Determine a and b so that the graph of f(x) =

Determine $a$ and $b$ so that the graph of $\displaystyle f(x) = \frac{\sin^{2}x}{a-b\cos x}$ has an extremum at the point $\left(\dfrac{\pi}{3},\,\dfrac14\right)$.

Step 1: The point condition

With $\sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$ and $\cos\dfrac{\pi}{3} = \dfrac12$, \[f\!\left(\frac{\pi}{3}\right) = \frac{3/4}{a-\dfrac{b}{2}} = \frac14 \;\Longrightarrow\; a-\frac{b}{2} = 3 \;\Longrightarrow\; 2a-b = 6. \tag{1}\]

Step 2: The derivative

With $u = \sin^{2}x$, $v = a-b\cos x$, so $u' = 2\sin x\cos x$ and $v' = b\sin x$, factoring $\sin x$ out of the numerator and using $\sin^{2}x = 1-\cos^{2}x$: \[f'(x) = \frac{\sin x\left[2\cos x\bigl(a-b\cos x\bigr)-b\sin^{2}x\right]}{\bigl(a-b\cos x\bigr)^{2}} = \frac{\sin x\left(2a\cos x-b\cos^{2}x-b\right)}{\bigl(a-b\cos x\bigr)^{2}}. \tag{2}\]

Step 3: The extremum condition

At $x = \dfrac{\pi}{3}$ neither $\sin x$ nor the denominator vanishes, so the bracket must: \[2a\cdot\frac12-b\cdot\frac14-b = 0 \;\Longrightarrow\; a-\frac{5b}{4} = 0 \;\Longrightarrow\; 4a = 5b. \tag{3}\]

Step 4: Solve

From (3), $a = \dfrac{5b}{4}$; substituting into (1) gives $\dfrac{5b}{2}-b = \dfrac{3b}{2} = 6$, so

\[ b = 4, \qquad a = 5, \qquad f(x) = \frac{\sin^{2}x}{5-4\cos x}. \]

Step 5: What kind of extremum?

With these values (2) reads

\[ f'(x) = \frac{-4\sin x\left(\cos^{2}x-\tfrac52\cos x+1\right)}{\bigl(5-4\cos x\bigr)^{2}} = \frac{-4\sin x\bigl(\cos x-2\bigr)\left(\cos x-\tfrac12\right)}{\bigl(5-4\cos x\bigr)^{2}}. \tag{4} \]

Near $x = \dfrac{\pi}{3}$ the factor $\cos x-2$ is negative, so $-4\bigl(\cos x-2\bigr)\gt 0$; with $\sin x\gt 0$ and the square positive, the sign of $f'$ is the sign of $\cos x-\tfrac12$. Since the cosine is decreasing there,

\[ x\lt \frac{\pi}{3}:\ \cos x\gt \tfrac12 \Rightarrow f'\gt 0; \qquad x\gt \frac{\pi}{3}:\ \cos x\lt \tfrac12 \Rightarrow f'\lt 0. \]

The derivative passes from $+$ to $-$: the point is a local maximum, not a minimum. (Indeed $f''\!\left(\tfrac{\pi}{3}\right) = -\tfrac12\lt 0$.)

Step 6: Verification

\[f(0.9) = 0.2441163; \quad f(1) = 0.2494278; \quad f\!\left(\tfrac{\pi}{3}\right) = 0.25; \quad f(1.1) = 0.2493241; \quad f(1.3) = 0.2362451 \;\;\checkmark\]

\[f'(1) = +0.0245711\gt 0, \qquad f'(1.1) = -0.0252075\lt 0 \;\;\checkmark\ \text{(a maximum)}\]

\[\text{The other root } \cos x = 2 \text{ is impossible, so } x = \pm\tfrac{\pi}{3}+2k\pi \text{ are the only critical points off } \sin x = 0 \;\;\checkmark\]