Let $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$. Then:

Let $f(x) = ax^2 - b|x|$,where $a$ and $b$ are constants. Then at $x = 0$,$f(x)$ has

Two functions $f$ and $g$ have first and second derivatives at $x = 0$ and satisfy the relations: $f(0) = \frac{2}{g(0)}$,$f'(0) = 2g'(0) = 4g(0)$,$g''(0) = 5f''(0) = 6f(0) = 3$. Then:

Let $f, g: R \rightarrow R$ be two real-valued functions defined as $f(x)=\begin{cases} -|x+3| & , x < 0 \\ e^{x} & , x \geq 0 \end{cases}$ and $g(x)=\begin{cases} x^{2}+k_{1} x & , x < 0 \\ 4 x+k_{2} & , x \geq 0 \end{cases}$,where $k_{1}$ and $k_{2}$ are real constants. If $(g \circ f)$ is differentiable at $x=0$,then $(g \circ f)(-4)+(g \circ f)(4)$ is equal to

Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x+1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$ is differentiable for all $x > 0$. Then $\frac{8f'(8)}{f'(\frac{1}{8})}$ is equal to $.............$.

Let $f : (0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$ for all $x \in (0, \pi)$. If $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,then which of the following statement$(s)$ is (are) $TRUE$?
$(A) f \left(\frac{\pi}{4}\right) = \frac{\pi}{4 \sqrt{2}}$
$(B) f(x) < \frac{x^4}{6} - x^2$ for all $x \in (0, \pi)$
$(C)$ There exists $\alpha \in (0, \pi)$ such that $f^{\prime}(\alpha) = 0$
$(D) f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0$