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

Revision as of 11:30, 12 September 2021

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 1

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

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

Solution 2 (Calculus)

Take the derivative of f(x) and f'(x) using the quotient rule. $\begin{align*} f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ f'(x) & = \frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} & = \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{(1 + x)^2} & = \frac{4x^2 + 8x - 5}{(1 + x)^2} f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4} \end{align*}$

Solution 3 (answer choices, no AM-GM or calculus)

We go from A through E and we look to find the smallest value so that $x \ge 0$ , so we start from A:

$\frac{4x^2 + 8x + 13}{6x+6} = 1$

$4x^2 + 8x + 13 = 6x + 6$

$4x^2 + 2x + 7 = 0$

However by the quadratic formula there are no real solutions of $x$ , so $x$ cannot be greater than 0. We move on to B:

$\frac{4x^2 + 8x + 13}{6x + 6} = 2$

$4x^2 + 8x + 13 = 12x + 12$

$4x^2 - 4x + 1 = 0$

$(2x-1)^2 = 0$

There is one solution: $x = \frac{1}{2}$ , which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be $\boxed{\textbf{B}}$

@@ Line 42: / Line 42: @@
 f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4}
 \end{align*}</cmath>
+==Solution 3 (answer choices, no AM-GM or calculus)==
+We go from A through E and we look to find the smallest value so that <math>x \ge 0</math>, so we start from A:
+<cmath>\frac{4x^2 + 8x + 13}{6x+6} = 1</cmath>
+<cmath>4x^2 + 8x + 13 = 6x + 6</cmath>
+<cmath>4x^2 + 2x + 7 = 0</cmath>
+However by the quadratic formula there are no real solutions of <math>x</math>, so <math>x</math> cannot be greater than 0. We move on to B:
+<cmath>\frac{4x^2 + 8x + 13}{6x + 6} = 2</cmath>
+<cmath>4x^2 + 8x + 13 = 12x + 12</cmath>
+<cmath>4x^2 - 4x + 1 = 0</cmath>
+<cmath>(2x-1)^2 = 0</cmath>
+There is one solution: <math>x = \frac{1}{2}</math>, which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be <math>\boxed{\textbf{B}}</math>

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