+30 votes
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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

(A) One, at $\dfrac{\pi}{2}$
(B) One, at $\dfrac{3\pi}{2}$
(C) Two, at $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$
(D) Two, at  $\dfrac{\pi}{4}$ and $\dfrac{3\pi}{2}$
asked in Calculus
edited | 2.4k views
+5

Local Minima :- $f(x)$ will have local minima at $x=a$ iff $f(a-h)>f(a)<f(a+h)$where $h->0$

Local Maxima :- $f(x)$ will have local maxima at $x=a$ iff  $f(a-h)<f(a)>f(a+h)$ where $h->0$

Global Minima :- $f(x)$ will have global minima at $x=a$ iff $f(x)≥f(a)$ ∀$x$∈ Domain of function.

Global Maxima :-$f(x)$ will have global maxima at $x=a$ iff $f(x)≤f(a)$ ∀$x$∈ Domain of function.

So overall we can say , when we talk about local minima/maxima we compare it with nearest points and then declare nature of function at that point but when we talk about Global minima/maxima we compare it with whole domain.

Global minima $⊆$ Local minima  and similarly Global maxima $⊆$ Local maxima

## 2 Answers

+21 votes
Best answer

answer is (d)

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

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

and 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 $\frac{3\pi}{2}$

answered by Active (4.9k points)
edited
0
+1 (y)
0

@Praveen Saini@Arjun @Bikram sir

I understand that if the endpoint is to the left of local maxima , then the endpoint is local minima and if the endpoint is to the left of local minima then the endpoint should be local maxima ... now what if the stationary point next to endpoint is an inflection point ??? Now how to find out whether endpoint is local maxima or minima ???

+3
Why pi/4 is local minima?
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how is pi/4 even a local mimima????

0
okay finally got it....
0
@Sushmita, Can you please explain in detail?
+2
+14 votes
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.
answered by Boss (17.8k points)
edited
+4

Answer is D only. 2012 onward official keys are available.

http://gatecse.in/w/images/b/b5/Key_CS_2012.pdf

+6

This is trick question. Local word matters !

0
Why are we considering pi/4 here. They have asked for local minima and not absoulte minina, so shouldnt we consider the open interval only?
+1
what has "local" to do with interval being closed or open? [ ] means it is closed interval and ( ) means it is open interval.
0
Sir, accroding to me there should be only one minima at 3pi/2. I am under the impression that "Local" means we should consider the open interval (pi/4,7pi/4). Please correct me if I am wrong.

Definition : http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx
0
@Arjun SIR

f(x) = sin x

f'(x) = cos x
solve for  cos x = 0  x= n (pi) / 2

f''(x) = - sin x
at  x= n (pi) / 2   f''(x) <0  implies local maxima

How did u find local minima ?
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@arjun sir

please explain why we have consider $\prod /4$ here ??
0
Simply visualize the graph of sinx between those interval.
+2
why is pi/4 considered?
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