Given,
f ( θ ) = 1 2 ∣ 1 a m p ; sin θ a m p ; 1 − sin θ a m p ; 1 a m p ; sin θ − 1 a m p ; − sin θ a m p ; 1 ∣ + ∣ sin π a m p ; cos ( θ + π 4 ) a m p ; tan ( θ − π 4 ) sin ( θ − π 4 ) a m p ; − cos π 2 a m p ; log e ( 4 π ) cot ( θ + π 4 ) a m p ; log e ( π 4 ) a m p ; tan π ∣ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\-\sin \theta & 1 & \sin \theta \\-1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\\sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\\cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right| f ( θ ) = 2 1 1 − sin θ − 1 am p ; sin θ am p ; 1 am p ; − sin θ am p ; 1 am p ; sin θ am p ; 1 + sin π sin ( θ − 4 π ) cot ( θ + 4 π ) am p ; cos ( θ + 4 π ) am p ; − cos 2 π am p ; log e ( 4 π ) am p ; tan ( θ − 4 π ) am p ; log e ( π 4 ) am p ; tan π
Here cos ( θ + π 4 ) = − sin ( θ − π 4 ) \cos \left(\theta+\frac{\pi}{4}\right)=-\sin \left(\theta-\frac{\pi}{4}\right) cos ( θ + 4 π ) = − sin ( θ − 4 π ) tan ( θ − π 4 ) = − cot ( θ + π 4 ) \tan \left(\theta-\frac{\pi}{4}\right)=-\cot \left(\theta+\frac{\pi}{4}\right) tan ( θ − 4 π ) = − cot ( θ + 4 π ) log e ( 4 π ) = − log e ( π 4 ) \log _e\left(\frac{4}{\pi}\right)=-\log _e\left(\frac{\pi}{4}\right) log e ( π 4 ) = − log e ( 4 π ) sin π = − cos π 2 = tan π = 0 \sin \pi=-\cos \frac{\pi}{2}=\tan \pi=0 sin π = − cos 2 π = tan π = 0
⇒ \Rightarrow ⇒ f ( θ ) = 1 2 ∣ 2 a m p ; sin θ a m p ; 1 0 a m p ; 1 a m p ; sin θ 0 a m p ; − sin θ a m p ; 1 ∣ + ∣ 0 a m p ; − sin ( θ − π 4 ) a m p ; tan ( θ − π 4 ) sin ( θ − π 4 ) a m p ; 0 a m p ; log e ( 4 π ) − tan ( θ − π 4 ) a m p ; − log e ( 4 π ) a m p ; 0 ∣ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}2 & \sin \theta & 1 \\0 & 1 & \sin \theta \\0 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\\sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{\mathrm{e}}\left(\frac{4}{\pi}\right) \\-\tan \left(\theta-\frac{\pi}{4}\right) & -\log _{\mathrm{e}}\left(\frac{4}{\pi}\right) & 0\end{array}\right| f ( θ ) = 2 1 2 0 0 am p ; sin θ am p ; 1 am p ; − sin θ am p ; 1 am p ; sin θ am p ; 1 + 0 sin ( θ − 4 π ) − tan ( θ − 4 π ) am p ; − sin ( θ − 4 π ) am p ; 0 am p ; − log e ( π 4 ) am p ; tan ( θ − 4 π ) am p ; log e ( π 4 ) am p ; 0
a m p ; f ( θ ) = ( 1 + sin 2 θ ) + 0 (skew symmetric) a m p ; g ( θ ) = f ( θ ) − 1 + f ( π 2 − θ ) − 1 a m p ; = ∣ sin θ ∣ + ∣ cos θ ∣ a m p ; g ( θ ) ∈ [ 1 , 2 ] for θ ∈ [ 0 , π 2 ] \begin{aligned}& f(\theta)=\left(1+\sin ^2 \theta\right)+0 \text { (skew symmetric) } \\\\& g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1} \\\\& =|\sin \theta|+|\cos \theta| \quad \\\\& g(\theta) \in[1, \sqrt{2}] \text { for } \theta \in\left[0, \frac{\pi}{2}\right]\end{aligned} am p ; f ( θ ) = ( 1 + sin 2 θ ) + 0 (skew symmetric) am p ; g ( θ ) = f ( θ ) − 1 + f ( 2 π − θ ) − 1 am p ; = ∣ sin θ ∣ + ∣ cos θ ∣ am p ; g ( θ ) ∈ [ 1 , 2 ] for θ ∈ [ 0 , 2 π ]
Again let P ( x ) = k ( x − 2 ) ( x − 1 ) \mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}-\sqrt{2})(\mathrm{x}-1) P ( x ) = k ( x − 2 ) ( x − 1 ) a m p ; 2 − 2 = k ( 2 − 2 ) ( 2 − 1 ) a m p ; ⇒ k = 1 ( P ( 2 ) = 2 − 2 given ) a m p ; ∴ P ( x ) = ( x − 2 ) ( x − 1 ) \begin{aligned}& 2-\sqrt{2}=\mathrm{k}(2-\sqrt{2})(2-1) \\\\& \Rightarrow \mathrm{k}=1 \quad(\mathrm{P}(2)=2-\sqrt{2} \text { given }) \\\\& \therefore \mathrm{P}(\mathrm{x})=(\mathrm{x}-\sqrt{2})(\mathrm{x}-1)\end{aligned} am p ; 2 − 2 = k ( 2 − 2 ) ( 2 − 1 ) am p ; ⇒ k = 1 ( P ( 2 ) = 2 − 2 given ) am p ; ∴ P ( x ) = ( x − 2 ) ( x − 1 )
For option (A) P\left(\frac{3+\sqrt{2}}{4}\right)<0 Correct. P\left(\frac{1+3 \sqrt{2}}{4}\right)<0 Incorrect. P\left(\frac{5 \sqrt{2}-1}{4}\right)>0 Correct. P\left(\frac{5-\sqrt{2}}{4}\right)>0 Incorrect.