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Q.
The set of all values of a for which the function $f(x)=\left(\frac{\sqrt{a+4}}{1-a}-1\right)x^5-3x+ \log\,5$ decreases for all real $x$ is
Application of Derivatives
Solution:
For $f(x)$ to be decreasing for all $x$, we must have $f'(x) < 0$ for all $x$.
$\therefore 5\left(\frac{\sqrt{a+4}}{1-a}-1\right)x^{4}-3 < 0$ for all $x$
or $\left(\frac{\sqrt{a+4}}{1-a}-1\right) x^{4} < \frac{3}{5}$ for all $x$
or $\left(\frac{\sqrt{a+4}}{1-a}-1\right)=0$
$\Rightarrow \frac{\sqrt{a+4}}{1-a}1$
This is trivially true for $a > 1$ i.e., when a $\in\left(1, \infty\right)$
But if $a < 1$, then $\because \sqrt{a+4}$ is real
$\therefore a \ge-4$
$\therefore -4 \le a < 1$
For these values of a $\le\frac{\sqrt{a+4}}{1-a} \le 1$
$\therefore \sqrt{a+4} \le1-a$
$\Rightarrow \left(a+4\right)=1+a^{2}-2a$
$\Rightarrow 0 \le a^{2}-3a-3$
$\Rightarrow 0 \le\left(a-\frac{3}{2}\right)^{2}-\left(\frac{\sqrt{21}}{2}\right)^{2}$
$\Rightarrow \frac{-3-\sqrt{21}}{2} \le a \le\frac{3-\sqrt{21}}{2}$
$\Rightarrow -4 \le a \le\frac{3-\sqrt{21}}{2}\quad\left[\because a=-4\right]$
Hence a $\in\left[-4, \frac{3-\sqrt{21}}{2}\right] \cup\left(1, \infty\right)$