parallelogram

D(–2,4) in order are the vertices of

C. The lines 7x 3y 33 0,3x 7y 19 0+ − = − + = R. a rectangle of area 10 sq. units

3x 7y 10 0 and 7x 3y 4 0− − = + − = form a

figure which is

D. Four lines 4y 3x 7 0,3y 4x 7 0,− − = − + = S. a square

4y 3x 21 0,3y 4x 14 0− − = − + = form a figure which is

Que. 3. Column - I Column - II (code-V1T17PBQ1)

A. The four lines 3x 4y 11 0;3x 4y 9 0;− + = − − = P. a quadrilateral which is neither a

4x 3y 3 0 and 4x 3y 17 0+ + = + − = enclose a parallelogram nor a trapezium nor

figure which is a kite.

B. The lines 2x y 1, x 2y 1, 2x y 3+ = + = + = and Q. a parallelogram which is neither a

x 2y 3+ = from a figure which is rectangle nor a rhombus

C. If ‘O’ is theorigin, P is the intersection of the R. a rhombus which is not a square.

lines 2 22x 7xy 3y 5x 10y 25 0,− + + + − = A and B

are the points in which these lines are cut by the line

x 2y 5 0+ − = , then the points O,A,P,B (in some order)

are the vertices of S. a square.

Teko Classes,



Que. 4. Set of family of lines are discribed in column - I and their mathematical equation are given in

column - II. Match the entry of column - I with suitable entry of column - II. (m and n are parameters).

Column - I Column - II (code-V1T20PBQ1)

A. having gradient 3 P. mx y 3 2m 0− + − =

B. having y intercept three times the x-intercept Q. mx y 3m 0− + =

C. having x-intercept (–3) R. 3x y 3a+ =

D. concurrent at (2,3) S. 3x y a 0− + =

Subjective Type ( Up to 4 digit)

Que. 1. A rhombus ABCD has sides of length 10. A circle with centre ‘A’ passes through C (the opposite

vertex) likewise, a circle with centre B passes through D.If the two circles are tangent to each other,

find the area of the rhombus. (code-V2T2PDQ1)

Que. 2. The circles, which cut the family of circles passing through the fixed points

A (2,1) and B (4,3)≡ = orthogonally, pass through two fixed points ( ) ( )1 1 2 2x , y and x , y , which may be

real or imaginary. Find the value of ( )3 3 3 3

1 2 1 2x x y y .+ + + (code-V2T8PDQ3)

Que. 3. Consider 3 lines (code-V1T7PDQ3)

1

2

3

L : 5x y 4 0

L : 3x y 5 0

L : x y 8 0

− + =

− + =

+ + =

If these lines enclose a triangle ABC and sum of the square of the tangent ofthe interior angles can be

expressed in form p/q where p and q are relatively prime numbers, compute the value of(p + q).

Que. 4. If the expression 2 2f (x, y) 9x 16y 777x 992y k= − − − + can be resolved into two linear factors then

find the value of k. (code-V1T12PDQ1)

Que. 5. Point P is 2/3 of the way from the point F( 5,3) to G(4,15).− Line L is perpendicular to the line

FG and passes through the point P. If the equation of the line L is ax + by = c, where a, b and c are

relatively prime integer and a > 0 then find the vlaue of ( )8a 9b 10c .+ + (code-V1T17PDQ1)

Que. 6. Consider the circle whose centre is in the first quadrant and which is tangent to both the co-

ordinate axes and the line L, whose equation is 3x 4y 120.+ = If the co-ordinates of the point of

tangency of the circle with the line L are ( )1 1p ,q and ( )2 2p ,q and (a,b) and (c,d) are the coordinates

of the centres of the two corcles. Find ( )1 2 1 2p p q q a b c d+ + + + + + + (code-V1T17PDQ2)

Que. 7. If the equation of the diagonals of the parallelogram formed by the lines

2x y 7 0;− + = 2x y 5 0;− − = 3x 2y 5 0+ − = and 3x 2y 4 0+ + = are ax by 5 0+ − = and

px qy 1 0.+ + = Find the value of ( )a b p q .+ + + (code-V1T17PDQ3)

Teko Classes,



[SOLUTION]Single Correct Type

Que. 1. A. Centre : ( )3, 4 and r 5− = perpendicular distance from (3, –4) on 3x 4y 25 0+ − = is

9 16 25 32 32 7p d 5 .

5 5 5 5

− −= = ⇒ = − =

Que. 2. C. ( ) ( ) ( ) ( )y x

k k 1 ; y x k y x y 1 k x 1 ky x

−= > − = + ⇒ − = +

+

1 ky x

1 k

+ ⇒ =

− where

1 k1

1 k

+< −

−.

Que. 3. A. Solving line and circle ( )22 2 2 2m x x m c+ δ − =l ( ) ( )2 2 2 2 2 2m x 2 x m c 0⇒ + − δ + δ − =l l

1x

2x

( )2 2m 1+ =l given ( )2 2 2 2 2 2

1 2

1 2

2 m c2x x m cH H

x x 2

δ − δ −= = ⇒ =

+ δ δl

l................ (1)

2 2 2c||| y mK

δ −=

δ

ll ..................... (2)

1 1

A

(x y )

2 2

B

(x y )

�x + my = δ

(1) + (2) ( )2 2 2 2 22 c m c

H mK 2δ − +

+ = = δ −δ δ

ll where ( )2 2m 1+ =l .

Que. 4. A. ( ) ( ) ( )( )22 2 2 1.1 1

x y 4y 4 0 x y 2 0 x y 2 x y 2 0 Area .2 2

− − + = ⇒ − − = ⇒ + − − + = = =

x + y = 2

x + y = 3

x

y

x y + 2 = 0−(0,3)

(1,2)

( 2,0)−

(0,2)

Que. 5. D. ( )3

2

0

x ax 12 dx 45− + + =∫ gives a = 4 Hence 2f (x) 12 4x x (2 x)(6 x)= + − = + − hence

m 2 and n 6= − = m n a 6 2 4 8.⇒ + + = − + =

Que. 6. D.

( )( )( )

2 2 2

2 2 2

2 2 2

p p 1 p q p q 0 p q 1 01 1 1

D q q 1 q r q r 0 p q q r q r 1 02 2 2

r r 1 r r 1 r r 1

− − − + − −

= = − + = + + −

− − −

( )( ) ( ) ( ) ( )( )( )1 1

p q q r p q q r p q q r p r .2 2

= + + + + − = + + −

Teko Classes,



Que. 7. D. ( ) ( )1 1sin sin12 sin sin 12 4 12 4

− −= − π = − π ( ) ( )1 1cos cos12 cos cos 4 12 4 12

− −⇒ = π − = π −

( ) 2k 2 x 8x k 4 0∴ − + + + > If k 2= then .8x 4 0+ > (not possible) then of k 2≠ then k 2 0− > and

( )( )64 4 k 2 k 4 0− − + < 216 k 2k 8⇒ < + − 2k 2k 24 0+ − > ( ) ( )k 6 k 4 0⇒ + − > k 5.=

−6 40

2

Que. 8. A.( )

