+ All Categories
Home > Education > CBSE Mathematics sample question paper with marking scheme

CBSE Mathematics sample question paper with marking scheme

Date post: 15-Jul-2015
Category:
Upload: mathematics-teacher-in-kendriya-vidyalaya
View: 2,028 times
Download: 1 times
Share this document with a friend
10
SAMPLE PAPER -2015 MATHEMATICS CLASS – XII Time allowed: 3 hours Maximum marks: 100 General Instructions: 1. All questions are compulsory. 2. The question paper consists of 26 questions divided into three sections-A, B and C. Section A comprises of 6 questions of one mark each, Section B comprises of 13 questions of four marks each and Section C comprises of 7 questions of six 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 4 questions of four marks each and 2 questions of six mark each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculators is not permitted. Section A Q1. Evaluate: tan –1 √3 – sec –1 (–2) Q2 Find gof if f(x) =8 x 3 , g(x)= √ 3 . Q3. If [ 3 βˆ’ 2 5 βˆ’2 ] = [ 3 5 βˆ’3 βˆ’2 ] , find the value of y . Q4. Evaluate: | 30 0 30 0 βˆ’60 0 60 0 | Q5. Find p such that p z y x 3 2 1 and 1 4 2 z y x are perpendicular to each other. Q6. Find the projection of on if . =8 and = 2Μ‚ +6Μ‚ + 3
Transcript

SAMPLE PAPER -2015

MATHEMATICS

CLASS – XII

Time allowed: 3 hours Maximum marks: 100

General Instructions:

1. All questions are compulsory.

2. The question paper consists of 26 questions divided into three sections-A, B and C. Section A

comprises of 6 questions of one mark each, Section B comprises of 13 questions of four marks

each and Section C comprises of 7 questions of six 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 4 questions of four

marks each and 2 questions of six mark each. You have to attempt only one of the alternatives in

all such questions.

5. Use of calculators is not permitted.

Section A

Q1. Evaluate: tan–1√3 – sec

–1 (–2)

Q2 Find gof if f(x) =8 x3

, g(x)= √π‘₯3

.

Q3. If [3π‘₯ βˆ’ 2𝑦 5

π‘₯ βˆ’2] = [

3 5βˆ’3 βˆ’2

] , find the value of y .

Q4. Evaluate: | 𝑠𝑖𝑛 300 π‘π‘œπ‘ 300

βˆ’π‘ π‘–π‘›600 π‘π‘œπ‘ 600|

Q5. Find p such that p

zyx

321 and

142

zyx

are perpendicular to each other.

Q6. Find the projection of οΏ½οΏ½ on οΏ½οΏ½ if οΏ½οΏ½ . οΏ½οΏ½ =8 and οΏ½οΏ½ = 2𝑖 +6𝑗 + 3οΏ½οΏ½

Section B

Q7. 𝐿𝑒𝑑 𝐴 = 𝑁𝑋𝑁, π‘Žπ‘›π‘‘ βˆ— 𝑏𝑒 π‘‘β„Žπ‘’ π‘π‘–π‘›π‘Žπ‘Ÿπ‘¦ π‘œπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘› 𝐴 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦

(π‘Ž, 𝑏) βˆ— (𝑐, 𝑑) = (π‘Ž + 𝑐, 𝑏 + 𝑑). Show that βˆ— is commutative and associative.

Find the identity element for βˆ— on A, if any.

Q8. Prove πΆπ‘œπ‘‘βˆ’1 (√1+sinπ‘₯+√1βˆ’sinπ‘₯

√1+sinπ‘₯βˆ’ √1βˆ’sinπ‘₯) =

π‘₯

2 , x∈ (0,

πœ‹

4)

OR

Solve for x . 2 π‘‘π‘Žπ‘›βˆ’1(cos π‘₯) = π‘‘π‘Žπ‘›βˆ’1(2 π‘π‘œπ‘ π‘’π‘ π‘₯)

