Question

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

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

Answer 1

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

Given $f\left(x\right)=(\frac{\sqrt{p+4}}{1-p}-1)x^5-3x+ln5$
$f^{\prime }\left(x\right)=(\frac{\sqrt{p+4}}{1-p}-1)5x^4-3$
Since, $f\left(x\right)$ is decreasing,
$f^{\prime }\left(x\right)<0$
$(\frac{\sqrt{p+4}}{1-p}-1)5x^4-3<0\forall xϵR$
$\Rightarrow \frac{\sqrt{p+4}}{1-p}-1\le 0$
If $-4\le p<1$ then
$\Rightarrow \sqrt{p+4}\le 1-p$
$\Rightarrow p+4\le 1-2p+p^2$
$\Rightarrow p^2-3p-3\ge 0$
$\Rightarrow p\le \frac{3-\sqrt{21}}{2}or\frac{3+\sqrt{21}}{2}\le p$
$\Rightarrow pϵ[-4,\frac{3-\sqrt{21}}{2}]$
If $p>1$ then $\sqrt{p+4}\ge 1-p$
$\Rightarrow$ Always true for $p>1$
$\Rightarrow pϵ[-4,\frac{3-\sqrt{21}}{2}]\cup (1,\infty )$