Matematika danas

8/19/2019 Matematika danas

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MATHEMATICS TODAY | JANUARY ’15 7

Maths and Science & Technology

Proficiency in Mathematics is always considered to be important inengineering. One does not know where suddenly one finds the needof mathematics till one is pushed to a corner. Earlier, students of agriculture

were not very much bothered about mathematics. Yet there is no research

without application of mathematics in one form or another.

How does one know whether a particular treatment is useful to get better

cotton, jute, silk or even artificial fibres? First we have to know what is a

good cotton. One studies all the physical properties such as fibre-length,

strength, ability to form strong threads and so on. Changes due to varioustreatments have to be significant statistically.

The minimum number of experiments that are needed to determine a

particular property – even energy levels of atoms, the wavelengths of the

various lines of spectra, the calculation of energy levels are all determined

by statistics. To determine the structure of molecules or crystals, group

theory is the main tool. There are no hard-bound barriers dividing any

two branches in science. To study in a systematic manner, we divide it

into Maths, Physics, Chemistry, Agriculture or Industrial research. ‘Science

is a single field’.

Anil Ahlawat

Editor

Vol. XXXIII No. 1 January 2015

Corporate OfficePlot 99, Sector 44 Institutional Area, Gurgaon, (HR).

Tel : 0124-4951200

e-mail : [email protected] website : www.mtg.in

Regd. Office406, Taj Apartment, Near Safdarjung Hospital,

Ring Road, New Delhi - 110 029.

Managing Editor : Mahabir Singh

Editor : Anil Ahlawat (BE, MBA)

CONTENTS Maths Musing Problem Set - 145 8

Practice Paper 10

JEE Main - 2015

Mock Test Paper 14

JEE Main - 2015

Enrich Your Concepts 17

Class XI (Series-8)

Junior Mathematical Olympiad 24 Maths Musing - Solutions 29

You Asked, We Answered 30

Concept Boosters (XI) 31

Concept Boosters (XII) 47

CBSE Board 2015 67

Chapterwise Practice Paper (Series-8)

Math Archives 75

Mock Test Paper 77

JEE (Main & Advanced) (Series-7)

Practice Paper 89

JEE (Main & Advanced) & Other PETs

rialedit

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8 MATHEMATICS TODAY | JANUARY ’15

JEE MAIN

1. 0 < x


4/89


5/89


1. lim sin sin ...n

n

n

n

n→∞ +

+

+

+

2 2 2 21 2

.... sin++

=

n

n n2 2

(a)p

(b)p

/2 (c)p

/4 (d)p

/6

2. Let f be defined by f x x

x ( ) sin ,=

≠1

0

= c, x = 0

where c ∈ [–1, 1]. For which value of c does there

exist an antiderivative of f ?(a) –1/2 (b) 1/2 (c) 0 (d) –1

3. Te function f dx

x ( )

cos,

/

λ λ

π=

−∫

1 20

2

l ∈ (0, 1) is(a) increasing (b) decreasing(c) increasing on (0, 1/2) and decreasing on (1/2, 1)(d) increasing on (1/2, 1) and decreasing on (0, 1/2)

4. lim (log ) logn k

n

k

n

nk

nk

→∞ = =−

=∑ ∑1 12

1 1

2

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

5. For positive a and b define,

B a b x x dx a b( , ) ( ) .= −− −∫ 1 10

1

1 For a ∈ (0, 1),

y

y dy

a−∞

+ =∫

1

01

(a) B(a, 1 – a) (b) B(a, 1 + a)

(c) B aa

12

−

, (d) Ba

a2

,

6. Suppose, f is continuous on [a, b], f (a) = 0 = f (b)

and f x dx

a

b2 1( ) .∫ = Ten xf x f x dx

a

b

( ) ( )′ =∫

(a) 0 (b) 1/2 (c) –1/2 (d) 1

7.

Suppose, f is continuous on [0, c] and strictlyincreasing on [0, c] with f (0) = 0, then for any

x ∈ [0, c], f t dt f t dt x f x

( ) ( )

( )

0

1

0∫ ∫ + =−

(a) x (b) f (x )(c) x f (x ) (d) x + f (x )

8. Te function f is twice differentiable on (0, ∞)

with f (5) = 3, f ′(5) = 2 and f x dx ( ) =∫ 51

5

, then

( ) ( )x f x dx − ′′ =∫ 1 21

5

(a) 15 (b) 18 (c) 20 (d) 0

9. For what value of a (> 1)1 1

32

2

x

x

a

a

⋅ −

∫ log is

minimum?(a) 2 (b) 2.5 (c) 3 (d) 3.5

10. Let f satisfy the equation x = f (x ) · e f (x ), then

f x dx e

( ) =∫ 0

(a) 1 (b) e – 1 (c) e + 1 (d) 0

11. Call a function g as a lower-approximation for f

on the interval [a, b] if for all x ∈ [a, b], f (x ) ≥ g (x ).

Te maximum possible value of g x dx ( ) ,

1

2

∫ where

By : ER Tapas Kr. Yogi, (BHUBANESWAR) Mob : 09778158718

JEE MainJEE Main Integral Calculus


6/89


g (x ) is a linear lower-approximation for f (x ) = x x on[1, 2] is

(a)3 6

4 (b)

3 6

3 (c)

3 6

2 (d) 3 6

12.

( ( ) ( ) )/ /x x x x dx − + + − − − − =∫ 1 1 1 13

0

1

2 3 3 2 23

(a) 1/2 (b) 1/3 (c) 1/4 (d) 1/5

13. Suppose f : R → R be a continuous functionwhich satisfies f (–5) = 8, f (0) = 2 and is even.

Let g x f x x

f x x g x dx ( )

( ),

( ),, ( )=

≤− >

=−∫

0

4 05

5

then

(a) 0 (b) 10 (c) 20 (d) 40

14.

Let D be the region in XY plane which isbounded by the parabola y = 1 – x 2 and the x -axis.For which +ve real no. c, does the parabola y = cx 2 divides D into three smaller regions of equal area?(a) 1 (b) 2 (c) 4 (d) 8

15.

dt

t

dt

t 1 14 40

1

0

1

−÷

+=∫ ∫

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

16. f (x ) = x , x ∈ [–1, 1] = x 3, x ∈ (1, 2) = x 2 + 4, x ∈ [2, 3)(a) primitive of f for x ∈ [–1, 1] is x 3.

(b) primitive of f for x ∈ (1, 2) is1

44x .

(c) primitive of f for x ∈ [2, 3) isx

x 3

34

77

12+ −

(d) none of these

17. Let f (x ) = 1, x ∈ (0, 1] = 0, x ∈ (1, 2](a) primitive of f (x ) for x ∈ (0, 1] is x + c1.(b) primitive of f (x ) for x ∈ (1, 2] is 1 + c1.(c) primitive of f (x ) for x ∈ (1, 2] is c2.(d) no primitive of f (x ) is possible.

18. Given e dx k x | | ,

−∞

∞

∫ = 1 then constant k =

(a) 1 (b) –1 (c) 2 (d) –2

19. lim

/

n

nn

n→∞

=

21

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

20. For every q ∈ (0, p], we have

(a) 1 2

0

2 2+ > +∫ cos sintdt θ

θ θ

(b) 1 2

0

2 2+ = +∫ cos sintdt θ

θ θ

(c) 1 2

0

2 2+ < +∫ cos sintdt θ

θ θ

(d) 11

2

2

0

2 2

+ = +∫ cos sintdt

θ

θ θ21. Let n ≥ 1 be an integer. Te real numbera ∈ (0, 1) that minimizes the integral

I a x a dx an n( ) | |= −∫ 0

1

is =

(a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3

22. I d I d I

I 18

26

0

21

20

2

= = =∫ ∫ cos , cos ,//

θ θ θ θππ

then

(a) 7/8 (b) 8/7 (c) 8/9 (d) 9/8

23. In the diagram l ( AB) = l 1, l ( AC ) = l 2, l (BP ) = x

and l (BC ) = l , then (cos )( )θ x l

dx =∫ 0

l 1

l 2

A

B x P C

(a) l 2 + l 1 (b) l 2 – l 1(c) 2l 1 + l 2 (d) 2l 1 – l 2

24. y (x ) satisfies the differential equation

dy

dx y y yex = +log . If y (0) = 1, then y (1) =

(a) ee (b) e–e (c) e1/e (d) e

25. (( ) log( ) )e ex x e dx x − + − + =∫ 1 12

0

1

(a) 1 (b) e (c) 1 + e (d) e – 1


7/89


26. Let I d x

= +

−∫ 3 21

17 80

cos

cos

θθ

θ . If q ∈ (0, p) and

tan ,I =2

3 then x =

(a) p/3 (b) 2p/3 (c) p/6 (d) p/2

27. (sin (sin ) cos (cos ))/

2 2

0

2

x x dx + =∫ π

(a) p/4 (b) p/2 (c) p (d) 2p

28. Te shortest possible length of an interval

[a, b]for which4

1 2+=∫

x dx

a

b

π is

(a) 2 (b) 2 2

(c) 2 2 2+ (d) 2 2 2−29. Suppose f and g are differentiable functions on

R, satisfying f (0) = g (0) = f ′(0) = g ′(0) = 1 and

f ′′(x ) g (x ) = 4 f (x ) g (x ) – f ′(x ) g ′(x )

g ′′(x ) f(x ) = 5 f (x ) g (x ) – f ′(x ) g ′(x ).

If f (1) = 1, then g (1) = ae3 + be–3, where a =(a) 5/6 (b) 2/3 (c) 6/5 (d) 3/2

30. Let f : R → R be a continuous non-constant

periodic function of period T and let F denote an

antiderivative of f . Let g : R→

R be such that

F x T

f t dt x g x T

( ) ( ) ( ),=

+∫ 1

0

then

(a) g (x ) is periodic with period = T .(b) g (x ) is periodic with period = 2T .(c) g (x ) is periodic with period = T /2.(d) g (x ) is not periodic.

