Difference between revisions of "1965 AHSME Problems/Problem 34"
(Created page with "== Problem 34== For <math>x \ge 0</math> the smallest value of <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> is: <math>\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \tex...") |
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\textbf{(E) }\ \frac{34}{5} </math> | \textbf{(E) }\ \frac{34}{5} </math> | ||
− | ==Solution== | + | ==Solution 1== |
To begin, lets denote the equation, <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> as <math>f(x)</math>. Let's notice that: | To begin, lets denote the equation, <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> as <math>f(x)</math>. Let's notice that: | ||
Line 28: | Line 28: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Therefore, <math>f(x) = \frac {4x^2 + 8x + 13}{6(1 + x)} \ge \boxed{\textbf{(B) } | + | 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> | <math>(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})</math> | ||
+ | |||
+ | ==Solution 2 (Calculus)== | ||
+ | Take the derivative of f(x) and f'(x) using the quotient rule. | ||
+ | <cmath>\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*}</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> |
Revision as of 12:30, 12 September 2021
Contents
Problem 34
For the smallest value of
is:
Solution 1
To begin, lets denote the equation, as
. Let's notice that:
After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because , which implies that both
are greater than zero. Continuing with AM-GM:
Therefore, ,
Solution 2 (Calculus)
Take the derivative of f(x) and f'(x) using the quotient rule.
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 , so we start from A:
However by the quadratic formula there are no real solutions of , so
cannot be greater than 0. We move on to B:
There is one solution: , 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