We know that the output of the GIF function would always be an integer irrespective of input. Here numerator is sin(π[x-π]). We know that [x-π] would always be an integer for every real value of x. We also no that sin(nπ) = 0 if n is an integer This means that the numerator of this function would always be zero. Let's check the denominator (to confirm that the denominator is not equal to zero for any x)
For denominator = 0 4 + [x]² = 0 [x]² = -4 but the square of any real number cannot be negative Hence denominator > 0 for all real values of x and numerator = 0 This means f(x) = 0 for all real values of x This means a is the correct answer
Answers & Comments
Answer:
A) Continuous as well as differentiable for all x ∈ R
Step-by-step explanation:
[tex]f(x) = \frac{\sin(\pi[x-\pi])}{4+[x]^2}[/tex]
We know that the output of the GIF function would always be an integer irrespective of input.
Here numerator is sin(π[x-π]). We know that [x-π] would always be an integer for every real value of x.
We also no that sin(nπ) = 0 if n is an integer
This means that the numerator of this function would always be zero.
Let's check the denominator (to confirm that the denominator is not equal to zero for any x)
For denominator = 0
4 + [x]² = 0
[x]² = -4
but the square of any real number cannot be negative
Hence denominator > 0 for all real values of x
and numerator = 0
This means f(x) = 0 for all real values of x
This means a is the correct answer