SOLUTIONS

1. (c) : In the inequality θ θ

θ θ− < <3

3!sin , replace

q byn

n kk n

2 21

+=, to .

and use limn k

n n

n k

dx

x →∞ = +=

+=∫ ∑ 2 2 2

0

1

1 1 4

π

2. (c) : Define

F x x x

t t

dt x x

( ) cos cos ,= 1

−

≠∫ 20

21

0

= 0, x = 0

Ten F ′(x ) = f (x ) for x ∈ R Tus F is an antiderivative

of f in the case when c = 0.

3. (a) : For 0 ≤ l1 < l2 < 1, x ∈ (0, p/2)

–l1cos2x > –l2cos

2x

So,1

1

1

11 2 2 2−

<

−λ λ cos cosx x

f (l1) < f (l2), increasing.

4. (b) : Rewrite the term as

1 12

1

12

1

1

n

k

n n

k

nk

n

k

n

log log

−

=

−

=

−

∑∑

So, given limit = −

=∫ ∫ (log ) logx dx xdx 20

12

0

1

1

5. (a) : Substitute x y y

=+1

and b = 1 – a in B(a, b).

6. (c) :

x f x f x dx x f x dx

a

b

a

b

( ) ( ) ( ( ))∫ ∫ ′ = ′1

22

= −

= −∫

1

2

1

22 2xf x f x dx

a

b

a

b

( ) ( )

7. (c) : Put u = f –1(t ) and integrate by parts,

f t dt f t dt f t dt uf u du

x f x x x

( ) ( ) ( ) ( )

( )

0

1

0 0 0∫ ∫ ∫ ∫ + = + ′−

= x f (x )

8. (b) : Use integration by parts to get

( ) ( ) ( ) ( ) ( ) ( )x f x dx x f x x f x dx − ′′ = − ′ − − ′∫ ∫ 1 1 2 12 2

1

5

1

5

1

5

= ′ − − + ∫ 16 5 2 1 215

1

5

f x f x f x dx ( ) ( ) ( ) ( )

= 32 – 24 + 10 = 18

9. (c) : Using Newton-Leibnitz formula,

′ =

+ − −I a

a a

a( )

( ) log( ) log1 1 322

So, for a ∈ (1, 3), I ′(a) < 0 and for a > 3. I ′(a) > 0.

So, minimum at a = 3.

10. (b) : Note that f is inverse function of

g ( y ) = ye y and f (e) = 1

f x dx e g y dy e ye dy ee

y ( ) ( )

0 0

1

0

1

1∫ ∫ ∫ = − = − = −


8/89


11. (a) : Because g is linear, the area is actually atrapezium, with area A = hm, where m is the lengtho mid-line. For g (x ) ≤ f (x ), x ∈ [1, 2]

m g f =

≤

=

=

3

2

3

2

3

2

3 6

4

3 2/

So, g x dx hm( ) ≤ =∫ 3 641

2

12. (d) : Notice that ( )x x dx − + − − =∫ 1 1 1 03 230

1

inverse unctions and shifed.

So, given integral = x x dx 2 3 3 2

0

1

11

5/ /( )− − =∫

13. (c) : Define h(x ) = g (x ) – 2 and notice that h(x )

is odd.

So, g x h x ( ) ( )= +−−∫ ∫ 25

5

5

5

= + =−−∫ ∫ h x dx ( ) 2 205

5

5

5

14. (d) : Area o D = 2 14

32

0

1

( ) .− =∫ x dx

Te two curves intersect at x c

2 1

1=

+. Hence

2 11

3

4

32 2

0

1

1(( ) )− − = ×+∫ x cx dx c

c gives = 8

15. (c) : Fordt

t 1 40

1

−∫ , put t 2 = sinq to simpliy to

1

20

2d θ

θ

π

sin.

/

∫

Fordt

t 1 40

1

+∫ , put t 2 = tan(b/2) to

simpliy to

1

2 20

2d β

β

π

sin .

/

∫ 16. (c) : Any primitive o f is o the ollowing

orm,

x x

c x

x c x

x x c x

( ) , [ , ]

, ( , )

, [ , )

= + ∈ −

= + ∈

= + + ∈

2

1

4

2

3

3

21 1

41 2

34 2 3

For F to be continuous,1

2

1

41 2+ = +c c and

16

4

1

4

8

381 3+ + = + +c c

17. (d) : Any primitive o f is o the orm

F (x ) = x + c1, x ∈ [0, 1] = c2, x ∈ (1, 2]So, F continuity gives 1 + c1 = c2So, F (x ) = x + c1, x ∈ [0, 1] = 1 + c2, x ∈ (1, 2]But this unction is not differentiable.

18. (d) :

e dx e dx k

ek

k x kx

x

kx | | lim

−∞

∞ ∞

→∞∫ ∫ = = −

=2

2 21

0

(given)So, k < 0 and0

21 2− = ⇒ = −

kk

19. (d) : Notice that

lim lim( )!

(( )!)

( !)

( )!nn

n n

a

a

n

n

n

n→∞+

→∞=

+

+×1

2

22 2

1 2

where an =2nC n

=

+ +

+=

→∞lim

( )( )

( )n

n n

n

2 1 2 2

14

2

and lim /

nn

na→∞

=1 4

20. (a) : Notice that 1 2

0

+∫ cos tdt θ

is the arc length

o the curve y = sinx rom (0, 0) to (q, sinq) andθ θ2 2+ sin is the normal distance between these

same points.

21. (a) : Since f (x ) = x n is an increasing unctionon [0, 1].

I a x a dx n n( ) | |= −∫ 0

1

= − + −∫ ∫ ( ) ( )a x dx x a dx n n

an n

a0

1

=

+ − ++

2

111

n

na an n

( )

Now,d I a

daa

( ( ))= =0

1

2 gives

22. (a) : I d = −∫ cos ( sin )/

6 2

0

2

1θ θ θπ

= − ∫ J d cos sin/

6

0

22θ θ θ

π and cos sin

/6

0

22θ θ θ

π

∫ d

can be evaluated by putting cos7q = t .Contd. on Page No. 16


9/89


1. I a1, a2, a3, …, a4001 are terms o an A.P. such

that1 1 1

101 2 2 3 4000 4001a a a a a a

+ + + =....

and a2 + a4000 = 50, then |a1 – a4001| is equal to

(a) 20 (b) 30 (c) 40 (d) 102. From a point P , perpendicular tangents PQ and PR are drawn to ellipse x 2 + 4 y 2 = 4. Locus ocircumcentre o triangle PQR is

(a) x y x y 2 2 2 2 216

54+ = +( )

(b) x y x y 2 2 2 2 25

164+ = +( )

(c) x y x y 2 2 2 2 2416

5+ = +( )

(d) x y x y 2 2 2 2 24 516+ = +

( )

3. Let F x x t dt t dt x x x x

( ) sin cos cos .= + + −∫ ∫ 0

2 2

0

2

Ten area bounded by xF (x ) and ordinate x = 0 andx = 5 with x -axis is

(a) 16 (b)25

2 (c)

35

2 (d) 25

4. Let a unction f (x ) be such that f ′′(x ) = f ′(x ) + ex

and f (0) = 0, f ′(0) = 1, then ln ( ( )) f 24

2

is equal to

(a)1

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

5. I the line y = x + 2 does not intersect anymember o amily o parabolas y 2 = ax , (a ∈ R+) attwo distinct points, then maximum value o latusrectum o parabola is(a) 4 (b) 8 (c) 16 (d) 32

6. Equation o circle inscribed in |x –a| + | y –b| = 1 is(a) (x + a)2 + ( y + b)2 = 2(b) (x – a)2 + ( y – b)2 = 1/2(c) (x – a)2 + ( y – b)2 = 1 2/ (d) (x – a)2 + ( y – b)2 = 1

7. Te number o terms in (a1 + a2 + a3 + a4)3 is(a) 64 (b) 81 (c) 30 (d) 20

8. Te number o unctions f rom the set A = {0, 1, 2} into the set B = {0, 1, 2, 3, 4, 5, 6, 7} suchthat f (i) ≤ f ( j) or i


10/89


13. A hyperbola passing through origin has3x – 4 y – 1 = 0 and 4x – 3 y – 6 = 0 as its asymptotes.Ten the equation of its transverse axis is(a) x – y – 5 = 0 (b) x + y + 1 = 0

(c) x + y – 5 = 0 (d) x – y – 1 = 0

14. Let f x y

f x f y +

= +21

2 ( ( ) ( )) for real x and

y . If f ′(x ) exists and equals to –1 and f (0) = 1, then

the value of f (2) is

(a) 1 (b) –1 (c)1

2 (d) 2

15. lim( !)

n

nn

n→∞

1

equals

(a) e (b) e–1 (c) e–2 (d) e2

16. If |tan A| < 1 and | A| is acute, then

1 2 1 2

1 2 1 2

+ + −

+ − −

sin sin

sin sin

A A

A A is equal to

(a) tan A (b) –tan A (c) cot A (d) –cot A

17. In D ABC , orthocentre is (6,10), circumcentre

is (2, 3) and equation of side BC

is 2x + y = 17.

