tn maths 10th ex 1.5

 Question 1

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

(i) f(x) = x – 6, g(x) = x2
(ii) f(x) = 2x, g(x) = 2x2 – 1
(iii) f(x) = x+63 g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4


answer:
(i) f(x) = x – 6, g(x) = x2
fog(x) = f(g(x)) = f(x2) = x2 – 6 …………….. (1)
gof(x) = g(f(x)) = g(x – 6) = (x – 6)2
= x2 + 36 – 12x = x2 – 12x + 36 ……………… (2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)


(ytt tvvrr) f(x) = 4x2 – 1, g(x) = 1 + x
fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1
= 4 (1 + x2 + 2x) – 1 = 4 + 4x2 + 8x – 1  

4x2 + 8x + 3 ……………. (1)
gof(x) = g(f(x)) = g(4x2 – 1)
= 1 + 4x2 – 1 = 4x2 …………….. (2)
(1) ≠ (2)jdjjifj9i

lpp(x) ≠ gof(x)






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tn10th maths ex:1.5

( v) f(x) = 4x 2  – 1, g(x) = 1 + x fog(x) = f(g(x)) = f(1 + x) = 4(1 + x) 2  – 1 = 4(1 + x 2  + 2x) – 1 = 4 + 4x 2  + 8x – 1 = 4x 2  + 8x + 3 ……………. (1) gof(x) = g(f(x)) = g(4x 2  – 1) = 1 + 4x 2  – 1 = 4x 2  …………….. (2) (1) ≠ (2) ∴ fog(x) ≠ gof(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 Answer: f(x) = 3x + 2 ;g(x) = 6x – k fog = f[g(x)] = f (6x – k) = 3(6x – k) + 2 = 18x – 3K + 2 g0f= g [f(x)] = g (3x + 2) = 6(3x + 2) – k = 18x + 12 – k But given fog = gof. 18x – 3x + 2 = 18x + 12 – k -3k + 2 = 12 – k -3 k + k = 12-2 -2k = 10 k =  − 10 2  = -5 The value of k = -5 (ii) f(x) = 2x – k, g(x) = 4x + 5 Answer: f(x) = 2x – k ; g(x) = 4x + 5 fog = f[g(x)] = f(4x + 5) = 2(4x + 5) – k = 8x + 10 – k gof = g [f(x)] = g(2x – k) = 4(2x – k) + 5 = 8x – 4k + 5 But fog = gof 8x + 10 – k = 8x – 4k + 5 -k + 4k = 5 – 10 3k = -5 k =  − 5 3 The value of k =  − 5 3 Q 4. (i) If f (x) = x 2  – 1,...
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ex:1.4

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  Q1. Determine whether the graph given below represent functions. Give reason for your answers concerning each graph. answer:(i) (ii)&(iii) (iv) (i) It is not a function. The graph meets the vertical line at more than one points. (ii) It is a function as the curve meets the vertical line at only one point. (iii) It is not a function as it meets the vertical line at more than one points. (iv) It is a function as it meets the vertical line at only one point. Q 2. Let f :A → B be a function defined by f(x) =  x 2  – 1, Where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}. Represent f by (i) set of ordered pairs; (ii) a table; (iii) an arrow diagram; (iv) a graph answer : f: A → B A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} f(x)=x/2-1 f(2)=2/2-1=0 f(4)=4/2-1=1 f(6)=6/2-1=2 f(10)=10/2-1=4 f(12)=12/2-1=5 (i) Set of ordered pairs = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)} (ii) a table (iii) an arrow diagram;
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tn 10th maths ex 1.2

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  Q1. Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B ? (i) R 1  = {(2, 1), (7, 1)} (ii) R 2  = {(-1, 1)} (iii) R 3  = {(2, -1), (7, 7), (1, 3)} (iv) R 4  = {(7,-1), (0, 3), (3, 3), (0, 7)} (i) A = {1, 2, 3, 7}, B = {3, 0,-1, 7} Solution: R 1  = {(2,1), (7,1)} It is not a relation there is no element as 1 in B. (ii) R 2  = {(-1, 1)} It is not [ ∵  -1  ∉  A, 1  ∉  B] (iii) R 3  = {(2, -1), (7, 7), (1, 3)} It is a relation. R 4  = {(7,-1), (0, 3), (3, 3), (0, 7)} It is also not a relation. [ ∵  0  ∉  A]   Q 2. Let A = {1, 2, 3, 4, ….., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R. Answer: A = {1,2, 3, 4 . . . . 45} The relation is defined as “is square of’ R = {(1,1) (2, 4) (3, 9) (4, 16) (5,25) (6, 36)} Domain of R = {1, 2, 3, 4, 5, 6} Range of R = {...
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