We are given the equation:
\[ 5f(x) + 4 \left(\frac{1}{x}\right) = x^2 - 2 \]
Solving for \(f(x)\):
\[ f(x) = \frac{x^2 - 2 - \frac{4}{x}}{5} \]
Now, substitute \(f(x)\) into the equation for \(y\):
\[ y = 9x^2 f(x) = 9x^2 \left(\frac{x^2 - 2 - \frac{4}{x}}{5}\right) \]
Simplifying:
\[ y = \frac{9x^4 - 18x^2 - 36x}{5} \]
Now, differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{1}{5} \left(36x^3 - 36x - 36\right) \]
Simplifying:
\[ \frac{dy}{dx} = \frac{36}{5} \left(x^3 - x - 1\right) \]
For \(y\) to be strictly increasing, we need \(\frac{dy}{dx} > 0\), which implies:
\[ x^3 - x - 1 > 0 \]
Solving the inequality \(x^3 - x - 1 > 0\), we find that the critical points are:
\[ x = \pm \frac{1}{\sqrt{5}} \]
Thus, \(y\) is strictly increasing in the intervals:
\[ x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right) \]
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)