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If  f(x) = [x] (sin \; kx)^p  is continuous for real x, then

Option: 1

k\; \epsilon \left \{ n\pi, \; n\;\epsilon \; I \right \},\; p>0


Option: 2

k\; \epsilon \left \{ 2n\pi, \; n\;\epsilon \; I \right \},\; p>0


Option: 3

k\; \epsilon \left \{ n\pi, \; n\;\epsilon \; I \right \},\; p \; \epsilon \; R- \left \{ 0\right \}


Option: 4

k\; \epsilon \left\{ n\pi, n \;\epsilon \;I , n \neq 0 \right \} p \;\epsilon \;R- \left\{0 \right \}


Answers (1)

best_answer

 

Continuity at a point -

A function f(x)  is said to be continuous at  x = a in its domain if 

1.  f(a) is defined  : at  x = a.

2. \lim_{x\rightarrow a}\:f(x)\:exists\:means\:limit\:x\rightarrow a

of  f(x) at  x = a exists from left and right.

3. \lim_{x\rightarrow a}\:f(x)=f(a)\:then\:the\:limit\:equals \:the\:value\:at\:x=a

-

 

 

Properties of Continuous function -

If   f,\:g   are two continuous functions at a point a of their common domain D.Then  f\pm g   fg are continuous at  a  and if   g(a)\neq 0  then 

 \frac{f}{g}      is also continuous at  x = a.

-

 

 

f(x)=[x](sin\; kx)^{p}

(sin kx)p is continuous and differentiable function x \epsilon R , k \epsilon R and  p > 0.

[x] is discoutinuous at x \epsilon I

For k = n \pi , n \epsilon I

f(x)=[x](sin (n\pi x))^{p}

f(x)=0,a\epsilon I

and f(a)=0

So   f (x) becomes coutinuous  for all x \epsilonR

Posted by

Devendra Khairwa

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