Question

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

• A. $(-\infty ,\infty )$
• B. $\left[-4,3-\frac{\sqrt{21}}{2}\right]\cup (1,\infty )$
• C. $\left[-3,5-\frac{\sqrt{21}}{2}\right]\cup (2,\infty )$
• D. $[1,\infty )$

Answer 1

Answer: (B)

For $f(x)$ to be decreasing for all $x$, we must have $f^{\prime }(x)<0$ 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 $\le \frac{\sqrt{a+4}}{1-a}\le 1$
$\therefore \sqrt{a+4}\le 1-a$
$\Rightarrow \left(a+4\right)=1+a^2-2a$
$\Rightarrow 0\le a^2-3a-3$
$\Rightarrow 0\le {\left(a-\frac{3}{2}\right)}^2-{\left(\frac{\sqrt{21}}{2}\right)}^2$
$\Rightarrow \frac{-3-\sqrt{21}}{2}\le a\le \frac{3-\sqrt{21}}{2}$
$\Rightarrow -4\le a\le \frac{3-\sqrt{21}}{2}\left[\because a=-4\right]$
Hence a $\in \left[-4,\frac{3-\sqrt{21}}{2}\right]\cup \left(1,\infty \right)$