Question

The set of all values of 0 for which the function $f\left(x\right)=(\frac{\sqrt{a+4}}{1-a}-1)x^5-3x+\log 5$ decreases for all real $x$ is

• A. $[-3,\frac{5-\sqrt{27}}{2}]\cup (2,\infty )$
• B. $[-4,\frac{3-\sqrt{21}}{2}]\cup (1,\infty )$
• C. $(-\infty ,\infty )$
• D. $(1,\infty )$

Answer 1

The correct option is B $[-4,\frac{3-\sqrt{21}}{2}]\cup (1,\infty )$

Solution Differentiating, we get

$f^{\prime }\left(x\right)=(\frac{\sqrt{a+4}}{1-a}-1)5x^4-3$

For j(x) to be decreasing for all x. we must have $f^{\prime }\left(x\right)<0$ for all $x$

$\Rightarrow \displaystyle \left ( \dfrac{\sqrt{a+4}}{1-a} -1\right )\: x^{4}< \dfrac{3}{5}\forall x.

$

This is possible only if

$\frac{\sqrt{a+4}}{1-a}-1\le 0$

This inequality is always true if $0>1$, i.e., $a\in (1,\infty )$. Moreover, we must have $a\ge -4$ for $\sqrt{a+4}$ to be real. Therefore, we have

$\frac{\sqrt{a+4}}{1-a}\le 1\Rightarrow \sqrt{a+4}\le 1-a$

[$\because$ we consider only 0 < 1 ]

$\Rightarrow a+4\le 1+a^2-2a\Rightarrow 0\le a^2-3a-3$

$\Rightarrow a\le \frac{3-\sqrt{21}}{2}$

Thus $a\in [-4,\frac{3-\sqrt{21}}{2}]\cup (1,\infty )$