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Question

f(x)=sinx has a local minima at x=3π2.


A

True

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B

False

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Solution

The correct option is A

True


From the definition of local minimum, we know that it is a point which has lower output compared to its neighborhood. In other words f(x) will have a local minimum at x=a if f(a)<f(ah) and f(a)<f(a+h) where h is an infinitesimal positive number.

Let’s find f(a),f(ah) and f(a+h).

f(3π2)=sin(3π2)=1f(3π2h)=sin(3π2h)=sin(π+π2h)=sin(π2h)

We know sin(π2)=1 and sin(π2h) will be slightly less than 1.

We can write it as 1+h

sin(π2h)=(1h)=1+h

So, f(3π2h) will be slightly greater than - 1.

Now, let’s find f(3π2+h)

sin(3π2+h)=sin(π+π2+h)=sin(π2+h)

Here, sine will again be slightly less than 1 as sine has local maximum at x=π2.

We can write it as 1+h again.

And in the same way sin(3π2+h)=1+h

Which is slightly greater than -1.

Here we find thatf(3π2)<f(3π2h) and f(3π2)<f(3π2+h)

Hence we can say that sin(x) has a minimum at x=3π2


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