Question

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

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