Maths

8. $f (x) =$ ${\begin{matrix} \frac{| x |}{x}, if x \neq 0 & 0, if x = 0 . \end{matrix}$

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8. Given, f(x) = $f x x x$

For x = c < 0,

f(c) = 1

$\underset{x \to 0}{l i m}$ f(x) = $\underset{x \to 0}{l i m}$ 1 = 1

∴f(c) = $\underset{x \to 0}{l i m}$ f (x)

f is continuous at x $| < |$ 0.

For x = c > 0,

F (c) = 1

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $=$ = 1.

∴f(c) = $\underset{x \to c}{l i m}$ f(x)

f is continuous at x > 0.

For x = c 0.

L.H.L. = $\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m}$ ( 1) = 1

R.H.L. $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ 1 = 1

∴ L.H.L. $=$ R.H.L.

is now continuous at x = 0, point of discontinuity of f is at x = 0.

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7. $f (x) =$ ${\begin{array}{l} | x | + 3 if x \leq - 3 & - 2 x if - 3 < x < 3 & 6 x + 2 if x \geq 3 \end{array}$

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7. Given, f(x) = ${\begin{array}{l} | x | + 3 if x \leq - 3 & - 2 x if - 3 < x < 3 & 6 x + 2 if x \geq 3 \end{array}$

For x = $? c < - 3,$

f ( 3) = e + 3 (∴x< 3, $| x | = - x$ )

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $| x | + 3 = - a + 3 .$

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So, f is continuous at x = c < 3.

For x = c > 3

f(3) = 6.3 + 2 = 18 + 2 = 20

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 6x + 2 = 18 + 2 = 20

∴ $\underset{x \to c}{l i m}$ f(x) = f(c).So f is continuous at x = c > 3.

For. C = 3,

f ( 3) = ( 3) + 3 = 6.

$\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ .x + 3 = ( 3) + 3 = 6.

$\underset{x \to c^{+}}{l i m}$ f(x) = $\underset{x \to c^{+}}{l i m}$ ( 2x) = 2 ( 3) = 6.

∴ $\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ f(x) = f( 3)

So, f is continuous at x = c = 3.

For c = 3,

f(3) = 6.3 + 2 = 18 + = 20.

$\underset{x \to 3^{-}}{l i m}$ f(x) = $\underset{x \to 3^{-}}{l i m}$ 2x = 2 (3) = 6

$\underset{x \to 3^{+}}{l i m}$ f(x) = $\underset{x \to 3^{+}}{l i m}$ (6x + 2) = 6.3 + 2 = 20

∴ $\underset{x \to 3^{-}}{l i m}$ f(x) $=$

...more

7. Given, f(x) = ${\begin{array}{l} | x | + 3 if x \leq - 3 & - 2 x if - 3 < x < 3 & 6 x + 2 if x \geq 3 \end{array}$

For x = $? c < - 3,$

f ( 3) = e + 3 (∴x< 3, $| x | = - x$ )

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $| x | + 3 = - a + 3 .$

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So, f is continuous at x = c < 3.

For x = c > 3

f(3) = 6.3 + 2 = 18 + 2 = 20

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 6x + 2 = 18 + 2 = 20

∴ $\underset{x \to c}{l i m}$ f(x) = f(c).So f is continuous at x = c > 3.

For. C = 3,

f ( 3) = ( 3) + 3 = 6.

$\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ .x + 3 = ( 3) + 3 = 6.

$\underset{x \to c^{+}}{l i m}$ f(x) = $\underset{x \to c^{+}}{l i m}$ ( 2x) = 2 ( 3) = 6.

∴ $\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ f(x) = f( 3)

So, f is continuous at x = c = 3.

For c = 3,

f(3) = 6.3 + 2 = 18 + = 20.

$\underset{x \to 3^{-}}{l i m}$ f(x) = $\underset{x \to 3^{-}}{l i m}$ 2x = 2 (3) = 6

$\underset{x \to 3^{+}}{l i m}$ f(x) = $\underset{x \to 3^{+}}{l i m}$ (6x + 2) = 6.3 + 2 = 20

∴ $\underset{x \to 3^{-}}{l i m}$ f(x) $=$ $\underset{x \to 3^{-}}{l i m}$ f(x).

f is not continuous at x = 3 point of discontinuity

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6. $f (x) =$ ${\begin{matrix} 2 x + 3 if x \leq 2 & 2 x - 3 if x > 2 . \end{matrix}$

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6. Given f(x) = ${\begin{matrix} 2 x + 3 if x \leq 2 & 2 x - 3 if x > 2 . \end{matrix}$

For x = c < 2,

F (c) = 2c + 3

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x + 3 = 2c + 3

∴ $\underset{x \to c}{l i m}$ f (x) = f(c)

So f is continuous at x $| < |$ 2.

For x = c > 2.

F (c) = 2c 3

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x 3 = 2c 3

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So f is continuous at x $| > |$ 2.

For x = c = 2,

L.H.L. = $\underset{x \to 2^{-}}{l i m}$ f(x) = $\underset{x \to 2^{-}}{l i m}$ .2x + 3 = 2. 2 + 3 = 4 + 3 = 7.

R.H.L. = $\underset{x \to 2^{+}}{l i m}$ f(x) = $\underset{x \to 2^{+}}{l i m}$ 2x 3 = 2. 2 3 = 4 3 = 1.

∴ LHL $=$ RHL

∴ f is not continuous at x = 2.i e, point of discontinuity

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5. Is the function $f$ defined by

$\begin{array}{l} f (x) = {\begin{cases} x, x \leq 1 \\ 5, x \end{cases} \\ x x \end{array}$

Find all points of discontinuity of $f$ , where $f$ is defined by

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5. Given, f (a) = ${\begin{array}{l} x, if x \leq 1 & 5, if x > 1 . \end{array}$

At x = 0,

(0) = 0

$\underset{x \to 0}{l i m}$ f (x) = $\underset{x \to 0}{l i m}$ x = 0

∴ $\underset{x \to 0}{l i m}$ f (x) = f (0)

So, f is continuous at x = 0.

At x = 1,

Left hand limit,

L.H.L = $\underset{x \to 1}{l i m}$ f (x) = $\underset{x \to 1}{l i m}$ x = 1.

Right hand limit,

R. H. L. = $\underset{x \to 1^{+}}{l i m}$ f (x) = $\underset{x \to 1^{+}}{l i m}$ 5 = 5.

L.H.L. $=$ R.H.L.

So, f is not continuous at x = 1.

At x = 2,

f (2) = 5.

$\underset{x \to 1}{l i m}$ f (x) = $\underset{x \to 2}{l i m}$ 5 = 5

$\underset{l i m \to 2}{l i m f}$ (x) = f (2)

So f is continuous at x = 2.

Find all points of discontinuity of f, where f is defined by

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31. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week, and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other types by at most 600 units. If the company makes a profit of ?. 12 and ?. 16 per doll on dolls A and B, respectively, how many of each should be produced weekly in order to maximise the profit?

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Let x and y be the number of dolls of type A and B, respectively, that are produced in a week.

The given problem can be formulated as given below:

$M a x i m i s e z = 12 x + 16 y \dots \dots \dots . . (i)$

Subject to the constraints,

$\begin{array}{l} \end{array} x + y \leq 1200 \dots \dots \dots \dots (i i)$

$y \leq x / 2 o r x \geq 2 y \dots \dots \dots . (i i i)$

$x- 3 y \leq 600 \dots \dots \dots \dots . (i v)$

$x, y \geq 0 \dots \dots \dots \dots \dots (v)$

The feasible region determined by the system of constraints is given below:

A (600, 0), B (1050, 150) and C (800, 400) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 12x + 16y
A (600, 0)	7200
B (1050, 150)	15000
C (800, 400)	16000	Maximum

The maximum value of z is 16000 at (800, 400).

