You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

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.

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

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

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.

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)

Question 6.

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

Solution:

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

Question 8.

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

Solution:

The range of cos x is – 1 to 1

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:

(i) Let f: R → R defined as f: x → \(-\frac{2}{x}\) then

f(x) = \(-\frac{2}{x}\) or y = \(-\frac{2}{x}\)

⇒ xy = – 2

f (x) is not a function since f(x) is not defined for x = 0

(ii) Let f: R – {0} → R defined as f(x) = \(-\frac{2}{x}\)

⇒ y = \(-\frac{2}{x}\) = xy = – 2

f is one – one but not onto because 0 has no preimage.

f : R – {0} → R {0} is a function which is one- one and onto

Domain = R – {0}

Range = R – {0}

Question 10.

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

Solution:

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

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

Question 13.

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

Solution:

Given W(x) = 0.35x

W(0) = W(1) = 0.35, W(2) = 0.7 ………….. W ( ∞ ) = ∞

Since x. denotes the bodyweight of a man, it will take only positive integers. That is x > 0.

W(x) : (0, ∞) → (0, ∞)

Domain = (0, ∞) , Range = (0, ∞)

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 |

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:

Given the cost of airfare function and fuel surcharge functions are as follows.

C(m) = 0.4 m+ 50 ———- (1)

S (m) = 0.03 m ———- (2)

Total cost of a ticket = C(m) + S(m)

f(x) = 0.4 m + 50 + 0.03 m

f(x) = 0.43 m + 50

Given m = 1600 miles

The cost of Airfare for flying 1600 miles

f( 1600 ) = 0.43 × 1600 + 50

= 688 + 50

= 738

∴ Airfare for flying 1600 miles is Rs. 738.

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

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 the 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

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:

Number of customers = 200 – x

Cost of one meal = Rs. 100

Cost of (200 – x) meals = (200 – x) × 100

Menu price of the meal = Rs. x

∴ Total menu price of (200 – x) meals = (200 – x) x

Profit = Menu price – Cost

= (200 – x) x – (200 – x) 100

Profit = (200 – x) (x – 100)

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: