Mathematics - Quadratic Equation Question with Solution | TestHub

(A) Let $y=\frac{x^2+2x+4}{x+2}$ $$\begin{array}{rl}& \Rightarrow x^2+(2-y)x+(4-2y)=0\\ & \Rightarrow (2-y{)}^2-4(4-2y)\ge 0\\ & \Rightarrow y^2+4y-12\ge 0\\ & \Rightarrow y\le -6,y\ge 2\\ & \end{array}$$ $\therefore$ Minimum value of $y$ is 2 . (B) Since, $(A+B)(A-B)=(A-B)(A+B)$ $$\begin{array}{rl}& \Rightarrow A^2-AB+BA-B^2=A^2+AB-BA-B^2\\ & \Rightarrow AB=BA\\ & \text{ and }(AB{)}^t=(-1{)}^{k}AB\\ & \Rightarrow B^t{A}^t=(-1{)}^{k}AB\\ & \Rightarrow -BA=(-1{)}^{k}AB\left[\because B^t=-B,A^t=A\right]\\ & \Rightarrow BA=(-1{)}^{k+1}AB\\ & \Rightarrow (-1{)}^{k+1}=1\\ & \end{array}$$ $\therefore k+1$ is even or $k$ is odd. (C) $1<2^{\left(-k+3^{-a}\right)}<2\Rightarrow 0<-k+3^{-a}<1$ Given, $a={\log }_3{\log }_{3}2\Rightarrow 3^a={\log }_{3}2$ $$\begin{array}{lc}\Rightarrow & 3^{-a}={\log }_{2}3\\ \therefore & k<{\log }_{2}3<2\\ \text{ and }& 1+k>{\log }_{2}3>1\Rightarrow k>0\end{array}$$ From Eqs. (ii) and (iii), $0<k<2\Rightarrow k=1$ $[\because k$ is an integer] $$\begin{array}{rlrl}\text{ (D) }& \sin \theta & =\cos \varphi & \\ \Rightarrow & \cos \left(\frac{\pi }{2}-\theta \right)& =\cos \varphi & \\ \Rightarrow & \frac{\pi }{2}-\theta & =2n\pi \pm \varphi ,n\in Z& \\ \Rightarrow & & \theta \pm \varphi -\frac{\pi }{2}& =-2n\pi ,n\in Z\\ \Rightarrow & & \frac{1}{\pi }\left(\theta \pm \varphi -\frac{\pi }{2}\right)& =-2n,n\in Z\end{array}$$