Question

The set of all values of $a$ for which the function $f\left(x\right)=-(\frac{b^2-ab-2}{b^2-3b+4}+1)x^5-3x+\log 5,b\in R$ decreases for all real $x$ is

• A. $[-1,2]$
• B. $(0,7)$
• C. $[-7,1]$
• D. $(2,7)$

Answer 1

The correct option is C $[-7,1]$

$f\left(x\right)=-(\frac{b^2-ab-2}{b^2-3b+4}+1)x^5-3x+\log 5$

$f^{\prime }\left(x\right)=-(\frac{b^2-ab-2}{b^2-3b+4}+1)\cdot 5\cdot x^4-3$

Since, $f\left(x\right)$ is decreasing function $\forall x\in R$
$f^{\prime }\left(x\right)<0$

$\Rightarrow -(\frac{b^2-ab-2}{b^2-3b+4}+1)x^4<\frac{3}{5}\forall x\in R$

This is possible only if
$-(\frac{b^2-ab-2}{b^2-3b+4}+1)\le 0$
$(\frac{b^2-ab-2}{b^2-3b+4}+1)\ge 0$
$\Rightarrow \frac{2b^2-b(3+a)+2}{b^2-3b+4}\ge 0$
$\Rightarrow b^2-3b+4>0$ and
$\Rightarrow 2b^2-b(3+a)+2\ge 0$
$\Rightarrow D\le 0$

$(3+a{)}^2-4(2\left)\right(2)\le 0$
$9+a^2+6a-16\le 0$

$\Rightarrow a^2+6a-7\le 0$
$\Rightarrow (a+7)(a-1)\le 0$

$\Rightarrow a\in [-7,1]$

Hence, option C.