Answer to Question #285619 in Algebra for Anonymous005

Question #285619

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.


1
Expert's answer
2022-01-11T17:39:30-0500

"\\displaystyle\\text{Suppose } f(x) = g(x) \\text{ then we have the following } \\\\\n 3\\tan^4x+2k = -\\tan^4x+8k\\tan^2 x + k \\\\\n\\implies 4\\tan ^4 x = k(8 \\tan ^2 x - 1) \\\\\n\\implies k = \\frac{4\\tan ^4 x}{8 \\tan ^2 x - 1}\\\\\n\\text{By futher simplification we have the following } 4\\tan^4 x - 8k\\tan ^2 x + k = 0\\\\\n\\text{set } y= 2\\tan^2 x\\text{ to obtain the following }\\\\\ny^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)} \\\\\n\\text{for the value of \\(y\\) to be valid we must have } k(4k - 1) \\ge 0 \\text{ since \\( 0 < k < 1\\) we have }\\\\\n4k \\ge 1 \\implies k \\ge \\frac14\\\\\n\\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.}"


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