Ten the radius of the circumcircle of D ABC is

(a) 4 (b) 5 (c) 7 (d) 3

18. Te inclination to the major axis of thediameter of an ellipse, the square of whose length

is the harmonic mean between the squares of the

major and minor axes is

(a)π4

(b)π3

(c)2

3

π (d)

π2

19. Te value of lim( ) /

x

x x e

x →

+ −0

11 is

(a)e

2 (b) −

e

2 (c)

3

2

e (d) −

2

3

e

20. If I x

x x

=−

∫

sin cos

,π3

then I equals

(a) 23

log sin sinx x C + −

+π

(b) 23

log sin secx x C −

+π

(c) 23

log sin sinx x C − −

+π

(d) None of these

21. Let A1, A2, A3, …, A40 are 40 sets each with 7elements and B1, B2, …, Bn are n sets each with 7

elements. If A B Sii

j j

n

= == =

1

40

1∪ ∪ and each element of

S belongs to exactly ten of Ai's and exactly 9 of B j's,then n equals(a) 42 (b) 35 (c) 28 (d) 36

22. If x , y , z are distinct positive numbers, thenx ln y – lnz + y lnz – lnx + z lnx – ln y ∈(a) (0, ∞) (b) (1, ∞) (c) (3, ∞) (d) (1, 3)

23.

Let PQ be a chord of the ellipse

x

a

y

b

2

2

2

2 1+ = , which subtends an angle of p/2 radians at thecentre. If L is the foot of perpendicular from (0, 0)to PQ, then(a) locus of L is an ellipse(b) locus of L is circle concentric with given

ellipse(c) locus of L is a hyperbola concentric with given

ellipse(d) a square concentric with given ellipse

24. A natural number is selected from 1 to 100so that the probability, if it satisfies the number

x x

x

2 60 800

300

− +−

< is

(a) 7/25 (b) 4/25 (c) 2/25 (d) 8/25

25. Te digit at unit place in the number171995 + 111995 – 71995 is(a) 0 (b) 1 (c) 2 (d) 3

26. Te range of values of the term independent

of x in the expansion of x x

sincos

,− −

+

1 1 10α α

a ∈ [–1, 1] is

(a)−

105

10

5

105

20

202 2

C C π π,

(b)10

52

20

105

2

202 2

C C π π,−

(c) [1, 2] (d) (1, 2)


11/89


27. Area of trapezium whose vertices lie on theparabola y 2 = 4x and its diagonals pass through(1, 0) and having length 25/4 units each , is

(a)75

4sq. units (b)

625

16sq. units

(c) 254

sq. units (d) 258

sq. units

28. If the sum of the slopes of the normal frompoint ‘P ’ to the hyperbola xy = c2 is equal tol(l ∈ R+), then locus of point ‘P ’ is(a) x 2 = lc2 (b) y 2 = lc2 (c) xy = lc2 (d) none of these

29. Te differential equation of all straightlines which are at a constant distance ‘ p’ from theorigin is

(a) y x dy

dx p

d y

dx − = +

22

21

(b) y x dy

dx p

dy

dx − = +

12

(c) y x dy

dx p

dy

dx −

= +

22

2

1

(d) y x dy

dx p

dy

dx −

= +

2 2

1

ANSWER KEYS

1. (b) 2. (b) 3. (b) 4. (d) 5. (b)

6. (b) 7. (d) 8. (c) 9. (d) 10. (a)

11. (b) 12. (a) 13. (c) 14. (b) 15. (b)

16. (c) 17. (b) 18. (a) 19. (b) 20. (b)

21. (d) 22. (c) 23. (b) 24. (a) 25. (b)

26. (b) 27. (a) 28. (a) 29. (c)nn

23. (b) : Let l ( AP ) = y , using cosine formula in boththe small triangles,

l 22 = (l – x )2 + y 2 + 2(l – x ) y cosq and

l 12 = x 2 + y 2 – 2xy cosq

Eliminating cosq from these two equations

y l x

l l lx l

l x 2 1

2 2 22

12 22

= − + − + −

Doingdy

dx amounts to

dy

dx = cos .θ

24. (a) : Put y = eu to convert the given differentialequation into a linear differential equation.

25. (b) : ransform the integral

( ) log( )e ex x dx − + −∫ 1 10

1

to

log ydy

e

1∫ whichis inverse of y = ex

2.

26. (b) : First put t = 2 y and then z = sin y .

27. (b) :

I x x dx = +∫ (sin (sin ) cos (cos ))/

2 2

0

2π

...(i)

I x x dx = +∫ (sin (cos ) cos (sin ))/

2 2

0

2π

...(ii)

Adding (i) and (ii), we get

2 1 1

0

2

I dx = +∫ ( )/π

. So, I = p/2

28. (d) : tan tan− −− =1 14

b a π

Simplifying, ba

a= +−

1

1

So, length of interval l = b – a =1

1

2+−a

a

Using,dl

daa= = −0 1 2gives suitably

29. (a) : Adding the two equations, give f ′′(x ) g (x ) + 2 f ′(x ) g ′(x ) + g ′′(x ) f (x ) = 9 f (x ) g (x )Define h(x ) = f (x ) g (x ), then the above equation

becomes ′′ =h x h x ( ) ( )930. (a) : Notice that F (x + T ) – F (x )

= = + −∫ f t dt h x T h x T

( ) ( ) ( ),

0

where h x T

f t dt x T

( ) ( )=

∫

1

0

So, g (x ) = F (x ) – h(x ) = F (x + T ) – h(x + T )is periodic with period = T .

nn

Contd. from Page No. 13

JEE MainJEE MainIntegral Calculus


12/89

MATHEMATICS TODAY

| JANUARY ‘15 17

DEFINITION

A type of mathematical analysis involving the

use of quantified representation, models and

summaries for a given set of empirical data or real

world observations. Statistical analysis involves the

process of collecting and analyzing data and thensummarizing the data into a numerical form.

MEASURES OF CENTRAL TENDENCY

It is a measure that tells us where the middle of

a bunch of data lies. Te three most common

measures of central tendency are mean, median

and mode.

MEASURES OF DISPERSION

It is possible to have two very different data sets

with the same means and medians. For that reason,measures of the middle are useful but limited.

So, dispersion or variability is a very important

attribute of a data set. It measures the degree of

scatteredness of the observations in a distribution

around the central value.

Te various measures of dispersion are

(a) Range (b) Quartile Deviation

(c) Mean Deviation (d) Standard Deviation

RANGE

It is the difference between two extreme values

i.e., Range = Maximum value – Minimum value.

MEAN DEVIATION

Te mean deviation from a central value ‘a’ is given by,

M.D. (a)

=Sum of absolute values of deviations from

Number of observati

a

oons

1. Mean deviation for ungrouped data

Let x 1, x 2, ....., x n be n observations, then

About mean, x 1

1n

x x ii

n

| |−=∑

About median, M 1

1n

x M ii

n | |−=∑

2. Mean deviation for grouped data

(i) For discrete frequency distribution : Let x 1, x 2,...., x n be a set of n observations occurring withfrequencies f 1, f 2, ......, f n respectively, then

About mean, x 1

1N

f x x i ii

n

| |−=∑

About Median, M 1

1N

f x M i ii

n| |−

=∑

where, N f i

i

n

= ==∑ Sum of frequencies

1

.

(ii) For continuous frequency distribution: Herex i are the mid-points of the classes.

About mean, x 1

1N

f x x i ii

n

| |−=∑

About Median, M 1

1N

f x M i ii

n| |−

=∑

Shortcut method for calculating mean deviation

About mean,• x A

f d

N h

i ii

n

= +

×=∑

1, where ‘ A’

is the assumed mean and d x A

hii=

−.

STATISTICS

Statistics & Probability

Series-8


13/89

18 MATHEMATICS TODAY | JANUARY ‘15

\

M.D.( )x

N f x x i i

i

n

= −=∑1

1

About median, M l

N C

f h= +

−×2 •

where median class is the class interval whose

cumulative frequency is just greater than or

equal to N

2, N is sum of frequencies, l , f , h

and C are respectively the lower limit, the

frequency, the width of the median class and

the cumulative frequency of the class just

preceding the median class.

M.D.( ) M

N f x M

i ii

n

= −=∑

1

1

VARIANCE

Te mean of the squares of the deviations from

mean is called the variance and is denoted by s2.

For ungrouped data σ2

1

21= −=∑

nx x i

i

n

( )

For grouped data σ2

1

21= −

=

∑N

f x x i ii

n

( )

STANDARD DEVIATION

Te proper measure of dispersion about the mean ofa set of observations is expressed as positive squareroot of the variance, called standard deviation and

is denoted by s.

For

ungrouped

data

σ = −

=

∑1

1

2

nx x i

i

n

( )

For a

discrete

frequency

distribution

σ = −=∑1

1

2

N f x x i i

i

n

( )

For

continuous

frequency

distribution

σ = −

= =∑ ∑1 2

1 1

2

N N f x f x i i

i

n

i ii

n

Shortcut Method for finding standard deviation

σ = −

= =∑ ∑h

N N f y f y i i

i

n

i ii

n2

1 1

2

,

where y x A

hi

i= −

, A = assumed mean, h = width

of class intervals.

COEFFICIENT OF VARIATION

Te measure of variability which is independent•of units, is called coefficient of variation andusually denoted by C.V.

C.V.= × ≠

σx

x 100 0, ,

where s = standard deviation, x = mean.

Comparing the variability or dispersion of two•series, we calculate the coefficient of variancefor each series. Te series having greater C.V. issaid to be more variable than the other.

Te series having lesser C.V. is said to be moreconsistent than the other.

For two series with equal means, the series•with greater standard deviation (or variance)is called more variable or dispersed than theother. Also, the series with lesser value ofstandard deviation (or variance) is said to bemore consistent than the other.

• Variance of two groups of observations takentogether

Let us consider two groups of variatescontaining n1 and n2 items with respective

means x x 1 2and . Let the standard deviationsfor the two groups be s1 and s2 respectively.Let s be the standard deviation of n1 + n2 items

taken together and x be the mean of two groupstaken together, then

x n x n x

n n=

++

1 1 2 2

1 2

∴ = + +σ σ σ

2

1 21 1

22 2

21

n nn n

++

−( )

n n

n nx x 1 2

1 21 2

2

( )


14/89


15/89


VERY SHORT ANSWER TYPE ( 1 MARK)

1. Find the range of the data, 35, 50, 48, 62, 27, 39,

43, 72, 56, 68.

