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CBSE Class 12 Mathematics Solved Question Paper 2015

Long Answer Type

31.
Minimum and maximum z = 5x + 2y subject to the following constraints:
x – 2y ≤ 2
3x + 2y ≤ 12
−3x + 2y ≤ 3
x ≥ 0, y ≥ 0

straight x minus 2 straight y less-than or slanted equal to 2 3 straight x plus 2 straight y less-than or slanted equal to 12 minus 3 straight x plus 2 straight y less-than or slanted equal to 3 straight x greater-than or slanted equal to 0 comma space straight y greater-than or slanted equal to 0

Converting the inequations into equations, we obtain the lines
x – 2y = 2…..(i)
3x + 2y = 12……(ii)
−3x + 2y = 3……(iii)
x = 0, y = 0

From the graph, we get the corner points as
A(0, 5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0, 0)
The values of the objective function are:

Point (x,y)	Values of the objective function Z= 5x+2y
A(0, 5)	5 × 0 + 2 × 5 = 10
B(3.5, 0.75)	5 x 3.5 +2 x 0.75 =19 (Maximum)
C(2, 0)	5 x 1.5 +2 x 3.75 =15
O(0,0)	5 x 0 + 2 x 0 = 0 (Minimum)

The maximum value of Z is 19 and its minimum value is 0.

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32.

Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.

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