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. $\left( - 3,\frac{5 - \sqrt{27}}{2} \right) \cup (2,\infty)$
• B. $\left\lbrack - 4,\frac{3 - \sqrt{21}}{2} \right\rbrack \cup (1,\infty)$
• C. $( - \infty,\infty)$
• D. $\lbrack 1,\infty)$

Answer 1

Answer: (B)

$\left\lbrack - 4,\frac{3 - \sqrt{21}}{2} \right\rbrack \cup (1,\infty)$

Answer 2

Differentiating, we get $f'(x) = \left( \frac{\sqrt{a + 4}}{1 - a} - 1 \right)5x^{4} - 3$

For $f(x)$ to be decreasing for all $x$, we must have

$f'(x)$ < 0 for all $x.$

⇒ $\left( \frac { \sqrt { a + 4 } } { 1 - a } - 1 \right)$ $x^{4} < \frac{3}{5}\forall x$

This is possible only if $\frac{\sqrt{a + 4}}{1 - a} - 1 \leq 0$

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

$\frac{\sqrt{a + 4}}{1 - a} \leq 1$⇒ $\sqrt{a + 4} \leq 1 - a$

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

⇒ $a + 4 \leq 1 + a^{2} - 2a$⇒ $a^{2} - 3a - 3$

⇒ $a \leq \frac{3 - \sqrt{21}}{2}$

Thus, $a \in \left\lbrack - 4,\frac{3 - \sqrt{21}}{2} \right\rbrack \cup (1,\infty)$