MCQ
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ 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 $\mathrm{f}(-10 \sqrt{2})=2 \sqrt{2}$, then $\mathrm{f}^{\prime \prime}(-10 \sqrt{2})=$

$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$

$2.$ The area of the region bounded by the curves $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty < \mathrm{a} < \mathrm{b} < -2$, is

$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$

$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$

$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$

$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$

$3.$ $\int_{-1}^1 g^{\prime}(x) d x=$

$(A)$ $2 g(-1)$ $(B)$ 0 $(C)$ $-2 g(1)$ $(D)$ $2 \mathrm{~g}(1)$

Give the answer question $1,2$ and $3.$

  • $B,A,D$
  • B
    $B,C,B$
  • C
    $A,D,B$
  • D
    $A,D,B$

Answer

Correct option: A.
$B,A,D$
a
$1.$  Differentiating the given equation, we get

$ 3 y^2 y^{\prime}-3 y^{\prime}+1=0 $

$ \Rightarrow y^{\prime}(-10 \sqrt{2})=-\frac{1}{21}$

Differentiation again we get $6 y y^{\prime 2}+3 y^2 y^{\prime \prime}-3 y^{\prime \prime}=0$

$\Rightarrow \mathrm{f}^{\prime \prime}(-10 \sqrt{2})=-\frac{6 \cdot 2 \sqrt{2}}{(21)^4}=-\frac{4 \sqrt{2}}{7^3 3^2} \text {. }$

$2.$  The required area $= $$ \int_a^b f(x) d x=\left.x f(x)\right|_a ^b-\int_a^b x f^{\prime}(x) d x$

$ =b f(b)-a f(a)+\int_a^b \frac{x}{3\left[\left(f(x)^2-1\right)\right]} d x .$

$3.$  We have $y^{\prime}=\frac{1}{3\left(1-\left(f(x)^2\right)\right)}$

which is even Hence $\int_{-1}^1 g^{\prime}(x)=g(1)-g(-1)=2 g(1)$.

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