The first derivative of the function $f(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right) + x^x$ with respect to $x$ at $x=1$ is

  • A
    $\frac{1}{4}$
  • B
    $\frac{5}{4}$
  • C
    $\frac{-1}{2}$
  • D
    $\frac{3}{4}$

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Let $f: R \rightarrow (0, \infty)$ and $g: R \rightarrow R$ be twice differentiable functions such that $f^{\prime \prime}$ and $g^{\prime \prime}$ are continuous functions on $R$. Suppose $f^{\prime}(2) = g(2) = 0$,$f^{\prime \prime}(2) \neq 0$ and $g^{\prime}(2) \neq 0$. If $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$,then:

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

Match each function in List-$I$ to its derivative given in List-$II$.
List-$I$List-$II$
$(A) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$$(I) \cos x-\sin x$
$(B) \tan ^{-1}\left(\frac{1-x}{1+x}\right)$$(II) \frac{-1}{1+x^2}$
$(C) e^{\log (\sin x+\cos x)}$$(III) \frac{2}{1+x^2}$
$(D) \sqrt{1-\sin 2 x} \text{ for } (0 < x < \frac{\pi}{4})$$(IV) \cos x+\sin x$
$(V) -\sin x-\cos x$

The correct match is:

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

If $f(x)=\sin \left(\cos ^{-1}\left(\frac{1-2^{2 x}}{1+2^{2 x}}\right)\right)$ and its first derivative with respect to $x$ is $-\frac{ b }{ a } \log _{ e } 2$ when $x =1,$ where $a$ and $b$ are integers,then the minimum value of $\left| a ^{2}- b ^{2}\right|$ is.........

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