Distinguish between the following: Fire Insurance & Marine Insurance

Distinguish between the following:

Fire Insurance & Marine Insurance

Basis of Difference	Fire Insurance	Marine Insurance
Compensation	Compensation is paid only in the event of loss due to fire. Nothing is paid otherwise.	Compensation is paid only in the event of loss due marine perils. Nothing is paid otherwise.
Duration	It does not generally exceed one year.	It is generally done for one year or for the period of voyage.
Subject matter	The goods and property of the person make the subject matter.	The goods in ship, cargo and freight make the subject matter.
Insurable interest	Insurable interest must exist at the time of taking the policy and also at the time of receiving the claim.	Insurable interest must exist at the time of receiving the claim or at the time of loss only.

UNIT EXERCISE 1, RELATIONS AND FUNCTIONS, 10TH MATHS NEW SYLLABUS, TAMIL NADU, TN,

UNIT EXERCISE 1, RELATIONS AND FUNCTIONS, 10TH MATHS NEW SYLLABUS, TAMIL NADU, TN,

Question 1 :

If the ordered pairs (x2 − 3x, y2 + 4y) and (-2,5) are equal, then find x and y.

Solution :

Question 2 :

The Cartesian product A×A has 9 elements among which (–1, 0) and (0,1) are found. Find the set A and the remaining elements of A×A.

Solution :

So, in this question since A X A has 9 elements, so the number of elements in the set  A must obviously be 3.

Since the given elements of the cross product have -1, 0 and 1 as part of the entries, clearly these only must be the elements of the set A.

So A = {-1, 0, 1}

A X A

  =  {(-1, -1),(-1, 0),(-1, 1),(0, -1),(0, 0),(0, 1),(1, -1),(1, 0),(1, 1)}

Question 3 :

Given that f(x)  =  

(i) f (0) (ii) f (3) (iii) f (a+1) in terms of a.(Given that a ≥  0)

Solution :

(i)  f(0)

Instead of x, we have 0, it is less than 1

f(0)  = 4

(ii)  f(3)

Instead of x, we have 3, it is greater than 1

f(3)  = √(3 - 1)

f(3)  = √2

(iii) f (a+1) 

Instead of x, we have a+1, it is greater than 1

f(a + 1)  = √(a + 1 - 1)

f(a + 1)  = √a

Question 4 :

Let A= {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f : A-> N be defined by f (n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.

Solution :

Set ordered pairs :

=  {(9,3) (10, 5)(11, 11)(12, 3)(13, 13) (14, 7) (15, 5) (16,2) (17, 17)}

Range  = {2, 3, 5, 7,  11, 13, 17}

Question 5 :

Find the domain of the function f(x) =

Solution :

Let t = √(1 - √(1 - x2))

f(x)  = √(1 - t)

1 - t ≥ 0

t ≤ 1

By applying the value of t, we get

√(1 - √(1 - x2)) ≤ 1

1 - √(1 - x2)) ≤ 1

Subtracting 1 through out the equation,

- √(1 - x2)) ≤ 0

√(1 - x2)) ≥0

Taking squares on both sides,

(1 - x2)  ≥ 0

 - x2 ≥ -1

x ≤ 1

Hence the domain is x ∈ [-1, 1]

Question 6 :

If f (x) = x2 , g(x) = 3x and h(x) = x −2 , Prove that (f ∘ g) ∘ h = f ∘ (g ∘ h) .

Solution :

L.H.S :

R.H.S :

Hence proved.

Question 7 :

Let A = {1, 2} and B = {1, 2, 3, 4} , C = {5, 6} and D = {5, 6 ,7, 8} . Verify whether A × C is a subset of B × D?

Solution :

A = {1, 2} and C = {5, 6}

A x C  = {(1, 5) (1, 6) (2, 5) (2, 6)}  ----(1)

B = {1, 2, 3, 4} and D = {5, 6 ,7, 8}

B x D  = { (1, 5) (1, 6) (1, 7) (1, 8) (2, 5) (2, 6) (2, 7) (2, 8) (3, 5) (3, 6) (3, 7) (3, 8) (4, 5) (4, 6) (4, 7) (4, 8) }

Hence A x C is the subset of B x D.

Question 8 :

If f(x)  = (x - 1)/(x + 1), x ≠ 1 show that f(f(x))  = -1/x, provided x ≠ 0

Solution :

f(x)  = (x - 1)/(x + 1) (Given)

f(f(x))  =  

Hence proved.

Question 9 :

The functions f and g are defined by f (x) = 6x + 8; g (x)  = (x - 2)/3 

(i) Calculate the value of gg (1/2)

(ii) Write an expression for gf (x) in its simplest form.