2. If the S.D. of the items x 1, x 2, x 3, ....., x n is s,

what is the S.D. of the data x 1 + a, x 2 + a, x 3 + a, ....., x n + a?

3. A fair die is rolled once. Find the probability

that a number less than 7 shows up.

4. A book contains 100 pages. A page is chosen at

random. What is the probability that the sum of

the digits on the page is equal to 9?

5. Tree unbiased coins are tossed once. What is

the probability of getting atleast 1 head?

SHORT ANSWER TYPE ( 4 MARKS)

6. wo cards are drawn at random from a pack of

52 cards. What is the probability that both the

cards are aces?

7. Te mean and variance of the heights and

weights of the students of a class are given

below.

Height Weight

Mean 160 cm 50.4 kg

Variance 116.64 cm2 17.64 kg2

Show that the weights are more variable than

heights.

8. Find the mean deviation about mean for the

following data.

x i 3 5 7 9 11 13 f i 6 8 15 25 8 4

9. A card is drawn at random from a well-shuffled

deck of 52 cards. Find the probability that it

being a spade or a king.

10. Find the mean deviation about the median for

the data given below.

45, 36, 50, 60, 53, 46, 51, 48, 72, 42

LONG ANSWER TYPE ( 6 MARKS)

11. Find the mean, variance and standard deviation

for the following data using short cut method.

x i 60 61 62 63 64 65 66 67 68

f i 2 1 12 29 25 12 10 4 5

12. Te following is the record of goals scored by

team A in football session.

Number of goals scored 0 1 2 3 4

Number of matches 1 9 7 5 3

For the team B, mean number of goals scored

per match was 2 with a standard deviation

1.25 goals. Find which team may be considered

more consistent.13. Four persons are to be chosen at random from a

group of 3 men, 2 women and 4 children. Find

the probability of selecting

(i) 1 man, 1 woman and 2 children.

(ii) exactly 2 children.

(iii) exactly 2 women.

14. Five marbles are drawn from a bag which

contains 7 blue marbles and 4 black marbles.

What is the probability that

(i) all will be blue?

(ii) 3 will be blue and 2 black?

15. Calculate the mean and standard deviation for

the following table given the age distribution of

a group of people.

Age 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of

persons

3 51 122 141 130 51 2

SOLUTIONS1. Range = 72 – 27 = 45.

2. Remains unchanged.

3. Sample space S = {1, 2, 3, 4, 5, 6,}.

Let E be the event that ‘the number less than 7

shows up’, then E = {1, 2, 3, 4, 5, 6} = S

\ P (E) =P (S) = 1

4. Sample space S = {1, 2, ....., 100}

\ n(S) = 100


16/89

MATHEMATICS TODAY

| JANUARY ‘15 21

Let E be the event that “the sum of the digits on

the page is equal to 9”.

⇒ E ={9, 18, 27, 36, 45, 54, 63, 72, 81, 90}

\ n(E) = 10

So, required probability = = =n E

n S

( )

( )

10

100

1

10

5. Sample space S = {HTT , THT , TTH , TTT , HHT ,

HTH , THH , HHH }

\ n(S) = 8

Let E = event of getting atleast 1 head. Ten,

E = {HTT , THT , TTH , HHT , HTH , THH , HHH }.

\ n(E) = 7.

\ P (getting atleast 1 head) = =n E

n S

( )

( )

7

8

6. Let S be the sample space. Ten,

n(S) = number of ways of selecting 2 cards out of 52

= =

××

=52 252 51

2 11326C

( )

( ).

Let E = event that both the cards are aces.

Ten,

n(E) = number of ways of drawing 2 aces out of 4

= =

××

=4 24 3

2 16C

( )

( ).

P (getting 2 aces) = P (E) = = =n E

n S

( )

( )

6

1326

1

2217. For height, we have

Variance, s2 = 116.64 cm2

⇒ S.D., σ = =116 64 10 8. .cm cm Also, mean = 160 cm

∴ = ×

C VS.D.

mean. .1 100 = ×

=

10 8

160100 6 75

..

For weight, we have Variance, s2 = 17.64 kg2

⇒ S.D., σ = =17 64 4 2. .kg kgAnd, mean = 50.4 kg

∴ = ×

= ×

=C.V.

S.D.

mean2100

4 2

50 4100 8 33

.

..

Clearly, (C.V. of weights) > (C.V. of heights).

Hence, weights are more variable than heights.

8. We prepare the table given below

x i f i f ix i | |x x i − f x x i i| |−

3 6 18 5 30

5 8 40 3 247 15 105 1 15

9 25 225 1 25

11 8 88 3 24

13 4 52 5 20

otal 66 528 - 138

∴ = = ==∑x f x

N

i ii 1

6

528

668

∴ =

−

= ==∑

M.D.( )

| |

.x

f x x

N

i ii 1

6

138

662 09

9. Let S be the sample space. Ten, n(S) = 52

Let E1 = event of getting a spade,

and E2 = event of getting a king.

Ten, E1∩ E2 = event of getting a king of spade.

Clearly, n(E1) = 13, n(E2) = 4 and n(E1 ∩ E2) = 1

∴ = = =P E n E

n S( )

( )

( ),1

1 13

52

1

4

P E

n E

n S( )

( )

( )22 4

52

1

13= = =

and P E E n E E

n S( )

( )

( )1 21 2 1

52∩ =

∩=

\ P (getting a spade or a king) = P (E1 or E2)

= P (E1 ∪ E2) =P (E1) + P (E2) – P (E1∩ E2)

= + −

= =

1

4

1

13

1

52

16

52

4

13

10. Arranging the given data in ascending order,

we get 36, 42, 45, 46, 48, 50, 51, 53, 60, 72

Here n = 10, which is even.


17/89


∴ =

+ +

Median( ) observation

observ

th

th

M 1

2 2

21

n

naation

⇒ = +

M

1

26(5 observation observation)

th th

⇒ = + = = M 1

248 50

98

249( )

\ Te values of (x i – M ) are

– 13, – 7, – 4, – 3, – 1, 1, 2, 4, 11, 23

∴ −=∑| |x M ii 1

10

= (13 + 7 + 4 + 3 + 1 + 1 + 2

+ 4 + 11 + 23) = 69

∴ = − = ==∑M.D.( ) | | . M x M n

ii 1

10

69

106 9

11. Let the assumed mean be A = 64.

Ten, we prepare the table given below.

x i f i d i = (x i – 64) d i2

f id i f id i2

60 2 – 4 16 – 8 32

61 1 –3 9 – 3 9

62 12 – 2 4 –24 48

63 29 – 1 1 – 29 2964 = A 25 0 0 0 0

65 12 1 1 12 12

66 10 2 4 20 40

67 4 3 9 12 36

68 5 4 16 20 80

otal 100 0 286

∴ = = = =∑∑∑N f f d f d i i i i i100 0 2862, and

∴ = + = +

=∑Mean A f d

N

i i 640

10064

Variance, σ22 2

= −

∑ ∑ f d N

f d

N

i i i i

= −

=

286

100

0

1002 86

2

.

\ Standard deviation, σ = =2 86 1 69. .

12. In order to determine the consistency of teams,

we will have to find the coefficients of variations

of two teams.

Computation of mean and standard deviation

of goals scored by team A.

x i f i f i x i f i x i2

0 1 0 0

1 9 9 9

2 7 14 28

3 5 15 45

4 3 12 48

otal 25 50 130

∴ = = = =∑ ∑∑N f f x f x i i i i i25 50 1302

, and

∴ = { } = =∑x N

f x A i i1 50

252

and σ A i i i iN f x

N f x 2 2

21 1

= −{ }

∑∑

= −

= − =

130

25

50

255 2 4 1 2

2

. .

⇒ = =σ A 1 2 1 095. .

It is given that x B B= =2 1 25and σ .Now,

Coefficient of variation of goals scored by team A

= × = × =σ A

Ax 100

1 095

2100 54 75

..

Coefficient of variation of goals scored by team B

= × = × =

σBBx 100

1 25

2 100 62 50

.

.

We observe that, the coefficient of variation

of goals scored by team A is lesser than that of

team B. Hence, team A is more consistent.

13. Tere are 9 persons i.e., 3 men, 2 women and 4

children. Out of these 9 persons, 4 persons can

be selected in 9C 4 = 126 ways.

\ otal number of elementary events = 126


18/89

MATHEMATICS TODAY

| JANUARY ‘15 23

(i) 1 man, 1 woman and 2 children can be

selected in 3C 1 ×2C 1 ×

4C 2 = 36 ways.

\ Favourable number of elementary events = 36

So, required probability = =36

126

2

7

(ii) Exactly 2 children means 2 children and

2 persons from 3 men and 2 women. Tis

can be done in 4C 2 ×5C 2 = 60 ways.


So, required probability = =

60

126

10

21

(iii) We have to select 4 persons of which 2 are

women and the remaining 2 are chosen

from 3 men and 4 children. Tis can bedone in 2C 2 ×

7C 2 = 21 ways.


So, required probability = =21

126

1

6

14. Tere are 7 + 4 = 11 marbles in the bag out of

which 5 marbles can be drawn in 11C 5 ways.

\ otal number of elementary events = 11C 5

(i) Tere are 7 blue marbles out of which 5blue marbles can be drawn in 7C 5 ways.

\ Favourable number of elementary

events = 7C 5

Hence, required probability =7

511

5

C

C

= × =7

2 5

5 6

11

1

22

!

! !

! !

!

(ii) 3 blue out of 7 blue marbles and 2 blackout of 4 black marbles can be drawn in7C 3 ×

4C 2 ways.

\ Favourable number of elementary events

= 7C 3 × 4C 2

Hence, required probability

= ×

= × × =7

34

2

115

7

3 4

4

2 2

5 6

11

5

11

C C

C

!

! !

!

! !

! !

!

