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Question: The function f defined by f(x) = \(f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty)\) is –...

The function f defined by f(x) = f:[1,∞)→[1,∞)f:\lbrack 1,\infty) \rightarrow \lbrack 1,\infty) is –

A

Every where continuous

B

Discontinuous at all integer values of x

C

Continuous at x = 0

D

None of these

Answer

Discontinuous at all integer values of x

Explanation

Solution

f(x) = Q sin πx > 0

2nπ < πx < (2n +1) π

2n < x < 2n + 1, n ∈ I and if sin πx < 0

(2n + 1) < πx < (2n + 2)π

2n + 1 < x < 2n + 2, n ∈ I and sin πx = 0

if x = 0, 1, 2, …..

f(x)={lim⁡t→∞1−1(1+sin⁡πx)t1+1(1+sin⁡πx)t,2n<x<2n+1lim⁡t→∞(1+sin⁡πx)t−1(1+sin⁡πx)t+1,2n+1<x<2n+2f ( x ) = \left\{ \begin{array} { l } \lim _ { t \rightarrow \infty } \frac { 1 - \frac { 1 } { ( 1 + \sin \pi x ) ^ { t } } } { 1 + \frac { 1 } { ( 1 + \sin \pi x ) ^ { t } } } , 2 n < x < 2 n + 1 \\ \lim _ { t \rightarrow \infty } \frac { ( 1 + \sin \pi x ) ^ { t } - 1 } { ( 1 + \sin \pi x ) ^ { t } + 1 } , 2 n + 1 < x < 2 n + 2 \end{array} \right.

=

k ∈ I, f(k) = 0, but f(x) = 1 or – 1

according as k ∈ (2n, 2n + 1) or k ∈ (2n + 1, 2n + 2)