Sample PaperCLASS: XII (New Pattern)

MATHEMATICS Max.Marks:100Time Allowed:

3 hours

General Instructions:1. All questions are compulsory.2. This question paper consists of 29 questions divided into Four sections A, B, C andD. Section A comprises of 4 questions of one mark each, section B comprises of 8 questions of 02 marks each and section C comprises of 11 questions of 04 marks each and section D consists of 06 questions of 06 marks each.3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.4. There is no overall choice. However, internal choice has been provided in 03 Questions of four marks each and 02 questions of six marks each. You have to attempt only one of the alternatives in all such questions.5. Use of calculators is not permitted. You may ask for logarithmic tables, if required.

________________________________________________________________________SECTION - A (1 mark questions)

1. Let * be a binary operation on Q given by a*b=a+ab where a,b ∈Q .Is * commutative?

2. Evaluate Sin{1

2cos−1( 4

5 )}3. Given that A, B are two symmetric matrices such that AB =BA .Is AB symmetric?

4. Find the values of x, y and z if (x+ y+z

x+zy+z )=(957 )

SECTION - B (2 mark questions)

5. Evaluate | Sin300 Cos300

−sin 600 Cos600|

6. Find a point on the curve y=x2-4x+5 where the tangent to the curve is parallel to the x axis

7. Evaluate ∫ x+cos6 x

3 x2+Sin 6 x dx

8. Find the projection of the vector a→= i

¿

−3k¿

on the vector b→=3 i

¿

+ j¿

−4k¿

9. If a→

is a unit vector and ( x→+a

→)(x

→−a

→)=15 find |x

→|

10. The Cartesian equation of a line AB is

2x−1√3

= y+22

= z−33 .Find the direction cosines of a

line parallel to AB

11. Let f: R→R be defined by f(x) = 3x+2.Show that f is invertible. Also find f−1

:R→R

12. Write in the simplest form : tan−1( 1

√ x2−1 ) where |x|>1

SECTION - C (4 marks questions)

13. Show that x = 2 is a root of the equation

|x −6 −12 −3x x−3

−3 2x x+2|=0

and solve it completely.

14. Discuss the continuity of the function at x = 0: f ( x )=¿¿

x≠0

x=0

15. Verify Rolle’s theorem for the function f(x)= Sin2x in [0 , π2 ]

16. If √1−x2+√1− y2=a( x− y ) prove that

dydx=√ 1− y2

1−x2

OR If y=x2 +4 and x changes from 2 to 2.1 find the approximate change in y

17. Evaluate ∫0

2

exdx as a limit of sum

OR

Evaluate ∫ 5

( x+1 ) (x2+4 )dx

18. Solve the differential equation Cos

2 x dydx

+ y= tan x

19. Find the differential equation for the family of circles which passes through the origin and have their centre on the x axis

20. Express the vector a→=5 i

¿

−2 j¿

+5k¿

as the sum of two vectors such that one is parallel to the

vector b→=3 i

¿

+k¿

and the other is perpendicular to b→

Or

If a→+b

→+c

→=0 and |a

→|=3 ; |b

→|=5 |c

→|=7 then show that the angle between a

→

and b→

is 600

21. Find the vector equation of the plane passing through the intersection of the planes r→

.

(2 i¿

−7 j¿

+4 k¿

)=3 and r→

. (3 i¿

−5 j¿

+4 k¿

)+11=0 and passing through the point (-2,1,3)22. A coin is tossed three times and all the possible outcomes are assumed to be equally likely. Let E

and F be two events given by E: “both tail and head have occurred” F: “at most one tail has occurred”. Show that E and F are independent.

23. The sum of three numbers is 6.Twice the third number when added to the first number gives 7.On adding the sum of the second and third numbers to thrice of the first number we get 12.Find the numbers using inverse of a matrix.

Or

Using elementary transformations find the inverse of the matrix [ 1 2 3

2 5 7−2 −4 −5 ]

SECTION - D (6 mark questions)

24. Let R be a relation on N¿N defined by (a, b)R (c, d) ⇔ ad=bc, for all (a,b) and (c,d)∈N . Show that R is an equivalence relation .

