# Let F and G Be Two Real Functions Defined by F ( X ) = √ X + 1 and G ( X ) = √ 9 − X 2 . Then, Describe Function: (Vi) 2 F − √ 5 G - Mathematics

Let f and g be two real functions defined by $f\left( x \right) = \sqrt{x + 1}$ and $g\left( x \right) = \sqrt{9 - x^2}$ . Then, describe function:

(vi)  $2f - \sqrt{5} g$

#### Solution

Given:

$f\left( x \right) = \sqrt{x + 1}\text{ and } g\left( x \right) = \sqrt{9 - x^2}$

Clearly,

$f\left( x \right) = \sqrt{x + 1}$  is defined for all x ≥ - 1.
Thus, domain (f) = [1, ∞]
Again,

$g\left( x \right) = \sqrt{9 - x^2}$   is defined for  9 -x2 ≥ 0 ⇒ x2 - 9 ≤ 0
⇒ x2 - 32 ≤ 0
⇒ (x + 3)(x - 3) ≤ 0
$x \in \left[ - 3, 3 \right]$
Thus, domain (g) = [ - 3, 3]
Now,
domain ( f ) ∩ domain( g ) = [ -1, ∞] ∩ [- 3, 3]    = [ -1, 3]
(vi) $2f - \sqrt{5}g: \left[ - 1, 3 \right] \to \text{ R is given by } \left( 2f - \sqrt{5}g \right)\left( x \right) = 2\sqrt{x + 1} - \sqrt{5}\left( \sqrt{9 - x^2} \right)$ $= 2\sqrt{x + 1} - \sqrt{45 - 5 x^2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.4 | Q 4.6 | Page 38