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
x | [x] | 0 x < 1 1 x < 2 2 x < 3 - 1 x < 0 - 2 x < - 1 | 0 1 2 - 1 - 2 |
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]
x | {x} | 0 x < 1 1 x < 2 2 x < 3 - 1 x < 0 - 2 x < - 1 | x x - 1 x - 2 x + 1 x + 2 |
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]
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