Sample Question Paper (With value based questions) Issued by CBSE for 2013 Examination Mathematics Class–XII Blue Print S. No.
Topics
VSA
SA
LA
Total
—
4(1)
—
—
2(2)
4(1)
—
10(4)
2(2)
—
6(1)
—
1(1)
4(1)
—
13(5)
1(1)
12(3)
—
—
(b) Applications of Derivatives
—
—
6(1)
44(11)
(c) Integration
—
12(3)
—
—
(d) Application of Integrals
—
—
6(1)
—
1(1)
—
6(1)
—
2(2)
4(1)
—
—
(b) 3-dimensional Geometry
1(1)
4(1)
6(1)
17(6)
5.
Linear Programming
—
—
6(1)
6(1)
6.
Probability
—
4(1)
6(1)
10(2)
10(10)
48(12)
42(7)
100(29)
1. (a) Relations and Functions (b) Inverse Trigonometric Functions 2. (a) Matrices (b) Determinants 3. (a) Continuity and Differentiability
(e) Differential Equations 4. (a) Vectors
Total
Note: — Number of questions are given within brackets and marks outside the t he brackets. — The Question Paper will include question(s) based on values to the extent of 5 marks.
CBSE Sample Question Paper–2013 (with Value Based Questions)
Time allowed: 3 hours
Maximum marks: 100
General Instructions: 1. All questions are compulsory. 2. The question paper consists of 29 questions divided into three Sections A, B and C. Section A comprises of 10 questions of one mark each; Section B comprises of 12 questions of four marks marks each; and Section C comprises of 7 questions 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. marks 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. permitted. You may ask for logarithmic tables, if required. 5. Use of calculator is not permitted.
SECTION–A
Question numbers 1 to 10 carry 1 mark each. 1. Using principal values, write the value of 2 cos 1
1 3 sin 1 1 . 2 2
1 . 2 1 0 0 x 1 3. Write the value of x y z, if 0 1 0 y 1 . 0 0 1 z 0 2. Evaluate tan 1 2 cos 2 sin 1
4. If A is a square matrix of order 3 such that |adj A| = 225, find | A’| cos sin . 5. Write the inverse of the matrix sin cos 6. The contentment obtained after eating x-units of a new dish at a trial function is given by the Function C(x) = x3 + 6x2 + 5x + 3. If the marginal contentment is defined as rate of change of (x) with respect to the number of units consumed at an instant, then find the marginal contentment when three units of dish are consumed.
d 2 y d 2 y dy 2 2 1 0. 7. Write the degree of the differential equation dx dx 2 dx
CBSE Sample Question Paper
(iii)
2 8. If a and b are two vectors of magnitude 3 and , respectively such that a b is a unit vector, 3 write the angle between a and b . 9. If a 7i j 4k and b 2i 6 j 3k, find the projection of a on b .
10. Write the distance between the parallel planes 2x y 3z 4 and 2x y 3z 18. SECTION–B
Question numbers from 11 to 22 carry 4 marks each. 11. Prove that the function f : N N , defined by f ( x) x 2
x 1 is one – one but not onto.
x2
1 x2 2
12. Show that sin [ cot 1 {cos(tan 1 x)}] OR
2 2x . 1 2x 1 1 x Solve for x: 3 sin 4 cos 2 tan 1 1 x 2 1 x 2 3 1 x 2
13. Two schools A and B decided to award prizes to their students for three values: honesty ( x), punctuality (y) and obedience (z). School A decided to award a total of ` 11,000 for these three values to 5, 4 and 3 students, respectively, while school B decided to award ` 10,700 for these three values to 4, 3 and 5 students, respectively. If all the three prizes together amount to ` 2,700 then (i) Represent the above situation by a matrix equation and form Linear equations by using matrix multiplication. (ii) Is it possible to solve the system of equations so obtained using matrices? (iii) Which value you prefer to be rewarded most and why? 14. If x a( sin ) and y a(1 cos ), find 15. If y
sin 2 x 1 x2
, show that (1 x 2 )
2
d y dx 2
d2y dx
3x
2
.
dy y 0. dx
x 2 ax b , 0 x 2 16. The function f ( x) is defined as f ( x) 3x 2 , 2 x 4 . If f ( x) is continuous on [0, 8], 2ax 5b , 4x8 find the values of a and b. OR 2 1 1x Differentiate tan
1 x 2 with respect to cos 1 x 2 . 1 x 2 1 x 2
(iv)
Xam idea Mathematics–XII
x3
17. Evaluate:
x1 . dx x 1 2
OR
(1 – sin x) dx. (1 cos x) 2x dx. 18. Evaluate: 2 2 (x 1)( x 2)
Evaluate: e x
4
19. Evaluate: log(1 tan x) dx , using properties of definite integrals. 0
20. Let a 4i 5j k, b i 4j 5k and c 3i j k. Find a vector d which is perpendicular to both a and b and satisfying d . c 21. 21. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, –1, 2), B(5, 2, 4), and C(–1, –1, 6) OR
Find the equation of the perpendicular drawn from the point P(2, 4, –1) to the line x 5 y 3 z 6 . Also, write down the coordinates of the foot of the perpendicular from 1 4 9 P to the line. 22. There is 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. SECTION–C
Question numbers from 23 to 29 carry 6 marks each. 1 2 3 23. If A 2 3 2 , find A 1 . Hence solve the following system of equations:
3 3 4 x 2 y 3z 4; 2x 3 y 2 z 2 and 3x 3y 4z 11 x7 24. Find the equations of tangent and normal to the curve y at the point where it ( x 2)( x 3) cuts the x-axis. OR Prove that the radius of the base of right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half that of the cone. 25. Find the area of the region which is enclosed between the two circles x 2
( x 1) 2 y 2 1.