( )4 y 6

x 9 ; put x 0, 4 y 6 27 y 3/ 4.3

−− = = − = − ⇒ = −

Que. 9. A. tan x 2sin 2 and y 2cos 2λ = θ⇒ = θ = θ 2 2E x xy y 4 4sin 2 cos 2 4 2sin 4⇒ = − + = − θ θ = − θ

[ ]E 2,6 a b 8.∈ ⇒ + =

Que. 10. C. Centres are (10,0) and (–15,0) 1 2r 6; r 9= = d 25⇒ = 1 2

r r d⇒ + < ⇒Circles are separted

( )22

1 2PQ d r r 625 225 20= = − + = − =l

Q

l

1C

2C

1rl

A (10,0)

A

( 15,0)−P1

r

d

2r

Que. 11. A. Let the equation of the variable line is ( )2 2

1 1 1 1xx yy x y 0+ − + =

( )2 22 21 1 1 21 1

1 22 2 2 2

1 1 1 1

2 2 2 2 2 2

1 1 1

ax x yx yp p k

x y x y

i.e., x y ax k locus x y ax k 0

− ++⇒ = =

+ +

+ − = + − ± =

y

x

( )1 1x , y

(a,0)O

]

( )2

2 2ar k

4∴ = − ± ⇒ +ve sign

22 2a

r k4

= − (not possible as 2r becomes –ve)

–ve sign2

2 2ar k .

4= −

Varible line 1

1

xm

y= -

2p1p

(0,0) (0,0)

1 1(x , y )

Que. 12. B.

Teko Classes,



Que. 13. B.

sin cos

D 1 cos sin 0

1 sin cos

λ α α

= α α =

− α α

[ ] [ ]2 2cos sin sin cos sin cos sin cos 0 = λ α + α − α α − α + α − α − α =

[ ]2 2D sin .cos sin cos .sin cos sin 2 cos 2= λ − α α + α − α α − α = λ − α + α

sin 2 cos 2 2, 2 . ∴ λ = α + α ∴ λ ∈ −

Que. 14. B. Solving 2

2

x 1mx x 1 or x 0

x 1 m= ⇒ + = =

+

2 1x 1 0

m⇒ = − > for a region

m 10

m

−<

( )m 0,1⇒ ∈ Note : form = 0 or 1 the line does not enclose a region.

Que. 15. A. Let ( ) ( )C cos ,sin ;H h,kθ θ is the orthocentre of the ABC∆

(h,k) (0,0)G

2 1 Circumcentre

cos 1 sin 1,

3 3

+ +θ θ ( ) ( )2 2

2 2

h 1 cos

k 1 sin

x 1 y 1 1

x y 2x 2y 1 0

= + θ

= + θ

− + − =

+ − − + =

Que .16. B.9!

n(S) 1264!.5!

n(A) 0 to F and F to P

5! 4!. 10.6 60

2!.3! 2!.2!

60 10P(A)

126 21

= =

=

= = =

= =1

0

2

3

4

5

1 2 3 4

P(4,5)

Que. 17. (C) a 1; b 1; c sec

1 3h 0; g ; f

2 2

= = = θ

= = = −

2 2 2using abc 2fgh af bg ch 0

9 1 4sec 0 sec 2 sec 2

4 4 3

+ − − − =

π− θ − + = ⇒ θ = − ⇒ θ = − ⇒ θ =

Que. 18. (B) Let x 3cos ; y 3sin z 2cos ; t 2sin= θ = θ = φ = φ

( ) o6cos .sin 6sin cos 6 6sin 1 90 90∴ θ φ − θ φ = θ − φ = θ = + θ φ − θ =

x 3cos ; y 3sin z 2sin ; t 2sin p xz 6sin cos 3sin 2∴ = θ = θ = − θ = θ = = − θ θ = − θ

maxp 3.∴ =

Teko Classes,



Que. 19. (D)

2y 8x x 15 (x 5)(3 x) y 0 (x 5)(3 x) (x 5)(x 3) 0 x 5 or x 3.= − − = − − ⇒ < ⇒ − − ⇒ − − > ∴ > <

Que. 20. (D) all values of ‘a’.

Que. 21. (D) Slope of the given line is – 1/3 let one line is x y b b 1

1 slopea b a a 3

+ = ∴ = − ⇒ − = −

3b a ........(1)⇒ = also given 2 2a b 100 ........(2)+ = (1) and (2) b 10 b 10;a⇒ = ± ⇒ = 3 10=

b 10;a 3 10⇒ = − = − ∴ None a and b must be of same sign ⇒ (b)

Que. 22. (C)10

C 4log b 2 r= = π 1010 10 10 a

10

log b4 log b 2 .2 log a (as r 2 log a) log b .

log a∴ = π = ⇒ = π ∴ = π

Que. 23. A. Let x r cos ; y r sin 2r cos 3r sin 6= θ = θ⇒ θ + θ = 6

r ;2 cos 3r sin

⇒ =θ + θ

to find

( )2 2

minmin

x y i.e. r+ for r to be minimum 2cos 3sinθ + θ must be maximum i.e., 13

min

6 6 13r .

1313∴ = =

Que. 24. C. 2y 2x 4ax k;− − + abscissa corresponding to the vertex is b

2a− i.e.

4a2 a 2

4

= − ⇒ =

−

now y( 2) 7 7 8 16 k k 1.− = ⇒ = − + + ⇒ = −

Que. 25. B. Verify each alternative.

Que. 26. A. ( )33 3x y 1 3(x)(y)( 1) 0+ + − − − =

2 2 2(x y 1) (x y) (y 1) (x 1) 0 ⇒ + − + + + + + =

x y 1 or x y 1⇒ + = = = − ⇒ a line x y 1+ = or a point ( )1, 1− − ⇒ (A).

Que. 27. D. ( ) ( )2 2

64 k 3 100 k 3 36 k 3 6 or 6 k 3 or 9.+ + = ⇒ + = ⇒ + = − ⇒ = + − A B(2, 3) (10, k)

10

Que. 28. C.

k 1 2 k 1 2k 2 0

D 1 k k 1 0; 1 2k 3 0 0

k 2 3 k 1 k 2 3 k 1

+ −

= − − = − − − =

+ − + −

[ ]1 6k 2 4k 0; 2k 2; l 1.− + + = + = = +

Que. 29. B. ( ) ( ) ( )2 2 2

x y x 1 y 1 2xy 2 2x 2y x y xy 1+ = − + − ⇒ = − − ⇒ + + =

( )( )x y xy 1 2 x 1 y 1 2.⇒ + + + = ⇒ + + =

Que. 30. D. 2 2f (z, y) (x 2) (y 1) 0 x 2 and y 1= − + − = ⇒ = =

( ) ( )( )

2 2

2 1 4 2 2 1 2 1E .

