Question

If $f\left(x\right)=\frac{sin\left(\pi \right[x-\pi \left]\right)}{1+\left[x^2\right]}$ (Where [.] denotes the G.I.F.,) then f(x) is

• A. Continuous at integral points
• B. Continuous everywhere but not differentiable
• C. Differentiable once but higher order derivatives do not exist
• D. Differentiable for all x

Answer 1

Answer: (A), (D)

The correct options are
A Continuous at integral points
D Differentiable for all x

$[x-\pi ]$ = integer =k (say) $\forall xϵR$
$\Rightarrow sin\left(\pi \right[x-\pi \left]\right)=sink\pi =0\forall xϵR$
Also denominator = $1+\left[x^2\right]\ne 0$
Hence $f\left(x\right)=0\forall xϵR$
Hence f(x) is a constant function and thus continuous and differentiable for all $xϵR$.