1. Given piecewise function

$$f(x)= \begin{cases} \dfrac{\pi}{4}+\tan^{-1}x, & |x|\le 1 \\ \dfrac12(|x|-1), & |x|>1 \end{cases}$$

We need to check continuity and differentiability on the real line.

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2. Points where trouble may occur

Each branch is individually continuous and differentiable on its own domain. So we only need to check the junction points:

$x=-1 \quad \text{and} \quad x=1.$

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3. Continuity at } x=1

For $|x|\le 1$,

$f(1)=\frac{\pi}{4}+\tan^{-1}(1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}.$

Left-hand limit:

$\lim_{x\to 1^-} f(x)=\frac{\pi}{4}+\tan^{-1}(1)=\frac{\pi}{2}.$

For $x>1$, $|x|=x$, so

$f(x)=\frac12(x-1).$

Hence right-hand limit:

$\lim_{x\to 1^+} f(x)=\frac12(1-1)=0.$

Since

$\frac{\pi}{2} \ne 0,$

$f$ is not continuous at $x=1$.

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4. Continuity at } x=-1

For $|x|\le 1$,

$f(-1)=\frac{\pi}{4}+\tan^{-1}(-1)=\frac{\pi}{4}-\frac{\pi}{4}=0.$

Left-hand side: for $x<-1$, $|x|=-x$, so

$f(x)=\frac12((-x)-1)=\frac12(-x-1).$

Thus,

$\lim_{x\to -1^-} f(x)=\frac12(-(-1)-1)=\frac12(1-1)=0.$

Right-hand limit:

$\lim_{x\to -1^+} f(x)=\frac{\pi}{4}+\tan^{-1}(-1)=0.$

All are equal, so $f$ is continuous at $x=-1$.

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5. Continuity elsewhere

Away from $x=\pm1$, each formula is continuous. Since it is also continuous at $x=-1$ but not at $x=1$, the function is:

$\text{continuous on } \mathbb{R}\setminus\{1\}.$

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6. Differentiability at } x=-1

For $-1<x<1$,

$f(x)=\frac{\pi}{4}+\tan^{-1}x \implies f'(x)=\frac{1}{1+x^2}.$

So right derivative at $x=-1$ is

$f'_+(-1)=\frac{1}{1+(-1)^2}=\frac12.$

For $x<-1$,

$f(x)=\frac12(-x-1),$ so

$f'(x)=-\frac12.$

Thus left derivative at $x=-1$ is

$f'_-( -1 )=-\frac12.$

Since

$f'_-( -1 ) \ne f'_+( -1 ),$

$f$ is not differentiable at $x=-1$.

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7. Differentiability at } x=1

Since the function is not continuous at $x=1$, it cannot be differentiable there.

So $f$ is not differentiable at $x=1$.

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8. Final classification

• Continuous everywhere except at $x=1$
• Differentiable everywhere except at $x=-1$ and $x=1$

Therefore,

$\boxed{\text{Option D: continuous on } \mathbb{R}\setminus\{1\} \text{ and differentiable on } \mathbb{R}\setminus\{-1,1\}}$

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9. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They match.