Solution :

g (x)  = (x - 2)/3

g(1/2)  = ((1/2) - 2)/3

  =  (-3/2)/3

g(1/2)   = -1/2

gg(1/2)  = ((-1/2) - 2)/3

=  (-5/2)/3

gg(1/2)  = -5/6

(ii) Write an expression for gf (x) in its simplest form.

gf (x)  = g[6x +8]

Now we have to apply 6x + 8 instead of x in g(x).

  =  [(6x + 8) - 2]/3

  =  (6x + 6)/3

  =  2x + 2  

=  2 (x + 1)

Question 10 :

Write the domain of the following real functions

(i) f (x)  = (2x + 1)/(x - 9)

Solution :

To find the domain, let us equate the denominator to 0

x - 9 = 0 

x  = 9

The function is defined for all real values of x except 9.

Hence the required domain is R - {9}.

(ii)  p(x) =  -5/(4x2 + 1)

Solution :

In the denominator, we have x2. For all real values of x, we get positive values. Hence the required domain is R. 

(iii)  g(x) =  √(x - 2)

Solution :

Since the given function is in radical sign, we should not get negative answer. For that we have to apply the values grater than 2.

Hence the required domain for he given function is [2, ∞)

(iv) h(x)  = x + 6

Solution :

For all real values of x, we get defined values of h(x). Hence the domain is R.

EX. NO. 1.5, RELATION AND FUNCTIONS, 10TH NEW SYLLABUS, Tamil Nadu, TN,

EX. NO. 1.5, RELATION AND FUNCTIONS, 10TH NEW SYLLABUS, Tamil Nadu, TN,

Question 1 : 

Using the functions f and g given below, find f o g and g o f . Check whether  f o g = g o f .

(i) f (x) = x −6, g(x) = x2

Solution : 

f o g (x)  = f[g(x)]

We apply the function given for g(x).

f o g (x)  = f[x2]

Instead of x, we have x2, so we apply  x2 instead of x in f(x).

f o g (x)  = f(x2) = x2 − 6   ---(1)

g o f (x)  = g[f(x)]

=  g [x- 6]

Instead of x, we have x - 6, so we apply  x - 6 instead of x in g(x).

g o f (x)  = g(x - 6)  = (x - 6)2  ---(2)

f o g (x) ≠ g o f (x)

(ii) f (x) = 2/x, g(x) = 2x2 - 1 

Solution : 

f o g (x)  = f[g(x)]

We apply the function given for g(x).

f o g (x)  = f[2x2 - 1]

Instead of x, we have 2x2 - 1, so we apply  2x2 - 1 instead of x in f(x).

f o g (x)  = f(2x2 - 1) = 2/(2x2 - 1)   ---(1)

g o f (x)  = g[f(x)]

=  g [2/x]

Instead of x, we have 2/x, so we apply  2/x instead of x in g(x).

g o f (x)  = g(2/x) = 2(2/x)2 - 1

=  2(4/x2) - 1

=  (8/x2) - 1   ---(2)

f o g (x) ≠ g o f (x)

(iii) f (x) = (x + 6)/3,  g(x) = 3 - x 

Solution : 

f o g (x)  = f[g(x)]

We apply the function given for g(x).

f o g (x)  = f[3 - x]

Instead of x, we have 3 - x, so we apply  3 - x instead of x in f(x).

f(3 - x)  = (3-x+6)/3

=  (9-x)/3  ---(1)

g o f (x)  = g[f(x)]

=  g [(x + 6)/3]

Instead of x, we have [(x + 6)/3], so we apply  [(x + 6)/3] instead of x in g(x).

 g [(x + 6)/3]  = 3 - [(x + 6)/3]

  =  (9 - x - 6)/3

=  (3 - x)/3  ---(2)

f o g (x) ≠ g o f (x)

(iv) f (x) = 3 + x,  g(x) = x - 4

Solution : 

f o g (x)  = f[g(x)]

We apply the function given for g(x).

f o g (x)  = f[x - 4]

Instead of x, we have x - 4, so we apply x - 4 instead of x in f(x).

f(x - 4)  = 3 + x - 4

=  x - 1  ---(1)

g o f (x)  = g[f(x)]

=  g [3 + x]

Instead of x, we have 3 + x, so we apply 3 + x instead of x in g(x).

 g [3 + x]  = 3 + x - 4 

  = x - 1  ---(2)

(1)  = (2)

f o g (x)  = g o f (x)

(v)  f (x) = 4x2 − 1, g(x) = 1 + x

Solution :

f o g (x)  = f[g(x)]