Q9. By using properties of determinants, show that:

|1 + π‘Ž2 βˆ’ 𝑏2 2π‘Žπ‘ βˆ’2𝑏

2π‘Žπ‘ 1 βˆ’ π‘Ž2 + 𝑏2 2π‘Ž2𝑏 βˆ’2π‘Ž 1 βˆ’ π‘Ž2 βˆ’ 𝑏2

| = (1 + π‘Ž2 + 𝑏2)3

Q10. If cos y = x cos(a + y) with cos a β‰  Β± 1, prove that 𝑑𝑦

𝑑π‘₯=

π‘π‘œπ‘ 2( π‘Ž+𝑦)

sinπ‘Ž

OR

Find 𝑑𝑦

𝑑π‘₯ of the function (cos π‘₯)𝑦 = (cos 𝑦)π‘₯

Q11. If = (π‘‘π‘Žπ‘›βˆ’1π‘₯)2 , show that (π‘₯2 + 1)2𝑦2 + 2π‘₯(π‘₯2 + 1)𝑦1 = 2

Q12

If f(x) =

{

1βˆ’cos4π‘₯

π‘₯2 π‘€β„Žπ‘’π‘› π‘₯ < 0

π‘Ž, π‘€β„Žπ‘’π‘› π‘₯ = 0 √π‘₯

√16+√π‘₯βˆ’4 , π‘€β„Žπ‘’π‘› π‘₯ > 0

and f is continuous at x = 0, find the value of a.

Q13. Find the intervals in which the function f given by f(x) = 2x3 βˆ’ 3x

2 βˆ’ 36x + 7 is

(a) strictly increasing (b) strictly decreasing

Q14. Show that [οΏ½οΏ½ + b οΏ½οΏ½ + οΏ½οΏ½ 𝑐 + οΏ½οΏ½ ] =2[οΏ½οΏ½ οΏ½οΏ½ 𝑐 ]

OR

Find a unit vector perpendicular to each of the vectors (οΏ½οΏ½+ οΏ½οΏ½) π‘Žπ‘›π‘‘ ( π‘Ž - οΏ½οΏ½) where οΏ½οΏ½ = 𝑖 + 𝑗 +

οΏ½οΏ½ and οΏ½οΏ½ = 𝑖 + 2 𝑗 + 3οΏ½οΏ½ .

Q15. Evaluate: ∫2π‘₯

(π‘₯2+1)(π‘₯2+3)𝑑π‘₯ dx

Q16. Evaluate: ∫ 𝑒π‘₯ (1+sinπ‘₯

1+cosπ‘₯) dx

Q17. Using properties of definite integrals, evaluate:

∫π‘₯

4 βˆ’ π‘π‘œπ‘ 2π‘₯𝑑π‘₯

πœ‹

0

OR

Using properties of definite integrals, evaluate:

∫ π‘™π‘œπ‘”(1 + tan π‘₯)𝑑π‘₯

πœ‹4⁄

0

Q18. . A man is known to speak truth 3 out of four times .He throw a die and report that it is a

six find the probability that it is actually six. Which value is discussed in this question?

Q19. Find the shortest distance between the lines

)k2j5-i(3k-ji2r

and )Λ†Λ†Λ†2(Λ†Λ†

kjijir

Section C

Q20. Two institutions decided to award their employees for the three values of resourcefulness,

competence and determination in the form of prizes at the rate of Rs. x , Rs.y and Rs.z

respectively per person. The first Institute decided to award respectively 4,3 and 2 employees

with a total prize money of Rs.37000 and the second Institute decided to award respectively 5, 3

and 4 employees with a total prize money of Rs.47000.If all the three prizes per person together

amount to Rs.12000, using matrix method find the value of x, y and z. Write the values described

in the question.

Q21. Solve the differential equation

𝑑𝑦

𝑑π‘₯+ 2 𝑦 tan π‘₯ = sin π‘₯ , given that y = 0 where x =

πœ‹

3

Q22. ) Find the equation of plane passing through the line of intersection of the planes

x + 2y + 3 z = 4 and 2 x + y – z + 5 = 0 and perpendicular to the plane 5 x + 3y – 6 z + 8 = 0.

Q23. The sum of the perimeter of a circle and square is k, where k is some constant. Prove that

the sum of their areas is least when the side of square is double the radius of the circle.