y 2 1 and
(v)
CBSE Sample Question Paper
26. Find the particular solution of the differential equation: ( x sin y) dy (tan y) dx 0: given that y 0 when x 0. 27. Find the vector and Cartesian equations of the plane containing the two lines r ( 2i j 3 k) (i 2 j 5k) and r ( 3i 3 j 2k) ( 3i 2 j 5k). 28. A dealer in rural area wishes to purchase a number of sewing machines. He has only ` 5,760.00 to invest and has space for at most 20 items. An electronic sewing machine costs him ` 360.00 and a manually operated sewing machine ` 240.00. He can sell an Electronic Sewing Machine at a profit of ` 22.00 and a manually operated sewing machine at a profit of ` 18.00. Assuming that he can sell all the items that he can buy how should he invest his money in order to maximize his profit. Make it as a linear programming problem and solve it graphically. Keeping the rural background in mind justify the ‘values’ to be promoted for the selection of the manually operated machine. 29. In answering a question on a MCQ test with 4 choices per question, a student knows the 1 1 answer, guesses or copies the answer. Let be the probability that he knows the answer, 2 4 1 be the probability that he guesses and that he copies it. Assuming that a student, who 4 3 copies the answer, will be correct with the probability , what is the probability that the 4 student knows the answer, given that he answered it correctly? Arjun does not know the answer to one of the questions in the test. The evaluation process has negative marking. Which value would Arjun violate if he resorts to unfair means? How would an act like the above hamper his character development in the coming years? OR An insurance company insured 2000 cyclists, 4000 scooter drivers and 6000 motorbike drivers. The probability of an accident involving a cyclist, scooter driver and a motorbike driver are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? Which mode of transport would you suggest to a student and why?
Solutions SECTION–A
1 3 sin 1 1 2 3 7 2 2 3 6 6 1 2. tan 1 2 cos 2 sin 1 tan 1 2 cos 2 3 tan 1 2 1 = tan 1 (1) 2 4 1. 2 cos 1
1
1
(vi)
Xam idea Mathematics–XII
1 0 0 x 1 0 1 0 y 1 0 0 1 z 0 x 1 , y 1 , z 0
3. Given
4. We know that |adj A| =| A|n 1 | A |3 1 225
and
x 1 y 1 z 0 x y z 1 1 0 0 | A||A |
1
| A |2 225 15
1
cos sin sin cos
5. Let A =
Then, |A| = cos 2 sin 2 1
and
cos sin sin cos
adjA
cos sin sin cos 6. M(X) C (X) 3 x 2 12 x 5 M(X)( at X 3) 27 36 5 = 68 units A 1
7. 2 8. a × b = | a | | b |sin n | a × b | = | a | | b ||sin |
sin 1
2
or
1
1 1
1 3 2 sin 3
b. a 9. Projection of a on b = | b|
1
6
14 6 12 2
2
2 6 3
2
8
10. Given parallel planes are 2x y 3z 4 ...(i) and 2x y 3z 18 Let P(x 1 , y 1 , z1 ) be a point on plane (i) 2x 1 y1 3z1 4 Now the distance 'd' from P( x 1 , y 1 , z1 ) to plane (ii) is given as 2x 1 y 1 3z1 18 d 41 9
d
1
7
4 18 14 14 14 14 14 14 14
...(ii)
1
SECTION–B
11. f ( x) x 2
x1 Let x 1 , y 1 N such that f ( x 1 ) f (y 1 )
1
CBSE Sample Question Paper
(vii)
x12
x 1 1 y12 y 1 1 x 12 y 12 x 1 y 1 0 ( x 1 y 1 )( x 1 y1 1) 0 ( x 1 y 1 ) 0 [As x 1 y 1 1 0 any x 1 , y 1 N] f is one-one function x1 y1 Clearly f ( x) x 2 x 1 3 for x N
1/2
But f ( x) does not assume values 1 and 2
1
f : N N is not onto function 1 1 12. cos(tan 1 x) cos cos 1 x 2 1 x2 1 2 1 x 1 1 1 sin 1 sin cot x 2 1 1 (x 2 1) 2 sin cot 1 1 sin sin 1 x 2 1 x 2 x 2 1
1
1 2
1
1 x 2 2 x2
x2
1 x2 2
1
1+1
OR
Let x tan LHS = 3 sin 1
1/2
2 tan 4 cos 1 1 tan 2 2 tan 1 2 tan 1 tan 2 1 tan 2 1 tan 2
= 3 sin 1 (sin 2) 4 cos 1 (cos 2 ) 2 tan 1 (tan 2 )
1
= 3 2 8 4 2 2 tan 1 x
1
2 tan 1 x tan 1 x 3
6
x 1
1
3
13.