OR Let * be the binary operation defined on QxQ by (a,b)*(c,d) = (ac,b+ad) where Q is the set of rational numbers. Determine whether * is commutative and associative . Find the

indentiy element for * and the inverttible elements of QxQ

25. An open box with a square base is to be made out of a given quantity of metal sheet of area

c2.Show that the maximum volume of the box is

c3

6√3

26. Evaluate ∫ 1

Sinx (2+Cosx )dx

Or

Evaluate∫0

π2

log Sinxdx

27. Find the area of region included between the parabolas y2=4ax and x2=4ay where a>028. Find the foot of the perpendicular drawn from the point A(1,0,3) to the line joining the points

B(4,7,1) C(3,5,3)29. Mona wants to invest at most Rs 12000 in Savings Certificates (SC) and National Saving Bond

(NSB). She has to invest at least Rs 2000 in SC and at least Rs 4000 in NSB. If the rate of interest on Sc is 8% and the rate of interest on NSB is 10% per annum, how much money should she invest to earn maximum yearly income?

ANSWER KEY WITH MARKING SCHEMEQ.No Value points Marks Total

Marks1 Not Commutative (as a*b=a+ab¿ b+ba=b*a) 1M 1M

2

101

(put cos−1

45 =θ ⇒ cosθ =

45 Given expression =Sin

θ2 =

√ 1−Cosθ2 =√ 1−4

52 =

1√10

1M 1M

3 AB is Symmetric. (AB)t

=Bt

At

=BA=AB 1M 1M

4 x=2 ; y= 4 ; z= 3 ( by equating corresponding entries we get x+y+z = 9 ; x+z =5 ; y+z = 7)

1M 1M

5 Sin300Cos600

+Cos300

Sin600

=Sin 900=1

1MIM

2M

6

(2,1) is the point .

dydx

=2x−4 =0 ⇒ x=2 putting x =2 in y=x2-

4x+5 we get y = 1

1M 2M

7

Put 3 x2+Sin6 x = u ⇒ Given integral =

16∫

1tdt

=16

log|3 x2+Sin6 x|+c

1M 2M

8projection of the vector a

→= i

¿

−3k¿

on the vector b→=

3 i¿

+ j¿

−4k¿

=

a→

.b→

|b→|

=

3+0+12√9+1+16 =

15√26

1M

1M

2M

9

|x→|2 -|a

→|2 =15⇒ |x

→|2

=15+1⇒ |x→|= 4

1M

1M

2M

10x−1

2√3

= y+24

= z−36 dr’s of AB are √3 ,4,6

Hence dc’s of AB are

√3√55 ,

4√55 ,

6√55

1M

1M

2M

11 f ( x1 )=f ( x2 )⇒3 x1+2=3 x2+2⇒ x1 =x2 Thus f is one to one

1M

2y=3x+2 ⇒ x=

y−23 Given any y ∈R there exists

y−23 =x∈R

s.t. f(x)=y showing that the function is onto.

1 M

f−1

(x)=

x−23 is the inverse of f

12 x=Cosecθ⇒θ =Cosec−1 x

2tan

−1( 1√ x2−1 )

= tan−1( 1

√Cosec2θ−1 ) 1M

= tan−1( 1

√Cot2θ )= tan

−1( 1Cot θ )

= tan−1 (Tanθ ) =θ

1 M

=Cosec−1 x

13

|x−2 3 x−6 −x+2

5 −5 x −5−3 2 x x+2

|=0 by performing R1→R1−R2 and

R2→R2−R3

1M

4

5(x-2)

|1 3 −11 −x −1

−3 2x x+2|=0

1M

5(x-2)

|0 3 −10 −x −1

x−1 2 x x+2|=0

C1→C1+C3

1M

5(x-2)(x-1)

|0 3 −10 −x −11 2x x+2

|=0

½M

= -5(x-2)(x-1)(x+3) = 0 ⇒ x=2,1,−3 ½M

14

Here f(0) = 5 and Ltx→0 f(x) =

Ltx→0

2Sin2 xx2

=2

2Ltx→0

Sinxx Lt

x→0

Sinxx =2¿1×1 =2

As Ltx→0 f(x) ¿ = f(0) ,f is not continuous

15

Consider the function f(x) = Sin2x in [0 , π2 ]