15. Here A = 55, h = 10.

Age x i f i ux

ii=

− 55

10 f i ui ui

2 f i ui2

20-30 25 3 –3 –9 9 27

30-40 35 51 –2 –102 4 204

40-50 45 122 –1 –122 1 122

50-60 55 141 0 0 0 0

60-70 65 130 1 130 1 130

70-80 75 51 2 102 4 204

80-90 85 2 3 6 9 18

otal 500 5 705

x A hN

f ui i= +

= +

=

155 10

5

50055 1Σ .

σ2 2 22

1 1= −

hN

f uN

f ui i i iΣ Σ

= −

100705

500

5

500

2

=14099

100

⇒ = = =σ14099

10

118 739

1011 8739

..

nn

JEE (Main-Ofine) : 4th April

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Important Exam Dates 2015


19/89


1. Prove that for no integer n, n6 + 3n5 – 5n4 –

15n3 + 4n2 + 12n + 3 is a perfect square.

2. Two dice are thrown once simultaneously.

Let E be the event “Sum of numbers appearing on

the dice.” What are the members of E? Can you

load these dice (not necessarily in the same way)

such that all members of E are equally likely? Give justification.

3. Let sinx + sin y = a and cosx + cos y = b, show

that tanx

2 and tan

y

2are two roots of the equation:

(a2 + b2 + 2b)t 2 – 4at + (a2 + b2 – 2b) = 0

4. In a triangle ABC , angle A is twice the angle

B. Show that a2 = b (b + c), where a, b and c are the

sides opposite to angle A, B and C respectively.

5. A, B, C, D are four points on a circle with radius

R such that AC is perpendicular to BD and meets

BD at E. Prove that : EA2 + EB2 + EC 2 + ED2 = 4R2.

6. Suppose A1 A2 A3 .......... An is an n-sided regular

polygon such that

1 1 1

1 2 1 3 1 4 A A A A A A= + . Determine the number of

sides of the polygon.

7. Find all positive integers a, b for which the

number2

3

+

+

a

b is a rational number.

8. If a, b, c are positive real numbers, prove that :

a b c a

b c

a b c

c aa b c c

a b a b c

+ + +

++

+ +

+

++ + +

+≥

+

+ +

9 3 3

2

9. Find integers a and b such that x 2 – x –1 divides

ax 17 + bx 16 + 1 = 0.

10. Consider the equation x 4 – 18x 3 + kx 2 + 174x

– 2015 = 0. If the product of two of the four roots of

the equation is –31, find the value of k.

SOLUTIONS

1. Case 1 : When n is even number ⇒ n = 2k for

some k ∈ I .Now, n6 + 3n5 – 15n3 + 4n2 + 12n + 3 becomes 64k6 + 96k5 – 80k4 – 120k3 + 16k2 + 24k + 3⇒ 4(l) + 3, where l ∈ I .Now an odd perfect square is always of the form4l + 1 (∵ (2m + 1)2 = 4(m2 + m) + 1), wherem ∈ I .The given expression can’t be perfect square for

any even integer.Case 2 : When n is odd number ⇒ n = 2k′ + 1for some k′ ∈ I .Now, n6 + 3n5 – 5n4 – 15n3 + 4n2 + 12n + 3becomes (2k′ + 1)6 + 3(2k′ + 1)5 – 5(2k′ + 1)4 –15(2k′ + 1)3 + 4(2k′ + 1)2 + 12(2k′ + 1) + 3Now, (2k′ + 1)6 ≡ (12k′ + 1) (mod 4) ≡ 1(mod 4)3(2k′ + 1)5 ≡ (30k′ + 3) (mod 4) ≡ (2k′ + 3) (mod 4)5(2k′ + 1)4 ≡ (40k′ + 5) (mod 4) ≡ 1 (mod 4)

Whizdom Educare, 50-C, Kalu Sarai, Sarvapriya Vihar, New Delhi-16


20/89


15(2k′ + 1)3 ≡ (90k′ + 15) (mod 4) ≡ (2k′ + 3) (mod 4)

4 (2k′ + 1)2 ≡ 0 (mod 4)

12 (2k′ + 1) ≡ 0 (mod 4)

or, n6 +3n5 – 5n4 – 15n3 + 4n2 + 12n + 3 ≡ (1

+ 2k′ + 3 + 1 + 2k′ + 3 + 3) (mod 4)

≡ (4 (k′ + 2) + 3) (mod 4)We have n6+ 3n5 – 5n4 – 15n3 + 4n2 + 12n + 3

= 4l′ + 3 (where l′ ∈ N )

Again the expression is not a perfect square for

all odd integers.

Consequently, the expression is not a perfect

square for any integer.

2. Two normal dice are thrown, possible outcomes

are

{( , ),( , ),( , ),......,( , ),

( , ),( , ),( , ), ......,(

1 1 1 2 1 3 1 6

2 1 2 2 2 3 22 6

3 1 3 2 3 3 3 6

, ),

( , ),( , ),( , ), .......,( , ),

.............................................

( , ),( , ),( , ), ......,( , )}6 1 6 2 6 3 6 6

total 36 outcomes

Now, E = sum of numbers on dice faces.

= {2, 3, 4, ….., 12}

n(E) = 11.

Loading the dice means tampering with thenumber on the faces of dice so that one result

is favoured or dismayed over other.

For a normal throw

P n

nn

nnn E

( )

,

( ),∈

=

−≤ ≤

− −< ≤

1

362 7

12 1

367 12

If all members of E are equally likely, then,

P n k

n E( )

∀ ∈=

36 for some k ∈ N .

Also, outcomes of E are mutually exclusive and

mutually exhaustive. So,

P nn E( )

∀ ∈=∑ 1 ⇒ × =11

361

k ⇒ 11 × k = 36

Now, k can’t be a natural number

⇒ such loading is not possible.

3. sinx + sin y = a

⇒ 22 2

sin cosx y x y

a+

−

= …(i)

cosx + cos y = b

⇒ 22 2

cos cosx y x y

b+

−

= …(ii)

to prove tan tanx y

2 2and are roots of the equation

(a2 + b2 + 2b)t 2 – 4at + (a2 + b2 – 2b) = 0

Let the roots of the equation be a and b.

Sum of roots α β+ =+ +

4

22 2a

a b b and product of

roots αβ = + −

+ +

a b b

a b b

2 2

2 2

2

2

Now, a2 + b2 = sin2x + sin2 y + 2sinx sin y + cos2x

+ cos2 y + 2cosx cos y .

= 2 (1 + sinx sin y + cosx cos y )

= 2 (1 + cos (x – y )) = 4 cos2 x y −

2

Clearly, a2 + b2 + 2b =

42 2 2

cos cos cosx y x y x y −

−

+

+

= − 4

22

2 2cos cos cosx y x y

….(iii)

and a2 + b2 – 2b

= −

−

−

+

4

2 2 2cos cos cos

x y x y x y

= −

4

22

2 2cos sin sin

x y x y …(iv)

Now, sum of roots =4

2

2 2

a

a b b+ +

=×

+

= +4 2

2

82 2

2 2

sin

sin .sin

tan tan

x y

x y

y x

& product of roots =a b b

a b b

x y 2 2

2 2

2

2 2 2

+ −

+ += tan . tan

Thus, roots are tan tan .x y

2 2

and


21/89


4. In D ABC , ∠ A is twice the ∠B.

Let ∠ A = 2x , so ∠B = x .

⇒ ∠C = p – 3x

By sine law, we havea

x

b

x

c

x sin sin sin( )2 3= = −π

⇒ = =a

x x

b

x

c

x 2 3sin cos sin sin

⇒ = =−

a

x x

b

x

c

x x 2 3 4 2sin cos sin sin ( sin )

⇒ = =−

a

x b

c

x 2 4 12cos cos ∵sin x x n≠ ≠0 as π

Now, we have cos cosx a

b

x a

b

= ⇒ =2 4

22

2

and 4 11

4

2 2cos cosx c

bx

c

b− = ⇒ =+

On comparing, we get a

b

b c

b

2

24 4=

+

⇒ a2 = b(b + c).

5. A, B, C and D are concyclic points with O as

centre of circle. AC

^

BD

and AC

meetsBD

atE

.

Construction : Extend BO to meet the circle at D′.In DDEC and D AEBWe have ∠DEC = ∠ AEB = 90°

∠CDB

=∠CAB

We have DDEC ~ D AEB (by AA similarity)

Also, ED

EC

EA

EB=

or ED

EC

EA

EB

ED EC

EC

EA EB

EB

2

2

2

2

2 2

2

2 2

2= ⇒

+=

+

(applying componendo)

Now, ED EC EC

EBEA EB

2 22

22 2+ = +( ) …(i)

L.H.S. : EA2 + EB2 + EC 2 + ED2

= + + +EA EB

EC

EBEA EB

2 22

22 2( )

=

+ +( )( )EA EB EB EC

EB

2 2 2 2

2

= ⋅ AB BC

EB

2 2

2 …(ii)

Now, in DEAB and DCD′B, we have∠BD′C = ∠BAC and ∠D′CB = ∠ AEB = 90°so D AEB ~ DD′CB (by AA similarity)

or BA

BE

BD

BC =

′

⇒ = ′BA

BEBC BD

2

22 2

…(iii)

Comparing (ii) & (iii), we get EA2 + EB2 + EC 2 + ED2 = (Diameter)2

= 4 × (radius)2

6. We have, n-sided regular polygon whose

vertices are A1, A2, A3, ….., An.

Clearly, ∠ A1OA2 =2πn

and let OA1 = r 1.

InD A1OA2, we have cos

( )2 2

2

21 2

2

2

πn

r A A

r =

−

⇒ = −

⇒ =

( ) cos sin A A r

n A A r

n1 2

2 21 22 1

22π π

Similarly, in D A1OA3 we have, A A r n1 32

2=

sin

π

and in D A1OA4, we have, A A r n

1 4 23

=

sin π

We have,1 1 1

1 2 1 3 1 4 A A A A A A= +


22/89


⇒

=

+

1

2

1

22

1

23

r n

r n

r n

sin sin sinπ π π

⇒

−

=

1 1

3

1

2sin sin sin

π π π

n n n

⇒

−

=

sin sin

sin .sin

sin

3

3

2

π ππ π

πn nn n

n

⇒

=

22 2 3

sin . cos . sin sin . sinπ π π π πn n n n n

∵ sin sin sinπ π πn n n

≠ ⇒

=

04 3

⇒ = − = +4 3 4

23π π π π π π

n n n nor .........

⇒ = =7

2π

π π

πn n

or .........

⇒ = =n n71

2or .........

∵ n is an integer.

⇒ Number of sides = 7.

7. Let N a

b

p

q= +

+ =

2

3, where p, q are non zero

integers.

or assumingN a

b

b

bb=

+

+ ×

−

− ≠

2

3

3

33( )

or N b a ab

b=

− + −−

6 2 3

3

Now 6 3 2+ − −a b ab is a non–zero integer.

Case1 : 2b = 6 ⇒ b = 3 (not possible)Case 2 : ab = 6 and 3 2a b− is an integer∵ a and b are positive integers.

\ Possible pairs of (a, b) are (1, 6), (3, 2), (6, 1)

(we can’t take b = 3)

Now, for (a, b) as (1, 6), we get

3 2 3a b− = −For (a, b) as (3, 2), we get

3 2 1a b− =

For (a, b) as (6, 1), we get

3 2 2 2a b− =Possible solution is (3, 2)

Case 3 : 6 2= +b ab and a = 3 12k , for some

k1 ∈ N .

⇒ 6 = 2b + ab + 2b 2a R.H.S. must be integer ⇒ a = 2k22 for some k2 ∈ I .We are contradicting our earlier specificationthat a = 3k1

2 & also a = 2k22

Thus, no solution exists.Possible solution of (a, b) is (3, 2).

8. a b c a

b c a b c a

+ + ++

=+ + −( )

1

Multiplying numerator and denominator by

a b c a+ + −( )Similarly for other two ratios, consequently

L.H.S. =+ + −

++ + −

++ + −

1 1

1

a b c a a b c b

a b c cFor three positive real numbers, we haveA.M. ≥ H.M.

⇒

+ + − +

+ + −+

+ + −

=

≥+ + − −

1 1

1

3

3

3

a b c a a b c b

a b c c

a b c a

α( )let

bb c−

⇒ ≥

+ + − + +

+ +

α9

3a b c a b c

a b c

Leta b c

a b c

+ +

+ + = β

Also, a b c a b c+ + ≥ + +3

∵

a a a

n

a a a a

n

m mnm

nm

1 2 1 2 3+ + + ≥ + + + +

.... ... ,

ai > 0, i ∈ N

We have, β ≥ 3 or 3 3 3− ≤ −β


23/89


or1

3

1

3 3− ≥

−β

or9

3

9

3 3a b c a b c+ + − ≥

− + +( ) ( )β

Also,( )

( )α β≥ + + − ≥ ++ +9

33 3 3

2a b c a b c

⇒+ + −

++ + −

++ + −

≥ +

+ +

1 1

1 9 3 3

2

a b c a a b c b

a b c c a b c

Hence proved.

9. Leta be roots of x 2 – x –1. Ten a2 –a –1 = 0.

or a2 = a +1 …(i)

Now, as (x 2 – x – 1) divides ax 17 + bx 16 + 1 = 0

⇒ a is a root of ax 17 + bx 16 + 1 = 0

⇒ + + =a bα α17 16 1 0 ⇒ +( ) + =α α16 1 0a b

⇒ + + =( ) ( )α α2 8 1 0a b

⇒ +( ) +( ) + = = +α α α α1 1 0 18 2a b [ ]putting

⇒ + + + + =( ) ( )α α α2 42 1 1 0a b

⇒ +( ) +( ) + = = +3 2 1 0 14 2α α α αa b [ ]putting

⇒ + + + + =( ) ( )9 12 4 1 02 2α α αa b

⇒ +( ) +( ) + = = +21 13 1 0 12 2α α α αa b [ ]putting

⇒ + + + + =( )( )441 546 169 1 02α α αa b

⇒ +( ) +( ) + = = +987 610 1 0 12α α α αa b [ ]putting

⇒ + +( ) + + =987 987 610 610 1 02a b a bα α

⇒ + +( ) + + =α 987 610 987 610 1 0b a a b

⇒ (a × k1)+ k2 = 0, where k1, k2 ∈ I

or (1597a + 987b)a + 987a + 610b + 1 = 0

Also, α = ±1 5

2

or, 1597 9875

22584 1597 1 0a b a b+( ) ×

±

+ + + =

or, 1597a + 987b = 0 …(i)

and 2584a + 1597b+ 1 = 0 …(ii)

We have a b= −987

1597, putting in (ii), we get

b 1597

987 584

15971 0−

×

+ =

⇒ = − ⇒ =b a1597 987

10. x x kx x 4 3 218 174 2015 0− + + − =

Let the roots be a, b, c and d ; and ab = –31We have a + b + c + d = 18 …(i)

ab ac ad bc bd cd k+ + + + + = …(ii)abc abd acd cbd + + + = −174 …(iii)abcd = −2015 …(iv)From (iv), we have cd = 65

From (iii), we have ab(c + d ) + cd (b + a) = –174

⇒ –31 (c + d ) + 65 (a + b) = –174

Let (c + d ) = p and a + b = q

We have, p + q = 18 and 65 q – 31 p = – 174⇒ p = 4 and q = 14

⇒ c + d = 4 & a + b = 14

From (ii),

ab + a (c + d ) + (c + d ) + cd = k

⇒ –31 + (c + d ) (a + b) + 65 = k

⇒ 34 + 56 = k ⇒ k = 90nn


24/89


SOLUTION SET-144

1. (b) : 2n + 1 = (n + 1)2 – n2

n n n n n32 21

2

1

2= +

−

−

( ) ( )

\ m contains all odd numbers and the even

numbers 23, 43, 63, 83, 103, 123.

\ m = 1007 + 6 = 1013 with digit sum 5.

2. (c) : AB b AD d

= =,

A M B

C

P N

DTe line AC is

r p b d = +( )

Te line MN is

r mb q nd mb= + −( )

P is a common point ⇒ p = m (1 – q) = qn

∴ =+

pmn

m n

3. (c) : z 18 = 1, w48 = 1

(z w)n =1, z nwn = 1 = 1 × 1

\ n is divisible by 18 and 48. L.C.M = 144 with

digit sum 9.

4. (a) : AA NN RR DE

2 2 13

2

3

15

2 2270+ + ⇒

=!

! !

2 1 1 13

1

4

35

2720+ + + ⇒

=!

!

1 + 1 + 1 + 1 + 1⇒ 5! = 120

\ n = 270 + 720 + 120 = 1110, digit sum 3

5. (b) :

xdx ydy

x y dx

+

+

=2 2

∴ + = + = +x y x c y c cx 2 2 2 2 2, , parabolas

6. (a,b,d) : A4n + A4n + 1 = (4x )2

A4n + 2 + A4n + 3 = – (4n + 2)2

\ N = –22 + 42 – 62 + 82 – ...... + 20122

– 20142

= – 4[12 – 22 + 32 – 42 + ..... –10062 + 10072]

= –4[ –503 · 1007 + 10072]

= – 4 · 1007 · 504 = – 25 · 32 · 7 ·19 · 53

7. (a) : Let BC subtend angle 2a at the centre. Tus AB subtend angle 2p – 10 at the centre.2r sina = 81, 2r sin 5a = 31

sin sin sin sin531

8116 204 2α α α= ⇒ −

Solving, sinα =11

6

r = =

81

2

243

11sinα8. (c) : BE = 2r sin 3a

= − − =2243

113 4 1443( sin sin )α α

9. (3):5

5

3 5

3 2

3 1 2

3 2

1a

a

n

n

n

n

n

n

+=

++

= + +

+( )

∴ = + = ⇒ = 15 3 2 11an n aα α( ),∴ = + = +5 3 2 3 25

an

n n a n, log ( )a41 = 3

10. (a) : P. f x x x ( ) (cos sin )= + ×− −1

21 1

π

(cos–1x – sin–1x )

= −− −1

21 1

π(cos sin ),x x decreasing function

∴ − = − − = + =b a f f ( ) ( )1 13

4

1

41

Q. f x x ( ) sin= −

+

−2

4 1621

2 2

π

π π

b a− = − =

5

4

1

8

9

8R. f (x ) decreases b – a = f (–1) – f (1)

= + =9

8

1

8

5

4

S.

f x x ( ) sin= −

+

−3

2 4 482

12 2

π

π π

b a− = − =

7

8

1

32

27

32

nn

Solution Sender of Maths Musing

SET-144

1. N. Jayanthi : Hyderabad

2. Khokon Kumar Nandi : W.B.

3. Gouri Sankar Adhikari Mayta : W.B.


25/89


Q1. Prove that the equation sin6x + cos6x = a is

solvable if1

1≤ ≤a .

- Sujatha, Trissur

Ans. We have, sin6x + cos6x = a⇒ (sin2x + cos2x )3 – 3sin2x cos2x

(sin2x + cos2x ) = a

⇒ 1 – 3 sin2x cos2x = a ⇒ − =13

422sin x a

⇒ − −

=1

3

4

1 4

2

cos x a

⇒ − + =1 38

38

4cos x a

⇒ + =

5

8

3

8 4cos x a ...(i)But –1 ≤ cos4x ≤ 1 ⇒ − ≤ ≤

3

8

3

84

3

8cos x

⇒ − ≤ + ≤ +5

8

3

8

5

8

3

84

3

8

5

8cos x

⇒ ≤ ≤1

41a (by (i))

Q2. Find the equation of the plane passing throughthe point (–1, 2, 1) and perpendicular to theline joining the points (–3, 1, 2) and (2, 3, 4).

Find also the perpendicular distance of theorigin from this plane. - Mahendra, Amritsar

Ans. Te required plane passes through the point

(–1, 2, 1) having position vector

a i j k= − + +^ ^ ^

2

and is perpendicular to the line joining thepoints A(–3, 1, 2) and B(2, 3, 4).\ A vector normal to the plane is given by

n AB i j k i j k= = + + − − + +( ) ( )^ ^ ^ ^ ^ ^

2 3 4 3 2

= + +5 2 2i j k^ ^ ^

Vector equation of a plane passing through

point having position vector

a and normal to vector

n is given by

( ) . .

r a n r n a n− ⋅ = ⇒ =0\ Equation of the required plane is

r i j k i j k i j k.( ) ( ) .( )^ ^ ^ ^ ^ ^ ^ ^ ^

5 2 2 2 5 2 2+ + = − + + + +

⇒ + + − + +

r i j k.( )^ ^ ^5 2 2 5 4 2=

⇒ + + =

r i j k.( )^ ^ ^

5 2 2 1

We have | |

n = + + =5 2 2 332 2 2

In normal form

r i j k.^ ^ ^5

33

2

33

2

33

1

33+ +

=

So, the perpendicular distance of the origin

from the plane is1

33.

Q3. On each evening a boy either watchesDoordarshan channel or en sports. Te

probability that he watches en sports is45

. If

he watches Doordarshan, there is a chance of34

that he will fall asleep, while it is 14

when

he watches en sports. On one day, the boy isfound to be asleep. Find the probability that

the boy watched Doordarshan.- Anjali Jha, Patna

Ans. Let E1 and E2 be the events of the boy watchingDoordarshan and en sports, respectively. Itis given that

P E P E( ) and ( )1 215

45

= =

Let E be the event of the boy falls asleep. Againby hypothesis

P E

E

P E

E1 2

3

4

1

4

=

=and

Now E = E ∩ (E1 ∪ E2) = (E1 ∩ E)∪ (E2 ∩ E)

so that P E P E P EE

P E P EE

( ) ( ) ( )=

+

1 1

22

By Bayes’ theorem

P E EP E P E E

P E P E E P E P E E( / )

( ) ( / )

( ) ( / ) ( ) ( / )11 1

1 1 2 2

=+

=

×× + ×

=( / ) ( / )

( / ) ( / ) ( / ) ( / )

1 5 3 4

1 5 3 4 4 5 1 437

nn

Do you have a question that you just can’t getanswered?

Use the vast expertise of our mtg team to get to thebottom of the question. From the serious to the silly,the controversial to the trivial, the team will tackle thequestions, easy and tough.

The best questions and their solutions will be printed inthis column each month.


26/89


STANDARD RESULTS

EQUATION OF A CIRCLE IN VARIOUS FORM

Te equation of circle with centre (z h, k) andradius ‘r ’ is (x – h)2 + ( y – k)2 = r 2.

Te general equation of a circle isz

x 2 + y 2 + 2 gx + 2 fy + c = 0

with centre (– g , – f ) & radius = g f c2 2+ − .

Remember that every second degree equation

in x and y in which coefficient of x 2 = coefficient

of y 2 & there is no xy term always represents

a circle.

If g 2

+ f 2

– c > 0⇒

real circle. g 2 + f 2 – c = 0 ⇒ point circle.

g 2 + f 2 – c < 0 ⇒ imaginary circle.

Note that the general equation of a circle contains

three arbitrary constants, g , f and c which

corresponds to the fact that a unique circle passes

through three non-collinear points.

Te equation of circle with (z x 1, y 1) and (x 2, y 2)as end points of its diameter is,

(x – x 1)(x – x 2) + ( y – y 1)( y – y 2) = 0.

Note that this will be the circle of least radiuspassing through (x 1, y 1) and (x 2, y 2).

INTERCEPTS MADE BY A CIRCLE ON THE

AXES

Te intercepts made by the circle

x 2 + y 2 + 2 gx + 2 fy + c = 0 with the coordinate

axes are 2 22 2 g c f c− −and respectively.

Note :

If g 2 – c > 0 ⇒ circle cuts the x -axis at two

distinct points.If g 2 = c ⇒ circle touches the x -axis.

If g 2 < c ⇒ circle lies completely above or

below the x -axis.

POSITION OF A POINT W.R.T. A CIRCLE

Te point (x 1, y 1) lies inside, on or outside the

circle x 2 + y 2 + 2 gx + 2 fy + c = 0, according

as x 12 + y 1

2 + 2 gx 1 + 2 fy 1 + c < = or > 0

respectively.

Note : Te greatest and the least distance of a

point A from a circle with centre C and radius

r is AC + r and AC – r respectively.

AC

P Q

( )x , y 1 1

LINE AND CIRCLE

Let L = 0 be a line and S = 0 be a circle. If r

is the radius of the circle and p is the length

of the perpendicular from the centre on the

line, then

pz > r ⇔ the line does not meet the circle i.e.the line passes outside the circle.

p = r z ⇔ the line touches the circle.

p < r z ⇔ the line is a secant of the circle.

pz = 0 ⇒ the line is a diameter of the circle.

* ALOK KUMAR, B.Tech, IIT Kanpur

This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced,

etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here

are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).

He trains IIT and Olympiad aspirants.


27/89


PARAMETRIC EQUATIONS OF A CIRCLE

Te parametric equations of a circle

(x – h)2 +( y – k)2 = r 2 are,

x = h + r cosq; y = k + r sinq; –p < q ≤ p where (h, k) is the centre, r is the radius and

q is a parameter.Note that equation of a straight line joining two

points a & b on the circle x 2 + y 2 = a2 is

x y acos sin cosa b a b a b+

+

+

=

−

2 2 2

TANGENT & NORMAL

he equation of the tangent to the circlezx 2 + y 2 = a2 at its point (x 1, y 1) is,

xx 1 + yy 1 = a2. Hence equation of a tangent

at (acosa,asina) is; x cosa + y sina = a. Tepoint of intersection of the tangents at the

points P (a)and Q(b) is

a acos

cos

,

sin

cos

a b

a b

a b

a b

+

−

+

−

2

2

2

2

he equation of the tangent to the circlezx 2 + y 2 + 2 gx + 2 fy + c = 0 at its point (x 1, y 1)

is xx 1 + yy 1 + g (x + x 1) + f ( y + y 1) + c = 0 y z = mx + c is always a tangent to the circlex 2 + y 2 = a2 if c2 = a2 (1 + m2) and the point

of contact is −

a m

c

a

c

2 2

, .

If a line is normal / orthogonal to a circle, thenzit must pass through the centre of the circle.Using this fact normal to the circle

x 2 + y 2 + 2 gx + 2 fy + c = 0 at (x 1, y 1) is

y y

y f

x g x x − = +

+ −1 111( )

FAMILY OF CIRCLES

Equation of circle circumscribing a trianglezwhose sides are L1 = 0; L2 = 0 & L3 = 0 isgiven by; L1L2 + l L2L3 + mL3L1 = 0 providedcoefficient of xy = 0 & coefficient of x 2 =coefficient of y 2.

Equation of circle circumscribing a quadrilateralzwhose sides in order are represented by the

lines L1 = 0, L2 = 0, L3 = 0 and L4 = 0 is L1L3 + l L2L4 = 0 provided coefficient of x

2 = coefficient

of y 2 and coefficient of xy = 0.

Te equation of the family of circles passingzthrough the point of intersection of two circlesS

1 = 0 & S

2 = 0 is, S

1 + KS

2 = 0 (K

≠ –1).

Te equation of the family of circles passingzthrough the point of intersection of a circleS = 0 & a line L = 0 is given by S + KL = 0.

he equation of a family of circles passingzthrough two given points (x 1, y 1) and(x 2, y 2) is (x – x 1) (x – x 2) + ( y – y 1) ( y – y 2)

+ K x y

x y

x y

1

1

11 1

2 2

= 0, where K is a parameter.

Te equation of a family of circles touchingza fixed line y – y 1 = m (x – x 1) at the fixedpoint (x 1, y 1) is

(x – x 1)2 + ( y – y 1)

2 + K [ y – y 1 – m (x – x 1)]

= 0, where K is a parameter.

In case, if the line through (x 1, y 1) is parallel

to y -axis then the equation of the family of

circles touching it at (x 1, y 1) becomes

(x – x 1)2 +( y – y 1)

2 + K (x – x 1) = 0.

Also, if line is parallel to x -axis the equationof the family of circles touching it at (x 1, y 1)

becomes (x – x 1)2 + ( y – y 1)

2 + K ( y – y 1) = 0.

LENGTH OF A TANGENT AND POWER OF

A POINT

Te length of a tangent from an external point

(x 1, y 1) to the circle

S ≡ x 2 + y 2 + 2 gx + 2 fy + c = 0 is given by

L x y g x fy c S= + + + + =12

12

1 1 12 2

Square of length of the tangent from a pointP is also called the power of point w.r.t. acircle. Power of a point remains constantw.r.t. a circle.Note that power of a point P is positive,negative or zero according as the point ‘P ’ isoutside, inside or on the circle respectively.

DIRECTOR CIRCLE

Te locus of the point of intersection of twoperpendicular tangents is called the directorcircle of the given circle. Te director circle of


28/89


a circle is the concentric circle having radius

equal to 2 times the original circle.

EQUATION OF THE CHORD WITH A GIVEN

MIDDLE POINT

he equation of the chord of the circlez

S ≡ x 2 + y 2 + 2 gx + 2 fy + c = 0 in terms of itsmid point M (x 1, y 1) is y – y 1 = –

x g

y f

1

1

+

+(x – x 1).

