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Let f(x)=xsinpix ,\ x >0 . Then for ...

Let `f(x)=xsinpix ,\ x >0` . Then for all natural numbers `n ,\ f^(prime)(x)` vanishes at (a) A unique point in the interval `(n ,\ n+1/2)` (b) a unique point in the interval `(n+1/2,\ n+1)` (c) a unique point in the interval `(n ,\ n+1)` (d) two points in the interval `(n ,\ n+1)`

A

`"a unique point in the interval "(n,n+(1)/(2))`

B

`"a unique point in the interval "(n+(1)/(2),n+1)`

C

`"a unique point in the interval "(n,n+1)`

D

`"two points in the interval "(n,n+1)`

Text Solution

Verified by Experts

The correct Answer is:
A and C
`"We have "f'(x)=sin pix+pi x cos pi x=0`
`"or "tan pi x=-pix`
The graph of `y=tan pi x and y= - pi x` is as shown in the following figure. Therefore,
`tan pix=-pix`

From the graph
`x in (n+(1)/(2),n+1)or (n, n+1)`
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