2 2 22 2 1

− + + +∴ = = =

+ +

Que. 31. D. Slope of 8 6h

APh 3

−=

+ hence

8 6h( 6) 1

h 3

− − = −

+ h 45 / 37⇒ = hence coordinate

9 6 (45 / 37) 63/ 37= − × = a b 100.⇒ + =

Teko Classes,



Que. 32. C. Substituting y = mx in the equation of circle we get

( ) ( )

( ) ( )

2 2 2

2 2

2 2

2 2 2 2

x m x ax bmx c 0

x 1 m a bm x c 0

D 0 a bm 4c 1 m 0

a b m 2abm 4c 4cm 0

+ + + + =

+ + + + =

= ⇒ + − + =

+ + − − =

y / x denotes the slope of

the tangent from the origin

on the circle

(0,0)

2y m x=

1y m x=

y

x

( )2 2 2m b 4c 2abm a 4c 0− + + − =1

2

m

m 1 2 2 2

2ab 2abm m .

b 4c 4c b∴ + = − =

− −

Que. 33. C. Equation of the line g is a

y xc

= as (h,k) lies on it,

hence a

k hc

= .......... (1)

Now equation of BC x y

1b c

+ = (h,k) lies on it

h k1

b c+ = ............... (2)

Substituting ah

ck

= in (2) form (1) 2 2h k

1hb ab

+ = locus of P is 2 2x y

xb a

+ = .

Que. 34. A.a 1 7 a

tan3 10 b

− −θ = =

− also

7 4 3tan

b 3 b 3

−θ = =

− −

hence 3 a 1 7 a

b 3 3 10 b

− −= =

− − form 1 st two relations

9 ab b 3a 3 3a 6 ab b= − − = ⇒ + = − .................. (1)

form last two 10a ab 10 b 21 3a− − + = −

13a ab b 31 or ab b 13a 31⇒ − + = − = − ............ (2)

hence from (1) and (2) 3a 6 13a 31 10a 37 a 3.7.+ = − ⇒ = ⇒ =

Que. 35. D. Homogeneous equation of degree 3 ⇒ 3 lines through ⇒ lines concurrent.

Que. 36. B. ( ) 2

1 2 3

1 3a p p p a

2 4∆ = + + =

1 2 3

3ap p p

2∴ + + = =

length of altitude o Altitude

sin 60a

= ∴ Length of altitude 3a

2=

Que. 37. A.x y

11

+ =λ − λ

Area of square is ( )22 21 17 1 17

d 12 4 2 4

= ⇒ λ − + λ =

22 2 2 1 17 ⇒ λ − λ + = 24 4 15 0⇒ λ − λ − = 24 10 6 15 0⇒ λ − λ + λ − = ( )( )2 5 2 3 0⇒ λ − λ + =

5 3or

2 2∴ λ = λ = − But

50

2λ > ⇒ λ = 2∴ vertics are

3 5,0 and 0,

2 2

Remaining 2 vertices

are AB CD

5 3m m

3 5= − ⇒ =

9 25 34AB

4 2

+⇒ = =

34BM

4∴ = using parametric coordinates

y

xA(a,0)

B(b,0)

m = c/a

(0,c)C

m = a/

c g

P(h,k)

(0,7)Q

y(b,7) 10 b R(10,7)

7 a−

(10,a)

a 1−

x

S(100)3

(7,1) 3

(3,4)

(0,0)P

θθ

θ

θo90 −θ

B CQ

A

h hp

1p

3p2p

Teko Classes,



of B and D are 3 34 5 5 34 3

. ; .4 4 34 4 4 34

± ± ( ) ( ) ( )1 1

D , B 2,22 2

⇒ ≡ − − ⇒ ≡ +

Que. 38. B. ( )n 0 1 2 3 nn

n 0

lim A A A A A ........... A∞

→∞=

= + + + + +∑ Now

2

2 2

0 0 1

2A r r 1

3

= π − = π −

4 6 8 10

2 2 4 2

2 2 3 3 2 5

2 2 2 2A r r A r r

3 3 3 3

= π − = π − ⇒ = π − = π − and so on

Hence ( )

2 4 6

n 2

2 2 2 9A 1 ............ .

3 3 3 131 2 / 3

π π = π − + − + ∞ = =

+ ∑

Que. 39. B.

2 2

2

1 1 5a 8

2 a 2

equation of line : y m(x 8)

6m2 9m 1 m

1 m

1m (reject +ve sign as slope is ve)

8

1m y intercept 8 2 2.

8

−= ⇒ =

−

∴ = −

= ⇒ = ++

= ± −

∴ = − ⇒ − = =

Que. 40. B.

0 0

2 2

2 2

The distance form a point to a line

Ax By CAx By c 0 is

A B

The centre of the circle is (5, 3),

so the distance from this point to

the line5x 12y 4 0 is

25 36 45.5 12.( 3) 4 15

311695 12

+ ++ + =

+

−

+ − =

− −+ − −= =

+

Teko Classes,



Que. 41. D.

( )

( ) ( ) ( )

( ) ( )

2 2

2

x y 73; xy 24

x y 73 48 121

x y 11

and x y 5

x, y 8,3 , 8, 3 ,]

3,8 , 3, 8 Area 110.

+ = =

+ = + =

⇒ + = ±

− = ±

≡ − −

− − ∴ =

Que. 42. C. Clearly, coordinates of circumcentre (0) is (0,0) OA OB OC= = and centroid is

8 5 4 2 5,

3 3

+ +

since centriod divides the line segment joining orthocentre and circum centre in

the ratio 2:1, hence coordinates of orthocentre are ( )8 5,4 2 5+ + .

Que. 43. B .

( )

( )

22 2

2

22 2

2

Let the point on the x-axis is (c,0)

Sum of the squares of travel times is

10 c 16c 16T

1 2

116 c 20c 5c 16 c 5c 45

4 4

5c 4c 36 T is minimum is c 2.

4

− ++ = +

+ −= + + = − +

= − + ∴ =

Comprehesion Type

# 1 Paragraph for Q. 1 to Q. 3

1. - A. 2. - B. 3 - C.

Point of intersection the line 3x 4y 12 0 x 2y 5 0+ − = ⇒ + − = is x = 2 and y = 3/2

(i). Equation of AB is

x y 2 31 1 4k 3h 4kh 3x 4y 4xy 0.

2h 2k 2h 4k+ = ⇒ + = ∴ + = ⇒ + − =

(ii).

2 2

k k (3/ 2) k 2k 3. 1 . 2

h h 2 h h 2

2h(h 2) k(2k 3) 0

2(x y ) 4x 3y 0.

− −= − ⇒ = −

− −

⇒ − + − =

⇒ + − − =

B P(2,3/2)

(h,k)

O Ax

y

B(0,2k)

P(2,3/2)

(h,k)

O (2a,0)Ax

y

Teko Classes,



(iii). Here, a 3h, and b 3k= = ∴ equation of AB is

x y 2 11 1 3x 4y 6xy 0.

3h 3h 3h 2k+ = ⇒ + = ⇒ + − =


4. - D. 5. - D. 6 - A.

(i). Locus of S is a part of circle with OP as diameter passing inside the circle ‘C’

N

Q

O

C

PM

RS(h,k)

(ii). ( )( ) ( )( ) ( )( )2 2 2P R PQ PT PN PM d r d r d r= = = − + = −

( )( ) ( ) ( )( ) ( )( ) ( )22 2 2PS SR PS SQ PS SQ SQ SR PS SQ SR PQ PR PS= − + = − ∴ = ⇒ − ∴ ≠

(iii). Using Ptolemy’s theorem

( )( ) ( )( ) ( )( ) ( )YD XZ XY ZD YZ XD XZ ZD XD= + = +

( ){ }XY YZ ZX= =Q ⇒ β = γ + α ⇒ (A).

