Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2-5 x+5=0$. If $b=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ and $t=x^2-4 x+3 b-\frac{1}{5}+\frac{1}{x^2-4 x+9}, x \in R$ then

• A. minimum value of $(b+t)$ is 8 .
• B. maximum value of $\log _{1 / 5}( t )$ is -1 .
• C. range of $y=\cot ^{-1}\left(\log _5 t\right)$ is $\left(0, \frac{\pi}{4}\right]$
• D. range of $y=\cot ^{-1}\left(\log _{1 / 5}(t)\right)$ is $\left[\frac{\pi}{4}, \pi\right)$.

Answer 1

Answer: (A), (B), (C)

$b =\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)-2 \alpha \beta}{\alpha \beta}=\frac{25-10}{5}=3$
$t=x^2-4 x+9-\frac{1}{5}+\frac{1}{x^2-4 x+9} $
$t=(x-2)^2+5-\frac{1}{5}+\frac{1}{(x-2)^2+5}$
$t_{\text {min. }}=5 (\text { at } x=2)$
(A) Minimum value of $b + t =3+5=8$
(B) $ \log _{1 / 5} 5=-1$, maximum
(C)$ y =\cot ^{-1}\left(\log _5 t \right), t \geq 5 $
$\Rightarrow \log _5 t \in[1, \infty) $
$\Rightarrow \cot ^{-1}\left(\log _{ t } 5\right) \in\left(0, \frac{\pi}{4}\right]$
(D)$y=\cot ^{-1}\left(\log _{1 / 5}(t)\right) $
$\log _{1 / 5} t \in(-\infty,-1]$
$\Rightarrow \cot ^{-1}\left(\log _{1 / 5}(t)\right) \in\left[\frac{3 \pi}{4}, \pi\right)$