Difference between revisions of "1978 AHSME Problems/Problem 17"

(Created page with "We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath> We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math> Thus, our function is now <cmath>[...")
 
 
Line 1: Line 1:
 +
== Problem 17 ==
 +
 +
If <math>k</math> is a positive number and <math>f</math> is a function such that, for every positive number <math>x</math>, <math>\left[f(x^2+1)\right]^{\sqrt{x}}=k</math>;
 +
then, for every positive number <math>y</math>, <math>\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}</math> is equal to
 +
 +
<math>\textbf{(A) }\sqrt{k}\qquad
 +
\textbf{(B) }2k\qquad
 +
\textbf{(C) }k\sqrt{k}\qquad
 +
\textbf{(D) }k^2\qquad
 +
\textbf{(E) }y\sqrt{k}  </math> 
 +
 
We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath>
 
We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath>
 
We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math>
 
We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math>
Line 10: Line 21:
  
 
~JustinLee2017
 
~JustinLee2017
 +
 +
 +
==See Also==
 +
{{AHSME box|year=1978|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 21:44, 13 February 2021

Problem 17

If $k$ is a positive number and $f$ is a function such that, for every positive number $x$, $\left[f(x^2+1)\right]^{\sqrt{x}}=k$; then, for every positive number $y$, $\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}$ is equal to

$\textbf{(A) }\sqrt{k}\qquad \textbf{(B) }2k\qquad \textbf{(C) }k\sqrt{k}\qquad \textbf{(D) }k^2\qquad  \textbf{(E) }y\sqrt{k}$

We are given that \[[f(x^2 + 1)]^{\sqrt(x)} = k\] We can rewrite $\frac{9+y^2}{y^2}$ as $\frac{9}{y^2} + 1$ Thus, our function is now \[[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k\] \[\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k\] \[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}\] \[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2\]

\[\boxed{D}\]

~JustinLee2017


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png