Let $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geqslant 1 \\ ax^2 + b, & |x| < 1 \end{cases}$ be continuous and differentiable everywhere. Then $a$ and $b$ are

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column $I$ with the properties in Column $II$.

Column $I$	Column $II$
$(A)$ $f(x) = x|x|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{|x|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = |x - 1| + |x + 1|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

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

Let $f: R \to R$ be a twice differentiable function such that the quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ in $m$ has two equal roots for every $x \in R$. If $f(0)=1$, $f^{\prime}(0)=2$ and $(\alpha, \beta)$ is the largest interval in which the function $g(x) = f(\log_{e}x-x)$ is increasing, then $\alpha+\beta$ is equal to: