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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3

**11th Maths Exercise 1.3 Answers Question 1.**

Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related toy if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer.

Solution:

(i) A = {set of students in 11^{th} standard}

B = {set of sections in 11sup>th standard}

R : A ➝ B ⇒ x related to y

⇒ Every students in eleventh Standard must in one section of the eleventh standard.

⇒ It is a function.

Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.

**11th Maths Exercise 1.3 Answers Samacheer Question 2.**

Write the values of f at – 4, 1, -2, 7, 0 if

Solution:

f(-4) = -(-4) + 4 = 8

f(1) = 1 – 1^{2} = 0

f(-2) = (-2)^{2} – (-2) = 4 + 2 = 6

f(7) = 0

f(0) = 0

**11th Maths Exercise 1.3 Question 3.**

Write the values of f at -3, 5, 2, -1, 0 if

Solution:

f(-3) = (-3)^{2} – 3 – 5 = 9 – 8 = 1

f(5) = (5)^{2} + 3(5) – 2 = 25 + 15 – 2 = 38

f(2) = 4 – 3 = 1

f(-1) = (-1)^{2} + (-1) – 5 = 1 – 6 = -5

f(0) = 0 – 3 = -3

**11th Maths Exercise 1.3 Solutions Question 4.**

State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?

(i) If A = {a, b, c] and/= {(a, c), (b, c), (c, b)};(f: A ➝ A).

(ii) If X = {x, y, z} and/= {(x, y), (x, z), (z, x)}; (f: X ➝ X).

Solution:

(i) f : A ➝ A

It is a function but it is not 1 – 1 and not onto function.

(ii) f : X ➝ X

x ∈ X (Domain) has two images in the co-domain x. It is not a function.

**Exercise 1.3 Class 11 Maths State Board Question 5.**

Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A ➝ B for each of the following:

(i) neither one-to-one nor onto.

(ii) not one-to-one but onto.

(iii) one-to-one but not onto.

(iv) one-to-one and onto.

Solution:

A = {1, 2, 3, 4}

B = {a, b, c, d}.

R = {(1, b) (2, b) (3, c) (4, d)} is not 1-1 and not onto

(iii) Not possible

(iv)

**11 Maths Exercise 1.3 Question 6.**

Find the domain of \(\frac{1}{1-2 \sin x}\)

Solution:

**11th Maths Exercise 1.3 In Tamil Question 7.**

Find the largest possible domain of the real valued function f(x) = \(\frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}}\)

Solution:

∴ No largest possible domain

The domain is null set

**Exercise 1.3 11th Maths Question 8.**

Find the range of the function \(\frac{1}{2 \cos x-1}\)

Solution:

The range of cos x is – 1 to 1

**11th Maths Exercise 1.3 10th Sum Question 9.**

Show that the relation xy = -2 is a function for a suitable domain. Find the domain and the range of the function.

Solution:

xy = – 2 ⇒ y = -2/x

which is a function

The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}

**Samacheer Kalvi Class 11 Maths Question 10.**

If f, g : R ➝ R are defined by f(x) = |x| + x and g(x) = |x| – x, find gof and fog.

Solution:

**Exercise 1.3 Class 11 Maths Question 11.**

If f, g, h are real valued functions defined on R, then prove that

(f + g)oh = foh + goh. What can you say about fo(g + h) ? Justify your answer.

Solution:

Let f + g = k

= (f + g((h(x))

= f[h(x)] + g [h(x)]

= foh + goh

(i.e.,)(f + g)(o)h = foh + goh

fo(g + h) is also a function

**Samacheer Kalvi Guru 11th Maths Question 12.**

If f: R ➝ R is defined by f(x) = 3x – 5, prove that f is a bijection and find its inverse.

Solution:

P(x) = 3x – 5

Let y = 3x – 5 ⇒ 3x = y + 5

**Samacheer Kalvi 11th Maths Question 13.**

The weight of the muscles of a man is a function of his body weight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.

Solution:

W(x) = 0.35x

Since body weight x is positive and if it increases then W(x) also increase.

Domain is (0, ∞) i.e.,x > 0

**Samacheer Kalvi.Guru 11th Maths Question 14.**

The distance of an object falling is a function of time t and can be expressed as s(t) = -16t^{2}. Graph the function and determine if it is one-to-one.

