Difference between revisions of "1965 AHSME Problems/Problem 34"

Revision as of 13:58, 29 January 2020

Problem 34

For $x \ge 0$ the smallest value of $\frac {4x^2 + 8x + 13}{6(1 + x)}$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ \frac {25}{12} \qquad \textbf{(D) }\ \frac{13}{6}\qquad \textbf{(E) }\ \frac{34}{5}$

Solution

To begin, lets denote the equation, $\frac {4x^2 + 8x + 13}{6(1 + x)}$ as $f(x)$ . Let's notice that:

$\begin{align*} f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ & = \frac{4(x^2+2x) + 13}{6(x+1)}\\\\ & = \frac{4(x^2+2x+1-1)+13}{6(x+1)}\\\\ & = \frac{4(x+1)^2+9}{6(x+1)}\\\\ & = \frac{4(x+1)^2}{6(x+1)} + \frac{9}{6(1+x)}\\\\ & = \frac{2(x+1)}{3} + \frac{3}{2(x+1)}\\\\ \end{align*}$

After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because $x\ge 0$ , which implies that both $\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}$ are greater than zero. Continuing with AM-GM:

$\begin{align*} \frac{\frac{2(x+1)}{3} + \frac{3}{2(x+1)}}{2} &\ge {\small \sqrt{\frac{2(x+1)}{3}\cdot \frac{3}{2(x+1)}}}\\\\ f(x) &\ge 2 \end{align*}$

Therefore, $f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge 2$ , $\boxed{\textbf{(B)}}$

$(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})$

@@ Line 28: / Line 28: @@
 \end{align*}</cmath>
-Therefore, <math>f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge \boxed{\textbf{(B) }2}</math>
+Therefore, <math>f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge 2</math>, <math>\boxed{\textbf{(B)}}</math>
 <math>(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})</math>

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Difference between revisions of "1965 AHSME Problems/Problem 34"

Revision as of 13:58, 29 January 2020

Problem 34

Solution