Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions 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 to y 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.

Answer:

Given A denotes the set of students and B denotes the set of sections. Aslo given there 120 students and 4 sections.

Let f be a relation from A to B as “x related to y if the student x belongs to the section y”

Two are more students in A may belong to same section in B. But one student in A cannot belong to two or more sections in B. Every student in A can belong to any one of the section in B. Therefore / is a function.

In B we can have sections without students. Every element in B need not have preimage in A.

∴ f need not be onto.

Thus, f is a function and inverse relation for f need not exist.

Question 2.

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

Answer:

When x = -4

f(x) = – x + 4

f(-4) = – (-4) + 4

= 4 + 4 = 8

When x = 1

f(x) = x – x^{2}

f(1) = 1 – 1^{2}

= 1 – 1 = 0

When x = -2

f(x) = x^{2} – x

f(-2) = (-2)^{2} – (-2)

= 4 + 2 = 6

When x – 7

f(x) = 0

⇒ f(7) = 0

When x = 0

f(x) = x^{2} – x

⇒ f(0) = 0^{2} – 0 = 0

Question 3.

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

Answer:

When x = – 3

f(x) = x^{2} + x – 5

f(-3) = (-3)^{2} + (-3) – 5

= 9 – 3 – 5

= 9 – 8 = 1

When x = 5

f(x) = x^{2} + 3x – 2

f(5) = 5^{2} + 3(5) – 2

= 25 + 15 – 2

= 40 – 2 = 38

When x = 2

f(x) = x^{2} – 3

f(2) = 2^{2} – 3 = 4 – 3 = 1

When x = – 1

f(x) = x^{2} + x – 5

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

When x = 0

f(x) = x^{2} – 3

f(0) = 0^{2} – 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 f = { (a, c), (b, c), (c, b) }; (f : A → A)

Answer:

A = { a, b, c }

f = {(a, c), (b, c), (c, b)}; f : A → A

f is a function since every element in the domain has a unique image in the codomain.

f is not one-one.

a, b belonging to the domain A has the same image in the codomain A. f is not onto since belonging to the codomain A does not have preimage in the domain A Thus the relation / is a function from A to A and it is neither one-one nor onto.

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

Answer:

X = { x, y, z }

f = {(x, y), (x, z), (z , x) } f : X → X

The relation f: X → X is not a function since the element x in the domain has two images in the co-domain.

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.

Answer:

f = { (1, b) , (2, c) , (3, d) , (4, d)

f is a function, it not one to one and not onto.

(ii) not one – to – one but onto

Answer:

Does not exists

(iii) one – to – one but not onto

Answer:

Does not exist

(iv) one – to – one and onto

Answer:

f = { (1, a) , (2, b) , (3, c) , (4, d) }

f is a function which is one – to – one and onto.

Question 6.

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

Answer:

Let f(x) = \(\frac{1}{1-2 \sin x}\)

When 1 – 2 sin x = 0

⇒ 1 = 2 sin x

sin x = \(\frac{1}{2}\)

⇒ sin x = sin \(\left(\frac{\pi}{6}\right)\)

x = nπ + (- 1)^{n}\(\frac{\pi}{6}\), n ∈ Z

sin x = sin α ⇒ x = nπ + (-1)^{n}d, n ∈ Z

∴ Domain of f(x) is

Question 7.

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

Answer:

∴ For no real values of x, f (x) is defined.

∴ Domain of f(x) = { }

Question 8.

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

Answer:

Let f(x) = \(\frac{1}{2 \cos x-1}\)

Range of cosine function is

– 1 ≤ cos x ≤ 1

– 2 ≤ 2 cos x ≤ 2

– 1 ≤ 2 cos x – 1 ≤ 2 – 1

– 3 ≤ 2 cos x – 1 ≤ 1

\(-\frac{1}{3}\) ≥ \(\frac{1}{2 \cos x-1}\) ≥ 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.

Answer:

xy = – 2 ⇒ y = -2/x

which is a function

The domain is (-∞, 0) ∪ (0, ∞) and range is 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.

Answer:

Given

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.

Answer:

Given f : R → R , g : R → R and h : R → R (f + g) oh: R → R and (f o h + g o h) : R → R for any x ∈ R.

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

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

= foh(x) + goh(x)

∴ (f + g)oh = foh + goh

Also fo(g + h)(x) = f((g + h)(x)) for any x ∈ R

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

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

= fog (x) + foh(x)

∴ fo(g + h) = fog +foh

Question 12.

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

Answer:

Given f(x) = 3x – 5

Let y = 3x – 5

y + 5 = 3x

⇒ \(\frac{y+5}{3}\) = x

Let g(y) = \(\frac{y+5}{3}\)

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

= g(3x – 5)

= \(\frac{3 x-5+5}{3}\) = \(\frac{3 x}{3}\) = x

gof (x) = x

fog (y) = f(g(y))

= f\(\left(\frac{y+5}{3}\right)\)

= 3\(\left(\frac{y+5}{3}\right)\) – 5

= y + 5 – 5

fog(y) = y

∴ gof = I_{x} and fog = I_{Y}

Hence f and g are bijections and inverses to each ot1er.

Hence f is a bijection and f^{-1}(y) = \(\frac{y+5}{3}\)

Replacing y by x we get f^{-1}(x) = \(\frac{x+5}{3}\)

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.

Answer:

W(x) = 0.35x

Since bodyweight x is positive and if it increases then W(x) also increases.

Domain is (0, ∞) i.e.,x > 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.

Answer:

Given s (t) = – 16t^{2}

s (t_{1}) = s (t_{2}) ⇒ – 16t_{1}^{2} = – 16t_{2}^{2}

⇒ t_{1}^{2} = t_{2}^{2}

⇒ ± t_{1} = ± t_{2}

Since s (t_{1}) = s (t_{1}) 14 t_{1} = t_{2}

∴ The function s(t) is not one-one

Graph of s(t) = – 16t^{2}

Take the time along x – axis and distance along y – axis.

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 ) = 0.4 m + 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.

Answer:

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

Question 16.

A salesperson whose annual earnings can be represented by the function A (x) = 30,000 + 0.04 x, 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 + 0.05 x. Find (A + S)(x) and determine the total family income if they each sell Rs. 1,50,00,000 worth of merchandise.

Answer:

Given A (x) = 30,000 + 0.04 x

S (x) = 25,000 + 0.05x

A(x) + S(x) = 30,000 + 0.04 x + 25,000 + 0.05x

(A + S)(x) = 55,000 + 0.09 x

Given x = 1,50,00,000

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

= 55000 + 1,35000000

Total family income = Rs. 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 Indian rupee.

Answer:

Given f(x) = 1.23x

where x represents the number of American dollars

g(y) = 50.50y

where y represents the number of Singapore dollars.

To convert American dollars to Indian rupees, we must find

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

= g (1.23x)

= 50.50 (1.23x)

= 62.115x

∴ The function for the exchange rate of American can dollars in terms of Indian rupees is

gof (x) = 62.1 15x

Question 18.

The owner of a small restaurant can prepare a particular meal at a cost of Rs. 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.

Answer:

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)

Question 19.

The formula for converting from Fahrenheit to Celsius temperature is y = \(\). Find the inverse of this function and determine whether the inverse is also a function?

Answer:

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)

Answer:

Given f(x) = 3x – 4

Let y = 3x – 4

⇒ y + 4 = 3x

⇒ x = \(\frac{y+4}{3}\)

Let g(y) = \(\frac{y+4}{3}\)

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

= g(3x – 4)

= \(\frac{3 x-4+4}{3}=\frac{3 x}{3}\)

gof(x) = x

and fog(y) = f(g(y))

= f\(\left(\frac{y+4}{3}\right)\)

= 3\(\left(\frac{y+4}{3}\right)\)

= y + 4 – 4 = y

fog (y) = y

Hence g of = I_{x} and fog = I_{y}

This shows that f and g are bijections and inverses of each other.

Hence f is bijection and f^{-1}(y) = \(\frac{y+4}{3}\)

Replacing y by x we get f^{-1}(x) = \(\frac{x+4}{3}\)

The line y = x

f(x) =

The line y =3x-4

When x = 0 ⇒ y = 3 × 0 – 4 = -4

When x = 1 ⇒ y = 3 × 1 – 4 = -1

When x = -1 ⇒ y = 3 × -1 – 4 = -7

When x = 2 ⇒ y = 3 × 2 – 4 = 2

When x = -2 ⇒ y = 3 × -2 – 4 = -10

When x = 3 ⇒ y = 3 × 3 – 4 = 5

When x = -3 ⇒ y = 3 × -3 – 4 = -13

The line y = \(\frac{x+4}{3}\)