FUNCTION & GRAPH |
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| Introduction
Absolute Value of Modulus Function: y = |x| =  Domain  R
Range (0,  ) Properties of modulus function: (i) |x|  a  - a x a (a  0)(ii) |x| a x - a or x a (a ) (iii) |x + y| |x| + |y|(iv) |x + y|  Illustration: Find domain of y =  Ans: y is defind if (|x| - x) > 0 |x| > x which hold for -ve x only Hence domain (- , 0)(7) Signum Function: y = Sgn(x) =   Domain x R Range {- 1, 0, 1}  (8) Greatest integer function: [x] indicates integral part of x which is nearest & smaller integer to x. It is also c/d floor of x or stepwise function. [2.3] = 2 [5] = 5, [- 0.6] = - 1 In general  n n < n + 1 (n Integer) [n] = n 
Properties of greatest integer function: (i) [x] = x holds if x I I  Integer. (ii) [x + I] = [x] + I if I is integer. (iii) [x + y] [x] + [y] (iv) If [  (x)] I, then (x) I. (v) If [ (x)] I, then (x) < I + 1 (vi) If [x] > n x n + 1, n I. (vii) [- x] = - [x] if x I& [- x] = - [x] - 1 if x  I. (viii) [x + y] = [x] + [x + y - [x]] for all x, y R. (ix) [x] +  = [n x], n N N natural no.Illustration: y = 2[x] + 3 & y = 3[x - 2] + 5 then find value of [x + y]. Ans: 2[x] + 3 = 3[x - 2] + 5 2[x] + 3 = 3[x] - 6 + 5 [x] = 4 4 < 5 x = 4 + f (f fraction)  y = 2[x] + 3 = 11 [x + y] = [4 + f + 11] = [15 + f] = 15 (9) Fractional part function: y = {x} Let x = I + f, I = [x] & f = {x} y = {x} = x - [x]
Domain x R Range (0, 1)  Properties of fraction part of x (i) {x} = x if 0 < 1 (ii) {x} = 0 if x I (iii) {- x} = 1 - {x} if x I Illustration: Prove that [x] + [y] [x + y] where x = [x] + {x} Ans: x + y = [x] + {x} + [y] + {y} [x + y] = [[x] + [y] + {x} + {y}] = [x] + [y] + [{x} + {y}] [By [x + I] = [x] + I] [x + y] [x] + [y] (10) Exponential Function: f(x) = ax, a > 0, a  1 Domain R Range (0, ) Case I: a > 1 f(x) = ax increase with increase in x. i.e. f(x) is increasing function ob R  Case II: 0 < a < 1 f(x) = ax decrease with increase in x  Dumb Question: How range comes out (0, ) ?Ans: y = ax Let a > 1 As x Since a > 1 So, ax and when x - ax 0 range (0, ) Logorithmic function: f(x) = logax (x, a > 0) & a 1 Domain (0, ) Range R  Properties: (i) Logaa = 1 (ii) logbm a =  logba {a, b > 0, b 1 m R} (iii) logab =  {a, b > 0, a, b 1 & m > 1} (iv)  {a, m > 0 & a 1} (v)  {a, b, c > 0 & c 1} (vi) logma < b  (vii) logma < b  Dumb Question: Why b 1 in logba ? Ans: logba = c bc = a if b = 1 1c a Now whatever value of c, 1c a, so, b 1 Illustration: Find domain of f(x) = log10(1 + x3) Ans: f(x) = log10(1 + x3) exists if 1 + x3 > 0 (1 + x)(1 - x + x2) > 0 But 1 - x2 + x2 > 0 as D < 0 & a > 0 So, 1 + x > 0 x > - 1 x (- 1, ) Trignometric Function: (1) Sine Function: f(x) = sinx Domain R Range [-1, 1]  (2) Cosione Function: f(x) = cos x Domain R Range [-1, 1]  |
