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13 Apr 2021, 02:05
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
40% (02:08) correct
60%
(02:13)
wrong
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Let M be the maximum value and N be the minimum value of the expression \(\frac{x^2 + y}{y}\) if \(a \leq x \leq b\) and \(c\leq y \leq d\), then what is the value of M - N ?
(1) d = 7
(2) a = -2, b = 5 and c = 3
(1) d = 7
(2) a = -2, b = 5 and c = 3
13 Apr 2021, 04:30
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Bunuel
Bunuel, please check the question. we will have to interchange the c and d, either in the main question or the statements.
—————x—————x—————x—————x
Answer to the edited question
We do not know whether any of a, b, c and d is negative or positive, and whether they are fraction or integers.
But, the first step would be to analyze the given expression.
\(\frac{x^2 + y}{y}\)\(=\frac{x^2 }{y}+\frac{y}{y}=\frac{x^2}{y}+1\)
Now, let the values of x and y when the expression is maximum be \(x_m \ \ and \ \ y_m\), and the values of x and y when the expression is minimum be \(x_n \ \ and \ \ y_n\).
A) M = \(\frac{(x_m)^2}{y_m}+1\).......Will depend on the values of \(x_m \ \ and \ \ y_m\), where x_m should be as high as possible and y_m should be as small as possible.
B) N = \(\frac{(x_n)^2}{y_n}+1\).......Will depend on the values of \(x_n \ \ and \ \ y_n\), but the minimum value of \(\frac{(x_n)^2}{y_n}+1\) will be 1, if \(x_n\) can take value 0. Otherwise \(x_n=|x|\) should be as low as possible and \(y_n\) as large as possible.
M-N=\(\frac{(x_m)^2}{y_m}+1\)\(-(\frac{(x_n)^2}{y_n}+1)\)=\(\frac{(x_m)^2}{y_m}-\frac{(x_n)^2}{y_n}\)
When x can take value 0, our equation M-N is \(\frac{(x_m)^2}{y_m}-\frac{(x_n)^2}{y_n}\)=\(\frac{(x_m)^2}{y_m}-0\)=\(\frac{(x_m)^2}{y_m}\)
(1) d=7
Nothing about other values
Insuff
(2) a = -2, b = 5 and c=3
So x can range from -2 to 5.
The minimum value of \(\frac{(x_n)^2}{y_n}+1\) = 1, as \(x_n\) can take value 0.
The maximum value of \(\frac{(x_m)^2}{y_m}+1\) will be when |x| is maximum, so \(x_m=b=5\), and when y is minimum but positive, so \(y_m=c=3\)
M-N=\(\frac{(x_m)^2}{y_m}=\frac{5^2}{3}\)
Sufficient
B
—————x—————x—————x—————
Question now edited.
Note: I believe that the question may have to interchange the value of c and d, that is \(c\leq y \leq d\) as then we will not require the value of c, and the answer would B, making the question 700+ level.
