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05 Mar 2019, 13:57
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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:
85%
(hard)
Question Stats:
52% (02:52) correct
48%
(02:23)
wrong
based on 161
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The figure above shows the representations of the graphs of the functions f and g, where f(x) = 2^(-0.5x) and g(x) = ax^2+b (a and b constants). What is the value of g(f(0)) ?
(A) 3
(B) 3.25
(C) 3.5
(D) 3.75
(E) 4
Attachment:
GMAT-Club-Forum-qkzehvps.png [ 27.94 KiB | Viewed 93 times ]
05 Mar 2019, 15:30
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fskilnik
\(? = g\left( {f\left( 0 \right)} \right)\)
\(\left[ {{\rm{blue}}} \right]\,\,\,\left( {0,f\left( 0 \right)} \right) \in {\rm{graph}}\left( f \right)\,\,\,\, \Rightarrow \,\,\,\,f\left( 0 \right) = {2^{ - {1 \over 2}\left( 0 \right)}} = 1\)
\(? = g\left( 1 \right) = a \cdot {1^2} + b = a + b\)
\(\left[ {{\rm{red}}} \right]\,\,\,\,\left\{ \matrix{\\
\left( { - 4,b} \right) \in {\rm{graph}}\left( f \right)\,\,\,\, \Rightarrow \,\,\,\,b = {2^{ - {1 \over 2}\left( { - 4} \right)}} = 4 \hfill \cr \\
\left( { - 4,0} \right) \in {\rm{graph}}\left( g \right)\,\,\,\,\,\mathop \Rightarrow \limits^{b = 4} \,\,\,\,\,0 = a \cdot {\left( { - 4} \right)^2} + 4\,\,\,\, \Rightarrow \,\,\,\,a = - {1 \over 4} \hfill \cr} \right.\)
\(? = - {1 \over 4} + 4 = 3.75\)
The correct answer is (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
05 Mar 2019, 15:32
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Okay, this strikes me as not entirely GMAT-authentic, but I'll bite.
We want \(g(f(0))\), which means plugging 0 into the equation \(f(x)\) and then putting the result into the equation \(g(x)\). The first part is straightforward enough: \(f(0)=2^{-0.5(0)}=2^0=1\). But plugging that result into \(g(x)\) is less satisfying: \(g(1)=a(1^2)+b=a+b\).
Let's go back to the graph and take a look... We see that \((-4,0)\) is on the graph of \(g(x)\). (The parabola must be \(g(x)\), even if you don't know the graphs of these equations automatically, because the equation for \(f(x)\) can never equal zero.) Plugging this in to \(g(x)\) gives \(a(-4^2)+b=0\), which is \(16a+b=0\). From the dotted lines, it also appears that \(f(-4)=g(0)\). Plugging all of this in to the two functions gives \(2^{-0.5(-4)}=a(0^2)+b\), which means \(2^2=4=b\). Now \(16a+b=0\) becomes \(16a+4=0\), so \(a=-\frac{1}{4}\).
Finally, we can return to the earlier result that \(g(f(0)=g(1)=a+b\). With \(a=-0.25\) and \(b=4\), we get \(g(f(0)=-0.25+4=3.75\).
We want \(g(f(0))\), which means plugging 0 into the equation \(f(x)\) and then putting the result into the equation \(g(x)\). The first part is straightforward enough: \(f(0)=2^{-0.5(0)}=2^0=1\). But plugging that result into \(g(x)\) is less satisfying: \(g(1)=a(1^2)+b=a+b\).
Let's go back to the graph and take a look... We see that \((-4,0)\) is on the graph of \(g(x)\). (The parabola must be \(g(x)\), even if you don't know the graphs of these equations automatically, because the equation for \(f(x)\) can never equal zero.) Plugging this in to \(g(x)\) gives \(a(-4^2)+b=0\), which is \(16a+b=0\). From the dotted lines, it also appears that \(f(-4)=g(0)\). Plugging all of this in to the two functions gives \(2^{-0.5(-4)}=a(0^2)+b\), which means \(2^2=4=b\). Now \(16a+b=0\) becomes \(16a+4=0\), so \(a=-\frac{1}{4}\).
Finally, we can return to the earlier result that \(g(f(0)=g(1)=a+b\). With \(a=-0.25\) and \(b=4\), we get \(g(f(0)=-0.25+4=3.75\).
