(agra region) session ending examination 2017-18 class … · find the ter m independent of x in...

KENDRIYA VIDYALAYA SANGATHAN(AGRA REGION)

SESSION ENDING EXAMINATION 2017-18 CLASS XI

SUBJECT- MATHEMATICS(SOLVED PAPER)

Time Allowed: 3 hrs. Max. Marks: 100

General Instruction: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided in four sections. A, B, C and D. Section − A comprises of 4 questions

of one mark each, Section−B Comprises of 8 Questions of 2 marks each, Section−C Comprises of 11 Questions of 4 marks each and Section − D is of the 6 questions of 6 marks each.

3. All questions in Section − A are to be answered in word, one sentence or as per the exact requirement of the question.

Section -A 1. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 5, 6}, B = {2, 3, 4, 7, 8}, find (A ∩ B)’ 2. Write the negation of the following statement: “Every natural number is greater than zero” 3. How many chords can be drawn through 21 points on a circle? 4. Find the centre and radius of circle (x − 2)2 + (y + 7)2 = 36

Seation -B 5. Draw appropriate Venn-diagram of (A ∪ B)’

6. Let A= {1, 2} and B= {3, 4}. Write A × B. How many subsets will A × B have?

7. Express ( 5 − 3i)3 in the form of a + i b.

8. Expand (2x − 3)4 using Binomial Theorem.

9. Find the sum of n ≠0 terms of the series whose nth term is tn = n2 +3n

10. Evaluate limsin

sinx

ax bx

ax bx®

++

æèç

öø÷0 , a, b, a+b≠0

11. Write the contrapositive and converse of the statement:- If you are born in India, then you are a citizen of India.

12. An experiment consists of tossing a coin and then throwing it second time if a head occurs . If a tail occurs on the first toss, then a die is rolled once. Find the sample space for this experiment.

Section -C 13. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the

locus of a point P on the rod, which is 3 cm from the end in contact with the x−axis.

14. Solve for x : 2cos2 x + 3 sin x =0

15. Convert the complex number Zi

i=

-+

1 7

2 2( ), into polar form.

16. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonant can be formed from the alphabet?

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Oswaal CBSE Solved Paper - 2018, MATHEMATICS, Class-XI [ 7


17. Find the term independent of x in the expansion of 3

2

1

32

6

xx

-æèç

öø÷ .

18. Prove that: tantan tan

tan tan4

4 1

1 6

2

2 4x

x x

x x=

-( )- +

OR Show that: tan 3x tan 2x tan x = tan3x − tan 2x − tan x 19. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show

that: 1 1 1

2 2 2p a b= +

OR Find the image of the point (3, 8) with respect to the line x+3y = 7 assuming the line to be a plane mirror.

20. Find the domain and range of the real function f x( ) = -x2 9 .

21. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, −1).

22. Find the derivative of following function

(i) y x= -4 2 (ii) y ax bn

= +( )

OR

Find the derivative of f(x) = x sin x from first principle.

23. Out of the 100 students, two sections of 40 and 60 are formed . If you and your friend are among the 100 students, what is the probability that

(a) You both enter the same section?

(b) You both enter the different section?

Section -D 24. In a survey of 100 students, the number of students studying the various languages was found to be as follows:

English only 18, English but not Hindi 23, English and Sanskrit 8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24. Find

(i) How many students were studying Hindi?

(ii) How many students were studying English and Hindi?

25. Prove that:

cos cos cos2 2 2

3 332

x x x+ +æèç

öø÷ + -æ

èç

öø÷ =

p p

OR

Prove that:

sin sin sin sin10 50 60 70

316

° ° ° ° =

26. Using the principle of mathematical induction, prove that 3 8 92 2n n+ - - is divisible by 8 for all n Î N.

OR

Using the principle of mathematical induction, prove that

11 4

14 7

17 10

13 2 3 1 3 1. . .

+ + + +-( ) +( )

=+( )

n n

nn

, for all n Î N

27. Solve the following system of linear equations graphically:

3 2 150 4 80 15 0x y x y x x y+ £ + £ £ ³, , , ,


8 ] Oswaal CBSE Solved Paper - 2018, MATHEMATICS, Class-XI


28. The ratio of the A.M. and G. M. of two positive numbers a and b is m:n, show that

a b m m n m m n: := + -( ) - -( )2 2 2 2

29. Find the mean and variance for the following date:

Classes 0−30 30−60 60−90 90−120 120−150 150−180 180−210Frequencies 2 3 5 10 3 5 2

OR The mean of 5 observations is 6 and the standard deviation is 2. If the three observations are 5, 7 and 9. Find the

other two observations.




Section-A 1. Given, U = 1, 2, 3, 4, 5, 6, 7, 8, 9

A = 1, 2,

{ } 3, 5, 6

B = 2, 3, 4, 7, 8

A B = 1, 2, 3, 5, 6

{ }{ }

\ {{ } { }{ }

( ) È ( )

2, 3, 4, 7, 8

A B = 2, 3

Now, A B '= A B

-

= 1, 2, 3, 4, 5, 6, 7, 8, 9 2,3

{ } -{ } = 1, 4, 5, 6, 7, 8, 9{ }

½

2. Negation: It is false that every natural number is greater than 0.

or There exists a natural number which is not greater

than 0. ½ 3. A chord is obtained by joining any two points on a

circle. Therefore, total no.of chords drawn through 21 points is same as the no.of ways of selecting 2 points out of 21 points. This can be done in 21C

2

ways.

Hence, total no.of chords 212C =

21!19! 2!

= 21 10

= 210

´ 1 4. We have,

x y

x y

-

- - -

2 + +7 = 36

2 + 7 = 6

2 2

2 2 2

( ) ( )Þ ( ) ( )( )

Comparing this equation with x y r- -a + b =2 2 2( ) ( ) ,

we find that given circle has its centre at (2, −7) and radius is 6. 1

Section B 5.

AB

∪

2

Shaded area is (A∪B)’ 6. Given,

A = 1, 2 and B = 3, 4

A B = 1, 3 , 1, 4 , 2, 3 , 2, 4

{ } { }\ ´ ( ) ( ) ( ) ( ){ }\nn A B = n A n B

= 2 2

´( ) ( )´ ( )´

= 4 1

No. of subsets of A×B = 2

= 2 =16

A×B

4

n( )

1

7. We have,

5 3

= 5 3 5 3

= 25 30 +9 5 3

= 16 30 5 3

= 80

3

2

2

-

- -

- -

- -

i

i i

i i i

i i

( )( ) ( )( )( )( )( )

-- -- -

- -

48 150 +90= 80 198 90= 10 198

2i i ii

i 1

8. Using binomial theorem, we have

2x 3

= C 2x 3 + C (2x) 3 + C 2x 3

4

40

4 0 4 3 1 4 2 2

-

- - -

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

+ C 2x 3 + C 2x 3

=16x +4 8x 3 +6 4x

4 1

34 0 4

4 3 2

3 4( ) ( ) ( ) ( )( )( ) (

- -

- ))( ) ( )( )9 +4 2x 27 +81

=16x 96x +216x 216x+814 3 2

-

- -

9. Given,

t = n +3nThe sum of n terms is

S = t = n +3n

n2

n n2

n=1

n

n=1

n

( )ååå

åå( ) ( ) = n +3 n

2

n=1

n

n=1

n

==n n+1 2n+1

6+3

n n+12

=16

n n+1 2n+1 +

( )( ) ( )

( )( ) 99n n+1

=16

n n+1 2n+1 +9

( )éë ùû

( ) ( )éë ùû

=16

n n+1 2n+10

=13

n n+1 n+5

T

( )( )

( )( )

hhus,required sum is 13

n n+1 n+5( )( )

