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SAMPLE MODEL PAPER BY GROUP IV

SUBJECT : MATHEMATICS

|M.M. 100 |CLASS XII |TIME : 3 HRS. |

| |

|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 calculator is not permitted. However, you can use log tables. |

| | |

| | |

| |SECTION A ( 1MARKS) |

| | |

|1 |If Sin (tan-11/5 + cot-1 x ) = 1, then find the value of x. |

|2 |For a given 3x3 squared matrix A, │A│ = -4, and A.( Adj. A) = ( Adj. A) . A = ƛ I. Find ƛ. |

|3 |Find the equation of plane through the point (1,4,-2) and parallel to the plane |

| |- 2x + y -3z = 7. |

|4 |Give an example of a relation which is symmetric but neither reflexive nor transitive. |

|5 |Find the vector in the direction of a→ = [pic]whose magnitude is 7. |

|6 |If A is a matrix of order 2 X 3 and B is a matrix of order 3 X 5. What is the order of matrix (AB)’. |

| |SECTION B(4MARKS) |

|7 |Solve for x: [pic] |

| |OR |

| |Show that tan-1 [pic]= sin-1 [pic]+ cos-1 [pic] |

|8 |Solve the differential equation y dx - (x+2y2) dy = 0 |

|9 |Show that the function f : W→W defined by |

| |[pic], is a bijective function. |

|10 |Using properties of determinants, Prove that: |

| |[pic] |

|11 |Show that function f(x) defined by |

| |[pic] is continuous at x=0 |

|12 |Find dy/dx if y = (Cos x)x + (Sin x)1/x. |

| |OR |

| |Find [pic] of functions (cos x)y = (cos y)x |

|13 |Evaluate ∫ (x – 4) ex/(x – 2)3 dx |

|14 |Find the intervals in which the function f given by |

| |f(x) = Sin x + Cos x, 0 ≤ x ≤ 2π is increasing or decreasing. |

| |OR |

| |Find intervals in which ƒ(x)=20-9x+6x2-x3 is strictly increasing or decreasing. |

|15 |Evaluate [pic] . |

|16 |Evaluate ∫(5x +3)/ [pic]) dx |

| |OR |

| |Integrate : [pic] |

|17 |If the sum of two units vectors is a unit vector. Show that the magnitude of their difference is √3. |

|18 |Find the shortest distance between the lines |

| |(x – 3)/1 = (y – 5)/ - 2 = (z – 7)/ 1 and (x + 1)/7 = (y + 1)/ - 6 = (z+ 1)/ 1. |

|19 |The probabilities of two students A and B coming to the school in time are 3/7 and 5/7 respectively. Assuming that the events, ‘ a coming |

| |in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time. |

| |Write at least one advantage of coming to school in time. |

| | |

| |SECTION C(6 MARKS) |

|20 |The cost of 2 cycles, one motorbike and one car is Rs. 2, 89,000: the cost of three cycles, two motorbikes and one car is Rs. 3, 43,500 and|

| |the cost of one cycle, one motorbike and two cars is Rs. 5, 14,500. |

| |i) Represent the following information as linear equations. |

| |ii) Solve the linear equations using matrices. |

| |iii) A Student has the option to buy a cycle, a motorbike or a car. Which mode of transport would you suggest? |

|21 |A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions |

| |of the window to admit maximum sunlight through the whole opening. |

| |Explain the importance of sunlight. |

| |OR |

| |Show that area of right triangle of given hypotenuse is maximum when triangle is isosceles triangle. Also find maximum area. |

| |******** |

|22 |In answering a question on a multiple choice questions test with four choices per question. A student knows the answer, guesses or copies |

| |the answer. If ½ be the probability that he knows the answer, ¼ be the probability that he guesses it and ¼ that he copies it. Assuming |

| |that a student who copies the answer will be correct with the probability ¾. What is the probability that the student knows the answer |

| |given that he answered it correctly? Mehul does not know the answer to one of the question in the test. The evaluation process has negative|

| |marking. Which value would Mehul ‘violate’ if he restores to unfair means? |

| |How would his personality would be hampered? |

|23 |Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point(1,3,4) from the plane 2x – y + z +3 = 0. |

| |Find also, the image of the point in the plane. |

| |OR |

| |Find equation of plane passing through point P(4,-1,2) and which is parallel to lines. |

| |[pic] and [pic] |

|24 |A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B |

| |are available at a cost of Rs.5 and Rs.4 per unit respectively. One unit of the food A contains 200 units of vitamins, 1 unit of minerals |

| |and 40 units of calories, while one unit of the food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Find |

| |what combination of the foods A and B should be used to have least cost, but it must satisfy the requirements of the sick person. Form the |

| |question as LPP and solve it graphically. Explain the importance of balanced diet. |

|25 |Using integration, find the area of triangle ABC, where A is (2,3), B is (4,7) and C is (6,2). |

|26 |Find the particular solution of the differential equation: |

| |(x + y) dy + (x – y) dx = 0 given that, when x = 1, y =1. |

SOLUTION TO SAMPLE PAPER BY GROUP IV

SECTION - A

1. Any correct relation

2. 1/5 0

3. ƛ= - 4

4. 2x – y + 3z +8 = 0

5. 7 ([pic])/ √5

6. 5 X 2

SECTION - B

7.

tan -1 + 2 tan -11/x = 2π/3 1 ½ m

tan-1 + tan-1 ((2/x) / (1 – 1/x2 ) = 2 π /3

tan-1 + tan-1 (2x)/ (x2 + 1) = 2 π /3 1 m

tan-1 (x + (2x/(x2 -1)) / (1 – (x.2x) /(x2 - 1) = 2 π /3 1m

Solving to get x = √3 1 ½

OR

Sin-1 ([pic]) ½ m

Cos-1([pic]=[pic]=[pic] ½ m

tan ([pic]+[pic]) =[pic] ½ m

[pic] 1 ½ m

[pic]+[pic]=tan-1 ([pic])

Sin-1 ([pic]) + cos-1([pic] 1 m

8. [pic] 1m

[pic] 1m

[pic] 1m

[pic] 1m

9. Proving one – one function for all three cases:

( One is odd & other is even)

( Both n1, n2 are even)

( Both n1, n2 are odd) 1 ½ m

Proving onto function 1 ½ m

f is 1 – 1 and onto

f(n) is bijective function 1m

10. Applying C1 → C1 - bC3, C2 → C2 + aC3 1m

Taking Common (1 + a2 + b2)2 from C1 and C2

[pic] 2m

Solving to get ( 1 + a2 + b2)3 1m

11. Solving LHL = 2. 1 m

Solving RHL = 2 1 m

Also f(0) = 2

LHL = RHL = f(0) 1m

f(x) is continous at x = 2. 1m

12. let u = (Cos x)x , v = (Sin x)1/x

y = u + v

dy/dx = du/dv + dv/dx 1m

u = (Cos x)x

log u = x log.cosx

finding du/dx = (Cos x)x [-x tanx + log(cosx) 1m

Similarly from v = (Sinx)1/x

finding dv/dx = (Sin x) 1/x [x cotx + log(Sinx)] 1m

getting dy/dx 1m

OR

Log (cos x)y =log (cos y)x

y log (cos x) =x log (cos y) 1m

y([pic])[pic]cos x + log (cos x)[pic]=x ([pic])[pic] 1m

-y tan x+ log (cos x) [pic] 1m

[pic] 1m

13. I = ∫ (x – 2 - 2 ) ex/(x – 2)3 dx.

∫ex [1/(x – 2 )2 – 2/(x – 2)3 ] dx 1m

Taking f(x) = 1/(x – 2 )2

f’(x) = – 2/(x – 2)3 1m

using formula ∫ex (f(x) + f’(x)) dx = ex f(x) + C

I = ex / (x – 2 )2+ C 1m

14. f’(x) = Cosx – Sin x

Put f’(x) = 0 tanx =1

x = [pic]/4 , 5[pic] /4 as 0 ≤ x ≤ 2[pic] 1m

Possible intervals are

[0, [pic] /4), ([pic] /4 , 5[pic] /4), (5[pic] /4 , 2[pic]] 1m

Finding increasing in [0, [pic] /4) and (5[pic] /4 , 2[pic]] 1m

And decreasing in ([pic] /4 , 5[pic] /4) 1m

OR

| |[pic]ƒ'[pic]= -9 + 12x-3x2 |

| |=-3 (x2-4x+3) |

| |=-3 (x-3) (x-1) 1m |

| |For stationary points |

| |[pic]ƒ’[pic]=0 |

| |[pic]x=3,1 1m |

| |F is decreasing in (-[pic])U [pic] 1m |

| |F is increasing in (1,3) 1m |

15. Simplifying to [pic] Sin 2 x dx 1m

I = [pic] 1m

2I = [pic] 1m

Solving to get I = [pic] 2 / 4 1m

16. 5x + 3 = A d/dx(x2 +4x +10) + B

A(2x + 4) + B

= A = 5/2 , B = -7 1m

I = 5/2 ∫ 2x + 4/ √(x2 +4x +10) dx – 7 ∫ dx/ √(x2 +4x +10) 1m

Getting I = ∫ 5/2 √(x2 +4x +10) - 7 ∫ dx/ √(x+2)2 +(√ 6)2 1m

Getting I = ∫ 5/2 √(x2 +4x +10) - 7 log|x + 2 + √(x2 +4x +10| + c 1m

OR

| |[pic] |1m |

| |=[pic][pic] | |

| |=[pic][pic] put sin x- cos x=t (cosx+sin x) dx =dt | |

| |=[pic][pic] | |

| |= [pic] sin-1 t | |

| |=[pic]sin-1 (sin x –cos x) +c | |

| | |1m |

| | | |

| | |1m |

| | | |

| | | |

| | | |

| | |1m |

17. Let [pic] , [pic], [pic]be three vectors

[pic] = [pic] [pic]

And a→ + b→ = c→ 1m

| a→ + b→| = | c→ | = 1

| a→ + b→|2 =1

( a→ + b→).( a→ + b→) =1 1m

To get 2 a→ . b→ = -1 1m

Finding | a→ - b→|2 = 3

And | a→ - b→| = √3. 1m

18. For writing the values of a1→ , a2→ , b1→ , b2→ 1m

Using correct formula of S.D. 1m

Finding S.D. = 2√29 2m

19. P(A )= 3/7, P(B) = 5/7, P( ‾A )= 4/7, P( ‾B) =2/7, 1m

P(only one of them coming in time) = P(A ) P( ‾B) + P( ‾A) P(B) 1m

Solving and get result = 26/49 1m

Any suitable justification for coming to school in time. 1m

SECTION C

20. Let cost of a cycle, motorbike, car = x, y, z respectively

2x + y + z = 289000 ,

3x + 2y + z = 343500,

x + y +2 z = 514500 1m

Writing in AX = B form and finding I A I = 2 and 1m

I Adj A I = [pic] 2m

Finding A-1 and solving the equations to get,

x = Rs. 4500, y = Rs. 50000, z = Rs. 230000. 2m

21. Correct figure 1m

Let length of rectangle =2 x, Let breadth of rectangle =y

2x + 2y + [pic] x = 10

Y = (10 – 2x - [pic] x )/2 1m

Finding area A, dA/dx, d2A/dx 1m

Put dA/dx = 0 we get x = 10/ ([pic] +4)

d2A/dx = - ([pic] +4) < 0 maximum area 1 ½ m

y = 10/ ([pic] +4) ½ m

Comments on importance of sunlight 1m

OR

[pic]

[pic]

22. A͢ : Knows the answer

B : Guesses the answer ,

C : Copies ,

D : Answered correctly

P(A) = 1/2 , P(B) = 1/4 , P(C) = 1/4,

P(E/A) = 1 , P(E/B) = 1/4, P(E/A) = 3/4, 2m

Applying Baye’s theorem & getting the result P(A/E) = 3 3m

Suitable remarks by Mehul 1m

23. Let P(1,3,4) & Q be the foot of perpendicular to the plane.

Equation of line through (1,3,4) is (x - 1)/2 = (y-3)/ -1 = (z – 4)/1 = ƛ 1m

Any point on the line Q (2 ƛ+1, - ƛ+3, ƛ-4)

Q lies on given plane 2x – y +z +3 =0and getting ƛ = -1 2m

Q(-1, 4, 3) and IPQI = √6 2m

Image = (-3, 5, 2) 1m

OR

[pic]

24. Let x units of food A, y units of food B are mixed Z = 5x + 4y

200x+100y ≥ 4000, 40x + 40y ≥ 1400, x ≥ 0, y ≥ 0 2m

Correct graph 2m

Points (50,0), (20,15), (5,30), (0, 40) and minimum cost is at (5, 30) 2m

25. Correct graph 1m

Finding the equations AB, BC, AC as y = 2x – 1, y = -5x/2 +17 and y = -x/4 + 7/2 2m

Finding area = 9 Sq. Units 3m

26. Put y = vx, dy/dx = v + x dv/dx On Solving we get 1m

∫ (v+1)/(v-1) .dv = - ∫ 1/x dx 1m

½ log I v2+1 I + tan-1 v = - log x +c 2m

At x =1, y =1, finding c = log 2 + π/2 1m

to get solution log I x2 + y2 I + 2 tan -1 y/x = log 2 + π/2 1m

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