Page 1 CBSE TEST PAPER-03 CLASS - XII MATHEMATICS (Relations and Functions) [ANSWERS] Topic:- Relations and Functions 1. the function f is one â€“ one, for f(x 1) = f(x 2) 2x 1 = 2x 2 x 1 = x 2 2. is x 1, x 2 ? R f(x 1) = f(x 2) 3 â€“ 4x1 = 3 â€“ 4x2 x 1 = x 2 Hence one â€“ one Y = 3 â€“ 4x 3 4 4 3 4 3 4 3 4 4 4 x f - ? ? = ? ? ? ? - - ? ? ? ? = - ? ? ? ? ? ? ? ? = y Hence onto also. 3. f is one â€“ one and onto, so that f is invertible with inverse f -1 = {(1, 1) (2, 2) (3, 3)} 4. fog (x) = f(g x) = f{|5x â€“ 2|) = |5x â€“ 2| 5. f = {(1, a) (2, b) (3, c)} f -1 = { (a, 1) (b, 2) (c, 3)} (f -1 ) -1 = {(1, a) (2, b) (3, c)} Hence (f -1 ) -1 = f. 6. (fog) (x) = f[g(x)] = f(x â€“ 7) = x â€“ 7 + 7 = x (fog) (7) = (7) Page 2 CBSE TEST PAPER-03 CLASS - XII MATHEMATICS (Relations and Functions) [ANSWERS] Topic:- Relations and Functions 1. the function f is one â€“ one, for f(x 1) = f(x 2) 2x 1 = 2x 2 x 1 = x 2 2. is x 1, x 2 ? R f(x 1) = f(x 2) 3 â€“ 4x1 = 3 â€“ 4x2 x 1 = x 2 Hence one â€“ one Y = 3 â€“ 4x 3 4 4 3 4 3 4 3 4 4 4 x f - ? ? = ? ? ? ? - - ? ? ? ? = - ? ? ? ? ? ? ? ? = y Hence onto also. 3. f is one â€“ one and onto, so that f is invertible with inverse f -1 = {(1, 1) (2, 2) (3, 3)} 4. fog (x) = f(g x) = f{|5x â€“ 2|) = |5x â€“ 2| 5. f = {(1, a) (2, b) (3, c)} f -1 = { (a, 1) (b, 2) (c, 3)} (f -1 ) -1 = {(1, a) (2, b) (3, c)} Hence (f -1 ) -1 = f. 6. (fog) (x) = f[g(x)] = f(x â€“ 7) = x â€“ 7 + 7 = x (fog) (7) = (7) 7. (i) (x, x) ? R, as x and x have the same no of pages for all x?R ? R is reflexive. (ii) (x, y) R x and y have the same no. of pages y and x have the same no. of pages ? (y, x) ?R ? (x, y) = (y, x) R is symmetric. (iii) if (x, y) ? R, (y, y) ? R (x, z) ?R ? R is transitive. 8. (i) a * b = a â€“ b + ab b * a = b â€“ a + ab a * b ? b * a (ii) a * (b * c) = a * (b â€“ c + bc) = a â€“ (b â€“ c + bc) + a. (b â€“ c + bc) = a â€“ b + c â€“ bc + ab â€“ ac + abc (a * b) * c = (a â€“ b + ab) * c = [ (a â€“ b + ab) â€“ c ] + ( a â€“ b + ab) = a- b + ab â€“ c + ac â€“ bc + abc a * (b * c) ? (a * b) * c. 9. (i) gof (x) = g[f(x)] = g (2x + 1) = (2x + 1) 2 â€“ 2 (ii) fog (x) = f (fx) = f (2x + 1) = 2(2x + 1) + 1 = 4x + 2 + 1 = 4x + 3 10. Let x 1 x 2 ? A Such that f(x1) = f(x2) 1 2 1 2 1 2 2 2 3 3 x x x x x x - - = - - = f is one â€“ one 2 1 3 2 2 1 y x x x y x y - = - - = - 3 2 1 y f y y ? ? - = ? ? - ? ? Hence ontoRead More

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