Mathematical & Statistical Techniques
Chapter 1
Functions and Theirs Applications
Mathematical Functions
1. Constant Functions: Here the value of the function remains constant for different value of x. It is designed as y = k or f(x) = k, where k is any constant.
Eg y = 6 or f(x) = 94, here the value of y is 6 and the value of f(x) is 94 irrespective of whichever value of x.
Thus a constant function assumes only one value for each and every value of x, the independent variable.
Note: Its graph is a Straight line Parallel to x-axis.
2. Linear Function: Here the value of the function is expressed as Linear expression in x, the independent variable i.e. it contains terms in x and a constant and it doesn't have terms containing higher powers of x. It can be express as y = 4x +7 or f(x) = 3x - 2
Note: Its graph is a Straight line
3. Power Function : Here the value of the function is expressed as some power of x like y = x^5 or f(x) = x^7
Note: Its graph is a Curve.
4. Step function: Here the value of the function remains same (or constant) in a particular class interval. Its defined as
y = [x] where [x] represent largest integer not greater than x i.e.
y = [x] = 0 , for 0≤ x < 1
= 1, for 1≤ x < 2
= 2, for 2 ≤ x < 3
And y = [x] = -1 for - 1 ≤ x < 0
= -2 for -2 ≤ x < -1
= -3 for -3 ≤ x < -2
Note : Its graph is a set of Straight line parallel to x-axis which look like Step.
Hence Its called Step function.
5. Exponential Function: Here the value of function contains x as the exponent with some constant as the base. It can be expressed as y=f(x) = a^x where "a" Is constant.
Eg y = 7^x or f(x) = (1/2) ^x
Note: Its graph is a Curve.
6. Logarithm Function: It is expressed as y or f(x) = logax where a>0 ( not = 1) if we replace constant a by e the base of logarithm, y = lgex = log x, usually when the base is e it is not written and vice versa. That is, if the base of logarithm is not written, it's is understood that the base is e.
Exercise
1. Find values of the following functions at x = 0, 2, -3, 1.6, -1
(i) f(x) = 7x + 9
Ans: we replace x with 0, 2, -3, 1.6, -1
(a) f(0) = 7 (0) + 9 = 0+ 9 = 9
(b) f(2) = 7 (2) + 9 = 14 + 9 = 23
(C) f(-3) = 7 (-3) + 9 = -21 + 9 = - 12
(d) f(1.6) = 7 (1.6) + 9 = 11.2 + 9 = 20.2
(e) f(-1) = 7 (-1) + 9 = -7 + 9 = 2
(ii) f(x) = 7
Ans: we replace x with 0, 2, -3, 1.6, -1
(a) f(0) = 7
(b) f(2) = 7
(C) f(-3) = 7
(d) f(1.6) = 7
(e) f(-1) = 7
(ii) f(x) = X2
Ans: we replace x with 0, 2, -3, 1.6, -1
(a) f(0) = (0)2 = 0
(b) f(2) = (2)2 = 4
(C) f(-3) = (-3)2 = 9
(d) f(1.6) = (1.6)2 = 2.56
(e) f(-1) = (-1)2 = 1
Q.2 f(x) = 4x-1 for 0 ≤ x < 4 find f(0), f(1), f(1.2), f(4), f(-1) if these exists.
f(0) = 4 (0)- 1 = -1
f(1) = 4 (1) -1 = 3
f(1.2) = 4 (1.2) - 1 = 4.8- 1 = 3.8
f(4) = 4 (4) - 1 = 16 -1 = 15
f(-1) = 4(-1) -1 = -4 -1 = -5 does not exist because it not fall into the interval 0 ≤ x < 4.
Q.3 If f(x) = 3x2 – 2x + 4, find x for which f(x) = f(x-1)
Ans: 3x2 – 2x + 4, f(x) = f(x-1)
f(x) = x2 – 2x + 4
f(x-1) = 3(x-1)2 – 2(x-1) + 4
3(x2
– 2x +1) – 2(x-1) + 4 ----- [(a-b)2
= (a2 – 2ab + b2)
= 3x2
-6x + 3 - 2x + 2 + 4
= 3x2
– 8x + 9
3x2
– 8x + 9 = 3x2 – 2x + 4
3x2
– 3x2 -8x + 2x = 4 – 9
- 6x = - 5
x = 5/6
Q.4 If f(x)
= 2 + 3x for 1 < x < 4
= 3 + x for
4 < x < 7
= 2x + 7 otherwise
Find f(0),
f(-3), f(2), f(6).
