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9TH MAHARASHTRA: MATHS I MATHS II SCIENCE 9TH TAMIL NADU: ENGLISH

CBSE 10TH CBSE 12TH: MATHS NCERT SOLUTION

ICSE BOARD ISC BOARD HOMI BHABHA JEE MAIN 2019 NEET - 2019 PRIVACY DISCLAIMER

### EX. NO. 1.3, RELATIONS AND FUNCTIONS, 10TH MATHS NEW SYLLABUS, TAMIL NADU, TN,

EX. NO. 1.3, RELATIONS AND FUNCTIONS, 10TH MATHS NEW SYLLABUS, TAMIL NADU, TN,

1. Let f = {(x, y) | x, y N and y = 2x} be a relation on ℕ. Find the domain, co-domain and range. Is this relation a function?
Solution :
Since x and y N,

 x = 1 y = 2(1) y = 2 x = 2 y = 2(2) y = 4 x = 3 y = 2(3) y = 6 x = 4 y = 2(4) y = 8

f = {(1, 2) (2, 4) (3, 6) (4, 8)................}
For each values of x, we get different values of y. So the given relation is a function.
Domain is the set of values of x
Domain  = {1, 2, 3, 4, ............}
Co domain is the set of value of y. Since y N
Co domain  = {1, 2, 3, 4, .............}
Range means the set of values of y, which are associated with x.
Range =  {2, 4, 6, 8, .....}

2. Let X = {3, 4, 6, 8}. Determine whether the relation  = {(x, f (x)) | x X, f (x) = x2 + 1} is a function from X to ?
Solution :
Given that :
f (x) = x2 + 1
x X

 if x = 3 f(3) = 32+1 f(3) = 10 if x = 4 f(4) = 42+1 f(4) = 17 if x = 6 f(6) = 62+1 f(6) = 37 if x = 8 f(8) = 82+1 f(8) = 65

R  = { (3, 10) (4, 17) (6, 37) (8, 65) }
For each values of x, we get different values of f(x).
Hence it is a function

3. Given the function f : x -> x2 −5x + 6 , evaluate
(i) f (-1)
(ii) f (2a)
(iii) f (2)
(iv) f (x −1)
Solution :
Given that :
f(x)  = x2 −5x + 6
(i) f (-1)
here we have -1 instead of x.
f(-1)  = (-1)2 −5(-1) + 6
= 1 + 5 + 6
f(-1)  = 12
(ii) f (2a)
here we have 2a instead of x.
f(2a)  = (2a)2 −5(2a) + 6
= 4a2 - 10a + 6
(iii) f (2)
here we have 2 instead of x.
f(2)  = (2)2 −5(2) + 6
= 4 - 10 + 6
=  0
(iv) f (x −1)
here we have x - 1 instead of x.
f(x-1)  = (x-1)2 −5(x-1) + 6
= x2 - 2x + 1 - 5x + 5 + 6
=  x2 - 7x + 12

4. A graph representing the function f (x) is given figure it is clear that f (9) = 2.
(i) Find the following values of the function
(a) f (0) (b) f (7) (c) f (2) (d) f (10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following (i) Domain (ii) Range.
(iv) What is the image of 6 under f ?
Solution :
(i)  (a) f(0)  = 9, (b) f(7)  = 6, (c) f (2) =  6, (d) f (10) = 0
(ii) For what value of x is f (x) = 1 ? For x = 9.5, we get 1.
(iii) Describe the following (i) Domain (ii) Range.
Domain  = { 0 ≤ x  ≤ 10 }
Range  = { 0 ≤ x  ≤ 9 }
(iv) What is the image of 6 under f ? The image of 6 under f is 5.

