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21 Apr 2021, 02:00
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D
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Difficulty:
95%
(hard)
Question Stats:
36% (03:09) correct
64%
(02:40)
wrong
based on 47
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If x and y belong to the set {2, 4}, and \(x^{ky} = x^{(ly^2−8)}\), is \(kl > 2\)?
(1) \(k = −6\)
(2) \(3l − k = 3\)
(1) \(k = −6\)
(2) \(3l − k = 3\)
21 Apr 2021, 20:42
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Bunuel
Since x is either 2 or 4, we can equate the powers in \(x^{ky} = x^{(ly^2−8)}\)
\(ky=ly^2-8.......ly^2-ky-8\)
If the quadratic equation is \(ax^2+bc+c=0\), then
Sum of the roots =\(x_1+x_2=\frac{-b}{a}\)
In \(ly^2-ky-8=0\), it will be \(\frac{k}{l}\)
Product of the roots =\(x_1*x_2=\frac{c}{a}\)
In \(ly^2-ky-8=0\), it will be \(\frac{-8}{l}\)
we know the roots \(x_1 \ \ and \ \ x_2\) can be 2 or 4.
Let us check the values that l can take
\(x_1* x_2\) can be 2*2 or 2*4 or 4*4
a) If it is 2*2, then \(4=\frac{-8}{l}.....l=-2\), so if k<-1, then kl>2.
b) If it is 2*4, then \(8=\frac{-8}{l}.....l=-1\), so if k<-2, then kl>2.
c) If it is 4*4, then \(16=\frac{-8}{l}.....l=-\frac{1}{2}\), so if k<-4, then kl>2.
Thus we can say if k<-4, then the answer is YES for sure.
(1) \(k = −6\)
k<-4, so kl>2
Sufficient
(2) \(3l − k = 3\)
We have 3 possible values of l
a) l=-2, then (3*-2)-k=3 or k=-9......kl=-2*-9=18>2
b) l=-1, then (3*-1)-k=3 or k=-6......kl=-1*-6=6>2
c) l=-\(\frac{1}{2}\), then \(3*\frac{-1}{2}-k=3.....k=\frac{-9}{2}.........kl=\frac{-1}{2}*\frac{-9}{2}=9/4>2\)
So in each case, the answer is yes
Sufficient
D
24 Apr 2021, 13:57
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Bunuel
Since x = 2 or 4, which are positive and not 1, we must have equal exponents on both sides. Hence \(ly^2 - 8 - ky = 0\). Since y can only be 2 or 4, we must have \(4l - 2k - 8 = 0\) or \(16l - 4k - 8 = 0\).
Statement 1:
Plugging \(k = -6\) in both equations give \(l = -1\). Hence \(kl = 6 > 2\) in either case. Sufficient.
Statement 2:
We can write this as \(k = 3l - 3\) to plug in. With the first equation, solving gives \(L = -1\) and \(k = -6\). Plugging in the second equation gives the same set of solutions, hence \(kl = 6 > 2\) in either case. Sufficient.
Ans: D
