Question

Let $âˆ\pounds Mâˆ\pounds$ denote the determinant of a square matrix $M$. Let $g:\left[0,\frac{Ï€}{2}\right]→R$ be the function defined by
$g(θ)=\sqrt{f(θ)−1}+\sqrt{f\left(\frac{π}{2}−θ\right)−1}$
where
$f(θ)=\frac{1}{2}â\begin{array}{ccc}1& \sin θ& 1\\ −\sin θ& 1& \sin θ\\ −1& −\sin θ& 1\end{array}â+â\begin{array}{ccc}\sin Ï€& \cos \left(θ+\frac{Ï€}{4}\right)& \tan \left(θ−\frac{Ï€}{4}\right)\\ \sin \left(θ−\frac{Ï€}{4}\right)& −\cos \frac{Ï€}{2}& {\log }_e\left(\frac{4}{Ï€}\right)\\ \cot \left(θ+\frac{Ï€}{4}\right)& {\log }_e\left(\frac{Ï€}{4}\right)& \tan Ï€\end{array}â$
Let $p(x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(θ)$, and $p(2)=2−\sqrt{2}$. Then, which of the following is/are TRUE?

• A. $p\left(\frac{3+\sqrt{2}}{4}\right)<0$
• B. $p\left(\frac{1+3\sqrt{2}}{4}\right)>0$
• C. $p\left(\frac{5\sqrt{2}−1}{4}\right)>0$
• D. $p\left(\frac{5−\sqrt{2}}{4}\right)<0$

Answer 1

Answer: (A), (C)

$f(θ)=\frac{1}{2}â\begin{array}{ccc}& \sin θ& 1\\ −\sin θ& 1& \sin θ\\ −1& −\sin θ& 1\end{array}â+â\begin{array}{ccc}\sin Ï€& \cos \left(θ+\frac{Ï€}{4}\right)& \tan \left(θ−\frac{Ï€}{4}\right)\\ \sin \left(θ−\frac{Ï€}{4}\right)& −\cos \frac{Ï€}{2}& {\log }_c\left(\frac{4}{Ï€}\right)\\ \cot \left(θ+\frac{Ï€}{4}\right)& {\log }_c\frac{Ï€}{4}& \tan Ï€\end{array}â$
$f(θ)=\frac{1}{2}â\begin{array}{ccc}2& \sin θ& 1\\ 0& 1& \sin θ\\ 0& −\sin θ& 1\end{array}â+â\begin{array}{ccc}0& −\sin \left(θ−\frac{Ï€}{4}\right)& \tan \left(θ−\frac{Ï€}{4}\right)\\ \sin \left(θ−\frac{Ï€}{4}\right)& 0& {\log }_c\left(\frac{4}{Ï€}\right)\\ −\tan \left(θ−\frac{Ï€}{4}\right)& −{\log }_c\left(\frac{4}{Ï€}\right)& 0\end{array}â$
$f(θ)=\left(1+{\sin }^2θ\right)+0\text{ (skew symmetric) }$
$g(θ)=\sqrt{f(θ)−1}+\sqrt{f\left(\frac{π}{2}−θ\right)−1}$
$=âˆ\pounds \sin θâˆ\pounds +âˆ\pounds \cos θâˆ\pounds \text{ for }θ∈\left[0,\frac{Ï€}{2}\right]$
$g(θ)∈[1,\sqrt{2}]$
$\text{ Again let }P(x)=k(x−\sqrt{2})(x−1)$
$2−\sqrt{2}=k(2−\sqrt{2})(2−1)$
$⇒k=1(P(2)=2−\sqrt{2}\text{ given) }$
$∴P(x)=(x−\sqrt{2})(x−1)$
$\text{ for option (A) }P\left(\frac{3+\sqrt{2}}{4}\right)<0\text{ correct }$ $\text{ option (B) }P\left(\frac{1+3\sqrt{2}}{4}\right)<0\text{ incorrect }$
option (C) $P\left(\frac{5\sqrt{2}−1}{4}\right)>0$ correct
option (D) $P\left(\frac{5−\sqrt{2}}{4}\right)>0$ incorrect