Let f(x)=x sin pix, x gt 0 Then for all ...

Let `f(x)=x sin pix`, `x gt 0` Then for all natural numbers n, f`(x) vanishes at

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

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

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

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

Text Solution

Verified by Experts

`"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)`

Topper's Solved these Questions

DIFFERENTIATION
CENGAGE|Exercise Matrix Match Type|1 Videos
DIFFERENTIATION
CENGAGE|Exercise Numerical Value Type|3 Videos
DIFFERENTIATION
CENGAGE|Exercise JEE Previous Year|15 Videos
DIFFERENTIAL EQUATIONS
CENGAGE|Exercise Question Bank|25 Videos
DOT PRODUCT
CENGAGE|Exercise DPP 2.1|15 Videos

Lft f(x)= x sin pi x, x gt 0 then for all natural number n, f'(x) vanishes at

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

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

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

two point in the interval `(n, n+1)`

Let f (x) = |sin x| then f (x) is

continous everwhere

non-differentiable at odd and even multiple of `pi`

everywhere continuous but not-differentiable at `x = npi, n in I`

All of these

Let f(x)=(sin x)/(x), x ne 0 . Then f(x) can be continous at x=0, if

`f(0)=0`

`f(0)=1`

`f(0)=2`

`f(0)=-2`