. Here f is

continuous in [0 , π2 ]

as f(x)= Sin x is continuous

½M

4f'

(x) = 2 Cos2x exists in (0 , π

2 ) thus f is differentiable on (0 , π

2 )

1M

f(0)=Sin (0) = 0 and f( π2 )

=Sinπ = 0

1M

Conditions for Rolle’s thm are satisfied. Hence there should be at

least one c ∈(0 , π2 )

such that f' (c ) = 0

½M

Let 2Cos2c = 0 ⇒ c =

π4 ∈(0 , π2 ) 1M

16 Let x=Sin A and y=Sin B ½M

4

Cos A + Cos B =a(Sin A –Sin B) ½M

2Cos

A+B2 Cos

A−B2 =a¿ 2Cos

A+B2 Sin

A−B2

1M

Cot

A−B2 =a ⇒ Sin-1x-Sin-1y=2Cot-1a

1M

Differentiating ,

1√1−x2

-

1√1− y2

y '

=0

½M

dydx=√ 1− y2

1−x2

½M

ORx= 2 Δx=0 .1 ½M

4Δy= f ' ( x )Δx =2xΔx =2¿2×0. 1=0.4 1½M

When x=2 y = 22+4=8 ½M

y+Δy =8+0.4=8.4 1M

y changes from 8 to 8.4 ½M17.

a=0 b=2 h=

b−an

=2n nh=2 f(x)=ex

½M

4

∫0

2

exdx=

Lth→ 0 h[ f (0)+ f (0+h)+ f (0+2h)+. .. . ..+ f (0+(n−1 )h )]

1M

=Lth→0h[1+eh+e2 h+e3 h+. .. . .+e (n−1 )h ] ½M

Lth→0h.

1. [ (eh )n−1 ]eh−1

1M

=Lth→ 0

heh−1

{e2−1 }=e

2-1

1M

Or5

( x+1 ) (x2+4 ) =

Ax+1

+Bx+Cx2+4

1M

4

To get A =1 B = -1 and C = 1 1M

∫ 5( x+1 ) (x2+4 )

dx=∫

1x+1

dx+∫ −x+1x2+4

dx½M

=log|x+1|-∫ x

x2+4dx+∫ 1

x2+4dx

1M

=log|x+1|-

12

log|x2+4|+ 12

tan−1 x2+c

½M

18

Dividing Cos

2 x dydx

+ y=tan x by Cos2x to get

dydx

+ ySec 2 x=tan xSec2 x

1M

4

I.F. =etanx ½M

Solution is y(IF)=∫ tan xSec 2 x etanxdx+ k 1M

⇒ Put tan x =u ⇒ y etanx =∫ueudu+k ½M

⇒ y etanx =eu (u−1 )+k ½M

⇒ y =tanx-1+ke−tan x ½M

19

Let a be the radius Then centre is (a,0)

1M

4

= n to the Circle is (x-a)2

+y2

=a2⇒ x2+ y2−2ax=0….(*)

1M

Differentiating ,2x+2yy’-2a=0⇒ a=x+yy'

=a 1M

Putting this value of a in (*) we get 2xyy'

-y2+x2

=0 1M

20 Any vector parallel to b

→

is of the form kb→

for some scalar k½M

4

Let a→

= kb→

+ c→

where c→

is perpendicular to b→ 1M

Then c→

=(a→

-k b→

)¿b→ ½M

⇔ (a→

-k b→

).b→⇔ (a

→

.b→

)-k(b→

.b→

)=0⇒ k=2

1M

kb→

=2b→

=6 i¿

+2k¿ ½M

c→

=(a→

-2b→

)=(−i¿

−2 j¿

+k¿

)½M

OR

4

a→+b

→+c

→=0⇒a

→+b

→=−c

→ ½M

⇒ (a→+b

→)2=(−c

→)2 ½M

⇒ (a→+b

→) .(a

→+b

→)=(−c

→) .(−c

→)

