Question

Let $f\left(x\right)$ be a function continuous $\forall x\in R-\left\{0\right\}$ such that $f^{\prime }\left(x\right)<0,\forall x\in (-\infty ,0)$ and $f^{\prime }\left(x\right)>0,\forall x\in (0,\infty ).$ If $\underset{x\to 0^{+}}{lim}f\left(x\right)=3,\underset{x\to 0^{-}}{lim}f\left(x\right)=4$ and $f\left(0\right)=5,$ then the image of the point $(0,1)$ about the line, $y\cdot \underset{x\to 0}{lim}f({\cos }^{3}x-{\cos }^{2}x)=x\cdot \underset{x\to 0}{lim}f({\sin }^{2}x-{\sin }^{3}x),$ is

• A. $(\frac{12}{25},\frac{9}{25})$
• B. $(\frac{12}{25},\frac{-9}{25})$
• C. $(\frac{16}{25},\frac{-8}{25})$
• D. $(\frac{24}{25},\frac{-7}{25})$

Answer 1

The correct option is D $(\frac{24}{25},\frac{-7}{25})$

With the given information on $f\left(x\right)$, we can consider the given sample graph for reference:

At $x\to 0,$ we have
${\cos }^{3}x-{\cos }^{2}x={\cos }^{2}x(\cos x-1)\to 0^{-}$
$\therefore \underset{x\to 0}{lim}f({\cos }^{3}x-{\cos }^{2}x)=4$

Similarly,
${\sin }^{2}x-{\sin }^{3}x={\sin }^{2}x(1-\sin x)\to 0^{+}$
$\therefore \underset{x\to 0}{lim}f({\sin }^{2}x-{\sin }^{3}x)=3$

$\therefore y\cdot \underset{x\to 0}{lim}f({\cos }^{3}x-{\cos }^{2}x)=x\cdot \underset{x\to 0}{lim}f({\sin }^{2}x-{\sin }^{3}x)$
$\Rightarrow 4y-3x=0$

Let image of point $(0,1)$ be $(h,k)$
$\Rightarrow \frac{h-0}{-3}=\frac{k-1}{4}=-2\left(\frac{4}{(-3{)}^2+4^2}\right)$
$\therefore h=\frac{24}{25},k=\frac{-7}{25}$