Alternatively :( ) ( )

( )

2 22 YD XY1

2 2 YD

α + −=

α form

XYD∆ ;( ) ( )

( )

2 22 YD YZ1

2 2 YD

γ + −=

γ form YZD∆

( ) ( ) ( ) ( ) ( ) ( )2 2 2 22 2YD YD XY .........(1); YD YD YZ .........(2)α = α − γ = γ + −

( ) 2 2(1) (2)− ⇒ α − γ β = α − γ ⇒ β = α + γ


7. - D. 8. - D. 9 - C.

Equation of line PM : ( )y 1 tan x 1− = θ − Intersection point ‘Q’

of AC and MP. ( )4 x 1 tan x 1− − = θ −

3 tan 1 3tanQ ,

1 tan 1 tan

+ θ + θ ≡

+ + θ Area of APQ∆ = modulus of

( )1 1 1

5 1 tan 212 2 1

2 1 tan3 tan 1 3tan

11 tan 1 tan

+ θ −=

+ θ+ θ + θ

+ θ + θ

B (a,b)

P(2,3/2)

G(h,k)

O (a,0)Ax

y

X

Y Z

β

o60o60 γ

αD

o30 o30

(1,1)

y = x

A(2,2)

y = 4 x−

Q

B (0,0)Mθ

C(4,0)

o90

o45 o45P

Teko Classes,



(i). Area of quadrilateral BPQC Area, ( )5 1 tan 21 1 tan 3 5tan

A 4 22 1 tan 1 tan 1 tan

+ θ −− θ + θ = × × − = =

+ θ + θ + θ

(ii).2

A 51 tan

= −+ θ

note that, ( ) ( )0, 1 tan 1,2 A 3, 4 .4

π θ∈ ⇒ + θ∈ ∴ ∈

(iii). ( )( )

2 22

2

3 tan 1 3tan 4 4PQ 1 1

1 tan 1 tan 1 sin 2cos sin

+ θ + θ = − + − = =

+ θ + θ + θ θ + θ

( ) ( ) ( ) ( )2sin 2 0,1 PQ 2, 4 PQ 2,2⇒ θ∈ ⇒ ∈ ⇒ ∈ .


10. - A. 11. - C. 12 - B.

(i).2

a bxy e ,+= passes through (1,1) a b1 e a b 0+⇒ = ⇒ + = also (1,1)

dy2

dx= −

2a bxe .2bx 2+⇒ = −

a be .2b(1) 2+⇒ = − b 1 and a 1⇒ = − = (a,b) (1, 1)⇒ = − 2 2a b 2⇒ + = .

(ii). Hence ( )

( ) ( ) ( )

22 2

11 2

22 2

2

C : (x 1) (y 1) 3 C & C

are separatedC : x 6 y 11 3 3

− + + =

− + − =

(1, 1)− (6,11)

A

B

l

1C2C

2r 3=

1r 3=

1r

13

( ) ( )222 2 2

1 2AB d r r 169 4 3 121 AB 11.= = − + = − = ⇒ =l

(iii). Again

( )

h 0

f x hf (x) 1

f (x)f (x h) f (x)f '(x) lim ; f (1) 1

h h→

+−

+ − = = ⇒ h 0

hf 1 1

f (x) xlim

hx

x

→

+ −

=

( )2f '(1).f (x) 2f (x)as f (1) f (1) but f (1) 0 f (1) 1

x x= = = ≠ ⇒ =

f '(x) 2

f (x) x⇒ = n(f (x)) 2 n x C= +l l x 1,⇒ =

2f (1) 1 C 0 f (x) x= ⇒ = ⇒ = ( ) ( )e

a 1 e 2 2 22

b 1 1/e 1/ e

x e eI f (x)d n x x d n x xdx

2 2

−

−

−∴ = = = = =

∫ ∫ ∫l l .


12. - C. 14. - D. 15 - B.

2

2

2

x 10x 8, x 2

f (x) ax bx c 2 x 0,a 0

x 2x x 0

+ + ≤ −

= + + − < < ≠

+ ≥

For continuous at x 0 c 0= ⇒ =

Continuous at x 2 4 20 8 4a 2b 8 4a 2b 2a b 4= − ⇒ − + = − ⇒ − = − ⇒ − = − ............... (1)

Teko Classes,



Now let the line y mx p= + is tangent to all the 3 cuves solving 2y mx p and y x 2x= + = +

2 2 2x 2x mx p x (2 m)x p 0 D 0 (2 m) 4p 0+ = + ⇒ − − = ⇒ = ⇒ − − = ................ (2)

again solving 2y mx p and y x 10x 8= + = + + 2 2x 10x 8 mx p x (10 m)x 8 p⇒ + + = + ⇒ + − + −

2 2 2 2(10 m) 4(8 p) 0 (10 m) 32 4p 0 (10 m) (2 m) 32− − − = ⇒ − − + = ⇒ − − − =

2(100 20m) (4 4m) 32 (m cancels out) 96 16m 32 64 16m m 4 and p 1⇒ − − − = ⇒ − = ⇒ = ⇒ = = −

hence equation of line tangent to 1st and last curves is y 4x 1= − ................ (3)

now solving this with 2y ax bx (as c 0)= + = 2 2ax bx 4x 1 ax (b 4)x 1 0⇒ + = − ⇒ + − + = D 0⇒ =

2(b 4) 4a⇒ − = Also b 2a 4 (form 1)= + 2

4a 4a a 0 a 1 and b 6∴ = ⇒ ≠ ⇒ = =

x 0f '(0 ) lim 2ax b b;−

→⇒ = + =

x 0f '(0 ) lim 2x 2 2+

→= = + = b 2⇒ = .


16. - C. 17. - D. 18 - B.

(i). Area of CDFE ( )21 1A 1 x 1 x

2 2= − − −

2 22 x 1 x 1 x x

2 2

− − + + −= =

max

1 11

5 12 4A at x2 8 2

+ −− = =

y

x

D(0,1)C(1,1)

(0,0)A

E(x,0) B(1,0)

F(0,x)

(ii). ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 2 2 2PA PB PC PD 0− + − = α + γ − α + δ + δ +β − γ + β =

y

x

D(0,1)C(1,1)

(0,0)A

B(1,0)

α

γ δ

β

(iii). ( ) ( )2

2

AQ

1 1 2 2 13y 1 (1) y L 1 .

2 3 3 3 3

= ⇒ = ⇒ = + =

y

x

D(0,1)C(1,1)

(0,0)A

B(1,0)

Q(1,y)

Teko Classes,




19. - A. 20. - B. 21 - C.

Triangle is right R = orthocentre ; M = circumcentre.

