Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If F(X) = Loge (1 − X) And G(X) = [X], Then Determine Function:(Iii) $\Frac{F}{G}$ - Mathematics

If f(x) = loge (1 − x) and g(x) = [x], then determine function:

(iii) $\frac{f}{g}$

#### Solution

Given:
f(x) = loge (1 − x) and g(x) = [x]
Clearly, f(x) = loge (1 − x)  is defined for all ( 1 -x)  > 0.
⇒ 1 > x
⇒ x < 1
⇒ x ∈ ( -∞, 1)
Thus, domain () = ( - ∞, 1)

Again,
g(x) = [x] is defined for all x ∈ R.
Thus, domain (g) = R
∴ Domain (f) ∩ Domain (g) = ( - ∞, 1) ∩ R      = ( -∞, 1)

Hence,

(iii) Given:
g(x) = [ x ]

If  [ ]  = 0,
x ∈ (0, 1)
Thus,

$\text{ domain } \left( \frac{f}{g} \right) = \text{ domain } \left( f \right) \cap \text{ domain } \left( g \right) - \left\{ x: g\left( x \right) = 0 \right\}$
$\frac{f}{g}: \left( - \infty , 0 \right) \to \text{ R is defined by } \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{\log_e \left( 1 - x \right)}{\left[ x \right]} .$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.4 | Q 5.3 | Page 38