TN 11th, Maths Chapter 1, Exercise 1.3
Welcome To
Class : XI
Subject : Maths
f(-3) = (-3)² – 3 – 5
Chapter : 1
Exercise : 1.3
Que 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 :
A = {set of students in 11th standard}
B = {set of sections in 11th 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.
Que 2 :
Write the values of f at – 4, 1, -2, 7, 0 if
Solution :
f(-4) = -(-4) + 4
= 4+4
= 8
f(1) = 1 – 1²
= 0
f(-2) = (-2)² – (-2)
= 4 + 2
= 6
f(7) = 0
f(0) = 0
Que 3 :
Write the values of f at -3, 5, 2, -1, 0 if
Solution :
= 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
Que 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
(ii) f : X ➝ X
x ∈ X (Domain) has two images in the co-domain x. It is not a function.
Que 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 :
(i) 
(ii) Not Possible
(iii) Not Possible
(iv) 
Que 6 :
Find the domain of 1/(1−2sinx)
Solution:
Denominator ≠ 0
Therefore 1-2sinx ≠ 0
-2sinx ≠ -1
sinx ≠ 1/2 (i.e) x≠ Ï€/6
So Domain = R - {nÏ€ +(-1)^n (Ï€/6)}, n ∈ Z
Que 7 :
Find the largest possible domain of the real valued function f(x) =√(4−ײ)/√(ײ−9)
Solution :
f(x) = [√(4−x²)] / [√(x²−9)]
Take [√(4-x²)] > 0
4-x² > 0
x² > 4
∴ x > -2 And x < 2
Take [√(x²-9)] > 0
x²-9 > 0
x² > 9
∴ x > -3 And x < 3
∴ No largest possible domain
The domain is null set.
Que 8 :
Find the range of the function 1/(2cosx−1).
Solution :
The range of cos x is – 1 to 1
-1 ≤ cos x ≤ 1
"Multiple By 2" -2 ≤ 2 cos x ≤ 2
"Add By (-1)" -2-1≤ 2 cos x -1 ≤ 2-1
-3 ≤ 2 cos x -1 ≤ 1
So 1 ≤ 1/(2cos x - 1) ≤ (-1/3)
The Range Is Outside -1/3 and 1
(i.e)., Range Is (-∞,-1/3] U [ 1, ∞)
Que 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, ∞)
Range is R – {0}
Que 10 :
If f, g : R ➝ R are defined by f(x) = |x| + x and g(x) = |x| – x, find gof and fog.
Solution :
f(x) = { 0 x < 0; g(x) = { -2x x < 0
f(x) = { 2x x > 0; g(x) = { 0 x > 0
Now fog (x) = 0
gof (x) = 0
Que 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
Now koh = k(h(x))
= (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.
Que 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
x = ( y + 5 ) / 3
Let g(y) = ( y + 5 ) / 3
g o f(x) = g [ f(x) ]
= g ( 3x-5 )
= (3x - 5+5)/3 = x
also fog (y) = f[g(y)] = f [(y+5)/3]
= 3([y+5]/3)-5
=y+5-5
=y
Thus g o f = Ix and fog = Iy
f and g are bijections and inverse to each other. Hence f is a bijection and f^-1(y) = (y+5)/3
Replacing y by x we get f^-1(x) = (x+5)/3
Que 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.
So Domain is (0, ∞) i.e.,x > 0.
Que 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)
-16t1² = - 16t2²
t1 = 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
Que 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
Que 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
= 14,05,000
∴ Total income of family = ₹ 14,05,000
Que 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
Que 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)
Que 19 :
The formula for converting from Fahrenheit to Celsius temperatures is y=(5x/9)−(160/9)
Find the inverse of this function and determine whether the inverse is also a function.
Solution :
y = (5x/9)-(160/9)
= ( 5x -160 )/9
9y= 5x - 160
5x= 9y + 160
x= (9y+160)/5
Therefore y = (9x + 160 ) / 5
Yes It Is Also A Function.
Que 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
Solution:
f(x) = 3x – 4
Let y = 3x – 4
y + 4 = 3x
(y+4)/3 = x 
x = ( y+4 ) / 3
f[(y+4)/3] = 3[(y+4)/3]-4 = y
So ( x + 4 ) / 3 Which Is Also A Function.
If Any Doubt Click Here To Contact In WhatsApp
Thank You For Refering On Our Page
Comments
Post a Comment