Q.Show that $f(x)=x^2$ is continuous and differentiable at $x=0$.

The left-hand derivative at $x=3$ is $1$ and the right-hand derivative is $-1$; since they differ, $f'(3)$ does not exist.

First check continuity at $x=3$: from the left piece, $\lim_{x\to3^-}(x+2) = 5$; the function value is $f(3)=5$; from the right piece, $\lim_{x\to3^+}(8-x) = 5$. So $f$ is continuous at $x=3$ -- continuity alone doesn't guarantee differentiability, though.

Left-hand derivative: for $x<3$, $f(x)=x+2$, so $f'(x)=1$ for $x$ near 3 from the left.

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The only non-differentiable 'corners' come from the two absolute-value terms, at $x=1$ and $x=3$.

$f(x)=|x-1|+|x-3|+\sin x$. The function $|x-1|$ has a sharp corner (a kink where the left and right derivatives disagree) exactly at $x=1$, and is smooth everywhere else. Similarly $|x-3|$ has a corner exactly at $x=3$. The term $\sin x$ is differentiable at every real number.

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At $x=a$ the left and right pieces give the same value (continuity holds) but different slopes ($-1$ and $3$), so $f$ is continuous but not differentiable at $x=a$; and since $f$ isn't even defined for $x\le -a$, option (a) is false.

Check continuity at $x=a$: left piece value $= 2a-a = a$; right piece value $=3a-2a=a$. They agree, so $f$ is continuous at $x=a$.

Check differentiability at $x=a$: derivative of $2a-x$ is $-1$; derivative of $3x-2a$ is $3$. Since $-1\ne 3$, the left and right derivatives at $x=a$ disagree, so $f$ is NOT differentiable at $x=a$ — this rules in option (c) and rules out (b).

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$\cot x$ is undefined (division by $\sin x = 0$) at $x=0$, which lies inside the stated interval, so option (c) fails to be differentiable — indeed not even defined — at that point.

1. $\sin x+\cos x$ is a sum of standard differentiable functions, differentiable on all of $(-\infty,\infty)$.
2. $\sin x$ is differentiable everywhere on $(-\infty,\infty)$. …