1

10. limsin ax+ bxax+sin bx

= lima sin ax

ax+ b

a+ b sin bxb

x 0

x 0

®

®

æèç

öø÷

xx

=a 1+ ba+ b 1

limsin

=a+ ba+ b

=1

0

æèç

öø÷

´´

=æ

èç

ö

ø÷®

q

qq

1

1

11. Contrapositive statement: If you are not born in India, then you are not a citizen of India. 1

1

SOLUTIONS

1

1

1

1

1

1+1




Converse Statement: If you are a citizen of India then you are born in India. 1

12. The sample space S for tossing a coin and then tossing it second time if a head occurs, if a tail occurs on the first toss, the dice is tossed once, then

S= { HH, HT, T1, T2, T3, T4, T5, T6 }

Here, T1 means tail T on the coin and the number 1 on the die. 2

Section C 13. Let AB be the rod where A touches the x−axis and

B touches the y−axis

B

Q�

�y

x

OX'

Y'

X

Y

R A

p(x,y)

3 cm

9 cm

1 Let point P(x, y) Given, AB = Length of rod = 12 cm and AP = 3cm PB = AB – AP = 12−3 = 9 cm Drawing PQ ^ BO and PR ^ OA Hence, PQ =x and PR = y Let ÐPAR = Q Now PQ & AQ are parallel lines and BA is the

transversal. So, ÐBPQ = Ð PAR = q (Corresponding angles) 1

In right triangle BPQ

cos =

cos = PQBP = x

9

Base

Hyp

D

q

q

In right triangle PAR

sin =

= PRAP = y

9

perp

Hyp

D

q

qsin 1

We know that, sin2 q + cos2 q = 1

Putting, sin =y3

and cos =x9

, we get

y3

+ x9

=1

y9

+x81

=

2 2

2 2

q q

æèç

öø÷

æèç

öø÷

Þ 11 x81

+y9

=12 2

Þ

Hence, it satisfies the equation of ellipse 2 2

2 2

x y+ =1

a b

Thus, locus of p is ellipse. 1

14. We have,

2 cos x+3sinx = 0

2 1 sin x +3sinx = 0

2sin x 3sinx 2 = 0

2s

2

2

2

-

- -

( )

iin x 4sinx+sinx 2 = 0

2sinx sinx 2 +1 sinx 2 = 0

sinx 2 2si

2 - -

- -

-

( ) ( )( ) nnx+1 = 0( )

1

2sin +1= 0 sin 2 sin 2 0

sin =12

sin = sin6

= n +

q q q -

q -

q-p

q p -

\ ¹ \ ¹[ ]

æèç

öø÷

116

,n Zn( ) æ

èç

öø÷ Î

-p 1

15.

1 7

2+=

1 74+4 +

=1 73+4

=1 73+4

×3 43 4

=3 21 4

2 2

- -

-

- --

- -

i

i

ii i

iiii

ii

i i

( )

++289 16

=25 25

25= 1

2

2

ii

i

i

-- -

- - ½ Put r cos q = −1 and r sin q = −1 ½ Squaring and adding

r =1+1= 2

r = 2

cos =12

and sin =12

2

q-

q-

½ sin q and cos q both are −ve, so q lies in III quadrant

\ æ

èç

öø÷q - p -

p-

p=

4=

34

½

Polar form of

1 7

2+= 2 cos

34

+ sin 34

= 2 cos34

2

--

p-

p

p

i

ii

( )æèç

öø÷

æèç

öø÷

ìíî

üýþ

ææèç

öø÷

æèç

öø÷

ìíî

üýþ

-p

i sin34 ½

16. There are 5 vowels and 21 consonants in English alphabets.

2 vowels can be chosen in 5C2 ways. 2 consonants can be chosen in 21C2 ways. 4 letters can be arranged in 4! ways. 2

½

½

½

½

½

1




The number of words consisting of 2 vowels and 2 consonants 1

= C × C ×4!

=5×41×2

× 21×201×2

× 24

=10×210×24

= 50400

52 2

21

1

17. General term in the expansion of 32

21

3

6

xx

-éëê

ùûú

T = C32

1

3

= 1 C3

3 2

r+16

r2

6-r r

r 6r

6-r

r 6

xx

æèç

öø÷

æèç

öø÷

( ) ( )( ) ( )

-

- --r12-2r

r

r 6r

6-2r

6-r12-3r

x1

= 1 C 3

2

x

x-( ) ( )( )

Thus, the term will be independent of x, if the index is zero, i.e.,

12 − 3r = 0 Þ r = 4 1 Hence, 5th term is independent of x and is given by,

T = 1 C3

2

=5

12

54 6

4

6-8

6-4-( ) ( )( )

1

18 We have,

LHS = tan4x = tan2 2x

LHS =2tan2x

1 tan 2x =

2 2tanx1 tan x

2

2

( )æèç

--

ööø÷

æèç

öø÷

( )( )

1 2tanx1 tan x

= 4tanx 1 tan x

1 tan x

2

2

2

2

--

-

-22 2

2

2 4

4tan x

=4tanx 1 tan x

1 6 tan x+tan x

-

-

-

( )( )

= RHS 1

OR We have,

3 = 2 +

tan3 = tan 2 +

tan3 =tan2 +tan

1 tan2 tant

x x x

x x x

xx x

x x

( )

Þ

Þ-

aan3 1 tan2 tan = tan2 +tan

tan3 tan3 tan2 tan = tan

x x x x x

x x x x

-

-

( )Þ 22 +tan

tan3 tan2 +tan = tan3 tan2 tan

tan3 tan2 tan

x x

x x x x x x

x x

Þ

Þ

-

xx x x x= tan3 tan2 tan- -

1

Hence proved

19. Equation of line where intercept on the axes are a

and b is x y+ =1

a b ½

Distance from origin (0,0) to the line x y

+ =1a b

in p

So, distance d=p & x1 = 0 and y1 = 0 Putting values

Perpendicular distance (d)=Ax +By +C

A +B

p = 0 1

a+0 1

b

1 1

2 2

æèç

öø÷

æèèç

öø÷

æèç

öø÷

æèç

öø÷

Þ

Þ

--

-

1

1a

+ 1b

p = 0+0 1

1a

+ 1b

p = 1

1a

+ 1b

2 2

2 2

2 22

2 2

2 2 2 2

p =1

1a

+ 1b

p =1a

+1b

1p

=1a

+1b

Þ

Þ

1

Squaring both sides, we get

1=

1a

+1b

1=

1a

+1b

2

2 2

2 2 2

p

p

æ

èç

ö

ø÷

æ

èçç

ö

ø÷÷

2

1 Hence proved

OR Let line AB be x+3y=7 & point P be (3, 8)

let Q (h, k) be the image of point p(3, 8) in the line AB x+3y=7

X' X

Y'

R

Q(h, k)B

x+y=

3

7

AP(3,8)

O

Y

1 Since line AB is mirror 1. Point P & Q are at equal distance from line AB, i.e.

PR = QR, i.e R is the mid point of PQ. 2. Image is formed perpendicular to mirror i.e. line

PQ is perpendicular to line AB Since, R is the mid point of PQ We know that, Mid point of a line joining

1

1

1

1

1

1

1

½




x , y & x , y =

x +x2

, y +y

21 1 2 21 2 1 2( ) ( ) æ

èç

öø÷

Mid point of PQ joining

3,8 & h, k is

3+h2

, 8+k

2( ) ( ) æ

èç

öø÷

Coordinate of point R =3+h

2,

8+k2

æèç

öø÷

Since, point R lies on the line AB, it will satisfy the equation of line AB.

Putting x =3+h

2 and y =

8+k2

in equation of line AB ½

3+

2+3

8+2

=7

3+ + 3 8+

2=7

h h

h h

h

æèç

öø÷

æèç

öø÷

( )

++3 = 13h - (i) ½

Also, PQ is perpendicular to AB we know that, If two lines are perpendicular then product of their

slopes is equal to −1

Slope of AB x slope of PQ = −1

Slope of PQ= 1

slope of AB

-

So, slope of PQ= 113

= 3--æ

èç

öø÷

½

\ Þé

ëêùûú

x+3y =7 y =13

x+73

-

½

Now, line PQ is joining point P(3, 8) & Q (h, k)

Slope of PQ=y yx x

3 =k 8h 3

2 1

2 1

--

--

3h k =1- (ii) ½ from eq. (1)

h+3k = 13or h = 13 3k

-- -

putting value of h in eq. (ii), we get

3h k =1

3 13 3k k =1

k = 4

-

- - -

-( )

½

putting k = −4 in eq.(i), we get

h+3k = 13h+3( 4)= 13 h = 1

-- -

- ½

Hence, Q (h, k) = Q (−1, −4) Therefore, image is Q (−1, −4).

20. Given function f x = x 92( ) - for the domain, we require

x 9 0

x 9x 3 or x 3

2

2

-

-

³

Þ ³Þ £ ³

Domain is -¥ -( ] +¥[ ), ,3 3 2

Range is y Î R, y≥0 2

21. Let A(1, 2, 3) & B (3, −2, −1)

Let point P be (x, y, z)

Since, it is given that point P (x, y, z) is equal distance from point A (1, 2, 3) & B (3, 2, 1)

i.e., PA = PB 1

1 2 3 3 2 12 2 2 2 2 2

-( ) + -( ) + -( ) = -( ) + -( ) + +( )x y z x y z

Squaring both sides

1 + 2 + 3 z = 3 + 2 + 1+z

1+ 2 + 4+ 4

2 2 2 2 2 2

2 2

- - - - -

- -

x y x y

x x y y

( ) ( ) ( ) ( ) ( ) ( )( ) (( ) ( )

( ) ( ) ( )+ 9+z 6z

= 9+ 6 + 4+ 4 + 1+z +2z

+ +z

2

2 2 2

2 2

-

- -x x y y

x y 22

2 2 2

2 4 6z+14

= + +z 6 4 +2z+142 4 6z = 6

- - -

- -- - - - -

x y

x y x yx y x 44 +2z

2 +6 6z 2z = 0

4 2z = 0

2z = 0,

yx x

x

x

- - -

-

-( )

which is the required eequation

22. (i) y = 4 2dyd

=d

d4 2

= 4×12

12

12

-1

x

x xx

x

-

-\ ( )éëê

ùûú

- 0

= 2d

d= n

=2

- 12 n n-1x

xx x

x

\éëê

ùûú

1

(ii) y = ax+ b

dydx

=d

dxax+ b

= n ax+ bd

dxax+ b

ddx

n

n

n-1

( )

\ ( )

( ) ( )

xx = nx

= n ax+ b ×a

= na ax+ b

n n-1

n-1

n-1

( )éëê

ùûú

( )( )

½

1

1

1

1

1

1

1




OR

Given function, f(x) = xsin x, then by first principle

f' = limf + f

= lim+ sin + sin

= lim

h 0

h 0

h 0

xx h x

h

x h x h x xh

( ) ( ) ( )

( ) ( )®

®

®

-

-

xx x h x

hx h

x x h x

sin + sin+sin +

= limsin + sin

h 0

( ){ } ( )é

ëêê

ù

ûúú

( ){ }®

-

-

hhx h

x x h h

h

é

ëêê

ù

ûúú

( )

æèç

öø÷

é

ë

êê

®

®

+lim sin +

= lim2 cos +

2sin

2

h 0

h 0 êêê

ù

û

úúúú

æèç

öø÷ ´

( )( )® ®

+sin

= ×lim cos +2

limsin 2

2+s

h 0 h 0

x

x xh h

hiin

= cos 1+sin

= cos +sin

x

x x x

x x x

´

1 23. My friend and I are among the 100 students.

Total no. of ways selecting 2 students out of 100 students = 100C2 1

(a) The two of us will enter the same section if both of us are among 40 students or among 60 students.

Number of ways in which both of us enter the same section = 40C2 + 60C2

Probability that both of us enter the same section

=C + C

C

=

40!2! 38!

+ 60!2! 58!

100!2! 98!

=(39×40)+(59×

402

602

1002

660)99×100

=1733

1

(b) P (we enter different section) = 1 –P (we enter the same section)

=11733

=1633

-

1

Section D 24. Let U denotes the set of surveyed students. Let E, H

and S denote the set of students who are studying English, Hindi and Sanskrit, respectively.

Then, n(È) = 100, n( E) = 26, n(S) = 48, n(E S) = 8 and n (S H) = 8 1

3 55

18

35

HE

SU=100 1 from the venn – Diagram, it is clear that n (E H S) =3 1 The number od students who study english only = 18 The number of students who study no language = 24 Number of students who study Hindi only =

100 18+5+3+5+35 24

=100 66 24=100 90=10

- -

- --

( )éë ùû

1 Number of students who study Hindi = 10+3+5= 18 1 and Number of students who study English and

Hindi = 3 1 25.

LHS = cos x+cos x+3

+cos x3

=1+cos2x

2+

1+c

2 2 2p-

pæèç

öø÷

æèç

öø÷

æèç

öø÷

oos2 x+3

2+

1+cos2 x3

2

p-

pæèç

öø÷

é

ë

êêêê

ù

û

úúúú

æèç

öø÷

é

ë

êêêê

ù

û

úúúú

cos2x = 2cos x 1 or

cos x =1+cos2x

2

2

2

-é

ë

êêê

ù

û

úúú

=12

1+cos2x+1+cos2 x+3

+1+cos2 xp

-æèç

öø÷

é

ëê

ù

ûú

pp

p-

p

3

=12

3+cos2x+cos2 x+3

+cos2 x3

æèç

öø÷

æèç

öø÷

æèç

öø÷

é

ëê

ù

ûú

Usinng cosx+cosy = 2cosx+y

2cos

x y2

Replace x by 2x+2

æèç

öø÷

æèç

öø÷

-

p33

and y by 2x23

=12

3+cos2x+2cos2x+ 2

3+2x 2

32

æèç

öø÷

æ

è

ççç

-p

p-

p

çç

ö

ø

÷÷÷÷

é

ë

êêêê

æèç

öø÷

æ

è

çççç

ö

cos 2x+ 2

32x 2

32

p- -

p

øø

÷÷÷÷

ù

û

úúúú

1

1

1

1

1

1

1




=12

3+cos2x+cos4x2

cos432

=12

3

æèç

öø÷

æ

è

çççç

ö

ø

÷÷÷÷

é

ë

êêêê

ù

û

úúúú

p

++cos2x+2cos2x cos23

=12

3+cos2x+2cos2x cos

p

p -p

æèç

öø÷

é

ëê

ù

ûú

33

=12

3+cos2x+2cos2x cos3

u

æèç

öø÷

é

ëê

ù

ûú

æèç

öø÷

æ

èç

ö

ø÷

é

ëê

ù

ûú-

p

ssing cos = cos

=12

3+cos2x+2cos2x12

=1

p - q - q

-

( )éë ùûæèç

öø÷

é

ëê

ù

ûú

223+cos2x cos2x

=32

= RHS

-[ ]

1 Hence, proved.

OR Given,

sin10° sin50° sin60° sin70°

= sin60° sin10° sin50° sin70°

=

( )33

2sin10° sin50° sin70°

=3

42sin10° sin50° sin70°

=3

4cos4

( )

( )

00° cos60° sin70°

2 sinA sinB = cos A B cos A+B

=3

4co

-

- -

( )

( ) ( )éë ùû

ss40° sin70° cos60° sin70°

=3

4cos40° sin70°

12

sin70°

-

-

( )

æèç

öøø÷

( )

=3

4 cos40° sin70°

38

sin70°

=3

82 cos40° sin70°

38

sin70°

-

-

==3

8sin110° +sin30°

38

sin70°

2 cosAsinB = sin A+B sin A B

( )

( ) ( )

-

- -ééë ùû

( )

( )

=3

8sin110° sin70° +

38

sin30°

=3

82 cos90° cos20° +

38

12

-

ææèç

öø÷

( ) ( )éë ùû

´ ´( )

2 cosA cosB = cos A+B +cos A B

=3

82 0 cos20° +

31

-

66

= 0+3

16=

316

= RHS

1

Hence Proved.

26. Let the given statement be P (n), i.e., P(n): 32x+2 −8n−9 is divisible by 8. It can be observed that P(n) is true for n = 1 since,

32x1+2 – 8 x 1 – 9 = 64, which is divisible by 8. Let P(k) be true for some positive integer k, i.e.,

32k+2 – 8k – 9 is divisible by 8 1

32k+2 – 8k – 9 = 8m, where m Î M (i) We shall now prove that P(k+1) is true whenever

P(k) is true. 1 Consider,

3 8 +1 9

= 3 . 3 8 8 9

= 3 3 8 9+8 +9

2 k+1 +2

2k+2 2

2 2k+2

( ) ( )

( )

- -

- - -

- - -

k

k

k k 88 17

= 3 3 8 - 9 +3 8 +9 8 17

= 3 .8 +9 8 +9 8 17

= 9

2 2k+2 2

2

k

k k k

m k k

-

- - -

- -

( ) ( )( )

.8 +72 +81 8 17= 9 .8 +64 +64

= 8 9 +8 +8 8r

where r = 9 +8

m k km k

m k

m k

- -

( ) =

++8 is a ( ) natural number.

Therefore, 32(k+1)+2 – 8 (k+1) – 9 is divisible by 8. Thus, P(k+1) is true whenever P(k) is true. 3 Hence, by the principle of mathematical induction,

statement P(n) is true for all natural numbers i.e. ‘n. 1

OR Let

P n :1

1.4+

14.7

+1

7.10+ +

13n - 2 . 3n+1

=n

3n+1

for n =1

LHS =1

1.4

( ) ( ) ( )

==14

RHS =1

3 1 +1=

13+1

=14( )( )

Hence, LHS = RHS P(n) is true for n = 1 1 Assume P(k) is true

11.4

+1

4.7+

17.10

+ +1

3k - 2 3k +1=

k3k +1

( )( )

(i) 1

We will prove that P (k+1) is true

R.H.S = +1

3 +1 +1=

+13 +4

L.H.S =1

1.4+

14.7

+1

7.10+ +

13 +

k

kkk

k

( )( )( )

11 2 3 +1 +1

=1

1.4+

14.7

+1

7.10+ +

13 +3 2 3 +3+1

=1

1.

( ){ } ( ){ }

( )( )

-

-

k

k k

44+

14.7

+1

7.10+ +

13 +1 3 +4

=1

1.4+

14.7

+1

7.10+ +

13 2 3 +

k k

k k

( )( )

( )- 11+

13 +1 3 +4

=3 +1

+1

3 +1 3 +4

=1

3 +1 1+

13 +4

( ) ( )( )

( )( )æèç

öø

k k

kk k k

kk

k ÷÷

( )( )é

ëê

ù

ûú

( )æ

èç

ö

ø÷

=1

3 +1k 3 +4 +1

3 +4

=1

3 +13 +4 +1

3 +4

=1

3

2

kkk

kk k

k

kkk k

k

kk

+13 +1 +1

3 +4

=+1

3 +4= R.H.S

( )( )( )

( )é

ëêê

ù

ûúú

1

1

1

1

1

1

1




R.H.S = +1

3 +1 +1=

+13 +4

L.H.S =1

1.4+

14.7

+1

7.10+ +

13 +

k

kkk

k

( )( )( )

11 2 3 +1 +1

=1

1.4+

14.7

+1

7.10+ +

13 +3 2 3 +3+1

=1

1.

( ){ } ( ){ }

( )( )

-

-

k

k k

44+

14.7

+1

7.10+ +

13 +1 3 +4

=1

1.4+

14.7

+1

7.10+ +

13 2 3 +

k k

k k

( )( )

( )- 11+

13 +1 3 +4

=3 +1

+1

3 +1 3 +4

=1

3 +1 1+

13 +4

( ) ( )( )

( )( )æèç

öø

k k

kk k k

kk

k ÷÷

( )( )é

ëê

ù

ûú

( )æ

èç

ö

ø÷

=1

3 +1k 3 +4 +1

3 +4

=1

3 +13 +4 +1

3 +4

=1

3

2

kkk

kk k

k

kkk k

k

kk

+13 +1 +1

3 +4

=+1

3 +4= R.H.S

( )( )( )

( )é

ëêê

ù

ûúú

Hence, LHS = RHS P(k+1) is true whenever P(k) is true. 3 By principle of mathematical induction, P(n) is

true for n, where nÎ N. 1 27. Given in all equations are 3x+2y≤ 150 (1) x + 2y ≤ 80 (2) x ≤ 15 (3) x ≥ 0 (4) y ≥ 0 (5) Graph of inequality (1). Let us draw the graph of

the line 3x+2y = 150 At y = 0 Þ 3x+2(0) =150 Þ x = 50 and at x = 0 Þ 3 (0) + 2y = 150 Þ y = 75 (50, 0) and (0, 75) are the points on the line 3x+2y = 150 putting x = y = 0 in (1), we have 0 ≤ 150, which is

true. Hence, half – plane region containing the origin is

the solution region of this inequalities. Graph of inequality (2). Let us draw the graph of

the line x+4y = 80. At y = 0 Þ x +4 (0) = 80 Þ x = 80 At x = 0 Þ 0+4y = 80 Þ y = 20 (80, 0) and (0, 20) are the points on the line

x+4y = 80. Putting x = y = 0 in (2), we have 0 ≤ 80, which is

true. Hence, half plane region containing the origin is

the solution region of this inequality. Graph of inequality (3). Let us draw the graph of

line x = 15, which is parallel to y−axis and is at a distance of 15 units from it.

Putting x = 0 in (3), we have 0 ≤ 15, which is true, the solution region of the given inequality lying on the left hand side of y−axis.

Graph of inequality (4). Clearly x ≥ 0 represents the region lying on the right hand side of y – axis.

Graph of inequality (5). Clearly y ≥ 0 represents the region lying on the upper side of x – axis.

The common region of the four regions represent. The solution set of the given linear system.

3 +2 =150x y

x y+4 =80x=15

X

Y

28. Let the two numbers be a and b

A.M.=a+ b

2and G.M.= ab 1

According to the given condition,

a+ b2 ab

=mn

a+ b4 ab

=mn

a+ b =4ab m

n

a+ b =2 abm

n

2 2

2

22

2

Þ( )

( )

Þ ( )

Þ (i) 1

Using this in the identity (a−b)2 = (a+b)2 – 4ab, we obtain

a b =4ab m

n4ab =

4ab m n

n

a b =2 ab m n

n

22

2

2 2

2

2 2

- --

--

( ) ( )

Þ ( ) (ii)

1

Adding (i) and (ii), we obtain

2a =2 ab

nm+ m n

a =abn

m+ m n

2 2

2 2

-

-

( )Þ ( )

Substituting the value of a in (i), we obtain

(from eq (i))

6

1




b =2 ab

n m

abn

m+ m n

=abn

mabn

m n

=abn

m m n

a : b =ab

=

a

2 2

2 2

2 2

- -

- -

- -

( )

( )

\

bbn

m+ m n

abn

m m n=

m+ m n

m m n

Thus, a : b = m+ m

2 2

2 2

2 2

2 2

2

-

- -

-

- -

( )( )

( )( )

-- - -n : m m n2 2 2 1

Hence proved 29.

