BLUE PRINT XII CBSE MATHEMATICS Time : 3 Hours Max. Marks :100. Page 2.
2. XII CBSE-MATHS. XII CBSE. SAMPLE QUESTION PAPER. MATHEMATICS.
XII CBSE-MATHS
BLUE PRINT XII CBSE MATHEMATICS S.No.
Max. Marks :100
VSA
SA
LA
Relations and functions
1 (1)
4 (1)
-
Inverse Trigonometric function
1 (1)
4 (1)
-
Matrices
2 (2)
-
6 (1)
Determinants
1 (1)
4 (1)
-
Continuity and Differentiability
-
8 (2)
-
(b)
Application of Derivatives
-
4 (1)
6 (1)
(c)
Integrals
2 (2)
4 (1)
6 (1)
(d)
Application of Integrals
-
-
6 (1)
(e)
Differential equations
-
8 (2)
-
Vectors
2 (2)
4 (1)
-
(b)
Three-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) (b) 2 (a) (b) 3 (a)
4 (a)
Units
Time : 3 Hours
Total
Total
10 (4)
13 (5)
44 (11)
1
XII CBSE-MATHS
XII CBSE SAMPLE QUESTION PAPER MATHEMATICS Time : - 3 Hours
Max. Marks:- 100
GENERAL INSTRUCTIONS : 1. 2.
All questions are compulsory. 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 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 questions. 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. 5. Use of calculators is not permitted. You may ask for logarithmic tables, if required. ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
SECTION - A Q.1
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one–one.
Q.2
Find the value of tan −1 2 cos 2sin −1
Q.3
Find the number of all possible matrices of order 3 × 3 with each entry 0 or 1.
Q.4
2 3 4 Compute the product : 3 4 5 4 5 6
Q.5
Find adjoint of the matrix
Q.6
Evaluate
Q.7
Evaluate [x]dx
1 2
.
1 −3 5 0 2 4 3 0 5
1 2 3 4
.
1 dx 1 + sin x 3 0
Q.8 Q.9
If a = ˆi + 2ˆj − kˆ and b = 3iˆ + ˆj − 5kˆ , find a unit vector in the direction of a − b . Find the projection of a on b if a . b = 8 and b = 2iˆ + 6ˆj + 3kˆ .
Q.10 Write the direction cosines of a line equally inclined to the three coordinate axes.
SECTION - B Q.11 Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this. (i) On Z+, define * by a * b = a – b (ii) On Z+, define * by a * b = ab 2 (iii) On R, define * by a * b = ab (iv) On Z+, define * by a * b = | a – b | Q.12 Prove that : tan −1
1+ x − 1− x 1+ x + 1− x
=
π 1 1 − cos −1 x , − ≤ x ≤1 4 2 2
OR 1 If sin sin −1 + cos −1 x = 1 , then find the value of x. 5
2
XII CBSE-MATHS Q.13 Solve system of linear equations, using matrix method 2x + y + z = 1 x – 2y – z = 3/2 3y – 5z = 9 Q.14 For what value of λ is the function defined by 2 f (x) = λ (x − 2x) if x ≤ 0 4x + 1 if x > 0
continuous at x = 0 ? What about continuity at x = 1 ? Q.15 Differentiate the function w.r.t. x :
x2 + 1 x x cos x + 2 x −1 OR Find the derivative of the function given by f (x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) and hence find f ' (1). Q.16 Find the equations of all lines having slope 0 which are tangent to the curve y =
1 2
x − 2x + 3
.
OR A box of maximum volume with top open is to be made by cutting out four equal squares from four corners of a square tin sheet of side length a ft and then folding up the flaps. Find the side of the square to be cut off. π /2
Q.17 Evaluate
log tan x dx 0
OR Evaluate
1 (x + 1) x 2 − 1)
dx
dy = cos (x + y). dx Q.19 From the differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to the x-axis. Q.20 If a = ˆi + ˆj + kˆ and b = ˆj − kˆ , find a vector c such that a × c = b and a.c = 3 .
Q.18 Find the solution of differential equation
x + 1 3y + 5 3 − z = = and the plane 10x + 2y – 11z = 3. 2 9 −6 Q.22 Three balls are drawn one by one without replacement from a bag containing 5 white and 4 green balls. Find the probability distribution of number of green balls drawn.
Q.21 Find the angle between the line
SECTION - C Q.23 A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below: Market Products I 10,000 2,000 18,000 II 6,000 20,000 8,000 (a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively. Find the gross profit. OR 1 3 −2 Using elementary transformations, find the inverse of the matrix, if it exists : −3 0 −5 2 5
0
3
XII CBSE-MATHS Q.24
Check whether the following functions satisfy the conditions of Rolle’s theorem (a)
Q.25 Evaluate
Evaluate
f (x) = x2 – 3 | x |, x ∈[ −1,1] sin −1 x − cos −1 x sin −1 x + cos −1 x OR 8 (x + 2)(x 2 + 4)
(b) f (x) = sec x, x ∈
−π π , 4 3
dx
dx
Q.26 Draw sketch of the region bounded between circle x2 + y2 = 1 and the curve | x | + | y | = 1 using integration find the area of the bounded region. Q.27 Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? Q.28 From the point P (1, 2, 4) a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and coordinates of the foot of the perpendicular. Q.29 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 units 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 persons. Form the question as L.P.P. and solve it graphically.
4
XII CBSE-MATHS
SOLUTIONS (1)
(2)
Here, A = {1, 2, 3} and B = {4, 5, 6, 7} f : A → B = {(1, 4), (2, 5), (3, 6)} ∴ f (1) = 4, f (2) = 5 and f (3) = 6 ⇒ f (x1) ≠ f (x2) for every x1 ≠ x2 so f is one–one function. −1 2 cos 2sin −1 Method 1 : tan
= tan −1 2 cos
π 3
1 =θ 2
Method 2 : sin −1
−1 2 cos 2 sin −1 ∴ tan
1 2
(3)
π 4
1 2
[∵ cos 2θ = 1 – 2sin2θ] = tan −1 2 1 −
1 2
= tan −1 (1) = tan −1 tan
π π = 4 4
There are 3 × 3 = 9 entries in 3 × 3 matrix. There are two options 0 or 1 for each entry. ∴ Number of possible matrices = 2 × 2 × 2 × ...... (9 times) = 29 = 512 2 + 0 + 12
1 −3 5
2 3 4
(4)
π 6
= tan–1 [2cos 2θ]
2
1 2
−1 2 cos 2 × = tan
= tan–1 (1) =
sin
= tan–1 [2 (1 – 2 sin2θ)] −1 2 1− 2 × = tan
1 2
Here, 3 4 5
0
2
4 5 6
3
0
−6 + 6 + 0
10 + 12 + 20
14
0
42
4 = 3 + 0 + 15 −9 + 8 + 0 15 + 16 + 25 = 18 −1 56 4 + 0 + 18 −12 + 10 + 0 20 + 20 + 30 22 −2 70 5
1 2
(5)
Let A =
3 4
∴ A11 = 4, A12 = –3, A21 = –2, A22 = 1 ∴ adj A =
(6)
4
−3 ′
−2
1
1 − sin x 2
cos x
dx =
−2
−3
1
1 2
cos x 1
3
dx −
sin x cos 2 x
2
dx = sec 2 x dx − tan x sec x dx = tan x − sec x + C
3
1
2
3
We have [x]dx = [x]dx + [x]dx + [x]dx = 0 dx + 1.dx + 2.dx 0
0 2
(8)
4
1 (1 − sin x) 1 − sin x 1 . dx = dx dx = 1 + sin x (1 − sin x) 1 + sin x 1 − sin 2 x =
(7)
=
1
2
0
1
2
3
= 0 + [ x ]1 + [ 2x ]2 = (2 − 1) + (6 − 4) = 3 Here, a = ˆi + 2ˆj − kˆ and b = 3iˆ + ˆj − 5kˆ
ˆ − (3iˆ + ˆj − 5k) ˆ ∴ a − b = (iˆ + 2ˆj − k) = ˆi + 2ˆj − kˆ − 3iˆ − ˆj + 5kˆ = −2iˆ + ˆj + 4kˆ
5
XII CBSE-MATHS
(9)
∴ | a − b | = (−2) 2 + (1)2 + (4) 2 = 4 + 1 + 16 = 21 The unit vector in the direction of a − b is a−b 1 −2 ˆ 1 ˆ 4 ˆ ˆ (−2iˆ + ˆj + 4k) i+ j+ k = = |a−b| 21 21 21 21 a.b = 8 and b = 2iˆ + 6ˆj + 3kˆ ∴ | b | = (2)2 + (6)2 + (3)2 = 4 + 36 + 9 = 49 = 7 a.b 8 ∴ Projection of a on b is = |b| 7
(10)
Let the line make angle α with the three coordinate axes. ∴ Direction cosines of the line are cos α, cos α, cos α ∴ cos2α + cos2α + cos2α = 1
⇒ 3 cos2α = 1 ⇒ cos α = ±
1 3
Thus direction cosines of line equally inclined to three coordinates axes are ±
1 3
(11)
(12)
,±
1 3
,±
(i) a * b = a – b Suppose a < b Then a – b < 0 ⇒ a – b ∉ Z+ Thus * is not a binary operation over Z+. (ii) The product of two positive integers is always a positive integer. i.e., a ∈ Z+ and b ∈ Z+ ⇒ ab ∈ Z+ Thus * is a binary operation over Z+. (iii) a ∈ R and b ∈ R ⇒ b2 ∈ R ⇒ ab2 ∈ R ⇒ a * b = ab2 ∈ R Thus * is a binary operation over R. (iv) a ∈ Z+, b ∈ Z+ ⇒a–b∈Z ⇒ | a – b | ∈ Z+ if a ≠ b But | a – b | = 0 if a = b Since 0 ∉ Z+, we can’t say that * is a binary operation over Z+. −1 L.H.S. = tan
1+ x − 1− x 1+ x + 1− x
Let x = 2 cos 2θ, then tan −1
−1 = tan
1 + cos 2θ − 1 − cos 2θ 1 + cos 2θ + 1 − cos 2θ
1 − tan θ π −θ = tan −1 tan 1 + tan θ 4
π 1 − cos −1 x 4 2 = R.H.S.
=
6
−1 = tan
(∵ x = cos 2θ)
2 cos2 θ − 2sin 2 θ 2 cos2 θ + 2sin 2 θ
= tan −1
cos θ − sin θ cos θ + sin θ
1 3
XII CBSE-MATHS OR 1 sin sin −1 + cos−1 x = 1 5 −1 1 + sin −1 1 − x 2 = 1 [∵ cos−1 x = sin −1 1 − x 2 ] ⇒ sin sin 5
⇒ sin sin −1
1 1 1 −1 + x2 + 1 − x2 1 − =1 5 25 1 24 x+ 1 − x2 5 25
⇒ sin sin −1
=1 ⇒
[∵ sin −1 x + sin −1 y = sin −1[x 1 − y 2 + y 1 − x 2 ]]
1 24 x+ 1 − x2 = 1 ⇒ 5 25
24 x 1− x2 = 1− 25 5
Squaring both sides, we have 24 x (1 − x 2 ) = 1 − 25 5
2
⇒ 24 − 24x 2 = 25 1 +
x 2 2x − 25 5
⇒ 24 – 24x2 = 25 + x2 – 10x ⇒ 25x2 – 10x + 1 = 0 ⇒ 25x2 – 5x – 5x + 1 = 0 ⇒ (5x – 1)2 = 0 ⇒ x = 1/5 1 is a root of given equation. 5 The system of equations can be written in the form AX = B,
Thus x =
(13)
2
1
1
x
1
where A = 2 −4 −2 , X = y and B = 3 9 z 0 3 −5 2
1
1
Now, | A | = 2 −4 −2 0 3 −5
= 2 (20 + 6) – 1 (–10 – 0) + 1 (6 – 0) = 52 + 10 + 6 = 68
A11 = (20 + 6) = 26, A12 = – (–10 – 0) = 10, A13 = (6 – 0) = 6 A21 = – (–5 – 3) = 8, A22 = (–10 – 0) = – 10, A23 = – (6 – 0) = – 6 A31 = (–2 + 4) = 2, A32 = – (–4 – 2) = 6, A33 = (–8 – 2) = –10 26
adj A =
8 −10 2
∴ A
−1
10
6 ′ −6
8
2
= 10 −10
26
6
6 −10
6
−6 −10
26 8 2 1 1 10 −10 6 = adj A = 68 |A| 6 −6 −10
∴ X = A–1B x 26 8 2 1 y = 10 −10 6 68 z 6 −6 −10 ∴ x = 1 and y =
1 3 9
26 + 24 + 18 68 1 1 1 10 − 30 + 54 = 34 = 1/ 2 = 68 68 6 − 18 − 90 −102 −3 / 2
1 3 and z = – 2 2
7
XII CBSE-MATHS (14)
2 Here f (x) = λ (x − 2x) if x ≤ 0 4x + 1 if x > 0
2 L.H.L. = lim f (x) = lim λ (x − 2x) x → 0−
x → 0− Put x = 0 – h as x → 0–, h → 0
2 ∴ lim λ [(0 − h)2 − 2 (0 − h)] = lim λ [(h + 2h)] = 0 h →0
h→0
R.H.L. = lim f (x) = lim (4x + 1) x → 0+
x → 0+ + Put x = 0 + h as x → 0 , h → 0
∴ lim [4 (0 + h) + 1] = lim [4h + 1] = 0 + 1 = 1 h →0
h →0
∴ L.H.L. ≠ R.H.L. Thus f (x) is not continuous at x = 0 for any value of λ. At x = 1. L.H.L. = lim f (x) = lim (4x + 1) x →1−
x →1− – Put x = 1 – h as x → 1 , h → 0
∴ lim [4 (1 − h) + 1] = lim [5 − 4h)] = 5 − 0 = 5 h →0
h→0
R.H.L. = lim f (x) = lim (4x + 1) x →1+
x →1+ Put x = 1 + h as x → 1+, h → 0
∴ lim [4 (1 + h) + 1] = lim (5 + 4h) = 5 + 0 = 5 h →0
h →0
f (1) = 4 × 1 + 1 = 5 [∵ f (x) = 4x + 1] ∴ L.H.L. ≠ R.H.L. Thus f (x) is not continuous at x = 1 for all values of λ.
(15)
x cos x + Let y = x
x2 + 1 x2 −1
Put u = x x cos x and v = Then, y = u + v ⇒
x2 + 1 x2 −1
dy du dv = + dx dx dx
........ (1)
Now, u = x x cos x Taking logarithm on both sides, we have log u = log (xxcos x) = x cos x log x Differentiating both sides w.r.t. x, we have d d (log u) = (x cos x log x) dx dx 1 du d d = x cos x (log x) + log x (x cos x) u dx dx dx
du 1 d d = u x cos x. + log x x cos x + cos x x dx x dx dx
8
XII CBSE-MATHS du = xx cos x [cos x + log x (– x sin x + cos x)] dx du = xx cos x [cos x (1 + log x) – x sin x log x] dx Also, v =
x2 + 1
x2 −1 Differentiating both sides w.r.t. x, we have d d (x 2 − 1) (x 2 + 1) − (x 2 + 1) (x 2 − 1) dv d x2 + 1 dx dx = dx dx x 2 − 1 = (x 2 − 1) 2 2x (x 2 − 1) − 2x (x 2 + 1)
=
(x 2 − 1)2
Putting value of
=
2x 3 − 2x − 2x 3 − 2x (x 2 − 1)2
=
−4x (x 2 − 1) 2
du dv and in (1), we get dx dx
4x dy = xx cos x [cos x (1 + log x) – x sin x log x] – 2 dx (x − 1) 2
OR Here f (x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) Taking logarithm on both sides, we have log f (x) = log [(1 + x) (1 + x2) (1 + x4) (1 + x8)] = log (1 + x) + log (1 + x2) + log (1 + x4) + log (1 + x8) Differentiating both sides w.r.t. x, we have d d [log f (x)] = [log (1 + x) + log (1 + x2) + log (1 + x4) + log (1 + x8)] dx dx
1 1 d 1 d 1 d 1 d .f ′(x) = . (1 + x) + . (1 + x 2 ) + . (1 + x 4 ) + . (1 + x8 ) 2 4 8 f (x) 1 + x dx 1 + x dx 1 + x dx 1 + x dx f (x) =
1 2x 4x 3 8x 7 + + + 1 + x 1 + x 2 1 + x 4 1 + x8
1 2x 4x 3 8x 7 + + f (x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) 1 + x + 1 + x 2 1 + x 4 1 + x8 1 2 × 1 4 × 13 8 × 17 + + ∴ f (1) = (1 + 1) (1 + 12) + (1 + 14) + (1 + 18) 1 + 1 + 1 + 12 1 + 14 1 + 18 =2×2× 2×2
(16)
Here y =
1 2 4 8 15 + + + = 16 × = 8 × 15 = 120 2 2 2 2 2
1 2
x − 2x + 3
dy d 1 1 −2 (x − 1) d = × (2x − 2) = 2 = [(x 2 − 2x + 3) −1 ] = − 2 2 2 dx dx x − 2x + 3 dx (x − 2x + 3) (x − 2x + 3) 2 It is given that slope of tangent is 0.
9
XII CBSE-MATHS ∴
−2 (x − 1) 2
(x − 2x + 3) 2
= 0 ⇒ – 2 (x – 1) = 0 ⇒ x = 1
When x = 1 Then y =
1
=
2
(1) − 2 × 1 + 3
Thus point on curve is 1,
1 1 = 1− 2 + 3 2
1 2
Equation of tangent at point 1, y−
1 2
1 1 = 0 (x − 1) ⇒ y − = 0 2 2
is 1 2
y
Thus required equation of line is y =
1 2
OR Volume of the box is V = (a – 2x)2x dV = (a – 2x)2 + x.2 (a – 2x).(– 2) = (a – 2x)(a – 6x) dx dV =0 dx
For V to be extremum
⇒x=
a a , 2 6
a ; V = 0 (minimum) and we know that minimum and maximum occurs alternately in a continuous function. Hence 2 V is maximum at x = a/6.
But when x =
π /2
(17)
log tan x dx
Let I =
....... (i)
0
π /2
Then, I =
log tan 0
π − x dx 2
π /2
⇒I=
log cot x dx
....... (ii)
0
Adding (i) and (ii), we get π /2
(log tan x + log cot x) dx
2I = 0
π /2
π /2
= 0
OR Let I =
1 (x + 1) x 2 − 1)
Putting x + 1 =
10
dx
1 1 and dx = – 2 dt, we get t t
π /2
0.dx = 0 ⇒ I = 0
log1.dx =
log (tan x.cot x) dx = 0
0
XII CBSE-MATHS 1
∴ I= 1 t
= −
(18)
1 −1 t
. −
2
−1
1 t
2
(1 − 2t)1/2 + C = 1 − 2t = 1 ( −2) 2
We are given that
dt
dt = −
1−
1 − 2t
= − (1 − 2t)−1/2 dt
2 +C= x +1
x −1 +C x +1
dy = cos (x + y) dx
dy dv = dx dx So, the given equation becomes
dy dx
Put x + y = v, so that 1 +
dy 1 dx
dv dv − 1 = cos v ⇒ = 1 + cos v dx dx 1 1 2v ⇒ 1 + cos v dv = dx ⇒ sec dv = dx 2 2 Integrating both sides, we get 1 2v sec dv = 1.dx 2 2
⇒ tan
(19)
v =x+C 2
tan
x+y 2
x C
which is the required solution. The general equation of a parabola with fixed latus rectum 4a and axis parallel to the x-axis is (y – k)2 = 4a (x – h) ......... (1) Differentiating (1) w.r.t. x, dy = 4a dx again differentiating
2 (y – k)
........... (2)
2
d2 y dy 2 + 2 (y – k) =0 ......... (3) dx 2 dx Eliminating (y – k) between equation (2) and (3), we have
(20)
2
dy 4a d2y . 2 = 0 i.e. dx dy dx dx ˆ ˆ ˆ Here, a = i + j + k and b = ˆj − kˆ 2
dy dx
+
3
+ 2a
d2 y =0 dx 2
Let, c = c1ˆi + c2 ˆj + c3kˆ Now, a × c = b ˆi ˆj 1 1 ∴ c1 c 2
kˆ 1 = ˆj − kˆ c3
⇒ (c3 − c2 ) ˆi − (c3 − c1 ) ˆj + (c2 − c1 ) kˆ = ˆj − kˆ 11
XII CBSE-MATHS Comparing coefficients of ˆi, ˆj & kˆ on both sides, we have c3 – c2 = 0, c1 – c3 = 1, c2 – c1 = –1 ⇒ c3 = c2 .......... (1) and c1 – c2 = 1 .......... (2) It is given that a.c = 3
ˆ ˆ ˆ ˆ ∴ (iˆ + ˆj + k).(c 1i + c 2 j + c3 k) = 3 ⇒ c1 + c2 + c3 = 3 ⇒ c1 + 2c2 = 3 .......... (3) Subtracting (2) from (3), we have (c1 + 2c2) – (c1 – c2) = 3 – 1 ⇒ c1 + 2c2 – c1 + c2 = 2
[∵ c3 = c2]
2 3
⇒ 3c2 = 2 ⇒ c2 =
2 3 Putting value of c2 in (2), we have
∴ c 2 = c3 =
c1 −
(21)
2 =1 3
c1
1
2 3
5 3
