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If f, g and h are real functions defined by f ( x ) = √ x + 1 , g ( x ) = 1 x and h(x) = 2x2 − 3, find the values of (2f + g − h) (1) and (2f + g − h) (0). - Mathematics

If fg and h are real functions defined by 

\[f\left( x \right) = \sqrt{x + 1}, g\left( x \right) = \frac{1}{x}\] and h(x) = 2x2 − 3, find the values of (2f + g − h) (1) and (2f + g − h) (0).
 
 
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Solution

Given:

\[f\left( x \right) = \sqrt{x + 1}, g\left( x \right) = \frac{1}{x}\text{ and } h\left( x \right) = 2 x^3 - 3\]
Clearly, f (x) is defined for x + 1 ≥ 0 .
⇒ x ≥-1
⇒ x ∈ [-1, ∞]
Thus, domain ( f ) = [-1, ∞] .
Clearly, g (x) is defined for x ≠ 0 .
⇒ x ∈ R – { 0} and h(x) is defined for all x such that  x ∈ R .
Thus,
domain ( ) ∩ domain (g) ∩ domain (h) = [ -1, ∞] – { 0}.
Hence,
(2f + g – h) : [ -1, ∞] – { 0} → R is given by:
(2f + g – h)(x) = 2f (x) + g (x) -h (x)
\[= 2\sqrt{x + 1} + \frac{1}{x} - 2 x^2 + 3\]
\[(2f + g - h)(1) = 2\sqrt{2} + 1 - 2 + 3 = 2\sqrt{2} + 4 - 2 = 2\sqrt{2} + 2\]

(2f + g – h) (0) does not exist because 0  does not lie in the domain x ∈[ - 1, ∞] – {0}.

 
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.4 | Q 6 | Page 38
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