Piecewise function analysis with transcendental components

The question involves a piecewise-defined function containing transcendental parts and asks about continuity, differentiability, or computing the derivative as a piecewise expression.

ap-calculus-ab 2011 Q6 View
Let $f$ be a function defined by $$f(x) = \begin{cases} 1 - 2\sin x & \text{for } x \leq 0 \\ e^{-4x} & \text{for } x > 0. \end{cases}$$
(a) Show that $f$ is continuous at $x = 0$.
(b) For $x \neq 0$, express $f'(x)$ as a piecewise-defined function. Find the value of $x$ for which $f'(x) = -3$.
(c) Find the average value of $f$ on the interval $[-1, 1]$.
isi-entrance 2010 Q7 View
Let $f(x) = |x|\sin x + |x - \pi|\cos x$ for $x \in \mathbb{R}$. Then
(a) $f$ is differentiable at $x = 0$ and $x = \pi$
(b) $f$ is not differentiable at $x = 0$ and $x = \pi$
(c) $f$ is differentiable at $x = 0$ but not differentiable at $x = \pi$
(d) $f$ is not differentiable at $x = 0$ but differentiable at $x = \pi$
isi-entrance 2020 Q2 View
For the function on the real line $\mathbb { R }$ given by $f ( x ) = | x | + | x + 1 | + e ^ { x }$, which of the following is true ?
(A) It is differentiable everywhere.
(B) It is differentiable everywhere except at $x = 0$ and $x = - 1$.
(C) It is differentiable everywhere except at $x = 1 / 2$.
(D) It is differentiable everywhere except at $x = - 1 / 2$.
isi-entrance 2021 Q26 View
Define $f : \mathbb { R } \rightarrow \mathbb { R }$ by $$f ( x ) = \begin{cases} ( 1 - \cos x ) \sin \left( \frac { 1 } { x } \right) , & x \neq 0 \\ 0 , & x = 0 \end{cases}$$ Then,
(A) $f$ is discontinuous.
(B) $f$ is continuous but not differentiable.
(C) $f$ is differentiable and its derivative is discontinuous.
(D) $f$ is differentiable and its derivative is continuous.
jee-advanced 2018 Q18 3 marks View
Let $f _ { 1 } : \mathbb { R } \rightarrow \mathbb { R } , f _ { 2 } : \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right) \rightarrow \mathbb { R } , f _ { 3 } : \left( - 1 , e ^ { \frac { \pi } { 2 } } - 2 \right) \rightarrow \mathbb { R }$ and $f _ { 4 } : \mathbb { R } \rightarrow \mathbb { R }$ be functions defined by
(i) $\quad f _ { 1 } ( x ) = \sin \left( \sqrt { 1 - e ^ { - x ^ { 2 } } } \right)$,
(ii) $\quad f _ { 2 } ( x ) = \left\{ \begin{array} { c c } \frac { | \sin x | } { \tan ^ { - 1 } x } & \text { if } x \neq 0 \\ 1 & \text { if } x = 0 \end{array} \right.$, where the inverse trigonometric function $\tan ^ { - 1 } x$ assumes values in $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$,
(iii) $\quad f _ { 3 } ( x ) = \left[ \sin \left( \log _ { e } ( x + 2 ) \right) \right]$, where, for $t \in \mathbb { R } , [ t ]$ denotes the greatest integer less than or equal to $t$,
(iv) $\quad f _ { 4 } ( x ) = \left\{ \begin{array} { c c } x ^ { 2 } \sin \left( \frac { 1 } { x } \right) & \text { if } x \neq 0 \\ 0 & \text { if } x = 0 \end{array} \right.$.
LIST-I P. The function $f _ { 1 }$ is Q. The function $f _ { 2 }$ is R. The function $f _ { 3 }$ is S. The function $f _ { 4 }$ is
LIST-II
  1. NOT continuous at $x = 0$
  2. continuous at $x = 0$ and NOT differentiable at $x = 0$
  3. differentiable at $x = 0$ and its derivative is NOT continuous at $x = 0$
  4. differentiable at $x = 0$ and its derivative is continuous at $x = 0$

The correct option is:
(A) $\mathbf { P } \rightarrow \mathbf { 2 ; } \mathbf { Q } \rightarrow \mathbf { 3 ; } \mathbf { R } \rightarrow \mathbf { 1 ; } \mathbf { S } \rightarrow \mathbf { 4 }$
(B) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 1 } ; \mathbf { R } \rightarrow \mathbf { 2 } ; \mathbf { S } \rightarrow \mathbf { 3 }$
(C) $\mathbf { P } \rightarrow \mathbf { 4 } ; \mathbf { Q } \rightarrow \mathbf { 2 } ; \mathbf { R } \rightarrow \mathbf { 1 } ; \mathbf { S } \rightarrow \mathbf { 3 }$
(D) $\mathbf { P } \rightarrow \mathbf { 2 } ; \mathbf { Q } \rightarrow \mathbf { 1 } ; \mathbf { R } \rightarrow \mathbf { 4 } ; \mathbf { S } \rightarrow \mathbf { 3 }$