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The following functions are continuous o...

The following functions are continuous on `(0, pi)` (a) `tan x`

A

tan x

B

`int_(0)^(x) t sin 1/t dt`

C

`{{:(1", "0 le x le 3pi//4),(2sin. 2/9 x", " (3pi)/4 lt xlt pi):}`

D

`{{:(xsin x", "0lt xle pi//2),(pi/2 sin(pi+x)"," pi/2 lt x lt pi):}`

Text Solution

Verified by Experts

The correct Answer is:
B, C
The function f(x) = tan x is not difined at ` x = pi/2`, so f(x) is not continuous on `(0, pi)`.
(b) Since, `g(x) = x sin 1/x` is continuous on `(0, pi)` and the integral function of a continuous function is continuous,
`:. F(x) = int_(0)^(x) t( sin 1/t) ` dt is continuous on `(0, pi)`.
(c ) Also, ` f(x) = {{:(" 1, " 0 lt xle (3pi)/4),(2sin((2x)/9)", " (3pi)/4 lt x lt pi):}`
We, have, `underset( x to (3pi^(-))/4) lim f(x) = 1`
`underset( x to (3pi^(+))/4) limf(x) = underset( x to (3 pi) /4 ) lim 2 sin ((2x)/9) = 1 `
So, f(x) is continuous at ` x = 3 pi//4`.
`rArr" " ` f(x) is continuous at all other points.
(d) Finally, ` f(x) = pi/2 sin (x +pi) rArr f(pi/2) = - pi/2`
`underset(x to(pi/2)^(-))lim f(x) = underset( h to 0) lim f(pi/2-h) = underset( h to 0) lim pi/2 sin ((3pi)/2 - h) = pi/2 and underset( x to(pi//2)^(+)) lim f(x) = underset( h to 0) lim f(pi/2 + h)`
` = underset( h to 0) lim pi/2 sin ((3pi)/2 + h) = pi/2 `
So, f(x) is not continuous at ` x = pi//2`.
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