View
235
Download
0
Category
Preview:
Citation preview
8/8/2019 Math Test 1
1/37
8/8/2019 Math Test 1
2/37
8/8/2019 Math Test 1
3/37
8/8/2019 Math Test 1
4/37
8/8/2019 Math Test 1
5/37
8/8/2019 Math Test 1
6/37
8/8/2019 Math Test 1
7/37
8/8/2019 Math Test 1
8/37
8/8/2019 Math Test 1
9/37
8/8/2019 Math Test 1
10/37
8/8/2019 Math Test 1
11/37
8/8/2019 Math Test 1
12/37
8/8/2019 Math Test 1
13/37
8/8/2019 Math Test 1
14/37
8/8/2019 Math Test 1
15/37
8/8/2019 Math Test 1
16/37
8/8/2019 Math Test 1
17/37
8/8/2019 Math Test 1
18/37
8/8/2019 Math Test 1
19/37
8/8/2019 Math Test 1
20/37
8/8/2019 Math Test 1
21/37
8/8/2019 Math Test 1
22/37
8/8/2019 Math Test 1
23/37
8/8/2019 Math Test 1
24/37
8/8/2019 Math Test 1
25/37
OR
8/8/2019 Math Test 1
26/37
Let
Thus, is a symmetric matrix.
8/8/2019 Math Test 1
27/37
Thus, is a skew symmetric matrix.
Thus, A is represented as the sum of a symmetric and a skew symmetric matrix.
Q uestion 24 ( 6.0 marks) Find the vector equation of the plane passing through the intersection of the planes,
, and the point (1, 3, 1).
Solution:
H ere,
Using the relation , we obtain
It is given that the plane passes through the point (1, 3, 1). Therefore, it must satisfy equation (ii).Therefore, we obtain
8/8/2019 Math Test 1
28/37
S ubstituting the value of in (i), we obtain
This is the required vector equation of the plane.
Q uestion 25 ( 6.0 marks) Use the following information to answer the next question.
A painting is in the form of a rectangle with a semi-circle along one of its lengths. The total perimeter of the painting is 5 m.
Find the dimensions of the rectangular part of the painting such that the area covered by the paintingis maximum.
Solution:Let x and y be the length and breadth of the rectangle respectively. Then, radius of the semi-circle
will be .
It is given that the perimeter of the painting is 5 m.
Therefore, we obtain
8/8/2019 Math Test 1
29/37
Area of the painting is given by,
Therefore, area of the painting is maximum when
8/8/2019 Math Test 1
30/37
Thus, the dimensions of the rectangular part are
Q uestion 26 ( 6.0 marks)
Find .
Solution:
Let I =
The integrand can be written as
Comparing the coefficients of x 2 and the constant terms, we obtain
3 A + B = 0 B = 3 A
4 A + 2 B + C = 1
5 A + 2 C = 3
S olving these equations, we obtain
Thus, the integrand is given by,
8/8/2019 Math Test 1
31/37
Put 3 x 2 + 4 x + 5 = t (6 x + 4) dx = dt
Put
From (2), (3), and (4), we obtain
Therefore, from (1), we obtain
8/8/2019 Math Test 1
32/37
Q uestion 27 ( 6.0 marks) Find the area of the region bounded by the ellipse, x 2 + 9 y 2 = 81, and the line, x = 3.
Solution:
The given ellipse, x 2 + 9 y 2 = 81 can be written as
The given ellipse is symmetrical about both x and y axis. Thus, the required area of the region BCDEis given by,
8/8/2019 Math Test 1
33/37
Q uestion 28 ( 6.0 marks) A factory manufactures two types of cloths, A and B. The factory has two machines, which can workfor at most 3 hours. It takes 5 minutes on first machine and 2 minutes on second machine to produce10 m of cloth A. It takes 3 minutes on first and 3 minutes on second machine to produce 10 m of cloth B. The manufacturer earns revenue of Rs 60 and Rs 50 on 10 m of cloth A and 10 m of cloth Brespectively. Determine the maximum revenue that the factory can earn.
Solution:Let x units and y units of cloths A and B be manufactured, where 1 unit = 10 m. Therefore, totalprofit (Rs) = 60 x + 50 y
Let z = 60 x + 50 y
The mathematical model for the given problem is as follows.
Maximize z = 60 x + 50 y (1)
S ubject to
The feasible region OABC determined by the inequalities (2), (3), (4) is represented as follows.
8/8/2019 Math Test 1
34/37
The objective function Z at each of the corner point of the shaded region is evaluated as
C orner point Z = 60 x + 50 y
O (0, 0)
A (0, 60)
B (20, 40)
C (36, 0)
0
3000
3200
2160
Thus, the maximum value of Z is 3200 at B (20, 40).
Therefore, the factory should produce 20 units of cloth A and 40 units of cloth B to earn maximumrevenue of Rs 3200.
That is the factory should produce 200 m of cloth A and 400 m of cloth B to earn maximum revenueof Rs 3200.
Q uestion 29 ( 6.0 marks) A pair of unbiased dice is thrown. A random variable X is the difference between the numbers thatappear on the top face of the two dice. Find the expectation of X. Also, find the variance of X.
OR
A die is thrown 10 times. If getting a number greater than or equal to 3 is a success, then find theprobability of
a. at most 3 successes
b . at least 8 successes
Solution:The sample space of the experiment consists of 36 elements.
8/8/2019 Math Test 1
35/37
The random variable X is the difference between the numbers on the two dice. Therefore, it can takethe values 0, 1, 2, 3, 4, or 5.
The probability distribution of X is
X 0 1 2 3 4 5
P ( X )
Thus, the expectation of X is .
8/8/2019 Math Test 1
36/37
Variance
OR
The event X getting a number greater than or equal to 3 is success.
Probability of success
It can be observed that X has a binomial distribution with n = 10 and
(a) P (at most 3 successes) = P(X 3)
8/8/2019 Math Test 1
37/37
(b) P (at least 8 successes) = P(X 8)
Recommended