1 2
5 4 3 x 11000 (i) The given situation can be represented as follows: 4 3 5 y 10700 1 1 1 2700 z or 5x 4y 3z 11000 4x 3y 5z 10700 x y z 2700
5 4 3 (ii) Let A 4 3 5 1 1 1
1
1 2
(viii)
Xam idea Mathematics–XII
| A| 5 ( 2) 4 (1) 3 (1) 10 4 3 3 0 1 1 1/2 A exists, so equations have a unique solution. (iii) Any answer of three values with proper reasoning which will be considered correct.1 [For example, I prefer the value "punctuality" for rewards, because a punctual students can study better.] dx a(1 cos ) 2a sin 2 14. x a ( sin ) 1 2 d dy a. sin 2a sin cos y a(1 cos ) 1 2 2 d
d2 y dx 2 15. y
2a sin . cos dy 2 2 cot 2 dx 2a sin 2 2
1 cosec 2 . d 2
2 dx
sin 1 x
1 2 sin 2
.
1
2a sin 2 2
1 cosec 4 4a
1
2
1/2
1x Differentiating both sides w.r.t x, we get y( 2 x) dy 1 1 x2 dx 2 1 x 2 1 x2
1
dy xy 1 0 dx Differentiating again w.r.t x, we get
(1 x 2 )
2
(1 x ) 2
(1 x ) 16.
d2y dx 2 d2y dx 2
1/2
dy y . 1 x 0 dx
2x
dy dx
3x
dy y 0 dx
1 2
2
1 x 2 . y sin 1 x
2
1/2
1 1
lim f ( x) lim x 2 ax b 2 a b 4
x 2
x 2
lim f ( x) lim ( 3x 2) 8
x 2
1
x 2
As f is continuous at x 2 2a b 4 8 2a b 4 Similarly as f is continuous at x 4, lim f ( x) lim f ( x)
...(i)
1
lim ( 3x 2) lim ( 2ax 5b), 14 8 a 5b
...(ii)
1
x 4
x 4
x 4
x 4
Solving (i) and (ii), we get a 3, b 2
1
CBSE Sample Question Paper
(ix)
OR
1 x2 1 x2 , Let x 2 cos 2 1 x 2 1 x 2 tan 1 1 cos 2 1 cos 2 tan 1 cos sin cos sin 1 cos 2 1 cos 2 tan 1 1 tan tan 1 tan 1 tan y 1 cos 1 x 2 : Let z cos 1 x 2 2 dy y 1 z 1
y tan 1
4
2
2 x3 x 1 2x 1 17. I = dx x dx x2 1 (x 1)( x 1) dz
Let
x2 3 3 dx 1 dx I = x dx log( x 1) 1 log( x 1) C 2 x1 2 x1 2 2 2 OR e x (1 2 sin x cos x ) 1 sin x 2 2 dx e x dx 1 cos x x 2 sin 2 2 1 2 sin x cos x 2 2 dx e x 1 cosec 2 x cot x dx = ex 2 2 2 2 sin 2 x 2 e x cot x dx 1 e x cosec 2 x dx 2 2 2 cot x . e x e x cosec 2 x . 1 dx 1 e x cosec 2 x dx 2 2 2 2 2 e x cot x 1 e x cosec 2 x dx 1 e x cosec 2 x dx C 2 2 2 2 2 e x cot x C 2 2x dx 18. I = ( x 2 1)( x 2 2) dt Let x 2 t, 2x dx dt I = (t 1)(t 2)
1 1
1 2
1
2 1
2x 1 A B A + B = 2, A – B = 1 A 3 , B 1 ( x 1)( x 1) x 1 x 1 2 2
1
1 1+1
1
1/2
1/2 1 1
1
(x)
Xam idea Mathematics–XII
Let
1 A B A + B = 0 (t 1)(t 2) t 1 t 2 2 A B 1 on solving, we get A 1, B 1
I = dt dt log|t 1| log|t 2| C t1 t2 log|x 2 1| log|x 2 2| C
1
1 2 1
1/2
4
19. Let I = log(1 tan x) dx
...(i)
0
4
4 1 tan x log1 tan 4 x dx = log1 1 tan x dx 0 0
1
4
2 dx 4 log 2 log(1 tan x)]dx 1 tan x 0
or I = log
0
...(ii)
1
Adding (i) and (ii), we get 4
2I = log 2 dx 0
log 2
1
4
I = log 2
1
8
20. Let d xi yj zk As d a and d b d . a 0 and d . b 0 d . a 0 4x 5 y z 0 and d . b 0 x 4 y 5z 0 d . c 21 3x y z 21
1
...(i)
1