C.I. x f f(x) x x-( )2f x xi -( )2

0-30 15 2 30 8684 17368

30-60 45 3 135 3844 11532

60-90 75 5 375 1024 5120

90-120 105 10 1050 4 40

120-150 135 3 405 784 2352

150-180 165 5 825 3364 16820

180-210 195 2 390 7744 15488

f = 30å fx = 3210å x xi -( )å 2 = 68720

3

\

( ) ( )

åå

åå

Mean, x =fxf

=321030

=107

variance, =f x - x

f

=68720

30=

2 i

2

s

22290.67 1

OR Let the other two observations be a and b Therefore, the series is 5, 7, 9, a, b Now, mean

x = 6

i.e 5+7+9+a+ b

5= 6

21+a+ b= 30

a+ b = 9

Þ

Þ

Also, S.D.=

2 =

(i)

variance

variance

vari\ aance

variance

= 4

Also, = 4 =1n

x x

4 =15

5 6 + 7 6 + 9 6

i

2

2 2

-

- - -

( )

Þ ( ) ( ) ( )

å22 2 2

2 2

+ a 6 + b 6

20 = 1+1+9+ a 6 + b 6

20 11= a 6

- -

- -

- -

( ) ( )éë

ùû

Þ ( ) ( )éë

ùû

Þ ( )) ( )Þ ( ) ( )Þ ( )Þ

2 2

2 2

2 2

+ b 6

9 = a +36 12a + b +36 12b

9 = a + b +72 12 a+ b

9 =

-

- -

-

aa + b +72 12 9

9 = a + b +72 108

9 = a + b 36

a + b = 45

2 2

2 2

2 2

2 2

-

-

-

( )Þ

Þ

Þ (ii)

But from (i)

a2+b2+2ab = 81 (iii) ½

From (ii) and (iii), we have

2ab = 36 (iv) ½

Subtracting (iv) from (ii), we get

a2 + b2–2ab = 45 – (36)

i.e., (a–b)2 = 9

or a–b = ±3 (v) ½

So, from (i) and (v), we get

a = 6, b = 3 when a–b = 3

or a = 3, b = 6 when a–b = –3

Thus, the remaining observations are 3 and 6. ½

1 1

1

1

1

1

½

½

[from (i)]


KENDRIYA VIDYALAYA SANGATHAN(JAMMU REGION)

SESSION ENDING EXAMINATION 2017-18 CLASS XI

SUBJECT- MATHEMATICS(SOLVED PAPER)

Time Allowed: 3 hrs. Max. Marks: 100

General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions. 3. Questions 1-4 in section A are very short - answer type questions carrying 1 mark each 4. Questions 5-12 in section B are short - answer type questions carrying 2 marks each 5. Questions 13-23 in section C are long - answer I type questions carrying 4 marks each 6. Questions 24-29 in section D are long - answer II type questions carrying 6 marks each 7. There is no overall choice, however, there are 3 internal choices in each of the sections C and D.

Section-A 1 mark

1. Write the quadratic equation whose one root is: 1 − i.

2. Write the negation of the statement: “For every real number x, x is less than x+1.”

3. Find eccentricity of the hyperbola: x y2 2

16 91− = .

4. Find the value of: cot .31

3π

Section-B 2 marks

5. Write all the subsets of the set {−1, 0, l}.

6. Find the domain of the real valued function: f x( ) = −3 2x .

7. If n nc c7 5= , find nc4

8. Find the derivative by first principle of the constant function: f(x) = a for a fixed real number a.

9. Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x = 0” is true by direct method.

10. Coefficient of variance of a distribution is 60 and standard deviation is 21, find the mean of the distribution.

11. In a lottery, 100 tickets are sold and 10 equal prizes are awarded. What is the probability of getting two prizes if you buy two tickets?

12. If E and F are events such that P E P F( ) = ( ) =14

12

, and P E and F� �( ) =18

, then find P(E or F).

Section-C 4 marks

13. Draw the graph of the following real-valued function and find its domain and range: f(x) = x2 + 2x + 2.

OR Define a relation R on Z, the set of integers by: R= {(x, y): y = x3− 2x + 1 and − 2 < x < 3 and x Î z}. Find the domain and range of R in roster form.





14. Prove that: cos cos cos .π π4 4

2+

+ −

=x x x

OR

Prove that: 1 21 2 4

2+−

= +

sinsin

.θθ

πθtan

15. Find the polar form of the complex number: − +4 4 3i .

16. The English alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed from the alphabet?

OR

In how many ways can the letters of the word, “MATUTIPERSON” be arranged if there are always 4 letters between R and O?

17. Let A and B be two finite sets such that n( A – B)= 10, n(AÈB)= 50, n(AÇB)=15. Find n(B).

18. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

19. A geometric progression consists of positive terms. If each term equals the sum of next two terms then find the common ratio.

20. Find the value of ‘p’ if the lines: p(p2 + l) x – y + q = 0 and (p2 + 1)2x + (p2 + 1)y + 2q=0 are perpendicular to a common line.

21. The foci of a hyperbola coincide with the foci of the ellipse:

x y2 2

25 91+ = ; find the equation of hyperbola if its eccentricity is 2.

22. If a triangle has the vertices (0, 0, 6), (0, 4, 0) and (6, 0, 0), find its centroid. Also prove that the triangle is isosceles.

23. Find lim ,x f x→ ( )0 where f xxx

x

x( ) =

≠

=

| |

,

,

0

0 0

Section-D 6 marks

24. Find Mean, Variance and standard deviation using short-cut method:

Classes 0−10 10−20 20−30 30−40 40−50

Frequencies 5 8 15 16 6

25. Solve the following system of inequalities graphically:

x + 2y ≤ 8, 2x + y ≤ 8, x ≥ 0 and y ≥ 0.

26. By using Principle of Mathematical Induction, prove that:

41n − 14n is a multiple of 27 for all n Î N.

OR

By using Principle of Mathematical Induction, prove that: xn − yn is divisible by x − y for all n Î N.

27. Find ‘n’ if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

213

44

+

n

is 6 1:

28. Prove that: cos cos cos2 2 2

3 332

x x x+ +

+ −

=

π π.




OR

Solve the equation: 2 sin2 x + 5 sin x − 3 = 0 in the interval [0, 2p].

29. In an examination, question number 1 was attempted by 67 students, question number 2 by 46 students and question number 3 by 40 students. 28 students attempted both questions 1 and 2, 8 attempted 2 and 3, 26 attempted 1 and 3 and 2 students attempted all the three questions. Find how many students attempted question number 1 but not 2 and 3 ?

OR

Let cos(A B)=+45

and sin ,A B−( ) =5

13 where 0

4≤ ≤A B, ,

π then find the value of tan(2A).




SOLUTIONS

Section –A 1. Since complex roots occurs in pair. The roots of

quadratic equation are 1 − i and 1 + i Sum of roots = 1 − i + 1 + i = 2 Product of roots = (1 − i) (1 + i) = 1−i2

= 1− (−1) = 2 Required quadratic equation: x2 − (Sum of roots)x + (Product of roots) = 0 Þ x2− 2x + 2 = 0 1 2. Negation: “ There exists a real number x such that

x is not less than x+1.” 1 3. Given equation of hyperbola,

−x16

y9

=1

Here, a =16 or a = 4

and b = 9 or b = 3

Now, c = a + b

2 2

2

2

2 2 22

2

=16+9

c = 25c = 5

Ecentricity, e =ca

=54

∴

1

4. cot31

3= cot 10 +

3

ππ

π

= × +

=

=

cot

cot

5 23

313

ππ

π

1

Section –B 5. Given set = {−1, 0, 1} Subsets of given set = f, {–1}, {0}, {1}, {–1, 0}, {0, 1}, {1, –1}, {–1, 0, 1}. 2

6. Given function f x x( ) = −3 2

Clearly, f(x) is defined for

3 0

3 0

3 0

3 0

3 3 0

2

2

2

2

− ≥

− ≥

− −( ) ≥

− ≤

−( ) +( ) ≤

x

x

x

x

x x

or

or

or

,

,

,

or, ,x ∈ −[ ]3 3 1

Hence, Domain (f) = [–3, 3] 1

7. Given, n = n

n = nn = n

r = n r

n 7 = 5

n =12

Now, n

C C

C CC C

C

7 5

n-7 5

n-rr⇒∴

⇒

⇒

−

−

44 4=12

=

C

124 12 4

!! !−( )

=112

4 8

12 11 10 9 84 3 2 1 8

!! !

