Suppose #f(x) = 3tan^4x + 2k# and #g(x) = -tan^4 x+8k tan^2 x+k#, where #0 <= k <= 1# and #f(x) = g(x)# has exactly one solution for #x in [0, 1]#. Then what is that solution?

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Answer 1 · 2018-01-13T10:31:32

#x = tan^(-1)(sqrt(2)/2)#

Given:

#f(x) = 3 tan^4 x+2k#

#g(x) = -tan^4 x+8k tan^2x+k#

#0 <= x <= 1#

#0 <= k <= 1#

#f(x) = g(x)# has exactly one solution for #x in [0, 1]#

That is:

#3 tan^4 x+2k = -tan^4 x+8k tan^2 x+k#

So:

#4 tan^4 x-8ktan^2 x+k = 0#

Let #u = tan^2 x#

Then we require one solution for #u in [0, tan^2 1]# for:

#0 = 4u^2-8ku+k#

#color(white)(0) = (2u)^2-2(2u)(2k)+(2k)^2-(2k^2)+k#

#color(white)(0) = 4(u-k)^2-k(2k-1)#

So in order to have real solutions, we require #k(2k-1) >= 0#

Since we also require #0 <= k <= 1#, that implies that we need #1/2 <= k <= 1#

Note that if #k=1/2# then we have exactly one repeated root, namely #u = 1/2#.

More generally, the roots are:

#u = k+-1/2sqrt(k(2k-1))#

Note that:

#1/2sqrt(k(2k-1)) < k#

So both of the roots are in the range #[0, 2k] sub [0, 2] sub [0, tan^2 1]#

So the only solution is given by:

#tan^2 x = u = 1/2#

Hence:

#x = tan^(-1)(sqrt(2)/2)#