½M

|a→|2 +|b

→|2 +2a

→.b→

=|c→|2

1M

a→

.b→

=

152

½M

Cosθ =

a→

.b→

|a→||b

→|=

152×3×5 =

12 ⇒θ=600

1M

21r→

. {(2+3 λ )i¿

+(−7−5 λ ) j¿

+(4+4 λ)k¿

¿−3−11 λ=0¿…(*)1½M

4

(*) Passes through the point with position vector (−2i¿

+ j¿

+3 k¿

)

⇒ λ=16

1½M

Putting the value of λ in (*) we get r→

. 1M

(15 i¿

−47 j¿

+28k¿

)−7=022. E: “both tail and head have occurred”

={HTT,THT,TTH,HHT,HTH,THH} F: “at most one tail has occurred”= {HHH,HHT,THH,HTH}

½M+½M

4E∩F ={HHT,HTH,THH} ½M

P(E) =

68 P(F)=

48 P(E∩F )=

38

1½M

As P(E∩F )=P(E) P(F)⇒

38=6

8×4

8 is true ,we conclude that E and F are independent.

1M

23 Let the first, second and third numbers be x, y and z respectively. Then x + y + z=6 : x + 2z =7 ; 3x + y + z=12

1M

4⇒AX=B where A=(1 1 11 0 23 1 1 )

X= ( xyz ) B=

( 6712)

½ M

|A|=4≠0⇒ A is invertible½M

adjA=(−2 0 2

5 −2 −11 2 −1 )

1M

X=A−1

B=

1A

(−2 0 25 −2 −11 2 −1 )( 6

712)

=(312 )

1M

The required numbers are 3 ,1 and 2 ½MOR

6Let A=[ 1 2 3

2 5 7−2 −4 −5 ]

½M

A=IA ⇒[ 1 2 3

2 5 7−2 −4 −5 ]

=[1 0 00 1 00 0 1 ]

A

1M

By performing elementary transformations ,to get[1 0 00 1 00 0 1 ]

=

[ 3 −2 −1−4 1 −12 0 1 ]

A

3½M

⇒A−1

=[ 3 −2 −1−4 1 −12 0 1 ]

1M

24(a,b) R (a,b) ⇔ ab=ba Reflexive 1.5M(a,b) R (c,d) ⇔ ad=bc ⇔ cb=da ⇔ (c,d) R (a,b) symmetric 1.5M 6(a,b) R (c,d) and (c,d) R (e,f) ⇔ ad=bc and cf=de ⇔ ad.cf=bc.de ⇔ af=be ⇔ (a,b) R (e,f) Transitive

2½ M

Thus R is an equivalence relation ½ M

25

Let x be the side of the square base and y be the height.

V=x2y ,Surface area S= x2+4xy=c2⇒ y= c2−x2

4 x

1½M

6

V=

14(c2 x−x3 )

dVdx

=14

(c2−3 x2 )

d2Vdx2

=−32

x1½M

dVdx

−0⇒ x= c√3

d2Vdx2

]x= c

√3

=−3c2√3

<01M

V is Maximum when x= c

√3 . Then y= c

2√31M

VMax=

c3

6√3

1M

26∫ Sinx

Sin 2 x (2+Cosx )dx

by multiplying the Nr and Dr by Sinx

½M

6

=∫ Sinx

(1−Cos2 x )(2+Cosx )dx

=∫ −1

(1−t2 )(2+t )dx

by letting Cosx=t

1M

=−∫ 1

(1−t )(1+t )(2+t )dx

½M

1(1−t )(1+t )(2+t ) =

A1−t

+ B1+t

+ C2+ t

1M

To get A=

16 B=

12 C=

−13

1½M

∫ 1(1−t )(1+ t )(2+ t )

dx=

16

log|1−t|+ 12

log|1+ t|−13

log|2+t|+k

1M

∫ 1sin x (2+Cosx )

dx =

−16

log|1−Cosx|−12

log|1+Cosx|+13

log|2+Cosx|+c½M

OR

6

I=∫0

π2

log Sinxdx=∫0

π2

log Sin ( π2 −x )dx=∫0

π2

logCosxdx

1M

2I=∫0

π2

log SinxCosxdx=∫0

π2

log 2SinxCosx2

dx

1M

=∫0

π2

log 2SinxCosx2

dx =

2log2

2log2

2

02log 1

∫ IxdxSin………(*)

2M

I1 =

12∫0

π

log S intdt=

12×2∫

0

π2

log Sinxdx=I

1M

(*) ⇒ 2I=I-2log

2

I=

−π2

log 21 M

27 2M

6

Solving the two equations given we get x=0 and x=4a 1M

Required area =∫0

4 a

√4ax dx−∫0

4 a x2

4adx

2M

=

32a2

3−16a2

3=16 a2

3 sq.units

1M

28 Let P be the foot of the perpendicular from A on BC. If P divides

BC in the ratio k:1 then P is ( 3k+4k+1

, 5k+7k+1

, 3k+1k+1 )

½M

1M

Dr’s of BC are 1,2,-2 ½M

6

Dr’s of AP are

3k+4k+1

−1 , 5k+7k+1

−0 , 3k+1k+1

−31M

Since AP ¿ BC ,dot product =0 ⇒ k=−7

42M

So the foot of the perpendicular is ( 5

3, 73,17

3 ) 1M

29 Suppose that she invests Rs x in SC and Rs y in NSB ½M

Then LPP is to maximize Z=

8 x100

+10 y100 subjecting to constraints x

¿2000 y¿4000 x+y¿12000

1½M

6

(To draw the graph, to identify the feasible region and to get coordinates)

2½M

To put the coordinates of the vertices in Z=

8 x100

+10 y100 to get values

Rs 560, Rs 1040 Rs 1160

1M

Rs 2000 should be invested in in savings certificates and Rs 10000 in National Savings Bonds to get a maximum yearly income of Rs 1160

½M

SESSION ENDING EXAMINATION

MATHEMATICS

CLASS XI

Time Allowed: - 3 Hrs. Max. Marks: - 100

General Instruction:I. All questions are compulsory.

II. The question paper consists of 29 questions divided into three sections A, B ,C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each ,Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.

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

Section A (1X6)Q1.For what value of k numbers −3

7, k ,−7

3 are in G.P .Q2.Write the compound statement by using connectives ‘or’ for the statements p: 25 is multiple of 5 q: 25 is multiple of 8Q3.Let A = {1,2} and B = {3,4}. How many subsets will AX B have?Q4.Find the slope of the line 3x+2 = 0

Section B (2X8)Q5.Write the power set of set A = {a , b, c}Q6.Write the contra positive and converse of the statement: If you are born in India, then you are a citizen of India.Q7.Find the value of tanπ

8 .

Q8.Evaluate : limx→0

cosx−1x2

Q9. Given that A (3,2,-4), B (5,4,-6) and C (9,8,-10) are collinear. Find the ratio in which B divides AC.

Q10. If p is length of perpendicular drawn from origin to the line whose intercepts on the axes are a and b, then show that

1p2=

1a2 +

1b2

Q11. Find the number of different 8 letters arrangements that can be made from the letters of the word ‘DAUGHTER’ so that all consonants occur together .

Q12. In a relay race there are 5 teams A, B, C, D & E . What is the probability that A, B , C finish first , second and third , respectively.

Section C (4X11)

Q13. Convert the complex number (1+7 i)(2−i)2 in polar form.

ORReduce the complex number in standard form a+ib

( 11−4 i

− 21+i )(3−4 i

5+ i )Q14. Solve the following system of in equalities graphically

4x+3y≤ 60 , y ≥ 2x , x≥ 3Q15. Prove that:

cos2 (x+π3 ) + cos2 (x−π

3 ) −¿sin2 x =12

Q16. Find (a + b)4 – ( a −¿b)4, hence evaluate (√5+√2)4 – (√5−√2)4 .

Q17. Find the coordinates of foci, vertices, eccentricity and length of latus rectum of hyperbola 49y2-16x2 = 784

Q18. Solve : sin2x+sin4x+sin6x=0

Q19. If an+bn

an−1+bn−1 is the arithmetic mean of a and b, then find the value of n.

OR If A.M. and G.M. of two positive numbers a & b are 17 and 15 respectively, find the numbers.

Q20. Find the domain and the range of real function f(x) =√16−x2 . OR

If A= {0,1,2,3,4,5} and define relation R from A to A, R= { (x,y) : x + y¿5: x , y∈A } .Write R in roster form also domain and range.

Q21. Find the derivative by first principle of f(x) = x−2x+2

Q22. A committee of 7 has to be formed from 8 men and 4 women for

“Swachhtta Abhiyan”. What is the probability that the committee consists of (i) exactly 3 women(ii) at least 3 women

What is the role of students in Swachhtta Abhiyan.

Q23. Find the co-ordinates of the foot of perpendicular from the point (1,0) to the line 3x-4y-16=0

Section D (6X6)

Q24. Prove by using the principle of mathematical induction for all n∈N ,

1.3+3.5+5.7+ ... + (2n−1) (2n+1) = n(4n2+6n−1)3

Q25. In ∆ABC ,prove that b2−c2

a2 sin2 A+ c2−a2

b2 sin 2B+ a2−b2

c2 sin 2C=0

OR In ∆ABC , prove that

(b2−¿c2) cotA + (c2−¿a2) cotB + (a2−¿ b2) cotC = 0

Q26. If a, b are the roots of x2−¿3x+¿p = 0 and c ,d are roots of x2−¿12x+¿q = 0 where a,b,c,d are in G.P. Prove that (q + p): (q −¿ p ) = 17:15

OR

Find the sum of series upto n terms

13

1+13+23

1+3+ 13+23+33

1+3+5+...

Q27. Find the mean, variance and standard deviation for following data

Marks 30-40 40-50 50-60 60-70 70-80 80-90 90-100No. Of students

3 7 12 15 8 3 2

Q28. Find the no. of ways in which a hand of 7 cards is drawn from a well shuffled deck of 52 cards such that it contains

(i) all queens ; (ii) 3 kings ; (iii) at least 3aces.

Q29. In a survey of 80 people it was found that 25 people read Dainik Jagran, 26 read Amar Ujala and 26 read Hindustan 9 read both Dainik Jagran & Amar Ujala 11

read Amar Ujala and Hindustan and 8 read Hindustan and Dainik Jagran 3 read all three newspapers find how many read

(i) none of them .(ii) at least one newspapers . (iii) exactly one news paper

SESSION ENDING EXAMINATION MATHEMATICS

CLASS – XIMARKING SCHEME

1. k = ± 1 12. 25 is multiple of 5 or 8 13. 24=16 14. slope is not defined 15. P(A) = { ∅ , {a}, {b},{c}, {a,b}, {b,c},{a,c}, {a,b,c}} 26. Contrapositive: If you are not a citizen of India then you are not born in IndiaConverse: If you are a citizen of India then you are born in India.

1+1

7. Applying identity of tan 2x

and getting 1 = 2 y

1− y 2

y = ± ¿ -1)

than tan π8 = √2 -1

1/21/2

1/2

1/2

8. limh→ 0

−2sin2 x /2x2

= - 1/2

1

1

9. ( 9k+3k+1 ,

8k+2k+1 ,

−10k+−4k+1 )

For comparing coordinate k = 1/2 For ratio k : 1 = 1 : 2 (internally )

1

1/2

1/2

10. Equation of line xa+ yb = 1

length of perpendicular p = 1

√ 1a2 +

1b2

1p2=

1a2 +

1b2

1/2

1

1/2

11. total no. of words = 8! 1

4! x 5! = 2880

1

12. Total Outcomes = 5*4*3 = 60 Probability = 1/60

11

13. Expressing Z= -1 + i r = √2 and

θ= 3π4

z=√2 (cos(3π4 ) + i sin(

3π4 ))

or

( 11−4 i

− 21+i )(3−4 i

5+ i ) = (−1+9 i

5−3 i )( 3−4 i5+i )

¿( 307442 )+( 599

442 ) i

111

1

2

2

14. For each correct graph (1X3)

and for correct shading

3

1

15. ½(- 2sin2 x + 2cos2 (x+π3 ) + 2cos2 (x

−π3 ))

= ½[1+(cos2x+cos(2x+2π3 )+cos(2x-2

π3 )

= ½ [1+cos2x+2cos2 π3 cos2x]

=1/2

1

111

16. (a + b)4 – ( a −¿b)4

= 2(4C1a3b+4C3 . ab3) = 8ab(a2+b2) (√5+√2)4 – (√5−√2)4 .= 56√10

2

11

17. a = 4, b = 7, c = √65 foci = (0,± √65) vertices = (0, ± 4)

e = √654

LR = 49/2

½ 11

1 ½

18. For reducing the equation in form and for factorizing sin 4x(1+ 2 cos2x) = 0

For general solution x = n π / 4, n ∈ Z or x = n π ± π3

21+1

19. an+bn

an−1+bn−1 =a+b

2 for cross multiplication

1111

separating a and b for getting n = 1

or

a+b

2 = 17

√ab = 15 for solving for getting a = 25, and b = 9

1111

20. Domain = [−4, 4] Range = [ 0, 4]

or For writing correct relation Domain = {1, 2, 3, 4, 5} Range = { 1, 2, 3, 4, 5}

22

211

21. a) f '( x) = limh→ 0

( x+h−2 ) / (x+h+2 )−( x−2 ) / (x+2 )❑h

Correct calculation = 4/(x+2)2

1

1

2

22. Total Outcomes = 12C7

i) Probability of 3 W & 4 M = 4C3 X 8C4 / 12C7

ii) 3W & 4 M = 4C3 X

8C4 = 280 4W & 3 M = 4C4 X

8C3 = 56 Total Probability = 336/ 12C7

For values

1

2

1

23. Equation of perpendicular line 4x+3y+k = 0 k= - 4 4x+3y-4 = 0 for getting point of intersection of perpendicular lines i.e. foot of perpendicular (64/25, - 52/25)

1½ ½

2

24. For n = 1 For n = k For solving n = k +1 For writing final statement for n ∈ N

11 ½3½

25. for using sin 2A = 2sin A cos A for using sine rule for using cosine rule For getting correct answer

or converting cotA, cot B and cot C in term of sine & cosine using cosine rule solving and getting correct answer

1 ½1 ½1 ½1 ½

222

26. a+ b = 3, ab = p c+d = 12, cd = q a b c d are in GP a+ar = 3, ar2 + ar3 = 12 getting r = ± 2 getting q / p = 16

2

12

1

applying Componendo and Dividendo (q + p) / (q – p) = 17/15

or

Tn = (n+1)2

4

Sn = ΣTn = Σ (n+1 )2

4 ¿

n(2n2+9n+13)24

3

12

27. Σfi = 50 Σf ixi = 3100 Σf i (xi - x)2 = 10050 x = 62 V = 201 SD = 14.18

½ 1211½

28. (i) 4C4 X 48C3

(ii) 4C3X

48C4

(iii) 4C4 X 48C3 +

4C3X 48C4

2

2

2

29. For correct Venn diagram (i ) none of them = 28 (ii) at least one newspaper = 52 (iii) exactly one newspaper = 30

1 ½1 ½1 ½1 ½