If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$,then $\left(\frac{d y}{d x}\right)_{(1,1)}=$

  • A
    $0$
  • B
    $\pi / 4$
  • C
    $1$
  • D
    $\pi / 2$

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Consider the functions defined implicitly by the equation $y^3-3y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$,the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$,the equation implicitly defines a unique real valued differentiable function $y=g(x)$ satisfying $g(0)=0$.
$1.$ If $f(-10 \sqrt{2})=2 \sqrt{2}$,then $f^{\prime \prime}(-10 \sqrt{2})=$
$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$
$2.$ The area of the region bounded by the curves $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$,where $-\infty < a < b < -2$,is
$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx+bf(b)-af(a)$
$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx+bf(b)-af(a)$
$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx-bf(b)+af(a)$
$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx-bf(b)+af(a)$
$3.$ $\int_{-1}^1 g^{\prime}(x) dx=$
$(A)$ $2g(-1)$ $(B)$ $0$ $(C)$ $-2g(1)$ $(D)$ $2g(1)$
Give the answer for questions $1, 2$ and $3.$

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