Consider the functions $f_{1}(x) = x$ and $f_{2}(x) = 2 + \ln x$ for $x > 0$. The graphs of these functions intersect:

The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

Let $f: R \to R$ be a differentiable function such that $f(2) = 6$ and $f'(2) = \frac{1}{48}.$ Then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{4t^3}{x - 2} dt$ equals

The function $f(x) = \begin{cases} e^{2x} - 1, & x \le 0 \\ ax + \frac{bx^2}{2} - 1, & x > 0 \end{cases}$ is continuous and differentiable for

Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1}, 0 < a < 2 .$
$1.$ Which of the following is true?
$(A)$ $(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$
$(B)$ $(2-a)^2 f^{\prime}(1)-(2+a)^2 f^{\prime \prime}(-1)=0$
$(C)$ $f^{\prime}(1) f^{\prime}(-1)=(2-a)^2$
$(D)$ $f^{\prime}(1) f^{\prime}(-1)=-(2+a)^2$
$2.$ Which of the following is true?
$(A)$ $f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
$(B)$ $f(x)$ is increasing on $(-1,1)$ and has a local maximum at $x=1$
$(C)$ $f(x)$ is increasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$(D)$ $f(x)$ is decreasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$3.$ Let $g(x)=\int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} d t$. Which of the following is true?
$(A)$ $g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$
$(B)$ $g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$
$(C)$ $g^{\prime}(x)$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$
$(D)$ $g^{\prime}(x)$ does not change sign on $(-\infty, \infty)$
Give the answer for questions $1, 2$ and $3.$

Match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A$. If $y = |x| + |x - 2|$,then at $x = 2$,$\frac{dy}{dx} =$	$I$. $2$
$B$. If $f(x) = |\cos 2x|$,then $f'(\frac{\pi}{4} +) =$	$II$. $0$
$C$. If $f(x) = \sin(\pi[x])$,where $[x]$ is the greatest integer function,then $f'(1-) =$	$III$. $-2$
$D$. If $f(x) = \log|x - 1|$,$x \neq 1$,then $f'(\frac{1}{2}) =$	$IV$. does not exist