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Consider the function $f(x) = \sin(x)$ in the interval $x =\left[\frac{\pi}{4},\frac{7\pi}{4}\right]$. The number and location(s) of the local minima of this function are
 

  1. One, at $\dfrac{\pi}{2}$
  2. One, at $\dfrac{3\pi}{2}$
  3. Two, at $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$
  4. Two, at  $\dfrac{\pi}{4}$ and $\dfrac{3\pi}{2}$
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Answer is $(D)$

$f '(s) = \cos x =0$ gives root $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ which lie between the given domain in question $\left[\frac{\pi}{4},\frac{7\pi}{4}\right]$

$f ''(x)= - \sin x$ at $\frac{\pi}{2}$ gives $-1<0$ which means it is local maxima and at $\frac{3\pi}{2}$ it gives $1>0$ which is local minima.

Since, at $\frac{\pi}{2}$ it is local maxima so, before it, graph is strictly increasing, so $\frac{\pi}{4}$ is also local minima.

So, there are two local minima $\frac{\pi}{4}$ and $\frac{3\pi}{2}.$
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Sine function increases till $\frac{\pi}{2}$ and so for the considered interval $\frac{\pi}{4}$ would be a local minimum. From $\frac{\pi}{2}$, value of sine keeps on decresing till $\frac{3\pi}{2}$ and hence $\frac{3\pi}{2}$ would be another local minima. So, (D) is the correct answer here.
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Simplest way possible! 

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Answer is B

NCERT 12th class maths book pg no:228 (chapter 6):

 

I don't think there is any doubt regarding local minima at point $c=\frac{\pi}{4}$.

QED: Answer is B

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