Question

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

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

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

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

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