...(ii)
1/2
Solving (i) and (ii), we get x 7 , y 7 , z 7
d i 7 j 7 k
1 1/2
21. The equation of plane passing through (3, –1, 2), (5, 2, 4) and (–1, –1, 6) is x 3 y1 z 2 x 3 y 1 z 2 3 5 1 2 2 4 0 or 2 3 2 0
3 1 1 1 6 2 ( x 3)(12) ( y 1)(16) (z 2)(12 ) 0 12x 16 y 12z 76 or 3x 4 y 3z 19 Length of from (6, 5, 9) to (i) is
4
4
0 ...(i)
2
1/2
CBSE Sample Question Paper
18 20 27 19
6
34
32 42 32
OR x 5 y 3 z 6 Any point on the line 9 1 4
is ( 5, 4 3; 9 6)
(xi)
1
1
1
1
2
2
P (2 , 4, 1)
A
x + 5 y + 3 z – 6 = = 1 4 –9 B
Q
Let ( , , ) is the foot of the perpendicular drawn from point P( 2, 4, 1) to the given line x 5 y 3 z 6 …(i) 9 1 4 Since, ( , ) be on line …(i) 5 3 6 (say) 9 1 4 5 , 4 3 , 9 6
PQ ( 5 2)i (4 3 4) j (9 6 1) k ( 7)i ( 4 7) j (9 7 )k PQ line (i) ( 7).1 (4 7 ).4 (9 7 ) 9 0 98 98 1 The pt. Q is (–4, 1, –3) y 4 z 1 and foot of is ( 4 , 1 , 3) Equation of PQ is x 2 6
3
2
1 1/2 1
22. Let X denote the number of non-violent persons out of selected two. X can take
values 0, 1, 2 non-violent 20: Violent patriotism: 30 30 29 87 P(X = 0) = 50 49 245 30 20 20 30 30 20 2 120 P(X = 1) = 50 49 50 49 50 49 245 20 19 38 P(X = 2) = 50 49 245
1/2 1/2 1/2 1/2
(xii)
Xam idea Mathematics–XII
Probability distribution is 0 87 245
X P(x)
Mean = 0
1 120 245
2 38 245
87 120 38 196 1 2 245 245 245 245
1
Importance: In order to have a peaceful environment both the values are required patriotism and non-violence because of patriotism with violence could be very dangerous. 1 SECTION–C
1 2 3 23. The given matrix is A 2 3 2 ,| A| = – 6 + 28 + 45 = 67 0 3 3 4 A 1 exists 6 17 13 6 17 13 1 1 14 5 8 Adj A = 14 5 8 , A 67 15 9 1 15 9 1
1
2
The given system of equations can be written as AX = B 1 2 3 x 4 Where A = 2 3 2 , X = y , B 2 3 3 4 z 11
1
6 17 13 4 201 3 14 5 8 2 1 134 2 67 67 15 9 1 11 67 1 x 3, y 2 , z 1 1 X A 1 B
1 1/2 1/2
24. The given curve cuts the x-axis at x = 7, and y = 0
y
x 7 x2
5x 6
dy dx
( x 2 5 x 6) ( x 7 )( 2x 5)
( 49 35 6) (0) dy ( at x 7) dx ( 49 35 6) 2
1
( x 2 5 x 6) 2
1
20
1 2
1 2
1 2
1
2
Equation of tangent to the curve at (7, 0) is 1 ( x 7) or x 20 y 7 0 20 Equation of normal to the curve at (7, 0) is y 0
1
1 2
y 0 20( x 7 ) 20 x y 140 0
1
1 2
CBSE Sample Question Paper
(xiii)
OR Let x and r be radius of base of cylinder and cone respectively Let OC = x , VOB ~ B DB h(r x) V VO OB BD h B D DB r Let S be the curved surface area of cylinder. h(r x) 2h A' O' [rx x 2 ] S 2xh 2x r r
dS 2h d2 s (r 2x), dx r dx 2 S is maximum dS 0 r 2x dx
1 1
B'
1
h
4 h0
h'
1
r
A
O
C r
D
B
x
1
S is is maximum when x r , i.e., when radius of base of cylinder is half the radius of base 2
of cone.
