Question

If the local maximum of f(x) = sin2x - xϵ(0,π) is at x = a, then find the value of 36 $\frac{a}{\pi }$

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Answer 1

If x=c is a local maximum of f(x), then we have $\pi$

1. f'(c) = 0

2. f''(c)<0

So, to find the the local maximum, we will find f’(x) and equate it to zero.

f’(x) = 2 cos2x - 1

f’(x) = 0 => 2cos2x - 1 = 0

In the interval (0, $\pi$ ), the solutions are x = $\frac{\pi }{6}$ and x = $\frac{5\pi }{6}$

Now we will use the second condition, that f”(c) < 0 when x =c is a maximum

f”(x) = -4 sin2x

f'' $\left(\frac{\pi }{6}\right)$ = -4sin(2. $\frac{5\pi }{6}$ < 0)

$\Rightarrow$ x = $\frac{\pi }{6}$ is a maximum

Now, f'' ($\frac{5\pi }{6}$)= $-4sin\left(2.\frac{5\pi }{6}\right)>0$

$\Rightarrow$ x = $\frac{5\pi }{6}$ is not a local maximum

So, we get a = $\frac{\pi }{6}$

$\Rightarrow 36\frac{a}{\pi }=36\times \frac{\pi }{6}\times \frac{1}{\pi }=6$