Hence, 800 and 400 dolls of type A and type B should be produced, respectively, to get the maximum profit of? 16000.

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30. Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

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Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$M a x i m i s e z = 3 x + 3.5 y \dots \dots \dots . . (i)$

Subject to the constraints,

$\begin{array}{l} \end{array} x + 2 y \geq 240 \dots \dots \dots . . (i i)$

$x + 0.5 y \geq 90 \dots \dots \dots . . (i i i)$

$1.5 x + 2 y \leq 310 \dots \dots \dots . . (i v)$

$x, y \geq 0 \dots \dots \dots \dots . (v)$

The feasible region determined by the system of constraints is given below:

A (140, 50), B (20, 140) and C (40, 100) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 3x + 3.5y
A (140, 50)	595	Maximum
B (20, 140)	550
C (40, 100)	470

The maximum value of z is 595 at (140, 50).

Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.

Thus, the maximum amount of nitrogen added to the garden is 595 kg.

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29. A fruit grower can use two types of fertilisers in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table, Tests indicate that the garden needs at least 240 kg of phosphoric acid, 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added to the garden?

Kg per Bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

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Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$M i n i m i s e z = 3 x + 3.5 y \dots \dots \dots \dots . (i)$

Subject to the constraints,

$\begin{array}{l} \end{array} x + 2 y \geq 240 \dots \dots \dots . . (i i)$

$x + 0.5 y \geq 90 \dots \dots . . (i i i)$

$1.5 x + 2 y \leq 310 \dots \dots \dots . . (i v)$

$x, y \geq 0 \dots \dots \dots \dots . (v)$

The feasible region determined by the system of constraints is given below:

A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 3x + 3.5y
A (140, 50)	595
B (20, 140)	550
C (40, 100)	470	Minimum

The maximum value of z is 470 at (40, 100).

Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.

Hence, the minimum amount of nitrogen added to the garden is 470 kg.

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28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has to supply oil to three petrol pumps, D, E and F, whose requirements are 4500L, 3000L and 3500L, respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is ?. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost

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Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $(7000 - x - y)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $, (3000 - y) L a n d 3500 - (7000 - x - y) = (x + y - 3500)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{l} \end{array} x \geq 0, y \geq 0, a n d (7000 - x - y) \geq 0$

$T h e n, x \geq 0, y \geq 0, a n d x + y \leq 7000$

$4500 - x \geq 0, 3000 - y \geq 0, a n d x + y - 3500 \geq 0$

$T h e n, x \leq 4500, y \leq 3000, a n d x + y \geq 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l} z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 - x - y) + 3 / 10 (4500 - x) + 4 / 10 (3000 - y) + 2 / 10 (x + y - 3500) \end{array}$

$\begin{array}{l} = 0.3 x + 0.1 y + 3950 \end{array}$

The problem can be formulated as given below:

$M i n i m i s e z = 0.3 x + 0.1 y + 3950 \dots \dots \dots . (i)$

Subject to cons

...more

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $(7000 - x - y)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $, (3000 - y) L a n d 3500 - (7000 - x - y) = (x + y - 3500)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{l} \end{array} x \geq 0, y \geq 0, a n d (7000 - x - y) \geq 0$

$T h e n, x \geq 0, y \geq 0, a n d x + y \leq 7000$

$4500 - x \geq 0, 3000 - y \geq 0, a n d x + y - 3500 \geq 0$

$T h e n, x \leq 4500, y \leq 3000, a n d x + y \geq 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l} z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 - x - y) + 3 / 10 (4500 - x) + 4 / 10 (3000 - y) + 2 / 10 (x + y - 3500) \end{array}$

$\begin{array}{l} = 0.3 x + 0.1 y + 3950 \end{array}$

The problem can be formulated as given below:

$M i n i m i s e z = 0.3 x + 0.1 y + 3950 \dots \dots \dots . (i)$

Subject to constraints,

$\begin{array}{l} \end{array} x + y \leq 7000 \dots \dots \dots . (i i)$

$x \leq 4500 \dots \dots \dots . . (i i i)$

$y \leq 3000 \dots \dots . (i v)$

$x + y \geq 3500 \dots \dots \dots (v)$

$x, y \geq 0 \dots \dots \dots \dots (v i)$

The feasible region determined by the constraints is given below:

A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 0.3x + 0.1y + 3950
A (3500, 0)	5000
B (4500, 0)	5300
C (4500, 2500)	5550
D (4000, 3000)	5450
E (500, 3000)	4400	Minimum

The minimum value of z is 4400 at (500, 3000).

Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F, respectively.

Therefore, the minimum transportation cost is ?. 4400.

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Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Q7. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has to supply oil to three petrol pumps, D, E and F, whose requirements are 4500L, 3000L and 3500L, respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is ?. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

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Contributor-Level 10

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $(7000 - x - y)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $, (3000 - y) L a n d 3500 - (7000 - x - y) = (x + y - 3500)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{l} \end{array} x ? 0, y ? 0, a n d (7000 - x - y) ? 0$

$T h e n, x ? 0, y ? 0, a n d x + y ? 7000$

$4500 - x ? 0, 3000 - y ? 0, a n d x + y - 3500 ? 0$

$T h e n, x ? 4500, y ? 3000, a n d x + y ? 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l} z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 - x - y) + 3 / 10 (4500 - x) + 4 / 10 (3000 - y) + 2 / 10 (x + y - 3500) \end{array}$

$\begin{array}{l} = 0.3 x + 0.1 y + 3950 \end{array}$

The problem can be formulated as given below:

$M i n i m i s e z = 0.3 x + 0.1 y + 3950 \dots \dots \dots . (i)$

Subject to cons

...more

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $(7000 - x - y)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $, (3000 - y) L a n d 3500 - (7000 - x - y) = (x + y - 3500)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{l} \end{array} x ? 0, y ? 0, a n d (7000 - x - y) ? 0$

$T h e n, x ? 0, y ? 0, a n d x + y ? 7000$

$4500 - x ? 0, 3000 - y ? 0, a n d x + y - 3500 ? 0$

$T h e n, x ? 4500, y ? 3000, a n d x + y ? 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l} z = (7 / 10) x + (6 / 10) y + 3 / 10 (7000 - x - y) + 3 / 10 (4500 - x) + 4 / 10 (3000 - y) + 2 / 10 (x + y - 3500) \end{array}$

$\begin{array}{l} = 0.3 x + 0.1 y + 3950 \end{array}$

The problem can be formulated as given below:

$M i n i m i s e z = 0.3 x + 0.1 y + 3950 \dots \dots \dots . (i)$

Subject to constraints,

$\begin{array}{l} \end{array} x + y ? 7000 \dots \dots \dots . (i i)$

$x ? 4500 \dots \dots \dots . . (i i i)$

$y ? 3000 \dots \dots . (i v)$

$x + y ? 3500 \dots \dots \dots (v)$

$x, y ? 0 \dots \dots \dots \dots (v i)$

The feasible region determined by the constraints is given below:

A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 0.3x + 0.1y + 3950
A (3500, 0)	5000
B (4500, 0)	5300
C (4500, 2500)	5550
D (4000, 3000)	5450
E (500, 3000)	4400	Minimum

The minimum value of z is 4400 at (500, 3000).

Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F, respectively.

Therefore, the minimum transportation cost is ?. 4400.

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27. Two godowns, A and B, have grain capacities of 100 quintals and 50 quintals, respectively. They supply to 3 ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops is given in the following table:

Transportation Cost per Quintal (in Rs)
From/To	A	B
D	6	4
E	3	2
F	2.50	3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

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