OR

Show that the volume of greatest cylinder that can be inscribed in a cone of height h and semi

vertical angle Ξ± is, 23 tan27

4h .

Q.24 There are a group of 50 people who are patriotic, out of which 20 believe in non-violence.

Two persons are selected at random out of them, write the probability distribution for the

selected persons who are non- violent. Also find the mean of the distribution. Explain the

importance of non- violence in patriotism.

Q25. Using integration Find the area lying above x-axis and included between the circle

π‘₯2 + 𝑦2 = 8 x and parabola 𝑦2 = 4 x

OR

Using the method of integration, find the area of the region bounded by the following lines

5x - 2y = 10, x + y – 9 =0 , 2x – 5y – 4 =0

Q26. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of

the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs

60/kg and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of

vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B.

Determine the minimum cost of the mixture? What is the importance of Vitamins in our body?

Pratima Nayak,KV Teacher

Marking Scheme Second Pre Board Examination Mathematics-2014 Kolkata Region

Q1. - πœ‹

3 Q2.2x Q3. y = -6 Q4. 1 Q5. p = - 2 Q6. 8/7 1 X 6

Q7. (π‘Ž, 𝑏) βˆ— (𝑐, 𝑑) = (𝑐, 𝑑) βˆ— (π‘Ž, 𝑏) for commutativity. 2

11

((π‘Ž, 𝑏) βˆ— (𝑐, 𝑑)) βˆ— (𝑒, 𝑓) = (π‘Ž, 𝑏) βˆ— ((𝑐, 𝑑) βˆ— (𝑒, 𝑓)) for associativity. 2

11

No identity element. 1

______________________________________________________________________ Q8.

1 1/2

2

11

2

11

Β½ _______________________________________________________________________ OR

1

1

2

11

1/2

_________________________________________________________________________________________

Ans 9.

Applying R1 β†’ R1 + bR3 and R2 β†’ R2 βˆ’ aR3, we have: 1

1

Expanding along R1, we have: (1 + π‘Ž2 + 𝑏2)3 1

Answer 1

____________________________________________________________________________________

Q10.π‘₯ = cos𝑦

cos (π‘Ž+𝑦) 1

𝑑π‘₯

𝑑𝑦 = =

sina

π‘π‘œπ‘ 2( π‘Ž+𝑦) 1+1

𝑑𝑦

𝑑π‘₯ =

π‘π‘œπ‘ 2( π‘Ž+𝑦)

sinπ‘Ž 1

________________________________________________________________________________________________

OR

Taking logarithm on both the sides,

Differentiating both sides

1

1

_______________________________________________________________________

Q11. 1

1

1

1

_____________________________________________________________________________

Q12

limπ‘₯β†’0βˆ’π‘“(π‘₯) =2𝑠𝑖𝑛22π‘₯

π‘₯2= 8 1Β½

RHL on rationalization limπ‘₯β†’+𝑓(π‘₯) =8 1Β½

a = 8 1

________________________________________________________________________

Q13.

1

x = βˆ’ 2, 3 1/2 Intervals: (- ∞,-2),( -2,3) and (3, ∞) 1

(f) is strictly increasing in (- ∞,-2) and (3, ∞) and strictly decreasing in interval ),( -2,3) 1½

___________________________________________________________________________

Q14. = {(οΏ½οΏ½ Γ— b)+(οΏ½οΏ½ Γ— c) + (οΏ½οΏ½ Γ— b) + (οΏ½οΏ½ Γ— c)} . (𝑐+οΏ½οΏ½ ) 1

= (οΏ½οΏ½ Γ— b) . 𝑐 + (οΏ½οΏ½ Γ— c). 𝑐+( οΏ½οΏ½ Γ— c). 𝑐 + (οΏ½οΏ½ Γ— b) . οΏ½οΏ½ + (οΏ½οΏ½ Γ— c). οΏ½οΏ½+( οΏ½οΏ½ Γ— c). οΏ½οΏ½ 2

