JoeAa wrote:
gmatophobia wrote:
Bunuel wrote:
The domain of the function\( f(x) = \sqrt{\sqrt{x+2} - \sqrt{4-x}}\) is the set of real numbers x such that
A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤4
A. -2 ≤ x ≤ 1
B. -2 ≤ x ≤ 4
C. 1 ≤ x ≤ 2
D. 1 ≤ x ≤ 4
E. 2 ≤ x ≤4
The value of an expression under root is always non -negative
- \(\sqrt{x+2}\)
\( x + 2 \geq0\)
\( x \geq-2\)
- \(\sqrt{4-x}\)
\( 4-x \geq0\)
\( x \leq4\)
- \( \sqrt{\sqrt{x+2} - \sqrt{4-x} }\)
\( \sqrt{x+2} - \sqrt{4-x} \geq 0\)
\( \sqrt{x+2} \geq\sqrt{4-x}\)
Applying the principle that\( \sqrt{n} =|n|\)
\( |x+2| \geq|4-x|\)
Squaring both sides weget
\( x^2 + 4x + 4 \geq 16 + x^2 -8x\)
\( x^2 + 4x + 8x - x ^2 \geq 16 - 4\)
\( 12x \geq 12\)
\( x \geq 1\)
Combining the information we get-
\( 1 \leq x \leq4\)
OptionD
I completely get the logics but can I understand from youwhy \( x \geq-2\), \( x \leq4\), \( x \geq 1\) at the end itis \( 1 \leq x \leq4\) , whynot \( -2 \leq x \leq4\)?
The conditions(\( x \geq-2\) ,\( x \leq4\) ,\( x \geq1\) ) represent restrictive conditions, meaning for the whole expression to be defined they must be truesimultaneously . You see,
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Statistics : Posted by Bunuel • on 13 Aug 2023, 05:00 • Replies 7 • Views 4231
