Cinemugam (Kollywood)

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, Question 1 : If the ordered pairs (x 2 − 3x, y 2 + 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 -

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) = x 2 Solution :   f o g (x)  = f[g(x)] We apply the function given for g(x). f o g (x)  = f[x 2 ] Instead of x, we have x 2 , so we apply  x 2 instead of x in f(x). f o g (x)  = f(x 2 ) = x 2 − 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) = 2x 2 - 1  Solution :   f o g (x)  = f[g(x)] We apply the function given for g(x). f o g (x)  = f[2x 2 - 1] Instead of x, we have 2x 2 - 1, so we apply  2x 2 - 1 instead of x in f(x). f o g (x)  = f(2x 2 - 1) = 2/(2x 2 - 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).