Answer to Question #285619 in Algebra for Anonymous005

Question

Let f(x) = 3tan^4x+2k and g(x) = -tan^4x+8ktan^2 x + k for 0 ≤ x ≤ 1, where 0 < k < 1. The graphs of f and g intersect at exactly one point. Find k. My book gets an answer of 1/4 where as I get k=4. Please explain.

Answer 1

$\displaystyle\text{Suppose } f(x) = g(x) \text{ then we have the following } \\ 3\tan^4x+2k = -\tan^4x+8k\tan^2 x + k \\ \implies 4\tan ^4 x = k(8 \tan ^2 x - 1) \\ \implies k = \frac{4\tan ^4 x}{8 \tan ^2 x - 1}\\ \text{By futher simplification we have the following } 4\tan^4 x - 8k\tan ^2 x + k = 0\\ \text{set } y= 2\tan^2 x\text{ to obtain the following }\\ y^2 - 4ky + k = 0 \implies \,\, y = \frac{4k -\pm \sqrt{16k^2 - 4k}}{2} = \frac{4k \pm 2\sqrt{4k^2 - k}}{2} = 2k \pm \sqrt{k(4k - 1)} \\ \text{for the value of \(y\) to be valid we must have } k(4k - 1) \ge 0 \text{ since \( 0 < k < 1\) we have }\\ 4k \ge 1 \implies k \ge \frac14\\ \text{Since the minimum value for $k$ is $\frac14$ then it suffices to take $k = \frac14$ since we have that $f$ and $g$ intersect at a point.}$