Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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If F(X) = Loge (1 − X) and G(X) = [X], Then Determine Function: (Iv) G F Also, Find (F + G) (−1), (Fg) (0), ( F G ) ( 1 2 ) , ( G F ) ( 1 2 ) - Mathematics

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

(iv) \[\frac{g}{f}\] Also, find (f + g) (−1), (fg) (0),

\[\left( \frac{f}{g} \right) \left( \frac{1}{2} \right), \left( \frac{g}{f} \right) \left( \frac{1}{2} \right)\]
 
 
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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,

(iv) Given:
f(x) = loge (1 − x)

\[\Rightarrow \frac{1}{f\left( x \right)} = \frac{1}{\log_e \left( 1 - x \right)}\]
\[\frac{1}{f\left( x \right)}\]   is defined if loge( 1 -x) is defined and loge(1 – x) ≠ 0.
⇒ (1 - x) > 0 and (1 - x) ≠ 0
⇒ x < 1 and x ≠ 0
⇒ x ∈ ( -∞, 0)∪ (0, 1)
Thus,
\[\text{ domain } \left( \frac{g}{f} \right) = \left( - \infty , 0 \right) \cup \left( 0, 1 \right)\]  = ( - ∞, 1)  .
\[\frac{g}{f}: \left( - \infty , 0 \right) \cup \left( 0, 1 \right) \to \text{ R defined by } \left( \frac{g}{f} \right)\left( x \right) = \frac{g\left( x \right)}{f\left( x \right)} = \frac{\left[ x \right]}{\log_e \left( 1 - x \right)}\]
(f + g)( -1) = f(-1) + g( -1)
 = loge{1 – (-1)}+ [ -1]
= loge  2 – 1
Hence, (f + g)( -1) = loge  2 – 1
(fg)(0) = loge ( 1 – 0) × [0] = 0
\[\left( \frac{f}{g} \right)\left( \frac{1}{2} \right) = \text{ does not exist}  . \]
\[\left( \frac{g}{f} \right)\left( \frac{1}{2} \right) = \frac{\left[ \frac{1}{2} \right]}{\log_e \left( 1 - \frac{1}{2} \right)} = 0\]
 

 

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.4 | Q 5.4 | Page 38
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