Ans: x = 0
lies between otherwise
F(0) = 2x +
7 = 2(0) + 7 =
7
X = -3 lies
between otherwise
F(-3) = 2x
+ 7 = 2 (-3) + 7 = -6+7 =1
X = 2 lies
between 1 < x < 4
F(2) = 2 +
3x = 2+ 3(2) = 2 +6 = 8
X = 6 lies
between 4 < x < 7
F(6) = 3 +
x = 3 + 6 = 9
Q.5 if
f(x) = 5 , for -1 < x < 1
= x + 4, for 1 < x < 5
= 2x – 1, for 5 < x < 10
Find f(0),
f(3.1), f(-0.5), f(6). Also solve the equation f(x) = 6
Ans: x = 0 lies between for -1 < x <
1
F(0) = 5
x = 3.1 lies
between 1 < x < 5
F(3.1) =
x+4 = 3.1 + 4 = 7.1
x = - 0.5 lies
between for -1 < x < 1
F(0) = 5
x = 6 lies between
5 < x < 10
F(6) = 2x -1 = 2(6) -1 =
12 -1 = 11
f(x) = 6
where we substitute
x = 2 than its lies 1 < x < 5
f(2) = x +
4 = 2+4 =6
Q.6 IF
f(x) = 3x , show that f(x+1) =3 f(x) = f(3) f(x-2)
Ans: f(x) = 3x
F(x+1) = 3(x+1)
= 3x 3 --- (i)
3 f(x) = 3 3x ---- (ii)
f(3) f(x-2)
= 3x 3(x-2) = (3x 3x ) 3-2
= 32x 3-2 = 3x ----- (iii)
From (i), (ii) and (iii)
f(x+1)
=3 f(x) is not equal f(3) f(x-2)
Q.7 If
f(x) = kx – 3 and f(3) = 6 , find the value of k hence find f(-1), f(2)
Ans: f(3) = 6
f(x) = kx –
3 when x = 3
f(3) = k(3)
– 3
6 +3 = 3k 9 = 3k k = 3
Q.8 if
f(x) = x2 – 2x + 7, find x for which f(x) = f(x-1) + 5
Ans: f(x) = x2 – 2x + 7
f(x-1) + 5
= (x-1)2 – 2(x-1) + 7 + 5
= x2 – 2x + 1 -2x +2
+7 + 5 = x2 -4x +
15
f(x) = f(x-1)
+ 5
x2
– 2x + 7 = x2 -4x + 15
x2
– 2x - x2 + 4x = 15 -7 2x
= 8 x = 4
Total
Revenue = p. D R = (20 – D) D
R = 20D – 3D2
= 20 (2)– 3(2)2 = 40 -3(4)
= 40 – 12 = 28
AR = P AR = 20- 3D = 20 – 3(2) = 20 – 6 = 14
D = 5
Total
Revenue = p. D R = (20 – D) D
R = 20D – 3D2
= 20 (5)– 3(5)2 = 100 -3(25)
=100– 75 = 25
AR = P AR = 20- 3D = 20 – 3(5) = 20 – 15 =
5
Ans: D =
2
Total
Revenue = p. D R = (30 – D2) D
R = 30D – D3
= 30 (2)– (2)3 = 60
-8 =
52
AR = P AR = 30- D2 = 30 – (2)2 = 30 – 4 = 26
D = 5
Total
Revenue = p. D R = (30 – D2) D
R = 30D – D3
= 30 (5)– (5)3 = 150
-125 = 25
AR = P AR = 30- D2 = 30 – (5)2 = 30 – 25 = 5
10. Find the values of total cost and average cost
at x = 1, 4 if the total cost functions is
(i) C = 2 + 5x + x2 (ii) C = 10 + 3x2
Ans: When
x =1
(i)
C = 2 + 5x + x2 =
2 + 5(1) + (1)2 =
2 + 5 + 1 = 8
Average
Cost = C/x
=
8/x =
8/1 = 8
(ii) C = 10 + 3x2 = 10
+ 3(1)2 =
10 + 3 = 13
Average
Cost = C/x
=
13/x =
13/1 = 13
When x =4
(i)
C = 2 + 5x + x2 =
2 + 5(4) + (4)2 =
2 + 20 + 16 = 38
Average
Cost = C/x
=
38/x =
38/4 = 9.5
(ii) C = 10 + 3x2 = 10
+ 3(4)2 =
10 + 48 = 58
Average
Cost = C/x
=
58/x =
58/1 = 14.5
11. The
demand D and price p are expressed by 5D + 8p= 50. Express demand function,
total revenue function in terms of D. Also find their values at D=2.
Ans:
5D + 8p= 50
8p = 50 – 5D
P = 5/8 (10 – D) ---- (Taking Common 5)
When D = 2 than,
P = 5/8 (10 – 2) = 5/8 (8) = 5
Total Revenue (R) = p.D
= 5/8 (10 – D) D
When D = 2 than,
R
= 5/8 (10 – 2) x 2 = 5/8 (8) x 2 =
5 x 2 =
10
Ans: 6D + 20p = 100
20 P = 100
– 6D
P = 2/20 (50 – 3D) = 1/10 (50 – 3D)
When D = 4,
P = 1/10 (50 – 3x4)
= 1/10 (50 – 12) = 1/10 (38) = 3.8
Total Revenue (R) = p.D
= 1/10 (50 – 3D) D
When D = 4,
=
1/10 (50 – 3x4) x 4
=
1/10 (50-12) x 4
=
1/10 (38) x 4
=
3.8 x 4 = 15.2
13. Find the equilibrium price and quantity for the following demand and supply functions.
(i) p= 15 -3D and p= 4 + 8D
(ii) p = 8 – D2 and p = D2 + 6D
Ans: (i) p= 15 -3D and p= 4 + 8D
The equilibrium price and quantity
15 -3D = 4 + 8D
-3D – 8D = 4 – 15
-11D = -11 D = -11
/ -11 D = 1
(ii) p = 8 – D2 and p = D2 + 6D
The equilibrium price and quantity
8 – D2 = D2 + 6D
-D2 – D2 – 6D + 8 = 0
-2D2 – 6D + 8 = 0
-D2 – 3D + 4 = 0
----(Divided by 2 both the side)
D2 + 3D – 4 = 0
D2 + 4D – D – 4 = 0
D (D + 4) – 1 (D + 4) = 0
(D + 4) or (D – 1) = 0
(D + 4) = 0 or (D – 1) = 0
D = - 4 or D = 1
Demand is not Negative. Therefore D = - 4 is not solution.
Hence D = 1
Q.14 The manufacturer produces x packets of biscuits per day. The fixed cost is Rs. 1,000 per day, the cost of manufacturing is 18 per packet and the distribution charges are 2 per packet. Find the total cost function. Also find average cost function and their values for x= 50.
Ans: The manufacturer produces x packets of
biscuits per day.
The fixed cost is Rs. 1,000 per day,
The cost of manufacturing is 18 per packet and the distribution charges
are 2 per packet.
Total Cost (C) = 1,000 + (18+2) x
C = 1000 + 20x
Average Cost = C/x =
1000/x + 20x /x
=
1000/x + 20
When x = 50
C = 1000 + 20x =
1000 + 20 (50) = 1000+1000 = 2000
Average Cost = 1000/x + 20
= 1000/5 + 20 = 200+20
= 220
Q.15 The total cost is C= 200 + 20x and revenue is R= 800 + 5x. Find the point of no profit, no loss.
Ans: The total cost is C= 200 + 20x Revenue
is R= 800 + 5x
The point of no profit, no loss (C = R)
200 + 20x = 800 + 5x
20x – 5x = 800 – 200
15x = 600
X = 600 / 15 x = 40
Q. 16. If the total revenue and total cost function for a product are as follows, find the minimum number of units which can be produced and sold to avoid loss.
(i) R= 400 + 25x – x , C= 50 + 50x (ii)
R= 49x- 2x, C=70+ 10x X= 10
Ans: (i) R= 400 + 25x – x2 , C= 50 + 50x
The minimum number of units which can be produced and sold to avoid
loss.
C = R
400 + 25x – x2 = 50 + 50x
-x2 + 25x – 50x + 400 – 50 = 0
-x2 - 25x + 350 = 0
x2 + 25x - 350 = 0
x2 + 35x – 10x – 350 = 0
x(x + 35) – 10 (x + 35) = 0
(x + 35) = 0 or (x – 10) = 0
X = - 35 or x = 10
No. of unit can not produce in Negative. x = - 35 is not solution.
X = 10
(ii) R= 49x- 2x2, C=70+ 10x
The minimum number of units which can be produced and sold to avoid
loss.
C = R
49x- 2x2 = 70+ 10x
- 2x2 + 49x – 10x – 70 =0
- 2x2 + 39x – 70 =0
2x2 - 39x + 70 =0
2x2 - 35x – 4x + 70 =0
X(2x – 35) – 2(2x – 35) =0
(2x – 35) =0 or (x – 2) =0
2x = 35 or x = 2
X = 35/2 or x =2
Q. 17. A manufacturer makes toys and the weekly total cost is given by
C= 1200 + 40x.
(i) If each toy is sold at 60, find no. of units to be produced
and sold for no loss.
(ü) If the selling price increases by 10%, find the no. of units
to be produced and sold to ensure no loss,
(iii) If it is known in advance that the weekly demand for goods will be
70, find selling price of a toy to ensure no loss.
Ans:
A manufacturer makes toys
and the weekly total cost is given by C= 1200 + 40x
(i) If each toy is sold at 60, find no. of units to be produced and sold
for no loss.
No. of unit to be produce be x unit
Total revenue = 60x
No loss (C = R)
60x = 1200 + 40x
60x – 40x = 1200
20x = 1200
X = 1200 / 20 x
= 60
No. of units to be produced
and sold for no loss is 60 units
(ü) If the selling price increases by 10%, find the no. of units
to be produced and sold to ensure no loss,
If selling price is RS. 60
If the selling price increases by 10% = 60 + 10% 0f 60
=
60 + 6 =RS. 66
NO Loss (C = R)
1200 + 40x = 66x
1200 = 66x -40x
1200 = 26x
X = 1200/ 26 x = 46.15 approx 47
No. of units to be produced
and sold for no loss is 47 units
(iii) If it is known in advance that the weekly demand for goods will be
70, find selling price of a toy to ensure no loss.
Let the selling price be Rs. P
Total Revenue = 70p
No Loss (R = C)
70p = 1200 + 40x
70p = 1200 + 40(70)
70p = 1200 + 2800
70p = 4000
P = 4000/ 70
P = 57.14
Selling price of a toy to ensure no loss is Rs. 57.14
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