Question

Two functions $f$ and $g$ have first and second derivatives at $x=0$ and satisfy the relations, $f\left(0\right)=\frac{2}{g\left(0\right)},$ $f^{\prime }\left(0\right)=2g^{\prime }\left(0\right)=4g\left(0\right),$ $g^{\prime \prime }\left(0\right)=5f^{\prime \prime }\left(0\right)=6f\left(0\right)=3$ then

• A. If $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ then $h^{\prime }\left(0\right)=\frac{15}{4}$
• B. If $k\left(x\right)=f\left(x\right).g\left(x\right)\sin x$ then $k^{\prime }\left(0\right)=2$
• C. $\underset{x\to 0}{lim}\frac{g^{\prime }\left(x\right)}{f^{\prime }\left(x\right)}=\frac{1}{2}$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A If $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ then $h^{\prime }\left(0\right)=\frac{15}{4}$
B If $k\left(x\right)=f\left(x\right).g\left(x\right)\sin x$ then $k^{\prime }\left(0\right)=2$
C $\underset{x\to 0}{lim}\frac{g^{\prime }\left(x\right)}{f^{\prime }\left(x\right)}=\frac{1}{2}$

$6f\left(0\right)=3;g\left(0\right)=\frac{2}{f\left(0\right)},f^{\prime }\left(0\right)=4g\left(0\right),2g^{\prime }\left(0\right)=4g\left(0\right)$

$\Rightarrow f\left(0\right)=\frac{1}{2},g\left(0\right)=4,f^{\prime }\left(0\right)=16,g^{\prime }\left(0\right)=8$

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

$h^{\prime }\left(x\right)=\frac{g\left(x\right)f^{\prime }\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{\left(g\right(x){)}^2}$

So, $h^{\prime }\left(0\right)=\frac{4.16-\frac{1}{2}.8}{4.4}=\frac{64-4}{16}=\frac{15}{4}$

$k\left(x\right)=f\left(x\right).g\left(x\right)\sin x$

$k^{\prime }\left(x\right)=f\left(x\right).g\left(x\right)cosx+sinx\left(f\right(x\left)g^{\prime }\right(x)+g(x\left)f^{\prime }\right(x\left)\right)$

$k^{\prime }\left(0\right)=\frac{1}{2}.4.1+0=2$

$\underset{x\to 0}{lim}=\frac{g^{\prime }\left(x\right)}{f^{\prime }\left(x\right)}=\frac{8}{16}=\frac{1}{2}$