1
1 3 25. On solving the equations of the two circles, we get points of intersection as A , and 2 2 1 D , 2
3 2
1 Y A
(
1 , 3 2 2
) ( x – 1) 2 + y 2 = 1
x 2 + y 2 = 1
’
O
X
B
D
(
1 , 3 2 2
)
Y’
Area of shaded region = 2 (Area OABCO)
1 1 2 2 1 ( x 1) 2 dx 1 x 2 dx 0 1 2
1
1
1 ( x 1) x 1 2 x 1 1 2 1 2 x x 1 ( x 1) 2 sin 1 1 sin 1 1 0 2 2 2 2 2
2
(xiv)
Xam idea Mathematics–XII
3 sin 1 1 sin 1 (1) sin 1 (1) 3 sin 1 1 2 4 2 4 3 3 4
6 2 2 4 2 3 sq. units 2 3
1
6
1
26. The given differential equation can be written as dx (cot y)x cos y dy
I.F. = e
cot y dy
e log sin y sin y
The solution is x sin y = sin y cos y dy C
1
1 sin 2 y dy C or x sin y
1 cos 2 y C
1
2
1
4 It is given that y 0, when x 0 1 C 1 C 0 4 4 x sin y 1 (1 cos 2y) 1 sin 2 y 4 2 2x sin y is the reqd. solution
27. Here a1 2i j 3k and a 2 3i 3j 2k b 1 i 2j 5k and b 2 3i 2 j 5k n b1 b2
1
1 1
1
i j k 1 2 5 20i 10j 8k 3 2 5
1
1 2
Vector equation of the required plane is ( r a1 ). n 0 or r . n a1 . n
or r .( 20i 10j 8k) (2i j 3 k).( 20i 10 j 8 k) 40 10 24 74
2
r .(10i 5j 4 k) 37 1 2 28. Suppose number of electronic operated machine = x and number of manually operated sewing machines = y 1/2
The cartesian equation of plane is 10x 5y 4z 37
1
CBSE Sample Question Paper
(xv)
Y 28
24
(0, 24)
20 A(0, 20) 16
2
P(8, 12)
12 8 4
B(16, 0) X’
O Y’
4
8
12
16
3 20 x + 2
y
= 4 8
24
28
X
x
+
y
= 2 0
...(i) x y 20 and, 360 x 240 y 5760 or 3x 2y 48 ...(ii) x 0, y 0 To maximise Z 22 x 18 y 2 Corners of feasible region are A ( 0, 20), P(8 , 12), B(16 , 0 ) Z A 18 20 360 1/2 Z P 22 8 18 12 392 Z B 352 Z is maximum at x 8 and y 12 The dealer should invest in 8 electric and 12 manually operated machines. Keeping the ‘save environment’ factor in mind the manually operated machine should be promoted so that energy could be saved. 1 29. Let A be the event that he knows the answer, B be the event that he guesses and C be the event that he copies. 1/2 1 1 1 Then P( A) , P( B) and P(C) 1/2 2 4 4 Let X be the event that he has answered correctly. X X 1 X 3 Also, P 1, P and P 1 A B 4 C 4
(xvi)
Xam idea Mathematics–XII
X P P( A) A A Thus, required probability = P 1 X X X X P P( A) P P( B) P P(C) A B C 1 1 1 2 2 1 2 1 1 1 1 3 1 1 3 3 1 2 4 4 4 4 2 16 16 If Arjun copies the answer, he will be violating the value of honesty in his character. He should not guess the answer as well as that may fetch him negative marking for a wrong guess. He should accept the question the way it is and leave it unanswered as cheating may get him marks in this exam but this habit may not let him develop an integrity of character in the long run. 2 OR Let the events defined are E1: Person chosen is a cyclist E2: Person chosen is a scooter-driver 1/2 E3: Person chosen is a motorbike driver A: Person meets with an accident 1/2 1 1 1 P(E1) = , P(E2) = , P(E3) = 1 6 3 2 A A A E P = 0.01, P = 0.03, P = 0.15, P 2 = Required 1 A E1 E2 E 3
A . P(E2 ) E E 2 P 2 A A A A P . P(E1 ) P . P(E2 ) P . P(E 3 ) E1 E2 E 3 P
1 3 3 100 600 6 3 1 1 1 3 1 15 100 52 52 26 6 100 3 100 2 100 Suggestion: Cycle should be promoted as it is good for (i) Health (ii) No pollution (iii) Saves energy( no petrol)
1 1/2 1/2 1/2 1/2
Value Based Questions Mathematics–XII 1. Everyone wants to be a perfect ideal human being. Let us assume that dishonesty is one of the factors that affects our perfectness and perfectness has an inverse square relationship with dishonesty. For any value x of level of dishonesty we have a unique value y of perfection. (i) Write down the equation that relates y with x. (ii) Is this relationship from x X ( 0,
) to y ( 0, ), forms a function?
1 (iii) For what level of dishonesty one can achieve th level of perfection? 4
Sol.
(iv) What will be the change in level of perfection when the level of dishonesty changes from 4 to 2? 1 (i) y , x0 x2
(ii) Yes (iii) When y
1 1 1 , we have 4 x 2 x 2, but x can not be –ve 4 4 x2
x 2
1 16 1 When x 2, y 4
(iv) When x 4, y
Change in level of perfection = 1 1 = 3
4 16 16 2. A trust fund has ` 30,000 that is to be invested in two different types of bonds. The first bond pays 5% p.a. interest which will be given to orphanage and second bond pays 7% interest p.a. which will be given to financial benefits of the trust. Using matrix multiplication, determine how to divide ` 30,000 among two types of bonds if the trust fund obtains an annual total interest of ` 1800. (i) What are the values reflected in the question? (ii) Why is it required to help orphan children? Sol. Let ` x be invested in Ist bond, then ` 30,000 – x will be invested in IInd bond.
