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14 Feb 2019, 12:11
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:47) correct
32%
(01:46)
wrong
based on 971
sessions
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Not Attempted Yet
GMATH practice exercise (Quant Class 11)
If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
If \(\,4 \le x \le 6\,\) and \(\,2 \le y \le 3\) , the minimum possible value of \(\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\) is:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
14 Feb 2019, 18:21
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fskilnik
\(?\,\, = \,\,\min \,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\, = \,\,\min \,\left| {{y^2} - {x^2}} \right|\)
\(2 \le y \le 3\,\,\,\,\, \Rightarrow \,\,\,\,\,4 \le {y^2} \le 9\)
\(4 \le x \le 6\,\,\,\,\, \Rightarrow \,\,\,\,\,16 \le {x^2} \le 36\,\,\,\,\, \Rightarrow \,\,\,\,\, - 36 \le - {x^2} \le - 16\)
\(\left. \matrix{\\
4 \le {y^2} \le 9 \hfill \cr \\
- 36 \le - {x^2} \le - 16\,\,\, \hfill \cr} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\, - 32 \le {y^2} - {x^2} \le - 7\,\,\,\,\, \Rightarrow \,\,\,\,\,7 \le \left| {{y^2} - {x^2}} \right| \le 32\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 7\)
The correct answer is (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
General Discussion
17 Feb 2019, 01:56
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fskilnik
\(4\leq x \leq6\)
\(2\leq y \leq3\)
Given,
\(|y^2 - x^2|\)
but we are asked to find out the minimum value of |y^2 - x^2| .
we must work with corner values / extremes values of x and y.
\(2^2 - 4^2\) = 4 - 16 = - 12 = | -12 | = 12
\(2^2 - 6^2\)= 4 - 36 = -32 = | -32| = 32
\(3^2 - 4^2\)= 9 - 16 = -7 = | - 7| = 7.
\(3^2 - 6^2\)= 9 - 36 = -27 = | -27| = 27.
D is the correct answer.