(i)

223 1

RM 1 122 2

= + =

M

15

R(1, 7)Q( 15, 19

20

35 D

P ( 8,15)

23

27

(ii) Incentre ( ) ( )20 8 15 15 25(1) 160 225 25 360

x 620 15 25 60 60

− + − + − − + −= = = = −

+ +

( ) ( ) ( )20 5 15 19 25 7 100 285 175 360y 6

60 60 60

+ − + − − − −= = = = − hence incentre (–6 , –6) (can be

used to determine the equation of PD) 20.15

r 5.s 2.30

∆= = =

(iii) Coordinates of D using section formulea are 23

5,2

− −

and PR

11m

2= −

∴ equation PD is 11x 2y 78 0 a c 89.+ + = ⇒ + =


22. - A. 23. - B. 24 - D.

Interection point is (–1, –1). Now proceed.


25. - D. 26. - B. 27 - D.

(i) Area of

9 3 11 1

DEF area( ABC) 7 1 1 3.4 4

1 1 1

∆ = ∆ = − =

−

(ii) Since P(a,b) is the circumcentre

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 2

2 2 2 2

a 7 b 1 a 9 b 3 .............(1)

a 7 b 1 a 1 b 1 .............(2)

∴ − + + = − + −

− + + = − + +

Solving (1) and (2) a 4 and b 3= =

( ) ( )2 2

R 4 7 3 1 5 a b c 4 3 5 12.∴ = − + + = ∴ + + = + + =

(iii)

H CG

12

( y)x , (17/3,1/3) (4,3)

( )

17 2 4 x 1 2 3 yx 17 8 9 y 5

3 3 3 3

point H 9, 5 .

× × × ×∴ = ⇒ = − = ⇒ = ⇒ = −

∴ −

Teko Classes,




28. - D. 29. - B. 30 - A.

Tangent ( )

( )

( )

2

2 2

2

y 2 m x 4

mx y 2 4m 0

2 4mp 2

1 m

1 2m 1 m

3m 4m 0

4m 0 or m

3

− = −

− + − =

−= =

+

− = +

− =

= =

O

2

B

A

Y

X

4α

θ

P(4,2)

2 5

Hence equation of tangent is y = 2 and (with infinite intercept on x - axis)

or ( )4

y 2 x 4 3y 6 4x 16 4x 3y 10 03

− = − ⇒ − = − ⇒ − − = x - intercept 10

4= Ans. (ii) ⇒ (B).

Variable line with mid point (h,k) x y

1,2h 2k

+ = it touches the circle 2 2x y 4+ =

2 2

2 2

1 1 1 12

4h 4k 41 1

4h 4k

−∴ = ⇒ + = ⇒

+

locus is 2 2x y 1− −+ = Ans. (iii) ⇒ (A)


31. - D. 32. - A. 33 - C.

Given ( ) ( ) ( )4a 3 x a 1 y 2a 1 0+ − + − + = ( ) ( )3x y 1 a 4x y 2 0⇒ − − + − − = family of lines passes

through the fixed point P which is the intersection of 3x y 1 and 4x y 2− = − = Solving P(1,2), now

(i)k k 2

. 1h h 2

−= −

−( ) ( ) 2 2locus is x x 1 y y 2 0 x y 2y x 0∴ − + − = ⇒ + − − =

( ) ( ) ( )2

2 2 21 5x y 1 2x 1 4 y 1 5

2 4

− + − = ⇒ − + − =

y

Ox

(h,k)

(1,2)

(ii) We have y 2m(x 1)− − ...........(1) this makes an angle of / 4π with 3x 4y 2− = with slope

3/ 4 ( )

( )

m 3/ 4 4m 31; 1 4m 3 4 4m

1 3m / 4 4 3m

− −∴ = = ± ⇒ − = +

+ + (with + ve sign) m 7=

with – ve sign ( )1

4m 3 4 3m 7m 1 m rejected7

− = − ⇒ = − ⇒ = − hence the line is

( )y 2 7 x 1 7x y 5 0.− = − ⇒ − − =

Teko Classes,



(iii) Again y 2 m(x 1)− = −2

x 0; y 2 m; y 0, x 1m

⇒ = = − = = − ( ) ( )2

2A 2 m 1 m 0m

∴ = − − <

4 42A 2 m 2 4 m

m m

= − − + = + − −

let ( )m M M 0− = >

42A 4 M

M⇒ = + +

22

4 M 4M

= − +

22

8 MM

= + −

area is minimum if M = 2 m 2⇒ = − ⇒ min min

2A | 8 A | 4.= ⇒ =


34. - C. 35. - D. 36 - C.

(i) ( )

( ) ( )

22 2 2 2

2 22 2

2

r a b 1 a 1 b 4 2a 4 0 a 2

and a 1 b 4 a b 1 4 2a 2b b 2

r 9 r 3.

= + + = + + + ∴ + = ⇒ = −

+ + + = + + + ⇒ = ⇒ = −

= ⇒ = Ans.

(ii)1 2 2 3

S S 0 x 1 S S y 1− = ⇒ = ⇒ − ⇒ = ∴ Radical centre ( )1,1= radius T 1

L S 1= =

∴ equation of circle is ( ) ( )2 2

x 1 y 1 1− + − = radius 1 and a 1; b 1 a b r 3.⇒ = = = ⇒ + + =

(iii)( )

( ) ( ) ( ) ( )2 2

2 2

family of circles touches the line x 1 0 at 1,0 is

x 1 y 0 x 1 0 passing through 3,2

4 4 2 0 4

x y 6x 5 0 radius 9 5 2.

− =

− + − + λ − =

⇒ + + λ = ⇒ λ −

∴ + − + = ∴ − =

Teko Classes,




37. - B. 38. - B. 39 - D.

( ) ( )

Angles BEC, ABD, ABE and BAC are in A.P.

let BEC 3 ABD

ABE and BAC 3

Now, 3 3

[using exterior angle theorem]

= α − β ⇒ = α −β

= α + β = α + β

α − β = α + β + α + β

( )

o o o

77 ,

24 24

and From ABD

3 2 22 2

B 2 ,4 2

A , C ABC si 30 90 60 triangle6 3

π π⇒ α − β ∴ β = − α =

∆

π πα −β + α + β = ⇒ β + β =

π πα + β = ∴ ∠ = α + β =

π π∠ = ∠ = ⇒ − −

(i) Area of circle circumscribing

21

ABC2 4

π ∆ = π =

(ii) BOC∆ is equilateral

23 1

14 2r

1 3s 4 3

2 2

∆ ⇒ = = =

(iii)1 3 3

BD OB sin sin BB' 2BD .3 2 3 4 2

π π= = = ∴ = =

Assertion & Reason Type

Que. 1. D. The given equation is ( ) ( )2 2

x 3 2 y 1 0− + − = hence it denotes only a point ( )P 3,1 or two

imaginary lines through ( )P 3,1 as 0.∆ =

Que. 2. A. Let x 20= and y 22= now, , x 20 X; y 22 Y− = − = if ( )20, 22 (0,0) (21,24) (1,2)→ ⇒ →

(22,23) (2,1)⇒ → .

Que. 3. B. 1 2C : centre (3,2); C : centre (4,3) radical axis of C1 and C

2 is 1 2C C 0− = 2x 2y 14 0⇒ + − =

x y 7 0+ − = ................(1) since (1) passes through the centre of 2C (4,3) hence S-1 is correct.

2C

1C also (3,2) lies on 2C hence S-2 is correct but that is not becorrect explanation S - 1.

Teko Classes,



Que. 4. D. Locus of P is polar of A (6,8) w.r.t. the circle S = 0 i.e., 2 2 2x y 6x y 6x 8y 5 0+ − + − − + =

( ) ( )1 1 1 1xx yy g x x f y y c 0 6x 8y 3(x 6) 4(y 8) 5 0+ + + + + + = ⇒ + − + − + + = 3x 4y 18 32 5 0⇒ + − − + =

3x 4y 45 0+ − = ⇒ S - 1 is incorrect. Also the point A (6,8) lies outside S. Hence S-1 is false and S-

2 is true.

Que. 5. D.

Que. 6. A.1 1

11 1

x y 1

3x2y dy dyC : 2x 0 m

3 dx dx y

− = ⇒ = =

1 1

2 3 12 2

x y 1

ydy dyC : 3xy y 0 m

dx dx 3x

+ = ⇒ = − = 1 2 1 2m .m 1 C and C∴ = − ⇒ are orthogonal.

Que. 7. (D) 2 2f (x, y) (2x y) (x y 3)= − + + − ⇒ S -1 is flase, It represent a point (1,2).

Que. 8. (A) D,E,F are the middle points of the sides of the triangle.

D A

E B 4

F C

∴ ∠ = ∠ ∠ = ∠ ⇒∠ = ∠

similiar triangles

Que. 9. (A.).The quadrilateral is obvioulsy a parallelogram and if the diagonals are at right angles, is must

be a rhombus. Hence, the distance between the pairs of opposite sides must be the same

i.e.( ) ( )

2 2 2 2

2 2 2 2

r r ' r r 'p q p ' q '

p q p ' q '

− −= ⇒ + = +

+ +

Que. 10. (D). The joint equation of y = x and y x= − is ( )( )x y x y 0− + = i.e. 2 2x y 0.− =

Que. 11. A.1 2 3 1 2 32x x 3x 2y y 3y

P ,6 6

+ + + += hence P lies inside the triangle

( )1 1

A

x , y ( )2 2

B

x , y

( )3 3C x , y

P (a,b)

3

3

1 D 2

1 2 1 22x x 2 y y,

3 3

+ +

∴ area of PBC∆ < area of ABC∆

Que. 12. A. ( )2 2 2

1 2 2 3 3 1 1 2 32cos cos 2cos cos 2cos cos cos cos cosθ θ + θ θ + θ θ + θ + θ + θ +

( )2 2 2

1 2 3 1 2 2 3 3 1sin sin sin 2sin sin 2sin sin 2sin sin 0θ + θ + θ + θ θ + θ θ + θ θ =

( ) ( )2

1 2 3 1 2 3cos cos cos sin sin sin 0⇒ θ + θ + θ + θ + θ + θ =

2 3 2 3cos cos cos 0 & sin sin sin 0⇒ θ + θ + θ = θ + θ + θ =

⇒ centroid and circumcentre of ABC∆ is at origin

⇒ ABC is equiliateral ∴ Orthocentre of ABC is also origin.

O

E

C

A

B

F

D

Teko Classes,



Que. 13. A.

Que. 14. A Concurrency of lines

1/ 3 2 / 3 1

1/ 2 3/ 4 1

5 3 1

−

⇒ − ⇒

−

S-2 is correct.

Que. 15. C. 1L : 3x 4y 2 0− − + = , 2

L : 5x 12y 7 0− + + = ( )0,0∴ containing bisector is

3x 4y 2 5x 12y 7

5 13

− − + + += 39x 52y 26⇒ − − + 25x 60y 35= − + + 14x 112y 9 0⇒ + + =

∴ None (0,0) containing bisector is 3x 4y 2 5x 12y 7

5 13

− − + − −=

Que. 16. B. The points A and B may be on same side also

More than One Correct TypeQue. 1. A,C,D. Case - I : If lines form a triangle then n = 4 i.e., 3 excircles and 1 incircle

O

y

x

Case - II : If lines are concurrent or all 3 parallel then n = 0

Case - III : If two are parallel and third cuts then n = 2 hence A,C,D.

Que. 2. A,C. S = Area of OAC∆ + area of BCD∆( )( )1 x y 11.x

0 x 12 2

− −= + < <

( )( )x 1 y 1xS

2 2

− −⇒ = − ....(1)

Now 's CBD and OCA∆ are similar

Teko Classes,



( ) ( )( ) ( )

( )

2

22 2

2

y 1 1 x 1 x 1y 1

1 x x x

x 1 1/ x 1 x 1x xS

2 2 2 2x

x x 1 2x 2x 1 1x 1

2x 2x 2x

1x 1 2

2x

− − −∴ = ⇒ = + =

− − −= − = +

+ − − += = = + −

= − − +

y

x(0,0)O

(0,1)A B(1,1)(x,1)

C

D(1,y)

∴ A is minimum if 1

x2x

= i.e, 1

x2

= which lies in (2/3, 1) and minA 2 1= − which lies in

( )1/ 3,1/ 2 A & C.⇒

Que. 3. B,C,D. (A). If 5 5

;6 2 12

π θ πθ = ⇒ = now,

1sin ;

2 c

θ=

( )o

1 2 2 4c 6 2

sin 75 6 23 1= = = = − ⇒

++ (A) is not correct.

(B).o o

3 1 1 4;c 5 1

2 10 sin 54 cos36 5 1

θ π= = = = = − ⇒

+ B is correct.

(C).( )1

,c 22 4 sin / 4

θ π= = = ⇒

π (C) is correct.

(D). o

1,c 2

2 6 sin 30

θ π= = = ⇒ (D) is correct.

Que. 4. C,D. a sec y x tan bsec x y tanθ = + θ ⇒ θ = − θ by squaring and adding

( ) ( ) ( )2 2 2 2 2 2 2a b sec x 1 tan y 1 tan⇒ + θ = + θ + + θ 2 2 2 2x y a b⇒ + = + ⇒ (C) and (D).

Que. 5. A,C,D.

Que.6. A. ( )( )2x y 1 x 3y 2 0− − + + = hence the lines are

2x y 1 0 1 5P ,

x 3y 2 0 7 7

− − = ⇒ − + + =

equation of the two lines lines join-

ing origin and the point of intersection of 3x 5y 2− = and f (x, y) 0=

is ( )( ) ( )

2

2 23x 5y 3x 5y 2 3x 5y

2x 5xy 3y 02 4

− − −+ − + − =

( ) ( )2 22 22x 5xy 3y 3x 4y 3x 5y 0⇒ + − + − − − =

2 22x 5xy 3y 0⇒ + − =

Y

XO

c = 1

(c,0)

θ/21

π

Teko Classes,



Que. 7. C,D.

Que. 8 . A,C,D. Using property of anlge bisector sec x

tan 1 x

θ=

θ −⇒ x tan sec xsecθ = θ − θ

secx

sec tan

θ=

θ + θ

1x

1 sin⇒ =

+ θ

Que. 9. A,B,C. ( ) ( )2f (x) Ax Bx C A a b 2c a c b c 0 A 0= + + ⇒ = + − = − + − > ⇒ > ⇒ mouth opens

upwards

now x= 1 is obvious solution terefore both roots are rational.

( ) ( )ve ve

b a c a 0 B 0;

− −

− + − < ⇒ <123 123

0 c b a

vertex B

02A

= − > hence abscissa ‘a’ of the vertex > 0

O

y

x1

(D) need not be correct as with a 5,b 4,c 2, P 0 and a 6,b 3,c 2,P 0= = = < = = = >

⇒ (A), (B) and (C) are correct.

Que. 10. C,D. Abscissa coresponding to the vertex is given by 1

x 1sin

= >α

is the vertex

the graph of ( ) 2f (x) sin x 2x b= α − + as shown x 1∀ ≤ ∴ minimum of 2f (x) (sin )x 2x b 2= α − + −

must be greater then zero but minimum is at x =1 i.e. ,

sin 2 b 2 0; b 4 sin , (0, );b 4 as sin 0 in (0, )α − + − ≥ ≥ − α α ∈ π ≥ α > π

O

y

x1 x=cosec �

vertex

Hence (C) and (D) are correct.

Alternatively : [ ]2f (x) (sin )x 2x b 2 f (x) 0 x 1= α − + − ≥ ∀ ≤ now vertex 1>

Case - I : (1) D 0, (2) F(1) 0≥ ≥

D 0 4 4sin (b 2) 2

cos ec 2 b

b cosec 2 b 3 ........(1)

≥ ⇒ − α − ≥

⇒ α + ≥

∴ ≤ α + ⇒ ≤

O

y

x1

and f (1) 0 sin b 4 0

b 4 sin

b 4 ..........(2)

≥ ⇒ α + − ≥

⇒ ≥ − α

⇒ ≥

Teko Classes,



Case - II : f (1) 0 and D 0 f (1) 0

b 4 D 0 b 3

b 4 (C) and (D) are the correct answers

> < ⇒ >

⇒ ≥ ⇒ < ⇒ >

∴ ≥ ⇒O

y

x1

Que. 11. B,C,D. 22x 2x a 3+ + + must be positive hence D 0<

i.e., 5

4 8(a 3) 0 1 2a 6 0 2a 5 a ..........(1)2

− + < ⇒ − − < ⇒ − < ⇒ > − Also base of the logrithm

7 a 0 and 7 a 1 a 7 & a 6 ...........(2)− > − ≠ ⇒ < ≠ from (1) and (2) ( )5

a ,6 6,72

∈ − ∪

.

Que. 12. A.

2 2ax 2cxy by d 0+ + − =2abd 0 0 0 dc 0⇒ ∆ = − + − − + = ( )2d ab c 0⇒ ∆ = − − = ( )2d c ab 0⇒ ∆ = − =

⇒ either d = 0 and 2c ab> or 2c ab=

Que. 13. B,D. 2 2 24a c 4ac b 2a c b Or b+ + = ⇒ + = − now ax by c 0 ax by b 2a 0+ + = ⇒ + + − =

( ) ( )a x 2 b y 1 0 (2, 1) (B)⇒ − + + = ⇒ − ⇒

or ( ) ( )ax by b 2a 0 a x 2 b y 1 0 (2,1) (D).+ − − = ⇒ − + − = ⇒ ⇒

Que. 14. B,D. For concurrency

2

2

1 1

1 1 0

1 1

2

λ − −

−λ =

−λ

( )

( )

2 4 2 2

2 4 2 2

2 4 2

1 1 1 1 0

1 1 1 0

1 2 2 0

⇒ λ λ − + −λ − − + λ =

⇒ λ λ − − λ − − − λ =

⇒ λ λ − − λ − =

let ( ) ( )( ) ( ) ( )( )2 2 2t t t 1 2t 2 0 t t 1 t 1 2 t 1 0 t t 2 0 t 2 t 1 0 t 2λ = ⇒ − − − = ⇒ − + − + = ⇒ − − = ⇒ − + = ⇒ =

2 2; 2.∴ λ = λ = ±

Que. 15. A,B. L must be angle. bisector of 1 2

L & L

D

1L

L

2L

A

P

∴ L is given by 1 2 1 2

3x 4y 1 5x 12y 2P DA P DA P AD P AD.

5 13

+ − − += ± ⇒ ∆ ≅ ∆ ⇒ ∠ = ∠

Teko Classes,



Que. 16. C,D.1 1

2 2 h 1 h2 2

∆ = = ⇒ = P wil lie on line parallel to AB at a perpendicular distanceof

1

2 from AB. locus of P is

5 3 3 5y x or y x y x 1 or y x 1

2 2 2 2− = − − = − ⇒ = + = −

P

B

(3,3)

(1,1)

A (2,2)

1 1 1 1 3 52 2

2 22 2 2 2+ + =

1 1 5 32 2

2 2 2 2+ =

Match Matrix Type

Que. 1. A - P,S. B - P. C - Q. D - P,Q,R.

A. Ovioulsy trapezium a 37

a bb 37

=⇒ =

= Here isoscelse trapazium, hence cyclic quadrilateral

also ⇒P,S.

B.

btan

b a cac bd

ac dtan

c

θ =

= ⇒ = ⇒ ⇒ θ = φφ =

hence cyclic quadrilateral ⇒ P..

Teko Classes,



C.c c

ax by c 0 if y 0, x if x 0, ya b

± ± = = = ± = = ± ⇒ rhombus ⇒ Q.

D. ( )( ) ( )( )2x 6 x 2 0 x 6 and x 2 y 14y 45 0 y 9 y 5 0− − = ⇒ = = ⇒ − + = ⇒ − − = ⇒ a square.

Que. 2. A - Q,R,S. B - P. C. - Q,S. D - Q.

A.1 2

10 10h 10 h 10

10 10= = ⇒ = = ⇒ a square of side 10 ⇒ are 10 Q,R,S.= ⇒

3x y 0- = 3x y5 0- =

x 3y 10 0+ - =

x 3y 20 0+ - =

B. AB DC

5 9m 5 m 9

1 1= = − ⇒ = = − ⇒

− − not a parallelogram ⇒ P..

||| y (C) and (D) can be checked.l

D( 2, 4)-

A(1, 2)

B(2, 3)-

C( 1, 5)- -

y

x

Teko Classes,



Que. 3. A - S. B - R. C - Q.

A.1

2

20d 4

5square

20d 4

5

= =

⇒= =

3x 4y+11=0−

3x 4y− −9=0

4x + 3y +3 = 04x + 3y 17 = 0−

B.

1

2

2d

5

2d

5

=

⇒

=

interior not 90o ⇒ rhombus

2x + y 1 = 0−

2x + y = 0−3

x + 2y = 0−1

x + 2y = 0−3

C. ( )( )2 22x 7xy 3y 5x 10y 25 0 x 3y 5 2x y 5− + + + − = ≡ − + − − the point of intersection is (4,3)

homogenising f(x,y) = 0 and x 2y 5 0+ − = we get the homogeneous equation 2 22x 7xy 3y 0− + =

hence OAPB is a parallelogram

O

y

x

A(3,1)

(1,2)(4,3)P x 3y+5=0−

x+2y 5=0−

2x y 5=0− −

(0,−5)

Que. 4. A - S. B - R. C - Q. D - P.

Can be easily analysed.

Subjective Type

Que. 1. (75 sq. unit) Let radius of circle with centre A = R and radius

of the circle with centre B = r now AC = R ; BD = r Area of the

rhombus 1 2d d R.r

2 2= = now R – r = 10 ................ (1)

also in right 2 2

2 2R r100 R r 400

2 2

+ = + =

................(2)

Squaring (1), 2 2 RrR r 2Rr 100; 2Rr 300 75 sq. units.

2+ − = ⇒ = ⇒ =

Teko Classes,



Que. 2. (40). T h e e q u a t i o n o f c i r c l e t a k i n g A B a s d i a m e t e r (x 2)(x 4) (y 1)(y 3) 0− − + − − = ........... (1)

The equation of the line joining the points A and B is x y 1 0− − = ..................... (2)

The equation of members of family of circle passing through A and B is given by

( )( ) ( )( ) ( )S x 2 x 4 y 1 y 3 x y 1 0≡ − − + − − + λ − − = where λ is parameter, , Rλ∈

( ) ( ) ( )2 2S x y 6 x 4 y 11 0≡ + λ − + −λ − + − λ = ............ (1), Let the circles which cuts the members of

circles be 2 2

1S x y 2gx 2fy c 0= + + + + = .................. (2), Applying condition of orthogonality for (1)

and (2), we get 6 4

2g 2f c 11 .2 2

λ − −λ − + = + − λ

i.e., ( )( 6g 4f c 11) g f 1 0− − − − + λ − + = . This will

also hold for all Rλ∈ ∴ we have 6g 4f c 11 0∴ − − − − = and g f 1 0− + = solving these equations

for g and f in terms of c, we get c 15 c 5

g and f10 10

− − − −= = substituting the values of g and f in terms of

c in (2), we get the circles cutting the circle of system (1) orthogonally as

2 2 2 2c 15 c 5 cx y 2 x 2 y c 0 or x y 3x y (x y 5) 0

10 10 5

− − − − + + + + = + − − − + − =

which represents

equatoin of family of circle passing through two fixed points whose coordinates obtained by solving

equation 2 2i.e., solving x y 3x y 0 and x y 5 0+ − − = + − = 2x 6x 10 0(D 0)⇒ − − = < 1 2x x 6;⇒ + =

1 2x x 10= ||| yl 2

1 2 1 2y 4y 5 0 y y 4; y y 5− + = ⇒ + = = ( )3 3 3 3

1 2 1 2x x y y⇒ + + + =

( ) ( ) ( )3 3

1 2 1 2 1 2 1 2 1 2 1 2x x 3x x x x y y 3y y (y y )+ − + + + − + 216 30(6) 64 60 36 4 40= − + − = + =

Que. 3. (465) Arranging the lines in descensing order 1 2 3m 5; m 3; and m 1= = = −

2 3 3 11 2

1 2 2 3 3 1

m m m mm m 2 1 3 1 1 5 3tan a ; tan B 2; tan C

1 m m 1 15 8 1 m m 1 3 1 m m 1 5 2

− −− + − −∴ = = = = = = − = = =

+ + + − + −

2 1 9 1 256 144 401tan A 4 p q 465.

64 4 64 64

+ += + + = = ⇒ + =∑

Que .4. (1394.25)777

a 9, b 16, h ;f 496;c k2

= = − = − = − =

condition; ( ) ( ) ( ) ( )2

2 2 22 2 777abc at bg 0 144k 9 496 16 0 or 144k 4 777 9 496

2

= − = ⇒ − − + = = − −

( ) ( ) ( ) ( )2 2 2 2

36k 777 9 248 36k 9 259 248 = − ⇒ = − ( )( )36k 9 507 11 K 1394.25.⇒ = ⇒ =

Que. 5. (0530.00) ( )FG 81 144 15= + = slope of 12 4FG

9 3= = slope of

3L

4= −

( )FP 2 5 8 3 30

P is 1; 11 P 1,11PG 1 3 3

− + +∴ = ∴ = = ∴ equation P is ( )

3y 11 x 1

4− = − −

15 L

GF 2 1P

10 5

(4,15)( 5,3)−

Teko Classes,



4y 44 3x 3 3x 4y 47 a 3; b 4; c 47 8a 9b 10c 24 36 470 530.− = − + ⇒ + = ⇒ = = = ⇒ + + = + + =

Que. 6. (0210.00) Circle touches both the coordinates axes and also the line L

( ) ( )

( ) ( ) ( ) ( )

( )

1 1 2 2

2 1

2 1

1 1 2 2

centre can be r , r or r , r

3r 4r 120r 7r 120 5r

5

2r 120 or 12r 120

r 60 or r 10

r , r 10,10 and r , r 60,60

Normal for circle1 with centre 10,10 is

∴

+ −∴ = ⇒ − =

= =

= =

⇒ ≡ ≡

∴

1 1r , r

1 1r , r

L: 3x + 4y = 120

x

y

( )4

y 10 x 10 3y 30 4x 40 4x 3y 10 or 16x 12y 403

− = − ⇒ − = − ⇒ − = − = ................(1)

and tangent line L is 3x 4y 120 or 9x 12y 360+ = + = ................(2)

solving (1) and (2) 16x 12y 40

9x 12y 360

− =

+ =

25x 400 x 16 and y 18= ⇒ = = Normal for circle 2 with centre (60,60) is

( )4

y 60 x 60 3y 180 4x 240 4x 3y 603

− = − ⇒ − = − ⇒ − =

or 16x 12y 240

9x 12y 360

− =

+ =

..................... (3) solving with equation (2)

( ) ( )25x 600 x 24 and y 12 16,18 ; 24,12= ⇒ = = ⇒

( )1 2 1 2p p q q a b c d 16 24 18 12 10 10 60 60 210⇒ + + + + + + + = + + + + + + + = .

Alternatively :600

r 10s 60

∆= = =

Que. 7. (40). Equation of the diagonal AC 1 3 2 4u u u u=

( )( ) ( )( )

( ) ( ) ( ) ( )

( )( )

( )( ) ( )1 4 2 3

2x y 7 3x 2y 5 3x 2y 4 2x y 5

5 2x y 7 3x 2y 5 3x 2y 4 2x y 20

11x 19y 35 7x 14y 20 18x 33y 15 0

6x 11y 5 0 a 6 and b 11

equation of the other diagonal ie. BD is

u u u u 2x y 7 3x 2y 4

2x y 5 3x 2y 5 4 2x y 7 3

− + + − = + + − −

− − + + = − + + − −

+ − = − − − ⇒ + − =

⇒ + − = ⇒ = =

= ⇒ − + + +

= − − + − ⇒ − + ( )

( ) ( )

( )

x 2y

5 2x y 5 3x 2y 25 29x 10y 28

25x 5y 25 54x 15y 3 0 18x 5y 1 0

p 18 and q 5

a b p q 6 11 18 5 40.

+

= − − − + + ⇒ + +

= − − + ⇒ + + = ⇒ + + =

⇒ = =

⇒ + + + = + + + =

A

B

C

D

1u 2x y 7 0= + =

2u 2x y 5 0= =

4u 3x

2y 4 0

= +

+ =

3u 3x

2y 5 0

=

+ =

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