=[οΏ½οΏ½ οΏ½οΏ½ 𝑐 ] + [οΏ½οΏ½ οΏ½οΏ½ 𝑐 ] 2

1

=2[οΏ½οΏ½ οΏ½οΏ½ 𝑐 ] 2

1

_________________________________________________________________________ OR

(π‘Ž + οΏ½οΏ½) = 2𝑖 + 3 𝑗 + 4οΏ½οΏ½. ( οΏ½οΏ½- οΏ½οΏ½) = 0οΏ½οΏ½ βˆ’ 𝑗 βˆ’ 2οΏ½οΏ½. 1

(π‘Ž + οΏ½οΏ½)𝑋 ( π‘Ž - οΏ½οΏ½) = βˆ’2𝑖 + 4 𝑗 βˆ’ 2οΏ½οΏ½. 2

11

|(π‘Ž + οΏ½οΏ½)𝑋 ( π‘Ž - οΏ½οΏ½ | =√24 1

2

1

1

√24(βˆ’2𝑖 + 4 𝑗 βˆ’ 2οΏ½οΏ½) Β½

_________________________________________________________________________

Q15. Let x2 = t β‡’ 2x dx = dt 1/2

1

A=1/2, B=-1/2 1/2

2

_____________________________________________________________________________________________________

Q16. 12

1

1

Β½+1 _____________________________________________________________________

Q17 Use of property

∫ 𝑓((π‘₯)𝑑π‘₯ = π‘Ž

0 ∫ 𝑓((π‘Ž βˆ’ π‘₯)𝑑π‘₯ = π‘Ž

0 , I = ∫

πœ‹βˆ’π‘₯

4βˆ’π‘π‘œπ‘ 2π‘₯𝑑π‘₯

πœ‹

0 1/2

2I = πœ‹ βˆ«π‘ π‘’π‘2π‘₯

3+4 π‘‘π‘Žπ‘›2π‘₯𝑑π‘₯

πœ‹

0 1/2

Use of property ∫ 𝑓((π‘₯)𝑑π‘₯ = 22π‘Ž

0 ∫ 𝑓((π‘Ž βˆ’ π‘₯)𝑑π‘₯ π‘Žπ‘  𝑓(2π‘Ž βˆ’ π‘₯) = 𝑓(π‘₯) π‘Ž

0

2I = 2πœ‹/4 βˆ«π‘ π‘’π‘2π‘₯

3+4 π‘‘π‘Žπ‘›2π‘₯𝑑π‘₯

πœ‹/2

0 1

tan x = t, sec2x dx =dt 1

& Correct result I =πœ‹2

4√3 1

_________________________________________________________________

OR 1

2

1

_________________________________________________________________________

Q18. P(T) =3/4, P(F) =1/4 1

E: getting a six,F: he is not getting a six

P(E/T)= 1/6, P(E/F)=5/6 1

By Bay,s Theorem P(T/E)= 𝑃(𝑇)𝑃(

𝐸

𝑇)

𝑃(𝑇)𝑃(𝐸

𝑇)+𝑃(𝐹)𝑃(

𝐸

𝐹) =3/8

2

11

Truthfulness Β½

Q19.

59bb

Λ†7Λ†Λ†31

,Λ†Λ†

21

2

12

kjibb

kiaa

1+1+1

shortest distance =

21

2221 )).((

bb

aabb

=

59

10 1

_____________________________________________________________________ Q20. 4 x + 3 y + 2z = 37000, 5 x + 3 y + 4z = 47000, x + y + z = 1200 1

|A| = - 3 β‰  0 so A-1 exists. X = A-1 B 1/2

Cofactors of A 2

[βˆ’1 βˆ’1 2βˆ’1 2 16 βˆ’6 3

]

Adjoint A 1/2 X = 4000 ,y = 5000, z = 3000 1Β½ Values Β½ _______________________________________________________________ Q21. P = 2tan x, Q = sin x

I.F = 𝑠𝑒𝑐2π‘₯ 1Β½

y 𝑠𝑒𝑐2π‘₯ = ∫ sin π‘₯ 𝑠𝑒𝑐2π‘₯ 𝑑π‘₯ + 𝐢 1

y 𝑠𝑒𝑐2π‘₯ = sec x + C 1Β½

y = 1

secπ‘₯ +

𝐢

𝑠𝑒𝑐2π‘₯= cos x + C π‘π‘œπ‘ 2π‘₯ -------------------(1) 1Β½

putting x = πœ‹

3 and y = 0 in eqn (1) C = -2 Β½

Y= cos x - 2π‘π‘œπ‘ 2π‘₯ 1 __________________________________________________________________________ Q22. Sol: The required plane is (x + 2 y + 3 z ) + k (2 x + y – z +5 )= 0 1

Or (1 + 2 k)x +(2 + k)y +(3 - k)z-4 + 5k = 0 1

5(1+2k) +3 (2+k) -6 (3-k)=0 1

Solving k = 7/19 1

The equation of the plane is : 33 x + 45y +50z = 41 . 2

________________________________________________________________________________

Q23.

Let r be the radius of the circle and a be the side

1+1 1/2 2 1/2 a =2 r 1 ___________________________________________________________________ OR fixed height (h) and semi-vertical angle (Ξ± )

relation of h and H 1+1/2 ( figure)

1

+ Β½ Result 1 _________________________________________________________________

24. Let X = The number of non -violent persons out of selected two. So, X = 0, 1, 2 1/2

P(X = 0) = 245

87

2

50

2

30

C

C P(X = 1) =

245

120

2

50

1

30

1

20

C

CCP(X = 0) =

245

38

2

50

2

20

C

C

3

X 0 1 2

P(X) 245

87

245

120

245

38

Mean = )(XPX = 245

196

245

382

245

1201

245

870 2

Importance of non- violence Β½

1

1

______________________________________________________________

25. (1) π‘₯2+𝑦2= 8x (π‘₯ βˆ’ 4)2+𝑦2= 16 represents a circle with centre (4,0) and radius 4 units 1/2

(2) 𝑦2 =4 x represents parabola with vertex at origin and axis as x-axis. Β½+ Β½ ( figure)

Point of intersection of the curves are (0,0) and (4,4)

= ∫ √4π‘₯4

0 dx + ∫ √8π‘₯ βˆ’ π‘₯2

8

4 dx 1+1/2

=2∫ π‘₯4

01/2 dx + ∫ √( 16 βˆ’ (π‘₯ βˆ’ 4)2

8

4 dx

= 2[2

3π‘₯32⁄ ]0

4

+ [π‘₯βˆ’4

2√16 βˆ’ (π‘₯ βˆ’ 4)2 +

16

2sinβˆ’1

π‘₯βˆ’4

4]4

8

2

= 32

3+ 4πœ‹ π‘ π‘ž 𝑒𝑛𝑖𝑑𝑠 1

_______________________________________ OR Solving (1) and (2) point of intersection is C(4,5) Solving (2) and (3) point of intersection is B(7,2)

Solving (1) and (3) point of intersection is A(2,0) 2

11

Area of triangle ABC= area of triangle ACD + area of CDEB + area of triangle ABE

= ∫5π‘₯βˆ’10

2

4

2𝑑π‘₯ + ∫ (9 βˆ’ π‘₯)𝑑π‘₯ βˆ’ ∫

2π‘₯βˆ’4

5

7

2

7

4𝑑π‘₯

2

11

= Β½ [ [5π‘₯2

2βˆ’ 10π‘₯]

2

4

+ [9π‘₯ βˆ’π‘₯2

2]4

7

βˆ’1

5[π‘₯2 βˆ’ 4π‘₯]

2

7 2

11

= 21

2 π‘ π‘ž 𝑒𝑛𝑖𝑑𝑠

2

11

-______________________________________________________________

Q26. Let the mixture contain x kg of food P and y kg of food Q. Minimise Z = 60x + 80y 1/2 subject to the constraints, 3x + 4y β‰₯ 8 … (2) 5x + 2y β‰₯ 11 … (3)

x, y β‰₯ 0 … (4) 2

11

Figure and shading 2

12

The corner points of the feasible region are .A(8/3,0),B(2,1/2),C(0,11/2)

minimum cost Rs 160 at the line segment A(8/3,0) & B(2,1/2) 2

11

Marking scheme can be for any alternative method by the evaluator. Pratima Nayak,KV Teacher


Recommended