Q.Find the values of $p$ and $q$ that make function $f(x)$ differentiable everywhere on $R$: $f(x)=3-x$, for $x<1$, $=px^2+qx$, for $x\ge 1$.

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.

Right-hand derivative: for $x>3$, $f(x)=8-x$, so $f'(x)=-1$ for $x$ near 3 from the right.

Since the left-hand derivative $(1)$ and right-hand derivative $(-1)$ are not equal, $f'(3)$ does not exist -- the graph has a sharp corner at $x=3$.

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.

A sum of differentiable functions is differentiable at any point where all the individual terms are differentiable, and can fail to be differentiable only where one of the terms itself has a corner. Since $x=1$ and $x=3$ are distinct points, $f$ is non-differentiable at exactly these 2 points.

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).

Option (d) is false since $f$ IS continuous at $x=a$ (shown above).

Option (a) is false too: $f$'s domain here is only $x>-a$ (the pieces cover $-a<x<a$ and $x\ge a$), so $f$ is not even defined — let alone continuous — for all $x\in\mathbb{R}$.

$\cot x$ 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 $(-\infty,\infty)$.
3. $\cot x = \dfrac{\cos x}{\sin x}$ requires $\sin x\ne 0$. But $x=0$ lies inside $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ and $\sin 0 = 0$, so $\cot x$ is not even defined there, let alone differentiable.
4. $\cos x$ is differentiable everywhere, including at the closed endpoints $\pm\pi$ (one-sided derivatives exist). So the function that fails to be differentiable is $\cot x$ on the given interval.