!!

=× × × ×× × × ×

=× × ×× × ×

12 11 10 94 3 2 1

n = n

n = nn = n

r = n r

n 7 = 5

n =12

Now, n

C C

C CC C

C

7 5

n-7 5

n-rr⇒∴

⇒

⇒

−

−

44 4=12

=

C

124 12 4

!! !−( )

=112

4 8

12 11 10 9 84 3 2 1 8

!! !

!!

=× × × ×× × × ×

=× × ×× × ×

12 11 10 94 3 2 1

1

8. Given function : f(x) = a (here ‘a’ is a constant)

∴ +( ) =

+( ) − ( )

=−

=

∴ (

f x h a

f x h f xh

a ah

f x

Now,

0

' )) =+( ) − ( )

=

=

→

→

lim

limh

h

f x h f xh0

00

0

Thereefore, ddx

a( ) = 0 1

9. Since, p: x is a real number such that x3 + 4x = 0 q: x is 0. In direct method, we assume p is true and prove q is true. 1

So, Let x be a real number such that x3 + 4x = 0, which

proves x = 0 Since, p is true Consider, x3 + 4x = 0, where x is real

x x

x x

x x

x x

x

2

2

2

4 0

0 4 0

0 4

0 4

4

+( ) =

= + =

= = −

= = −

= −

either or

or

or

,

is not possible because it is given

x is real.

Hence, x =0 Þ q is true 1

Hence proved.

1

1




10. Given, C.V. = 60 and σ = 21

C.V. (where x is mean)

60

or,

= ×

∴ = ×

=

σx

xx

100

21100

35 1

11. Since, 2 tickets are chosen out of 100 tickets

n s = C

1002( )

= =× ×× ×

= × =100

2 98100 99 98

2 1 9850 99 4950

!! !

!!

½

Now, out of the 100 tickets only 10 have a prize choosing 2 tickets out of 10 tickets = C

=10!

2! 8!= 45

102

½

\ Probability of getting two prizes =CC

=45

4950

=1

110

102

1002

1

12. Given,

P E P F P E F P E F( ) = ( ) = ( ) = ∩( ) =

14

12

18

, , and

We know that, P E F P E F

P E P F P E F

or( ) = ∪( )= ( ) + ( ) − ∩( )

= + −

=+

−

=

14

12

18

1 24

18

334

18

6 18

58

−

=−

=

1

Section –C 13. Let function f(x) = y = x2 + 2x + 2 (i) Step 1: On comparing y = x2 + 2x + 2 with y = ax2+bx+c, we get a = 1, b = 2 and c = 2 Since, the given equation is quadratic, it represents

a parabola. Step 2: Now, the axis of symmetry

xba

x

= −

∴ = −21

Step 3: Vertex of parabola put x = −1 in eq (i), we get y = 1 \ Vertex (–1, 1)

Step 4: Some other points on parabola are

x y= x2+2x+2

x = 0 y = 2

x = 1 y = 5

x = –2 y = 2

x = –3 y = 5

Now, plot all these points on the rectangular axes.

xx'

axis of

(-3,5)

(-2,2) (0,2)

y'

y

(1,5)

symmetry

2 Domain of y = (−∞,∞) 1 To find range of y, we first write the given quadratic

function in vertex by completing the square

y x x

y x

= + +

⇒ = +( ) +

2

2

2 2

1 1

As the domain of the given function is R with x taking any value in the interval, (x+1)2 is either zero or positive.

Hence, x

x

y

+( ) ≥

+( ) + ≥

≥

1 0

1 1 1

1

2

2or

or

Hence, the range of y = [1, ∞) 1OR

Given,

R x y y x x x x Z

R x y y x x

= ( ) = − + − < < ∈{ }= ( ) = − +

, :

, :

3

3

2 1 2 3

2

and and

or 11 1 0 1 2

1 1 2 1 1 13

and and

For

x x Z

x y

= − ∈{ }= − = −( ) − −( ) + = − +

, , ,

, 22 1 2

0 0 2 0 1 1

1 1 2 1 1 1 2

3

3

+ =

= = ( ) − ( ) + =

= = − ( ) + = − +

x y

x y

,

, 11 0

2 2 2 2 1 8 4 1 53

=

= = − ( ) + = − + =and x y,

∴ ( ) ( ) ( ) ( ){ }R = 1, 2 , 0, 1 , 1, 0 , 2, 5

Domain of R = 1, 0, 1,

−

− 22

Range of R = 2, 1, 0, 5

{ }{ } 1

1

1

1

2




14. L.H.S

Using

= +

+ −

+( ) + −( ) =

cos cos

cos cos

π π4 4

2

x x

A B A B ccos cos

cos cos

cos

A B

x

x

=

=

24

212

π

cos

cos

π4

12

2

=

= x

1

= R.H.S Hence proved. 1

OR

L.H.S =+−

=+ ++

1 21 2

22 2

2

sinsin

cos sin sincos sin

θ

θ θ θ

θ

θ 22

2 2

2 2

21

2

θ θθ θ

θ θ θ θ

−+ =( )

=+ +

sincos sin

cos sin sin cos

ccos sin sin cos

sin sin cos

cos s

2 2 2

2 2

θ θ θ θ

θ θ θ

θ

+ −

=( )

=+

iin

cos sin

tan

tan

θ

θ θ

θ

θ

( )−( )

=+( )−( )

2

2

2

2

1

1

=

+

−

tan tan

tan tan

tan

πθ

πθ

π4

14

4

2

2 ==

= +

∴ +( ) =

+−

1

4 12 tan tan

tan tanta

πθ A B

A Bnn tanA B+

= R.H.S = R.H.S.

15. Given Complex number

Z i

i r i

r r

= − +

− + = +( )∴ = − =

4 4 3

4 4 3

4

Let cos sin

cos , sin

θ θ

θ θ 44 3 1

On squaring and adding, we get

r

r

rr

2 2 2

2

2

4 4 3

16 48

648

= −( ) + ( )= +

=∴ = 1

On dividing, we get

tansincos

θθθ

=

rr

=4 3

4

= 3 = tan3

= tan3

−

− −π

−π 1

Since, q lies in II quadrant, as cos q is –ve and sin q is + ve.

i.e., θ π −π π

=3

=23

Required polar form is 8 cos

∴223

+ sin23

π πi

θ π −π π

=3

=23

Required polar form is 8 cos

∴223

+ sin23

π πi

1

16. There are 5 vowels and 21 consonants in English alphabets.

2 vowels can be chosen in 5C2 ways

2 consonants can be chosen in 21C2 ways

4 letters can be arranged in 4! ways 2

\ The number of words consisting of 2 vowels and 2 consonants = × ×

=××

×××

×

= × ×=

5 21 45 41 2

21 201 2

24

10 210 2450400

C C2 2 ! 1

OR Totally there are 12 letters

R and O occupy 7 positions like (1, 6), (2, 7) , (3, 8), (4, 9), (5, 10), (6, 11), (7, 12). 1

In this, R and O can interchange among themselves.

In the remaining positions the other 10 letter can arrange themselves in 10! ways. 1

So, the total number of ways where R and O have 4 letters between them considering T occurs two times

=× ×

= ×

= ×

=

7 2 102

7 10

7 3 628 800

25 401 600

!!!

