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Let $f(x)=\sin x^2, \: x \in \mathbb{R}$. Then

  1. $f$ has no local minima
  2. $f$ has no local maxima
  3. $f$ has local minima at $x=0$ and $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for odd integers $k$ and local maxima at $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for even integers $k$
  4. None of the above
in Calculus by Veteran (431k points)
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$f(x) = \sin x^2$

$f'(x) = \cos x^2 2x = 2x \cos x^2 = 0$

$\Rightarrow \cos x^2 = 0 \Rightarrow \frac{\pi}{2}, 3\frac{\pi}{2}$

$f''(x) = -4x^2\sin x^2 + 2\cos x^2$

Now, put $x^2 = \frac{\pi}{2} \; \text{in}\; f''(x)$, we get:

$f(x) < 0$,

$\Rightarrow f(x) \text{ has maxima at x  = $\pm\sqrt{\frac{\pi}{2}}$} $
0

@ankitgupta.1729

Help me to proceed ahead in this please.

+2

@`JEET

when $f'(x)= 0$ i.e. $2xcosx^2 = 0$ then either $x=0$ or $cos\;x^2 = 0 $

Now, $cos\;x^2 = 0$ means $x^2 = \frac{ \pi }{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2} ,.......$

(or) $x = \pm \sqrt \frac{ \pi }{2} ,\pm \sqrt \frac{3 \pi}{2} , \pm \sqrt \frac{5 \pi}{2} ,.......$

We can also write it as :

$x = \pm \sqrt {0 \pi + \frac{ \pi }{2}} ,\pm \sqrt{ 1 \pi + \frac{ \pi}{2}} , \pm \sqrt {2\pi + \frac{ \pi}{2}} ,.......$

Now, $f''(x) = -4x^2sin\;x^2 + 2cos\;x^2$

1) at $x=0,$ $f''(0) = 2$ . So, $f''(0) > 0$ , So, at $x=0$ , $f(x)$ has local minima.

2) at $x= \pm \sqrt{ 1 \pi + \frac{ \pi}{2}}$,  $f''(\pm \sqrt{ 1 \pi + \frac{ \pi}{2}}) = 6 \pi $ . So, $f''(0) > 0$ , So, at $x=\pm \sqrt{ 1 \pi + \frac{ \pi}{2}}$ , $f(x)$ has local minima.

3) at $x= \pm \sqrt {2\pi + \frac{ \pi}{2}}$,  $f''(\pm \sqrt {2\pi + \frac{ \pi}{2}}) = -10 \pi $ . So, $f''(0) < 0$ , So, at $x=\pm \sqrt {2\pi + \frac{ \pi}{2}}$ , $f(x)$ has local maxima.

So, observe the pattern or we can prove it for even $k$ and odd $k$, answer should be $C$

+1
Thanks.

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