Solution:

s(t) = -16t^{2}

Suppose S(t_{1}) = S(t_{2})

since time cannot be negative, we to take t_{1} = t_{2}

Hence it is one-one.

t | 0 | 1 | 2 | 3 |

s | 0 | -16 | -64 | -144 |

**Samacheer Kalvi Class 11 Maths Solutions Question 15.**

The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.

Solution:

C – base cost,

S = fuel surcharge,

m = mileage

C(m) = 0.4 m + 50

S(m) = 0.03 m

Total cost = C(m) + S(m)

= 0.4 m + 50 + 0.03 m

= 0.43 m + 50

for 1600 miles

T(c) = 0.43 (1600) + 50 = 688 + 50 = ₹ 738

**Samacheer Kalvi 11th Maths Solution Book Question 16.**

A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1,50,00,000 worth of merchandise.

Solution:

A(x) = 30, 000 + 0.04x, where x is merchandise rupee value

S(x) = 25000 + 0.05 x

(A + S) (x) = A(x) + S(x)

= 30000 + 0.04x + 25000 + 0.05 x

= 55000 + 0.09x

(A + S) (x) = 55000+ 0.09x

They each sell x = 1,50,00,000 worth of merchandise

(A + S) x = 55000 + 0.09 (1,50,00,000)

= 55000 + 13,50,000

∴ Total income of family = ₹ 14,05,000

**11th Maths Samacheer Kalvi Guru Question 17.**

The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.

Solution:

f(x) = 1. 23x where x is number of American dollars.

g(y) = 50.50y where y is number of Singapore dollars.

gof(x) = g(f(x))

= g(1. 23x)

= 50.50 (1.23x)

= 62.115 x

**Maths Class 11 Samacheer Kalvi Question 18.**

The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200 – x. Express his day revenue, total cost and profit on this meal as functions of x.

Solution:

cost of one meal = ₹ 100

Total cost = ₹ 100 (200 – x)

Number of customers = 200 – x

Day revenue = ₹ (200 – x) x

Total profit = day revenue – total cost

= (200 – x) x – (100) (200 – x)

**Samacheer Kalvi Guru 11th Maths Solution Question 19.**

The formula for converting from Fahrenheit to Celsius temperatures is \(y=\frac{5 x}{9}-\frac{160}{9}\)

Find the inverse of this function and determine whether the inverse is also a function.

Solution:

Question 20.

A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).

Solution:

f(x) = 3x – 4

Let y = 3x – 4

### Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3 Additional Questions

Question 1.

Find the domain and range of the function \(f(x)=\frac{1}{\sqrt{x-5}}\)

Solution:

Given that : f(x) \(f(x)=\frac{1}{\sqrt{x-5}}\)

Here, it is clear that / (x) is real when x – 5 > 0 ⇒ x > 5

Hence, the domain = (5, ∞)

Now to find the range put

For x ∈ (5, ∞), y ∈ R^{+}.

Hence, the range of f = R^{+}.

Question 2.

If \(f(x)=\frac{x-1}{x+1}\), then show that

Solution:

Question 3.

Find the domain of each of the following functions given by:

\(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)

Solution:

Here, f(x) is not defined if x^{2} – 1 ≠ 0

(x – 1) (x + 1) ≠ 0

x ≠ 1, x ≠ -1

Hence, the domain of f = R – {-1, 1}

Question 4.

Find the range of the following functions given by

f(x) = 1 + 3 cos 2x

Solution:

Given that: f(x) = 1 + 3 cos 2x

We know that -1 ≤ cos 2x ≤ 1

⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1

⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4

Hence the range of f = [-2, 4]

Question 5.

Find the domain and range of the function \(f(x)=\frac{x^{2}-9}{x-3}\)

Solution:

Domain off: Clearly f(x) is not defined for x – 3 = 0 i.e. x = 3.

Therefore, Domain (f) = R – {3}

Range off: Let f(x) = y. Then,

It follows from the above relation that y takes all real values except 6 when x takes values in the set R – {3}. Therefore, Range (f) = R {6}.

Question 6.

Find the range of the following functions given by f(x) = \(\frac{1}{2-\sin 3 x}\)

Solution: