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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 11th 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 – 12 = 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) = -16t2. Graph the function and determine if it is one-to-one.
Solution:
s(t) = -16t2
Suppose S(t1) = S(t2)
since time cannot be negative, we to take t1 = t2
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 x2 – 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:
