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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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