Let k and m be positive real numbers such that the function f(x)=3x^2+kx+1, & 00.

Question

Let $k$ and $m$ be positive real numbers such that the function $f(x)=3{x}^{2}+k\sqrt{x+1}, \& 00$. Then $\frac{8{f}^{'}(8)}{{f}^{'}(\frac{1}{8})}$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 309

Accepted Answer

Here, $f(x)= $$3{x}^{2}+k\sqrt{x+1}, \& 00$$$

So, $f(x)$ is differentiable at $x=1$

$\begin{gathered} f({1}^{-})=f(1)=f({1}^{+}) \ 3+k\sqrt{2}=m+{k}^{2}......(i) \end{gathered}$

$\begin{gathered} {f}^{'}({1}^{-})={f}^{'}({1}^{+}) \ 6(1)+\frac{k}{2\sqrt{1+1}}=2m(1) \ \Rightarrow 6+\frac{k}{2\sqrt{2}}=2m......(ii) \end{gathered}$

Using (i) and (ii),

$\begin{gathered} 3+k\sqrt{2}=3+\frac{k}{4\sqrt{2}}+{k}^{2} \ \Rightarrow {k}^{2}+k[\frac{1}{4\sqrt{2}}-\sqrt{2}]=0 \end{gathered}$

$\Rightarrow k[k+\frac{1-8}{4\sqrt{2}}]=0\Rightarrow k=0,\frac{7}{4\sqrt{2}}$

As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = $\frac{7}{4\sqrt{2}}$

$\begin{gathered} k=\frac{7}{4\sqrt{2}},m=3+\frac{\frac{7}{4\sqrt{2}}}{4\sqrt{2}} \ =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32} \end{gathered}$

$\frac{8{f}^{'}(8)}{{f}^{'}(\frac{1}{8})}=\frac{8\times [2\times \frac{103}{32}\times 8]}{6\times \frac{1}{8}+\frac{7}{4\sqrt{2}}\times 2\sqrt{918}}$

$=\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412\times 12}{16}=309$

Concept :

$f(x)$ is differentiable at $x=a$, if ${f}^{'}({a}^{-})={f}^{'}({a}^{+})$

Solutions · 1