We apply the function given for g(x).

f o g (x)  = f[1 + x]

Instead of x, we have 1 + x, so we apply 1 + x instead of x in f(x).

f(1 + x)  = 4(1+x)2 - 1

  =  4(1 + x2 + 2x) - 1

  =  4 + 4x2 + 8x - 1

f o g (x)  = 4x2 + 8x + 3   ---(1)

g o f (x)  = g[f(x)]

=  g [4x2 − 1]

Instead of x, we have 4x2 − 1, so we apply 4x2 − 1 instead of x in g(x).

 g [4x2 − 1]  = 1 + 4x2 − 1

  = 4x2  ---(2)

f o g (x) ≠ g o f (x)

Question 2 :

Find the value of k, such that f o g = g o f

(i) f (x) = 3x +2, g(x) = 6x −k

Solution :

Given that 

f o g = g o f

f o g  = f[g(x)]

= f[6x - k] 

Now we are going to apply 6x - k instead of x in f(x).

f o g  = 3(6x - k) + 2 ----(1)

g o f  = g[f(x)]

=  g[3x + 2]

Apply 3x + 2 instead of x in the function g(x).

g o f  = 6(3x + 2) - k  -----(2)

(1)  = (2)

3(6x - k) + 2  = 6(3x + 2) - k

18x - 3k + 2  = 18x + 12 - k

-3k + k  = -2 + 12

-2k  = 10

k  = -5

Hence the value of k is -5.

(ii) f (x) = 2x −k, g(x) = 4x + 5

Given that 

f o g = g o f

f o g  = f[g(x)]

= f[4x + 5] 

Now we are going to apply 4x + 5 instead of x in f(x).

f o g  = 2(4x + 5) - k ----(1)

g o f  = g[f(x)]

=  g[2x - k]

Apply 2x - k instead of x in the function g(x).

g o f  = 4(2x - k) + 5  -----(2)

(1)  = (2)

2(4x + 5) - k  = 4(2x - k) + 5

8x + 10 - k  = 8x - 4k + 5

-k + 4k  = 5 - 10

3k  = -5

k  = -5/3

Hence the value of k is -5/3.

Question 3 :

If f (x) = 2x −1, g (x) = (x + 1)/2, show that f o g = g o f = x

Solution :

f o g  = f[g(x)]

=  f[(x + 1)/2]

Now we apply (x + 1)/2 instead of x in f(x).

  =  2((x + 1)/2) - 1

  =  x + 1 - 1

  =  x --(1)

g o f  = g[f(x)]

=  g[2x - 1]

=  (2x - 1 + 1)/2

=  2x/2

  =  x --(2)

(1)  = (2)

Question 4 :

(i) If f (x) = x2 −1, g(x) = x −2 find a, if g o f (a) = 1 .

Solution :

f(a)  = a2 −1 and g(a) = a −2

g o f (a)  = g[f(a)]

  =  g[a2 −1]

  =  (a2 −1) - 2

g o f (a)  = a2 −3  

Given that 

 g o f (a) = 1 

a2 −3 = 1

a2 = 4

a  = ±2

(ii) Find k, if f (k) = 2k −1 and f o f (k) = 5.

Solution :

f o f (k)  = f[f(k)]

  =  f[2k - 1]

Now we apply 2k - 1 instead of k in f(k)

  =  2(2k - 1) - 1

  =  4k - 2 - 1

f o f (k)  = 4k - 3

Given that :

f o f (k)  = 5

4k - 3  = 5

4k  = 5 + 3

4k  = 8

k  = 2 

Hence the value of k is 2.

Question 5 :

Let A, B, C ⊆ N and a function f : A -> B be defined by f(x) = 2x + 1 and g : B -> C be defined by g(x) = x2 . Find the range of f o g and g o f .

Solution :

f o g  = f[g(x)]

=  f[x2]

Now we apply x2 instead of x in f(x).

f o g  = 2 x2 + 1

y = 2x2 + 1

Range :

{y | y = 2x2 + 1 and x ∊ N}

g o f  = g[f(x)]

=  g[2x + 1]

Now we apply 2x + 1 instead of x in g(x).

g o f  = (2x + 1)2

Range :

{y | y = (2x + 1)2 and x ∊ N}

Question 6 :

Let f (x) = x2 −1 . Find (i) f o f (ii) f o f o f

Solution :

(i) f o f  

  =  f[f(x)]

  =  f[x2 −1]

Now we apply x2 −1 instead of x in f(x).

  =  (x2 −1)2 - 1

  =  x4 - 2x2 + 1 - 1

f o f  = x4 - 2x2 

(ii) f o f o f

f o f  = x4 - 2x2 

f o f o f  = f [f o f]

  =  f[x4 - 2x2]

Now we apply x4 - 2x2 instead of x in f(x).

  =  (x4 - 2x2)2 - 1

Question 7 :

If f : R -> R and g : R -> R are defined by f(x) = x5 and g(x) = x4 then check if f, g are one-one and f o g is one-one?

Solution :

 f(x) = x5

For every positive and negative values of x, we get positive and negative values of y.

Every element in x is associated with different elements of y. Hence it is one to one function.

g(x) = x4 

For every positive and negative values of x, we get only positive values of y.

Negative values of y is not associated with any elements of x. Hence it is not one to one function.

fog(x)  = f[g(x)]

  =  f[x4]

now, we apply x4 instead of x in f(x)

f[x4]  = (x5)4

fog(x)  = x20

fog is not one to one function.

Question 8 :

Consider the functions f (x), g(x), h(x) as given below. Show that (f o g) o h = f o (g o h) in each case.

(i) f(x) = x −1, g(x) = 3x +1 and h(x) = x2

Solution :

(f o g) o h

fog(x)  = f[g(x)]

  =  f[3x + 1]

  =  3x + 1 - 1

fog(x)  = 3x

(f o g) o h  = (f o g) [h(x)]

  =  (f o g) [x2]

(f o g) o h  = 3x2     --------(1)

f o (g o h)

(g o h)  = g[h(x)]

  =  g[x2]

  =  3x2 +1 

f o (g o h)  = f [goh]

  =  f[3x2 +1] 

=  3x2 + 1 - 1

f o (g o h)  = 3x2 -------(2)

(1)  = (2)

(f o g) o h  = f o (g o h)

(ii) f (x) = x2, g(x) = 2x and h(x) = x + 4

Solution :

(f o g) o h

fog(x)  = f[g(x)]

  =  f[2x]

  =  (2x)2

fog(x)  = 4x2

(f o g) o h  = (f o g) [h(x)]

  =  (f o g) [x + 4]

(f o g) o h  = 4(x+4)2

=  4(x2 + 8x + 16)

=  4x2 + 32x + 64     --------(1)

f o (g o h)

(g o h)  = g[h(x)]

  =  g[x + 4]

  =  2(x + 4)

(g o h)  = 2x + 8

f o (g o h)  = f [goh]

  =  f[2x + 8] 

f o (g o h)  = (2x + 8)2

 =  (2x)2 + 2(2x)(8) + 82

=  4x2 + 32 x + 64-------(2)

(1)  = (2)

(f o g) o h  = f o (g o h)

(iii) f (x) = x −4, g(x) = x2 and h(x) = 3x −5

Solution :

(f o g) o h

fog(x)  = f[g(x)]

  =  f[x2]

  =  x2 −4

fog(x)  = x2 −4

(f o g) o h  = (f o g) [h(x)]

  =  (f o g) [3x −5]

(f o g) o h  = (3x - 5)2 - 4

=  (3x)2 - 2 (3x)(5) + 52 - 4

=  9x2 - 30x + 25 - 4

=  9x2 - 30x + 21   --------(1)

f o (g o h)

(g o h)  = g[h(x)]

  =  g[3x −5]

  =  (3x −5)2

=  (3x)2 - 2 (3x)(5) + 52

(g o h)  = 9x2 - 30x + 25

f o (g o h)  = f [goh]

  =  f[9x2 - 30x + 25] 

f o (g o h)  = 9x2 - 30x + 25 - 4

=  9x2 - 30x + 21   --------(2)

(1)  = (2)

Hence proved.

Question 9 :

Let f = {(−1, 3),(0,−1),(2,−9)} be a linear function from Z into Z . Find f (x).

Solution :

Let the linear function be "y = ax + b"

By applying the value of b in (1)

-a + (-1)  = 3

-a  = 3 + 1

-a  = 4

a  = -4

By applying the value of "a" and "b", we get 

y = -4x - 1

Hence the required linear equation is -4x - 1.

Question 10 :

In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a,b are constants. Show that the circuit C(t) = 3t is linear.

Solution :

Let's take two points  t1 and t2 from domain of C(t).

now,  c(at1 + bt2)  = 3(at1 + bt2)

c(at1) =  3at1

ac(t1) =  3at1

bc(t2) =  3bt2

3(at1 + bt2)  = 3at1 + 3at2 

(or) C(at1 + bt2) = aC(t1) + bC(t2),

Hence c(t) is linear.