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

The Domain of Definition of F ( X ) = √ X − 3 − 2 √ X − 4 − √ X − 3 + 2 √ X − 4 is (A) [4, ∞) (B) (−∞, 4] (C) (4, ∞) (D) (−∞, 4) - Mathematics

MCQ

The domain of definition of  $f\left( x \right) = \sqrt{x - 3 - 2\sqrt{x - 4}} - \sqrt{x - 3 + 2\sqrt{x - 4}}$ is

• (a) [4, ∞)

• (b) (−∞, 4]

• (c) (4, ∞)

• (d) (−∞, 4)

Solution

(a) [4, ∞)  $f\left( x \right) = \sqrt{x - 3 - 2\sqrt{x - 4}} - \sqrt{x - 3 + 2\sqrt{x - 4}}$

$\text{ For f(x) to be defined } , x - 4 \geq 0$

$\Rightarrow x - 4 \geq 0$

$\Rightarrow x \geq 4 . . . . (1)$

$\text{ Also} , x - 3 - 2\sqrt{x - 4} \geq 0$

$\Rightarrow x - 3 - 2\sqrt{x - 4} \geq 0$

$\Rightarrow x - 3 \geq 2\sqrt{x - 4}$

$\Rightarrow (x - 3 )^2 \geq \left( 2\sqrt{x - 4} \right)^2$

$\Rightarrow x^2 + 9 - 6x \geq 4\left( x - 4 \right)$

$\Rightarrow x^2 - 10x + 25 \geq 0$

$\Rightarrow (x - 5) {}^2 \geq 0, \text{ which is always true .}$

$\text{ Similarly,} x - 3 + 2\sqrt{x - 4} \geq 0 \text{ is always true } .$

$\text{ Thus, dom } (f(x)) = [4, \infty )$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Q 38 | Page 45