2 2 5 ∴ c = ˆi + ˆj + kˆ 3 3 3 Here the equation of given line is x + 1 3y + 5 3 − z x +1 y + 5 / 3 z − 3 = = = = ⇒ 2 9 −6 2 3 6 and given plane is 10x + 2y – 11z = 3 Let θ be the angle between the line and plane then
2 × 10 + 3 × 2 + 6 × −11
sin θ =
2
(2) + (3)2 + (6)2
(10)2 + (2)2 + (−11)2
20 + 6 − 66
=
4 + 9 + 36 100 + 4 + 121
−40
=
49 225
=
40 8 = 7 × 15 21
8 8 ⇒ θ = sin–1 21 21 X = Number of green balls, balls are drawn without replacement. X can take values 0, 1, 2, 3; bag contains 5 white, 4 green balls ⇒ sin θ =
(22)
5
5 C3 × 4 C0 5 × 4 × 3 × 1 5 C2 × 4 C1 10 × 4 × 6 10 = = = = ; P (1) = 9 9 9×8×7 42 9 × 8 × 7 21 C3 C3
5
C1 × 4 C2
P (0) =
P (2) =
9
C3
=
5 C 0 × 4 C3 1 × 4 × 3 × 2 1 5 × 6 × 6 15 = = = ; P (3) = 9 9 × 8 × 7 42 9×8× 7 21 C3
∴ Probability distribution is –––––––––––––––––––––––––––––––––––––– X 0 1 2 3
5 20 15 2 1 42 42 42 42 ––––––––––––––––––––––––––––––––––––––
P (X)
12
XII CBSE-MATHS
(23)
Here matrix of products =
10000 2000 18000 6000 20000 8000 2.50
(a) Matrix of sale prices = 1.50 1.00 2.50 1.50
10000 2000 18000 ∴ Total revenue of each market = 6000 20000 8000
25000 + 3000 + 18000
=
=
15000 + 30000 + 8000
1.00
46000 53000
Total revenue of Ist market = Rs. 46000 Total revenue of IInd market = Rs. 53000 2.00 (b) Matrix of costs = 1.00 0.50
∴ Total costs of each market =
10000
2000
18000
6000
20000
8000
1.00 0.50
20000 + 2000 + 9000
=
2.00
31000 36000
12000 + 20000 + 4000
Total cost of Ist market = Rs. 31000 Total cost of IInd market = Rs. 36000 Gross profit of Ist market = Rs. (46000 – 31000) = Rs. 15000 Gross profit of IInd market = Rs. (53000 – 36000) = Rs. 17000 OR 1 3 −2 Let A = −3 0 −5 2 5 0 Now, A = IA 1 3 −2 1 0 0 ⇒ −3 0 −5 = 0 1 0 A 2 5 0 0 0 1
Applying R2 → R2 + 3R1 and R3 → R3 – 2R1 ⇒
1 0
3 −2 9 −11 =
0 −1
4
1 0 0 3 1 0 A −2 0 1
13
XII CBSE-MATHS 1 3 −2 1 0 0 Applying R2 → R2 + 8R3 ⇒ 0 1 21 = −13 1 8 A 0 −1 4 −2 0 1 1 3 −2
1 0 0
Applying R3 → R3 + R2 ⇒ 0 1 21 = −13 1 8 A 0 0 25 −15 1 9 1 3 −2 1 0 0 1 Applying R3 → R ⇒ 0 1 21 = −13 1 8 A 25 3 0 0 1 −3 / 5 1/ 25 9 / 25 1 0 −65 Applying R1 → R1 – 3R2 ⇒ 0 1 21 = 0 0 1 0
−3
1
−24
8 A
1
−3 / 5 1/ 25 9 / 25
0
1 −2 / 5 −3 / 5
Applying R1 → R1 + 65R3 ⇒ 0 1 21 = 0 0
40 −13
1
−13 −3 / 5
1
8 A
1/ 25 9 / 25
1 0 0 1 −2 / 5 −3 / 5 Applying R2 → R2 – 21R3 ⇒ 0 1 0 = −2 / 5 4 / 25 11/ 25 A 0 0 1 −3 / 5 1/ 25 9 / 25 1 −2 / 5 Thus,
A–1
−3 / 5
= −2 / 5 4 / 25 11/ 25 −3 / 5
1/ 25
9 / 25
x 2 + 3 x , −1 ≤ x < 0
(24)
(a) We have f (x) =
x2 − 3 x ,
0 < x ≤1
which is continuous in [–1, 1]. Also, we have f (–1) = 0 = f (1)
1 1 , –1 ≤ x < 0, = 2x − , 0