Tis on simplication can be put in the formxx 1 + yy 1 + g (x + x 1) + f ( y + y 1) + c = x 1

2 + y 1

2 + 2 gx 1 + 2 fy 1 + c which is designated byT = S1.Note that the shortest chord of a circle passingthrough a point ‘ M ’ inside the circle, is onechord whose middle point is M .

CHORD OF CONTACTIf two tangents PT 1 & PT 2 are drawn from thepoint P (x 1, y 1) to the circle S ≡ x

2 + y 2 + 2 gx + 2 fy + c = 0, then the equation of the chordof contact of T 1T 2 isxx 1 + yy 1 + g (x + x 1) + f ( y + y 1) + c = 0.

REMEMBER

Equation of the circle circumscribing thezDPT 1T 2 is (x – x 1)(x + g ) + ( y – y 1)( y + f ) = 0.Te joint equation of a pair of tangents drawnz

from the point A(x 1, y 1) to the circlex 2 + y 2 + 2 gx + 2 fy + c = 0 is SS1 = T

2.

where, S ≡ x 2 + y 2 + 2 gx + 2 fy + c;S1 ≡ x 1

2 + y 12 + 2 gx 1 + 2 fy 1 + c

T ≡ xx 1 + yy 1 + g (x + x 1) + f ( y + y 1) + c.Area of the triangle formed by the pair of thez

tangents & its chord of contact =R L

R L

3

2 2+

where R is the radius of the circle & L is the

length of the tangent from (x 1, y 1) on S = 0Chord of contact exists only if the point ‘z P ’is not inside.

Length of chord of contactz T 1T 2 =2

2 2

L R

R L+.Angle between the pair of tangents fromz

(x 1, y 1) = tan–1

2

2 2

R L

L R−

where R = radius;

L = length of tangent.

POLE AND POLAR

If through a pointz P in the plane of the circle,two straight lines are drawn to meet the circleat points Q and R, the locus of the point ofintersection of the tangents at Q and R is calledthe polar of the point P ; also P is called the

pole of the polar.he equation of the polar of a pointz P (x 1, y 1)w.r.t. the circle x 2 + y 2 = a2 is given byxx 1 + yy 1 = a

2, and if the circle is general,then the equation of the polar becomesxx 1 + yy 1 + g (x + x 1) + f ( y + y 1) + c = 0.

Note that if the point (x 1, y 1) be on the circle,

then the chord of contact, tangent and polar

will be represented by the same equation.

Pole of a given linez Ax + By + C = 0 w.r.t. any

circle x 2 + y 2 = a2 is − −

Aa

C

Ba

C

2 2

, .

If the polar of a pointz P pass through a pointQ, then the polar of Q passes through P .

wo linesz L1 and L2 are conjugate of each otherif pole of L1 lies on L2 and vice-versa. Similarlytwo points P and Q are said to be conjugateof each other if the polar of P passes throughQ and vice-versa.

COMMON TANGENTS TO TWO CIRCLESTe direct common tangents meet at a pointzwhich divides the line joining centre of circlesexternally in the ratio of their radii.

ransverse common tangents meet at a pointzwhich divides the line joining centre of circlesinternally in the ratio of their radii.

When they touch each other :z

(a) Externally : here are three common

tangents, two direct and one is the tangentat the point of contact .

(b) Internally : Only one common tangent

possible at their point of contact.

When two circles neither intersect nor touchzeach other , there are four common tangents,two of them are transverse and the others aredirect common tangents.

When they intersect there are two commonztangents, both of them being direct.


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Length of an external common tangent andzinternal common tangent to the two circlesis given by:

L d r r L d r r ext = − − = − +2

1 22 2

1 22( ) & ( )int

where d = distance between the centres of the

two circles, r 1 and r 2 are the radii of the twocircles.

RADICAL AXIS & RADICAL CENTRE

Te radical axis of two circles is the locus ofzpoints whose powers w.r.t. the two circles areequal. Te equation of radical axis of the twocircles S1 = 0 and S2 = 0 is S1 – S2 = 0

i.e. 2( g 1 – g 2)x + 2( f 1 – f 2) y + (c1 – c2) = 0.

REMARKS

Radical axis bisects the common tangentzbetween the two circles.

Te common point of intersection of the radicalzaxes of three circles taken two at a time is calledthe radical centre of three circles.

A system of circles , every two of which havezthe same radical axis, is called a co-axialsystem.

Pairs of circles which do not have radical axiszare concentric.

If two circles intersect, then the radical axis iszthe common chord of the two circles.

If two circles touch each other, then the radicalzaxis is the common tangent of the two circlesat the common point of contact.

Radical axis is always perpendicular to the linez joining the centres of the two circles.

Radical axis need not always pass through thezmid point of the line joining the centres of thetwo circles.

ORTHOGONALITY OF TWO CIRCLES

wo circlesz S1= 0 & S2 = 0 are said to beorthogonal or said to intersect orthogonallyif the tangents at their point of intersectioninclude a right angle. Te condition for twocircles to be orthogonal is 2 g 1 g 2 + 2 f 1 f 2 = c1 + c2.

REMARKS

If two circles are orthogonal, then the polar ofza point ‘P ’ on first circle w.r.t. the second circle

passes through the point Q which is the otherend of the diameter through P . Hence locusof a point which moves such that its polarsw.r.t. the circles S1 = 0, S2 = 0 and S3 = 0 areconcurrent in a circle which is orthogonal toall the three circles.

Locus of the centre of a variable circlezorthogonal to two fixed circles is the radicalaxis between the two fixed circles .

PROBLEMS

SECTION-I

Single Correct Answer Type

1. Te locus of the mid points of the chords of the

circle x 2 + y 2 – ax – by = 0 which subtend a right

angle at a b2 2

, is

(a) ax + by = 0 (b) ax + by = a2 + b2

(c) x 2 + y 2 – ax – by +a b2 2

8

+ = 0

(d) x 2 + y 2 – ax – by –a b2 2

8

+ = 0

2. A rhombus is inscribed in the region common

to the two circles x 2 + y 2 – 4x – 12 = 0 and

x 2 + y 2 + 4x – 12 = 0 with two of its vertices on the

line joining the centres of the circles. Te area of the

rhombus is

(a) 8 3 sq.units (b) 4 3 sq.units

(c) 16 3 sq.units (d) none of these

3. In a right triangle ABC , right angled at A, on

the leg AC as diameter, a semicircle is described.

Te chord joining A with the point of intersection

D of the hypotenuse and the semicircle, then the

length AC equals to

(a) AB AD

AB AD

⋅

+2 2 (b)

AB AD

AB AD

⋅+

(c) AB AD⋅ (d) AB AD

AB AD

⋅

−2 2

4. Te locus of the mid points of the chords of the

circle x 2 + y 2 + 4x – 6 y – 12 = 0 which subtend an

angle ofp

3

radians at its circumference is


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(a) (x – 2)2 + ( y + 3)2 = 6.25(b) (x + 2)2 + ( y – 3)2 = 6.25(c) (x + 2)2 + ( y – 3)2 = 18.75(d) (x + 2)2 + ( y + 3)2 = 18.75

5. Te angle at which the circles (x – 1)2 + y 2 = 10

and x

2

+ ( y – 2)

2

= 5 intersect is(a)

p6

(b)p4

(c)p3

(d)p2

6. Te value of 'c' for which the set,{(x , y )|x 2 + y 2 + 2x ≤ 1} ∩ {(x , y )| x – y + c ≥ 0}contains only one point in common is(a) (–∞, – 1] ∪ [3, ∞) (b) {– 1, 3}(c) {– 3} (d) {– 1 }

7. P is a point (a, b) in the first quadrant. If thetwo circles which pass through P and touch both

the coordinate axes cut at right angles, then(a) a2 – 6ab + b2 = 0 (b) a2 + 2ab – b2 = 0(c) a2 – 4ab + b2 = 0 (d) a2 – 8ab + b2 = 0

8. Tree concentric circles of which the biggestis x 2 + y 2 = 1, have their radii in A.P. If the line

y = x + 1 cuts all the circles in real and distinct points.Te interval in which the common difference of theA.P. will lie is

(a) 01

4

,

(b) 0

1

2 2

,

(c) 02 2

4,

−

(d) none of these

9. B and C are fixed points having coordinates(3, 0) and (–3, 0) respectively . If the vertical angleBAC is 90º, then the locus of the centroid of theD ABC has the equation(a) x 2 + y 2 = 1 (b) x 2 + y 2 = 2(c) 9(x 2 + y 2) = 1 (d) 9(x 2 + y 2) = 4

10. If two chords, each bisected by the x -axis canbe drawn to the circle, 2(x 2 + y 2) – 2ax – by = 0(a ≠ 0, b ≠ 0) from the point (a, b/2), then(a) a2 > 8b2 (b) b2 > 2a2 (c) a2 > 2b2 (d) a2 = 2b2

11. angents are drawn to a unit circle with centreat the origin from a point on the line 2x + y = 4.Ten the equation of the locus of the middle pointof the chord of contact is

(a) 2(x 2 + y 2) = x + y (b) 2(x 2 + y 2) = x + 2 y (c) 4(x 2 + y 2) = 2x + y (d) none of these

12. Te common chord of two intersecting circlesC 1 and C 2 can be seen from their centres at theangles of 30° and 60° respectively . If the distance

between their centres is equal to 3 1+ , then theradii of C 1 and C 2 are(a) 3 3, (b) 2 2 2, (c) 2 2, (d) 2 2 4,

13. If the line x cosq + y sinq = 2 is the equationof a transverse common tangent to the circles

x 2 + y 2 = 4 and x 2 + y 2 – 6 3 x – 6 y + 20 = 0, thenthe value of q is(a) 5p/6 (b) 2p/3 (c) p/3 (d) p/6

14. Te locus of the mid-points of the chords of

the circle x 2

+ y 2

�

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