Find the relation between the coefficients k and m

Find the relation between the coefficients $k$ and $m$ for which the function $\displaystyle f(x) = \sqrt2\,\frac{\sin^{2}x}{k\sqrt2-m\sin x}$ has an extremum at $x = \dfrac{\pi}{4}$.

Step 1: The derivative

Put $u = \sqrt2\sin^{2}x$ and $v = k\sqrt2-m\sin x$, so that

\[ u' = 2\sqrt2\sin x\cos x, \qquad v' = -m\cos x, \]

\[ f' = \frac{u'v-uv'}{v^{2}} = \frac{2\sqrt2\sin x\cos x\bigl(k\sqrt2-m\sin x\bigr)+\sqrt2\,m\sin^{2}x\cos x}{\bigl(k\sqrt2-m\sin x\bigr)^{2}}. \tag{1} \]

Step 2: Factor the numerator

Taking out $\sqrt2\cos x$ and collecting what remains: \[2\sin x\bigl(k\sqrt2-m\sin x\bigr)+m\sin^{2}x = 2k\sqrt2\sin x-2m\sin^{2}x+m\sin^{2}x = \sin x\bigl(2k\sqrt2-m\sin x\bigr),\]

so the derivative takes the compact form

\[ f'(x) = \frac{\sqrt2\sin x\cos x\bigl(2k\sqrt2-m\sin x\bigr)}{\bigl(k\sqrt2-m\sin x\bigr)^{2}}. \tag{2} \]

Step 3: Impose $f'\!\left(\tfrac{\pi}{4}\right) = 0$

At $x = \dfrac{\pi}{4}$ both $\sin\dfrac{\pi}{4} = \dfrac{\sqrt2}{2}$ and $\cos\dfrac{\pi}{4} = \dfrac{\sqrt2}{2}$ are non-zero, so the zero must come from the last factor:

\[ 2k\sqrt2-m\cdot\frac{\sqrt2}{2} = 0 \;\Longrightarrow\; 2k-\frac{m}{2} = 0 \;\Longrightarrow\; m = 4k. \tag{3} \]

Step 4: Check that the point is admissible and is a genuine extremum

The denominator must not vanish there: $k\sqrt2-m\dfrac{\sqrt2}{2} = \sqrt2\left(k-\dfrac{m}{2}\right) = -\sqrt2\,k$, which is non-zero precisely when $k\neq0$ (and then $m = 4k\neq2k$).

With $m = 4k$ the critical factor becomes $2k\bigl(\sqrt2-2\sin x\bigr)$, which changes sign at $x = \dfrac{\pi}{4}$ because $\sin x$ is increasing there; meanwhile $\sqrt2\sin x\cos x\gt 0$ near $\dfrac{\pi}{4}$ and the denominator is positive. Hence $f'$ really does change sign: for $k\gt 0$ from $+$ to $-$, a maximum, with

\[ f\!\left(\frac{\pi}{4}\right) = \frac{\sqrt2\cdot\tfrac12}{k\sqrt2-4k\cdot\tfrac{\sqrt2}{2}} = \frac{\tfrac{\sqrt2}{2}}{-\sqrt2\,k} = -\frac{1}{2k}. \]

Step 5: Verification

\[k = 1,\ m = 4:\quad f\!\left(\tfrac{\pi}{4}\right) = -0.5; \qquad f(0.7) = -0.5048108; \qquad f(0.87) = -0.5028183 \;\;\checkmark\]

\[f(0.6) = -0.5339939; \qquad f(0.95) = -0.5086889 \;\;\checkmark\ \text{(all below the value at } \tfrac{\pi}{4})\]

\[\text{The numerator of } f' \text{ is } \sin x\cos x\bigl(4k-\sqrt2\,m\sin x\bigr),\ \text{zero at } \tfrac{\pi}{4} \text{ iff } m = 4k \;\;\checkmark\]