Total interest = ` 1800
(xviii)
V
Xam idea Mathematics–XII
Now,
A L U E
B A S E D
Q U E S T I O N S
5 [1800] [x 30000 x] 100 7 100 5x 7 ( 30000 x) 1800 100 100 5x + 210000 – 7x = 1800 × 100
30000 15000 2 For investment in IInd bond, amount = 30000 15000 = 15000. So, equal amount is invested in both of the bonds. (i) Values reflected are helping poor and needy children. Provided that the interest rate in financial benefits (IInd bond) is more than the Ist bond (money given to orphanage) trust decides to invest fund equally. It reflects that the motive of the trust is not to only to earn the interest but also to help the needy orphan children. This charity should be a concern of every one. (ii) The children living in orphanage are also talented and possess potential. If they are given the proper brought up and opportunity, they can contribute to the development of the society and country and will become good citizens.
–2x + 210000 = 180000
x
3. Of the students in a school; it is known that 30% has 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who has 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he has A grade. What is the probability that the student has 100% attendance? (i) Write any two values reflected in this question. (ii) Is regularity required only in school? Justify your answer Sol. Let E1 : Student has 100% attendance E2 : Student is irregular
A : Student attains A grade A : Student attains A grade given that she has 100% attendance E1 A : Student attains A grade given that she is irregular E2 E1
: Student has 100% attendance given that she attains A grade A Using Bayes’ theorem
Value Based Questions
A E1
E1
A
(xix)
P
P(E1 ). P
V
A P(E2 ). P A E1 E2
A
P(E 1 ). P
L 30 70 100 100 U 30 70 70 10 100 100 100 100 E 30 70 30 3 30 70 70 10 30 10 4 (i) Regularity and intelligence (ii) Regularity is the value which is required at every stage of our life. In our B childhood, during school education we can inculcate this value in our personality. Regularity increases our capabilities and makes us able to put the best of our A potential. We are able to achieve certain targets due to regular efforts. 4. In a survey of 20 richest person of three residential society A, B, C it is found that in society A, 5 believes in honesty, 10 in hard work, 5 in unfair means while in B, 5 in honesty, 8 in hard in work, 7 in unfair means and in C, 6 in honesty, 8 in hard work, 6 in unfair means. If the per day income of 20 richest persons of society A, B, C are ` 32,500, 30,500, 31,000 respectively, then find the per day income of each type of people by matrix method. (i) Which type of people has more per day income. (ii) According to you, which type of person is better for country. Sol. Let x, y, z be the per day income of person believing in honesty, hard work and unfair means, respectively. The given situation can be written in matrix form as
A X = B, 5 10 A 5 8 6 8 AX B Now for A 1
Where 5 x 32500 7 . X y , B 30500 6 31000 z
X A 1 B
C 11
(1) 1 1
8 7 (48 56) 8 8 6
E D
Q U E S
...(i)
5 10 5 | A| = 5 8 7 = 5(48 – 56) – 10(30 – 42) + 5(40 – 48) = 40 0 6 8 6 Also,
S
T I O N S
(xx)
Xam idea Mathematics–XII
V A
C 12
( 1) 1 2
5 7 ( 30 42) 12 6 6
C 13
( 1) 1 3
5 8 ( 40 48) 8 6 8
C 21
( 1) 2 1
10 5 ( 60 40) 20 8 6
C 22
( 1) 2 2
5 5 ( 30 30) 0 6 6
C 23
( 1) 2 3
5 10 ( 40 60) 20 6 8
C 31
( 1) 3 1
10 5 (70 40) 30 8 7
C 32
( 1) 3 2
5 5 ( 35 25) 10 5 7
C 33
( 1) 3 3
5 10 (40 50) 10 5 8
L U E
B A S E D
Q U E S T I O N S
T 8 12 8 8 20 30 Adj ( A) = 20 0 20 12 0 10 30 10 10 8 20 10
A 1
adj( A) | A|
8 20 30 1 1 3 40 40 40 5 2 4 12 1 3 10 = 0 = 0 40 408 20 410 10 1 1 1 4 40 40 40 5 2 Putting the value of X, A 1 , B in (i) we get 1 1 3 x 5 2 4 32500 y 3 0 1 . 30500 10 4 z 1 1 1 31000 5 2 4 x 1500 y 2000 z 1000
x 2000 y 1000 z 3000
Value Based Questions
x 1500 , y 2000 , z 1000 Hence, per day income of person who believe in honesty = ` 1,500 Per day income of person who believe in hard work = ` 2,000 Per day income of person who believe in unfair means = ` 1,000 (i) A person, who believe in hard work has more per day income. (ii) A person, who believe in hard work and honesty, are better for country. 5. The male-female ratio of a village increases continuously at the rate proportional to the ratio at any time. If the ratio of male-female of the villages was 1000 : 980 in 1999 and 1000 : 950 in 2009, what will be the ratio in 2019?