5. Let f (x) = 2x + 5. If x 0 then find [f(x + 2) - f(2)] / x
Solution :
f (x) = 2x + 5
f(x + 2)  = 2(x + 2) + 5
f(x + 2)  = 2x + 4 + 5   -------(1)
f(x + 2)  = 2x + 9
f(2)  = 2(2) + 5
= 4 + 5
f(2)  = 9 -------(2)
(1) - (2)   = 2x + 9 - 9
=  2x
[f(x + 2)  - f(2) ]/x =  2x / x = 2

6. A function f is defined by f (x) = 2x – 3
(i) find [f(0) +  f(1)]/2
(ii) find x such that f (x) = 0.
(iii) find x such that f (x) = x .
(iv) find x such that f (x) = f (1−x) .
Solution :
f(x) = 2x – 3
(i)  [f(0) +  f(1)]/2
f(0)  = 2(0) - 3  = -3
f(1)  = 2(1) - 3  = -1
[f(0) +  f(1)]/2 =  [-3 + (-1)] / 2
= -4/2
= -2
(ii) find x such that f (x) = 0.
f(x) = 2x – 3
2x - 3  = 0
2x  = 3
x  = 3/2
(iii) find x such that f (x) = x .
f(x) = 2x – 3
2x - 3  = x
2x - x  = 3
x  = 3
(iv) find x such that f (x) = f (1−x) .
f(1 - x)  = 2(1- x) - 3
= 2 - 2x - 3
= -2x - 1
2x - 3  = -2x - 1
2x + 2x  = - 1 + 3
4x  = 2
x  = 2/4  = 1/2

7. An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x. Solution :
Since the original shape is square, length of all sides will be equal.
length  = width  = 24 - 2x
height  = x
Volume of cuboid  = length ⋅ width ⋅ height
= (24 - 2x) ⋅ (24 - 2x) ⋅ x
=  x(24 - 2x)2
=  x [242 - 2(24) (2x) + (2x)2]
=  x [576 - 96x + 4x2]
V (x)  = 4x3 - 96x2 + 576x
Hence the volume of the cuboid is 4x3 - 96x2 + 576x.

8. A function f is defined by f (x) = 3−2x . Find x such that f (x2) = (f (x))2 .
Solution :
Given that :
f (x)  = 3 − 2x
f (x2)  = 3 − 2x2 ------(1)
(f (x))2  = (3 − 2x)2
=  32 - 2(3)(2x) + (2x)2
(f (x))2  = 9 - 12x + 4x2  ----(2)
(1)  = (2)
3 − 2x2  =  9 - 12x + 4x2
4x2 + 2x2 - 12x + 9 - 3  = 0
6x2 - 12x + 6  = 0
x2 - 2x + 1  = 0
(x - 1)2  = 0
x - 1  = 0 (or)  x - 1 = 0
x  = 1 (or) x  = 1
9. A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.
Solution :
Time  = Distance / speed
Speed of plane  = 500 km per hour
Distance travelled  = d
time taken  = t in hours
t  = d/500
d  = 500t
Hence the required distance is 500 t.
10. The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax +b , where a, b are constants.

 Length 'x' of forehand (in cm) Height 'y' (in inches) 35 56 45 65 50 69.5 55 74

(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.
Solution :
y = ax + b
(i)  For every values of x, we get different values of y.Hence it is a function.
(ii)
By solving these two equation, we get a and b
(1) - (2)
(45.5 a + b) - (35a + b)  = 65.5 - 56
45.5 a - 35a + b - b  = 9.5
10.5a  = 9.5
a  = 9.5/10.5
a  = 0.90
Substitute a = 0.90 in (1),
45.5(0.90) + b  = 65.5
b  = 65.5 - 40.95
b  = 24.5
y = 0.9x + 24.5
(iii) Find the height of a woman whose forehand length is 40 cm.
y = ? if x = 40
y = 0.9(40) + 24.5
y = 60.5
Height of woman is 60.5 inches.
(iv)  x = ? if y = 53.3
53.3 = 0.9x + 24.5
53.3 - 24.5  = 0.9x
28.8/0.9  = x
x  = 32
Length of forehand is 32 cm.