prepared by m.srinivasan, ziet, mumbai · web view5. linear progrmming. 6 (1) 6 (1) 6....

PREPARED BY M.SRINIVASAN, ZIET, MUMBAI

SAMPLE PAPER

CLASS XII : MATHEMATICS

BLUE PRINT

S.No. Topics VSA SA LA TOTAL1 a) RELATIONS AND FUNCTIONS 1 (1) 4 (1)

10 (4)b) INVERSE TRIGONOMETRIC

FUNCTIONS

1 (1) 4 (1)

2 a) MATRICES 2 (2) 6(1)13 (5)

b) DETERMINANTS 1 (1) 4 (1)

3 a) CONTINUITY &

DIFFERENTIABILITY

8 (2)

44 (11)

b) APPLICATION OF

DERIVATIVES

4(1) 6(1)

c) INTEGRATION 2(2) 4(1) 6(1)

d) APPLICATION OF INTEGRALS 6 (1)

e) DIFFERENTIAL EQUATIONS 8(2)

4 a) VECTORS 2 (2) 4 (1)17 (6)

b) 3-DIMENTIONAL GEOMETRY 1 (1) 4 (1) 6 (1)

5 LINEAR PROGRMMING 6 (1) 6 (1)

6 PROBABILITY 4 (1) 6 (1) 10 (2)

TOTAL 10

(10)

48

(12)

42(7) 100(29)

SAMPLE PAPER : CLASS XII : MATHEMATICS


Time:03 hours Max Marks: 100

General Instructions All questions are compulsory The question paper consists of 29 questions divided into three sections

A, B and C Section A contains 10 questions of 1 mark each, Section B is of 12

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

There is no overall choice. However, internal choice has been provided in 4 questions of four marks and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.

Use of calculators is not permitted

SECTION – A

1. Find gf(-3) if f(x) =|x| and g(x) = |5 x−2|

2. Find the principal value of sin−1[sin 3π5 ]3. If A = (3 1

2 −3), then find |adj A|

4. If A=( sinα cosα−cosα sin α ), verify that A’ A = I.

5. If A and B are symmetric matrices of the same order, then show that AB – BA is a skew symmetric matrix.

6. ∫ 1+cot xx+ log(sinx)

dx|7. Evaluate: ∫

−12

12

cos x log [ 1+sin x1−sinx ]dx

8. If a=i+2 j−3 k and b=2 i+4 j+9 k, find a unit vector parallel to a + b9. If the vectors, a=2 i−5 j+k and b=2 i+ j+4 k are parallel find the value of

and 10. Using direction ratios, show that the points A(2, 3, 4), B(-1, -2, 1) and

C(5, 8, 7) are collinear.


SECTION – B 11. Show that the relation R, defined on the set A of all triangles as:

R = {(T1 , T2): T1 is similar to T2} is an equivalence relation.(OR)

Let f: N {0} N {0} defined by : f ( x )={n−1if nis evenn+1if nis odd .

Show that f(x) is a bijective function.

12. Evaluate: tan−1 15 +tan−1 17 + tan−1 1

3+tan−1 1

8

13. Using properties of determinants prove that: |1 x x2

x2 1 xx x2 1 | = (1 – x3)2

14. If the function f(x) given by: f ( x )={3ax+b , if x>111x=15ax−2b if x<1

is continuous at x = 1, find the value of a and b.

15. Find dydx for y = tan−1[ √1+sinx+√1−sinx

√1+sinx−√1−sinx ](OR)

If y = log {x+√ x2+a2 } prove that (x2 + a2) y2 + x y1 = 0 16. Water is running in to a conical vessel, 15 cm deep and 5 cm in radius, at

the rate of 0.1 cm3/sec. When the water is 6 cm deep, find at what rate is the level of water increasing.

17. Evaluate: ∫ (x2+1 ) (x2+4 )(x2+3 ) (x2−5 )

dx

(OR)Evaluate: ∫ 1

sinx ¿¿¿

18. Solve the differential equation: (tan-1y – x)dy = (1 + y2) dx19. Form the differential equation of the family of hyperbolas having foci on x-

axis and center at origin.20. A bag A contains 8 white and 7 black balls while the other bag B contains 5

white and 4 black balls. One ball is randomly picked up from bag A and mixed up with the balls in the bag B. Then a ball is randomly drawn from it. Find the probability the ball drawn is white.

(OR)Find the mean and variance of the number of heads in a two tosses of a coin.


21. Find the area of the triangle with vertices A(1 , 1, 2), B(2, 3, 5) and C(1, 5, 3)22. Find the equation of the plane passing through the line of intersection of

the planes 2x + 6y + 12 = 0 and 3x – y + 4z = 0 which are at a unit distance from the origin.

SECTION C

23. Using Matrices, solve the following system of equations: 2x – 3y + 5z = 11; 3x + 2y – 4z = -5 ; x + y – 2z = -3

24. A window is in the form of a rectangle surrounded by a semi-circular opening. The total perimeter of window is 10 meters. Find the dimensions of the window so as to admit maximum light through the whole opening.

(OR)An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when depth of the tank is half of its width.

25. Evaluate: ∫0

π xa2 cos2 x+b2 sin2 x

dx

26. Find the area of the region: {(x , y) : x2 + y2 ≤ 1 ≤ x + y}(OR)

Compute the area bounded by the lines x +2y = 2 ; y – x = 1 and 2x + y = 727. Find the distance of the point (1, 2, 3) from the plane x – y + z = 5 measured

parallel to the line x2=y3= z

−6

28. A diet is to contain at least 90 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 cost Rs.4/- per unit and F2 costs Rs.6/- per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost of the diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

29. In a bolt factory, machines A, B and C manufacture 25%, 35% and 40% of the total. Of their output 5%, 4%, 2% are defective. A bolt is drawn at random from the product. a) What is the probability that the bolt drawn is defective? b) If the bolt drawn is found to be defective, find the probability that it is a product of machine B?


SAMPLE PAPER : CLASS XII : MATHEMATICSSOLUTIONS

Q.No SOLUTION MARKS AWARDED1 gf(x) = |5|x|−2| gf(-3) = 13 1

2sin 3π

5=sin [π−2 π

5 ] = sin 2π5 , 2π5 ∈[−π2 , π2 ]

PRINCIPAL VALUE = 2π5

1

3 adj A=(−3 −1−1 3 ) |adj A| = -11 1

4 A' A=(sinα −cosαcos α sin α )( sinα cos α

−cosα sin α )=(1 00 1)=I 1

5(AB – BA)’ = (AB)’ – (BA)’ = B’A’ – A’B’ = BA – AB(AB – BA)’ = - (AB – BA)

1

6

Put t = x + log (sin x)

∫ 1+cot xx+ log(sinx)

dx=∫ dtt = log t = log [ x + log (sinx)]

1

7

f(x) = cos x log [ 1+sin1−sinx ] ; f(-x) = -cos x log [ 1+sin x1−sinx ]f(-x) = - f(x) f(x) is a odd function

∫−12

12

cos x log [ 1+sin x1−sin x ]dx=0

1

8

a + b =3 i+6 j+6 k

Parallel unit vector = a+b

|a+b|=3 i+6 j+6 k

9

1

9The vectors are parallel if a b = 0 = 4 ; = -5

1

10

Direction ratios of AB are -3 : -5 : -3 = 3 : 5 : 3Direction ratios of BC = 6 : 10: 6 = 3 : 5: 3 AB parallel to BC. B is common. The points are collinear

1


Q.No SOLUTION MARKS AWARDED

11

For T1 A, T1 is similar to T1.(T1 , T1) R R is Reflexive

For T1 , T2 A, Let T1 is similar to T2. Then T2 is also similar to T2. Hence (T1 , T2) RR is symmetricFor T1 , T2 , T3 A, Let T1 is similar to T2 and T2 similar to T3. Then T1 similar to T3

Hence (T1 , T3) R R is transitive As R is reflexive, symmetric and transitive, R is an equivalence relation

1

1

1

1

11

Let m and n are even number.For f(m) = f(n), we have m – 1 = n – 1 m = nLet m and n are odd numbersFor f(m) = f (n), we have m+1 = n+ 1 m = nIf m is even and n is odd then nmThen f(m) is odd and f(n) is even f(m) f(n)Hence f is injective (1 – 1)If n is odd natural number then there exists an even natural number n+1 such that f(n+1) = n.If n is even natural number then there exists an even natural number n-1 such that f(n-1) = n.Hence f is surjective. (ONTO)Hence f is bijective

2

2

12

tan−1 15+tan−1 1

7=tan−1 12

34

tan−1 13+tan−1 1

8=tan−1 11

23

tan−1 15+tan−1 1

7 + tan−1 13+tan−1 1

8 = tan−11 =π4

1 ½

1 ½

1`

13

(1 – x3)2 = [(1 – x) (1 + x + x2)]2

Taking (1 – x) as common Taking (1 – x ) as commonTaking (1 + x + x2) as commonExpanding the determinant and getting the answer

111

1



14Using LHL, 5a – 2 b = f(1) = 11using RHL, 3a + b = f(1) = 11Using the above equations, a = 3 and b = 2

1 ½1 ½

1

15

√1+sinx = √sin2 x2+cos2 x2+2sin x2 cos x2=(cos x2 +sin x2 )

√1−sinx = √sin2 x2+cos2 x2−2sin x2 cos x2=(cos x2−sin x2 )y=tan−1[ √1+sinx+√1−sinx

√1+sinx−√1−sinx ]= tan−1{cot x2 }y=tan−1 {( tan( π2− x

2 ))} = π2− x2

dydx

=−12

1

1

1

1

15Find the first derivative and cross multiplying : y1√ x2+a2=1

Find the second derivate and getting the answer

2

2

16

V = Volume of water in the cone

tan = rh

= 515

h = 3r

V = π27h3

dhdt

= 273π h2

= 140 π

11

2

17

PUT y = x2

(x2+1 ) (x2+4 )(x2+3 ) (x2−5 )

= y2+5 y+4y2−2 y−15

= 1 + 7 y+19( y+3 )( y−5)

1+ 7 y+19( y+3 )( y−5) = 1 +

14 (x2+3)

+ 27

4 (x2−5)

∫ (x2+1 ) (x2+4 )(x2+3 ) (x2−5 )

=x+ 14√3

tan−1 1√3

+ 278√5

log|x−√5x+√5 |

1

2

1

17 Multiply and divide by sinx Put cos x = t and sin x dx = - dt

∫ sinxsin2 x¿¿

¿ = ∫ −dt(1−t ) (1+ t )(2+t )

1

1

1


1(1−t ) (1+t )(2+t)

= 16(1−t)

+ 12(1+t)

− 13(2+t )

∫ 1sinx ¿¿

¿ =

16log|1−cosx|−1

2log|1+cosx|+ 1

3log|2+cosx|

1


18The given DE is dxdy

+ x1+ y2

= tan−1 y

1+ y2

I.F = e tan−1 y

Solution: x = (tan-1y – 1 ) + c e tan−1 y

11

2

19The equation of family of hyperbolas: x

2

a2− y2

b2 = 1

Differentiate twice to eliminate a and bThe required D.E is x y y’’ + x (y’)2 – y y’ = 0

11 +1

1

20

A = event of transferring a white ballB = event of transferring a black ball

P(A) = 815 and P(B) =

715

E = event of selecting white ball from II bagP(E) = P(AE or BE) = P(AE) + P(BE)

= 815× 610

+ 715

× 510 =

83150

1

1

2

20

X 0 1 2P(X) 1

412

14

Mean = 32

Variance = 34

2

1

1

21

Area of triangle = 12|BC×BA|

BC=− I+2 J+0 k ; BA=− I−2 J−3 kBC× BA=−6 I−3 J +4 k

Area of the triangle = √612

1

2

1

22

Required plane : (2x + 6y + 12) + (3x – y + 4z) = 0Use perpendicular distance from (0,0,0) = 1 = 2The required planes are 2x + y + 2z + 3 = 0 and x – 2y +2 z – 3 = 0

1

2

1

23 Expressing : AX = B 1


Getting A−1=( 0 1 −2−2 9 −23−1 5 −13)

x = 1 ; y = 2 ; z = 3

4

1



24

Let x and y are length and breadth of the rectangle

Area of the figure = 20−(2+π ) x2

4+π x2

8When A’(x) 0 gives x = 20

4+πSecond derivative is negative at this value of x

Area is maximum when

Length = 204+π

Breadth = 104+π

Radius of semi-circle = 104+π

1

11

1

11

24

For the tank: length = x ; breadth = x and height = y

S= x2+ 4Vx

dSdx = 0 2 x3 = 4V x = 2y

The second derivative is positive at this valueS is minimum when x = 2yThe depth of the tank is half of its width

2

21

1

25

Using the property of definite integrals:I =∫0

π 1a2 cos2 x+b2 sin2 x

dxMultiply and divide by Sec2x, put t = tan x Change the limits

I = π2

2ab

1

11

3

26

points of intersection (1,0)

Area = ∫0

1

¿¿

Area = π4−12

1

2

3



26

Getting Points of intersection

Area =

∫0

2

{( x+1 )− 2−x2 }dx+∫

2

4

{(7−2 x )− 2−x2 }dx

Area = 6 square units

1

3

2

27

The equaiton of the line through A and parallel to given line:

x−12

= y+23

= z−3−6

Coordinate of the point P is

( 97 ,−117 , 157 )

Distnce of AP = 1

2

2

2

28

The LPP is Min Z 4x + 6y subject to3x + 6y 80 ; 4x + 3y 100, x,y0

Mimimum cost = Rs.104 when 24

units of F1 and 43 units of F2 are mixed.

1 + 1

3

1


29

P (A )= 25100

; P (B )= 35100

; P (B )= 40100

a) P (D )=P (A )P ¿) + P(B) P (D / B) + P(C) P(D / C) = 0.0345

b) P(B / D) = 2869 using baye’s theorem

1

2

3

prepared by m.srinivasan, ziet, mumbai · web view5. linear progrmming. 6 (1) 6 (1) 6....

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