(xxi)
V A L U E
(i) Why gender equality is value for society? (ii) What should society do to reduce the male-female ratio to 1? Sol. Let male-female ratio at any time be r. dr dr Given where k is the constant of proportionality. r kr dt dt dr We have k dt r
Integrating both sides we get log r kt log c where log c is the constant of integration. r log r log c kt log kt c
r c e kt
B A S E D
...(i)
Let us start reckoning time from the year 1999 for this problem. 1000 50 So in 1999, t 0 and r 980 49 Substituting in (i) we get 50 50 c c . e 0 49 49 (i) becomes 50 kt ...(ii) r e 49 1000 20 Also in the year 2009, t 10 and r 950 19 Substituting in (ii) we get 20 50 10 k 98 e 10 k e 19 49 95 Substituting in (ii) we get
Q U E S T I O N S
(xxii)
Xam idea Mathematics–XII
V A
In the year 2019, t 20
L
20 50 98 10
U E
B A S E D
Q U E S T I O N S
t
t
50 10 k 10 50 98 10 r .( e ) 49 49 95
r
50
98
...(iii)
98
49 95 49 95 95 100 98 ~ 1. 085 ~ 1085 : 1000 95 95
Thus in the year 2019, the male-female ratio will be 1085 : 1000 (i) Gender equity promotes economic growth, reduce fertility, child mortality and under nutrition. (ii) (a) Stop female-foeticide. (b) Empower women to realise their rights. (c) Provide special opportunities to women to come at par with men in all walks of life. 6. A window is in the form of rectangle surmounted by a semi-circular opening. Total perimeter of the window is 10 m. What will be the dimensions of the whole opening to admit maximum light and air? (i) How having large windows help us in saving electricity and conserving environment? (ii) Why optimum use of energy is required in the Indian context? Sol. Let ABCED be required window having length 2x and width y. If A is the area of window. Then 1 A 2 xy x 2 2 1 x(10 2x x) x 2 2 1 10x 2x 2 x 2 x 2 2 1 Given, Perimeter = 10 2x y y 2x 10 2
E
D
C
y
A
2 x
2 y 10 2 x x 1 1 10x 2x 2 x 2 10x 2 x 2 2 2 Obviously, window will admit maximum light and air if its area A is maximum. dA 1 Now, 10 2x 2 2 dx
B
Value Based Questions
(xxiii)
For maxima or minima of A dA 0 dx 1 10 2x 2 0 10 x(4 ) 0 2
x
10 4
d2 A dx 2
V A L
( 4 ) < 0
For maximum value of A, x
10
U 10
and thus y 4 4 Therefore, for maximum area i.e., for admitting maximum light and air, 20 Length of rectangular part of window = 2x 4 10 Width = 4 (i) Large windows allow more light during daytime and hence will reduce the use of electricity. Saving energy (electricity) helps in conserving environment as electricity is produced by using natural resources which we should conserve for the sake of future generation. (ii) India is the 2nd most populated country in the world so have more consumers of energy but less sources of its production. Therefore, in Indian context energy saving is like energy production. 7. In a competition, a brave child tries to inflate a huge spherical balloon bearing slogans against child labour at the rate of 900 cubic centimeter of gas per second. Find the rate at which the radius of the balloon is increasing when its radius is 15 cm.
E
B A S E D
Q
(i) Which values have been reflected in this question? (ii) Why child labour is not good for society? Sol. Let r be the radius and V be the volume of the balloon. Then dV 900 cm3/sec dt dr ?, when r = 15 dt 4 V r 3 3
Differentiating both sides w.r.t ‘t’ dV 4 dr 3r 2 3 dt dt 900 1 dr dr 900 4(15) 2 cm/sec dt dt 225 4
U E S T I O N S
(xxiv)
V A L U E
B
Xam idea Mathematics–XII
(i) Three values reflected are bravery, sympathy for child labour and raising voice against child labour. (ii) We know that child labour is illegal and harmful to both society and country. We should spread awareness in society so that child labour should be abolished. In the childhood they should be sent to school for their education so that they can contribute for the development of the society. 8. A manufacturing company makes two type of teaching aids A and B of mathematics of class XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30, respectively. The company makes a profit of ` 80 on each piece of type A and ` 120 on each piece of type B. How many pieces of type A and B should be manufactured per week to get a maximum profit? What is the maximum profit per week? Is teaching aid necessary for teaching learning process? If yes, justify your answer.
A S E D
Q U
Sol. Let x and y be the number of pieces of type A and B manufactured per week respectively. If Z be the profit then,
Objective function, Z = 80x 120y We have to maximize Z, subject to the constraints 9x 12 y 180 3x 4y 60 ...(i) x 3y 30 ...(ii) ...(iii) x 0, y 0 The graph of constraints are drawn and feasible region OABC is obtained, which is bounded having corner points O ( 0, 0), A (20 , 0), B (12 , 6) and C ( 0, 10) Y 3
x
E
+ 4 y
= 6 0
S T
25
20 x
+ 3 y = 15 3 0
I
10
O
5
N S
C(0, 10)
B(12, 6) A(20, 0) X’
O Y’
10
20
30
40
50
60
X
Value Based Questions
(xxv)
Now the value of objective function is obtained at corner point as Corner point
Z = 80 x 120y
O (0, 0)
0
A (20, 0)
1600
B (12, 6)
1680
C (0, 10)
1200
V A
Maximum
Hence, the company will get the maximum profit of ` 1,680 by making 12 pieces of type A and 6 pieces of type B of teaching aid. Yes, Teaching aid is necessary for teaching learning process as (i) it makes learning very easy. (ii) it provides active learning. (iii) students are able to grasp and understand concept more easily and in active manner. 9. A village has 500 hectares of land to grow two types of plants, X and Y. The contribution of total amount of oxygen produced by plant X and plant Y are 60% and 40% per hectare respectively. To control weeds, a liquid herbicide has to be used for X and Y at rates of 20 litres and 10 litres per hectare, respectively. Further no more than 8000 litres of herbicides should be used in order to protect aquatic animals in a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total production of oxygen?