, ,

, , 2 17. We know that,

n A B = n A +n B n A B

and n A B = n A n A B

∪( ) ( ) ( ) ∩( )( ) ( ) ∩( )

−

− −

(i) 1

Now, according to question, 1

n A B = 50, n A B =15 and n A B =10∪( ) ∩( ) ( )−

Using (ii), we get

n A B = n A n A B

10 = n A 15

n A = 25

− −

−

( ) ( ) ( )⇒ ( )

⇒ ( )

(iii) 1

Now, using (i), we get

n A B = n A +n B n A B

50 = 25+n B 10

n B = 50 25+10

∪ ∩( ) ( ) ( ) ( )⇒ ( )

⇒ ( )⇒

−

−

−

nn B = 35( ) [using (iii)]

1

1

1

1

1

1

1

(ii)

1

1




18. Let the G.P. be A1, A2, A3, A4, … A2n. The number of terms in the G.P. is 2n. (Since, it is

given that the G.P. has even number of terms) Now, the sum of all the terms is 5 times the sum of

terms occupying odd places, hence we have

A + A + A + + A

= 5 A + A + A + + A

A + A + A +

1 2 3 2n

1 3 5 2n-1

1 2 3

( )⇒ ++ A

5 A + A + A + + A = 0

A + A + A + + A

2n

1 3 5 2n-1

2 4 6 2n

−

( )⇒

= 4 A + A + A + + A1 3 5 2n-1( ) 1

Now, assume that the G.P. to be a, ar, ar2, … where ‘a’ is the first term and ‘r’ is the common ratio. 1

Then, ar r

r

a r

r

ar ar

n n

−( )−( )

=−( )

−

⇒ =⇒ =

1

14

1

1

44

1

Hence, the common ratio is 4. 1 20. Given lines are:

p p x y q2 1 0+( ) − + = (i))

and p x p y q2 2 21 1 2 0+( ) + +( ) + = (ii)

\ Lines perpendicular to same line are parallel to each other

i.e., slope of lines should be equal 1 Now, slope of line

p p x y q p p 2 21 0 1+( ) − + = − +( )is 1 and slope of line

p x p y q p2 2 2 21 1 2 0 1+( ) + +( ) + = +( ) is 1

Hence, − +( ) = +

⇒ = −

p p p

p

2 21 1

1

1

21. The equation of the ellipse is

x25

+y9

=12 2

(i)

Therefore, a2 = 25 and b2 = 9 Let ‘e’ be the eccentricity of the ellipse

b = a 1 e

9 = 25 1 e

or, 1 e =925

or, e =19

2 2 2

2

2

2

−

−

−

−

( )( )

225

or, e =1625

or, e =45

2

1

So, the co ordinates of the foci of the ellipse are (± ae, 0) i.e., (± 4, 0) ½ Let e’ be the eccentricity of the required hyperbola

and its equation be

xa'

yb'

=12

2

2

2− (ii)

The co-ordinates of the foci of the hyperbola are (± a'e', 0) or (± 4, 0) ½

It is given that e'= 2 Therefore, a'× 2 = 4 i.e., a'= 2 or a'2 = 22 = 4 ½

b a e' ' '2 2 2

2 2

1

2 2 1

4 312

= −( )= −( )= ×=

1

Therefore, the required equation of the hyperbola is

x4

y12

=12 2

− ½

22. We know that, the co-ordinates of the centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) are

x +x +x3

, y +y +y

3,

z +z +z3

1 2 3 1 2 3 1 2 3

Here, x , y , z 0, 0, 6

x , y , z 0, 4, 0

x , y , z 6

1 1 1

2 2 2

3 3 3

( ) ≡ ( )( ) ≡ ( )( ) ≡ ,, 0, 0( )

\ Coordinates of centroid

=0+0+6

3,

0+4+03

, 6+0+0

3= 2,

43

, 2

2

Now, let A(0, 0, 6), B(0, 4, 0), C(6, 0, 0) denote the vertices of a ∆ ABC, then

AB = 0 0 + 0 4 + 6 0

= 0+16+36 = 52

BC = 0 6 + 4 0 + 0

2 2

2 2

− − −

− −

( ) ( ) ( )

( ) ( )

2

−− 0

= 36+16+0 = 52AB = BC

2( )

∴

Thus, the ∆ ABC is isosceles. 2 23. We have,

f xxx

x

x

f x x

( ) =≠

=

( ) =

=

,

,

li

0

0 0

0LHL of at

mm

lim lim

lim

-x

h h

h

f x

f hhh

→

→ →

→

( )

= −( ) =−−

=−

0

0 0

0

0

hhh

f x x

h

x

= −( ) = −

( ) =

=

→

→ +

lim

lim

0

0

1 1

0RHL of at

ff x

f hh

( )= +( )

→ lim

00




=

=

= ( ) =

→ → →lim lim limh h h

hh

hh0 0 0

1 1

Clearlyy

So, does not exist.

, lim lim

lim

x x

x

f x f x

f x→ →

→

− +( ) ≠ ( )

( )0 0

0

Section –D 24.

Class Interval

Mid- value

(xi)

frequencies (fi)

y =x - 2510ii

fiyi yi2 fiyi

2

0−10 5 5 −2 −10 4 20

10−20 15 8 −1 −8 1 8

20−30 25 15 0 0 0 0

30−40 35 16 1 16 1 16

40−50 45 6 2 12 4 24

Total 50 10 68

y =x 25

10ii − { h = 10, A (assumed mean) = 25}

1

∴ ( ) = + ×

= + ×

∑∑

Mean Axf yf

hi i

i

251050

10

=

∴ ( ) = − ( ) ∑ ∑

27

22

22 2

Variance σhN

N f y f yi i i i

= × − ( )

1002500

50 68 102

= −[ ]

=

125

3400 100

330025

= 132 2

\ Standard deviation (S. D.), σ= 132 =11.489 1 25. Given, system of inequalities is x + 2y ≤ 8, 2x + y ≤ 8, x ≥ 0, y≥ 0 First we solve x + 2y ≤ 8 (i) Let first draw graph of x + 2y = 8

Putting x = 0 in (i) Putting y = 0 in (i)

0 + 2y = 8 x + 2(0) = 8

2y = 8 x + 0 = 8

y = 4 x = 8

Drawing graph

x 0 8

y 4 0 1

Points to be plotted are (0, 4) and (8, 0) Checking for (0, 0) Putting x=0, y = 0 x + 2y ≤ 8 0 + 2(0) ≤8 0 ≤ 8, which is true So, we shade left side of the line Hence origin lies in the plane x + 2y ≤ 8 Now, we solve 2x + y ≤ 8 (ii) Lets first draw graph of 2x + y =8

Putting x = 0 in (ii) Putting y = 0 in (ii)

2(0) + y = 6 2x + (0) = 8

0 + y = 6 2x = 8

y = 6 x = 4

Drawing graph

x 0 4

y 8 0 1

Points to be plotted are (0, 8) and (4, 0) Checking for (0, 0) Putting x = 0, y = 0 2x + y ≤ 8 2(0) + 0 ≤ 8 0 ≤ 8, which is true So, we shade left side of line. Hence, origin lies in the plane 2x + y ≤ 8. Also, give that x ≥ 0, y ≥ 0 Hence, the shaded region will lie in the first

quadrant. Hence, the shaded region OABC represents the

given inequalities.

(4,0) (8,0)

(0,4)

(0,8)

Line2

Line1

XX'

Y'

Y

A

B

CO

x+2y ≤ 8

2x+y ≤ 8

4

26. Let, P n = 41 14 = 27dn n( ) − where d Î N 1

For n =1

L.H.S = 41 14 = 41 14

1 1−−

= 27

= R.H.S= ×27 1

2

1

(ii)




n =1

L.H.S = 41 14 = 41 14

1 1−−

= 27

= R.H.S= ×27 1

\ P(n) is true for n =1 1 Assume P(k) is true 41k – 14k = 27m, where mÎN (i) 1 We will prove that P(k+1) is true

L.H.S

= −

= −

+ +41 14

41 41 14 14

1 1

1

k k

k k. .