L U E
B A S E D
(i) How do you think excess use of herbicides affect our environment? (ii) What are the general implications of this question towards planting trees around us? Sol. Let plants X and Y be grown in x and y hectares.
So,
x 0 and y 0 x y 500
...(i) Contribution of oxygen by the plants = 60% of x + 40% of y 6 x 4y z 0 .6 x 0 . 4 y 10 10 Also, Amount of liquid herbicides required = ( 20 x 10 y) litres Given 20 x 10 y 8000 2x y 800 ...(ii) The LPP for given problem is Maximum, Z 0. 6x 0. 4 y S.t. ...(iii) x y 500 and 2x y 800 ...(iv) x, y 0
Q U E S T I O N S
(xxvi)
V
Xam idea Mathematics–XII
Sketching a graph for the above LPP, we get the region shown in the figure Y
A
900
L
800
U
700
E
600
(0, 800)
A(0, 500) 500
B
400
A
300
S
200
E
100
B
D
O
100
(500, 0)
(400, 0) C
(0, 0) 200
300
400
2 500
600 700
X
x
+ y
Q U
= 8 0 0
x
+
y
= 5 0 0
Solving x y 500 and 2x y 800, we get B ( 300 , 200)
E
Corner point
Value of the optimizing function
(0, 0)
Z = 0 + 0 = 0
S
(400, 0)
Z = 0.6 × 400 + 0.4 × 0 = 240
(300, 200)
Z = 0.6 × 300 + 0.4 × 200 = 180 + 80 = 260
(0, 500)
Z = 0.6 × 0 + 0.4 × 500 = 200
T I O N S
Maximum Production of oxygen will be achieved when plant X is planted in 300 hectare and plant Y is planted in 200 hectare. (i) Excess herbicide will get absorbed in the soil and may contaminate the water source also. Thus it can affect the whole ecosystem. (ii) Care should be taken while planting trees that the variety of the plants is such that they provide more oxygen for our environment.
Value Based Questions
(xxvii)
10. In shop A, 30 tin pure ghee and 40 tin adulterated ghee are kept for sale while in shop B, 50 tin pure ghee and 60 tin adulterated ghee are there. One tin of ghee is purchased from one of the shop randomly and it is found to be adulterated. Find the probability that it is purchased from shop B. (i) How adulteration is dangerous for humanity? (ii) What you can do against adulteration? Sol. Let the event defined be as
E1 = Selection of shop A.
V A L U E
E2 = Selection of shop B. A = Purchasing of a tin having adulterated ghee. 1 1 P(E 1 ) , P(E2 ) 2 2
A 40 4 , E 1 70 7 E P 2 required A P
A 60 6 E2 110 11
P
A P(E2 ). P E2 E P 2 A A A P(E 1 ). P P(E 2 ). P E1 E2 1 6 3 . 2 11 11 21 1 4 1 6 2 3 43 . . 2 7 2 11 7 11 (i) Adulteration is dangerous as it is harmful for user’s health. (ii) To prevent adulteration, we should spread awareness against it in society. 11. In a self-assessment survey 60% persons claimed that they never indulged in corruption, 40% persons claimed that they always speak truth and 20% say that they neither indulged in corruption nor tell lies. A person is selected at random out of this group. (i) What is the probability that the person is either corrupt or tells lie?
B A S E D
Q U E S T I
(ii) If the person never indulged in corruption, find the probability that she/he tells, truth.
O
(iii) If the person always speaks truth find the probability that she/he claims to have never indulge in corruption.
N
(iv) What values have been discussed in this question?
S
(xxviii)
Xam idea Mathematics–XII
(v) Why is it must for all to practice values in our life?
V Sol. Let A : Set of persons never indulged in corruption A L U E
B A S E D
Q U E S T I O N S
B : Set of persons always speak truth 60 40 Then, , P( A) P( B) 100 100 20 and P( A B) 100
(i) P(Either A or B) = P( A B) P( A) P( B) P( A B) 60 40 20 80 4 100 100 100 100 5 20 P( A B) 100 1 B (ii) P 60 A 3 P( A) 100 20 P A B ( ) A 1 (iii) P 100 40 2 B P( B) 100 (iv) The following values have been discussed (a) We should never indulge in any type of corruption. (b) We should never tell lies i.e., we should always speak truth. (v) Values contribute to intellectual development, use of abilities, achieve creativity, personal development and development of society.