(

= +( ) −27 14 41 14 141m from ik k. ( ))

= × + × −

= × + −

41 27 41 14 14 14

41 27 14 41 14

m

m

k k

k

.

(( )= × + ( )= +( )

41 27 14 27

27 41 14

m

m

k

k

= 27 r; where r = (41m+14k) is a natural number

\ P(k+1) is true whenever P(k) is true. 2 \ By the principle of mathematical induction, P(n) is true, where n is a natural number, 1

OR Let P(n): xn – yn is divisible by (x – y) for nÎN For n = 1 1 L.H.S = x1 – y1

= x−y which is divisible by (x – y) Hence, P(n) is true for n = 1 1 Assume P(k) is true. xk – yk is divisible by (x – y), where kÎN (i) 1 We will prove that P(k+1) is true

L.H.S = x y

= x x y+x y y

= x

k+1 k+1

k+1 k k k+1

k

−

− −++1 k k k+1

k k k k

x y + x y y

= x .x x y + x y y .y

− −

− −

( ) ( )( ) ( )

= x x y +y x yk k k− −( ) ( ) By statement (i), we have xk – yk is divisible by x – y. Hence, xk(x−y) + y(xk−yk) will be divisible by (x−y) Hence, P(n) is true for n = k+1 2 Thus, P(1) is true and P(k+1) is true whenever P(k)

is true. Hence, by principle of mathematical induction P(n)

is true for all nÎN. Hence, xn – yn is divisible by x−y for nÎN is always

true. 1 27. In the expansion,

a+ b = C a + C a b+ C a b

n n0

n n n-1 n n-2 2( ) 1 2

+ + C ab + C bn n-1 n n n n−1

Fifth term from the beginning

= C a bn

4n-4 4

Fifth term from the end

= C a bn 4 n-4n−4

Therefore, it is evident that in the expansion of

2 +

13

44

n

The fifth term from the beginning is

n4

4n-4

4

4

C 213

( )

1

and fifth term from the end is

n 44

4

n-4

n4

4n-4

4

4n

4

4n

44

C 213

C 213

= C2

2.1

n− ( )

( )

( )( )

4

33

= C2

213

n4

4n( )

⋅

=n!

6 4! n 4 !2

C 213

4n

n 44

4

⋅ ( ) ( )

( )

−

−

n 4

nn-4n

44

4n

n

= C 23

3

= C

n− ⋅( )( )4

nn− ⋅( )

⋅( )

4 23

3

=6 n!

n 4 !

4n

− 4!1

34n⋅

( ) It is given that the ratio of the fifth term from the

beginning to the fifth term from the end is 6 : 1 Therefore, from (1) and (2), we obtain

nn

nn

n

n

n

n

!! - !

:!

- ! !:

:

6 4 42

64 4

1

36 1

2

66

3

4

4

4

4

⋅ ( ) ( ) ( ) ( )=

⇒( )

( )=

66 1

2

6

3

66

6 36 6

6 6

452

452

10

4 4

4

45

2

:

⇒( )

×( )

=

⇒ ( ) =

⇒ =

⇒ =

⇒ = × =

n n

n

n

n

n

1

28. Let first calculate all 3 terms separately we know that

(i) 1

1

1

1




cos 2x = 2cos x 1

cos 2x+1= 2 cos xcos2x+1

2= cos x

So, cos x =cos

2

2

2

2

−

22x+12

Replacing x with x+3p is about

cos x+3

=cos2 x+

3+1

2

=cos

2 ππ

22x+ 23

+1

2

π

½ Similarly,

Replacing x with x+3p

in cos x =cos2x+1

22

cos2

3x -

p

=

cos23

1

2

x - p

+

=

cos 23

1

2

x - 2p

+

½

Now, L.H.S = cos cos cos2 2 2

3 3x x x+ +

+

p-p

=

1 22

1 223

2

1 223

2+

++ +

++

cos

cos cosx

x xp - p

=12

1 2 1 223

1 223

+ + + +

+ +

cos cos cosx x x

p-

p

=

12

1 1 1 2 223

223

+ + + + +

+

cos cos cosx x x

p-

p

=

12

3 2 223

223

+ + +

+

cos cos cosx x x

p-

p

Using cos x + cos y = 2

2 2cos cos

x y x y+

-

Replacing x by 2

23

x +p

and y by

2

23

x -p

=

12

3 2 22

23

223

2+ +

+ +

cos cosxx x

p - p

cos2

23

223

2

x x+ −

p - p

= + + + + +

12

3 2 2 2 223

23

cos cos

co

x x xπ π

ss

cos cos cos

2 223

23

2

12

3 2 242

43

x x

xx

− + −

= + +

π π

π

22

12

3 2 2 223

= + +

cos cos cosx x

π

= + + −

= + +

12

3 2 2 23

12

3 2 2 2

cos cos cos

cos cos

x x

x x

ππ

−−

cos

π3 1

[using cos (p - q) = – cos q]

= + + −

= + − × ×

12

3 2 2 212

12

3 2 212

2

cos cos

cos cos

x x

x x

= + −[ ]

= +

=

=

12

3 2 2

12

3 0

32

cos cos

[ ]

x x

RHS

Hence ProvedOR

2 5 3 0 0 2

2 5

2

2

sin sin int ,

, sin s

x x in the erval

We have x

+ − = [ ]+

π

iin

sin sin sin

sin sin sin

x

x x x

x x x

− =

⇒ + − − =

⇒ +( ) − +( ) =

⇒

3 0

2 6 3 0

2 3 1 3 0

2

ssin sin

sin sin sin

sin

x x

x x x

x

+( ) −( ) =

⇒ −( ) = ≠ − ∴ + ≠[ ]⇒ =

3 2 1 0

2 1 0 3 3 0

112

6

16

⇒ =

⇒ = + −( )

∈

sin sin

,

x

x n n Zn

π

ππ

1

29. Let q1 denotes the no. of students attempted question no. 1

q2 denotes the no. of students attempted question no. 2

1

½

1

1

1

1

1

1

2




q3 denotes the no. of students attempted question no. 3 1

According to question,

n q = 67; n q = 46; n q = 40; n q q = 28;

n q q = 8; n q1 2 3 1 2

3 1

( ) ( ) ( ) ( )( )

2 qq = 26 and n q q q = 23 1 2 3( ) ( )

2 The no. of students who attempted question 1, but

not question number 2 and 3

= n q n q q n q q +n q q q

= 67 28 26+2= 69 54=15

1 1 2 1 3 1 2 3( ) ( ) ( ) ( )− −

− −−

1

OR

cos A+B =45

and sin A B =5

13

cos A+B =45

=BaseHyp.

and sin

( ) ( )

⇒ ( )

−

AA B =5

13

sin A+B =35

=Parp.Hyp.

Use pythagoras theorem t

−

( )

⇒ ( )

oo find perpendicular

A+B = sin35

-1

[ ]

⇒ ( )

−

(i)

and A B = sin5

13-1

(ii) 1

On adding (i) and (ii), we get 1

235

513

235

15

135

13

1 1

12

A

A

=

+

⇒ = −

+

sin sin

sin

- -

- 1135

235

144169

513

1625

2

1

−

⇒ = +

⇒

A sin-

2235

1213

513

45

25665

256

1

1

A

A

A

= × + ×

⇒ =

⇒ =

sin

sin

sin

-

-

665

=

=

PH

Now, Base = Hyp - Perp. = 65 - 56

2 2 2 2 2

44225 3136− 1

= 1089

Þ Base = 1089 = 33

Now tan 2A = PerpHyp

.

.=

5633

1

1

½

½


(agra region) session ending examination 2017-18 class … · find the ter m independent of x in...

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