Mathematics - Matrix Question with Solution | TestHub

(P) $A^2+B=1$

$B{A}^2+B^2=B$

$\Rightarrow {\mathrm{B}}^2=\mathrm{B}$

$\therefore {\mathrm{A}}^2+{\mathrm{B}}^2=1\Rightarrow \mathrm{tr}\left({\mathrm{A}}^2+{\mathrm{B}}^2\right)=3$

(Q) $x\frac{dy}{dx}+y(\ln y)=0\Rightarrow \int \frac{dx}{x}+\int \frac{dy}{y(\ln y)}=C$;

$\ln (x\ln y)=C$. If $x=1$ then $y=e$

Hence, $\Rightarrow \ln (\mathrm{lne})=\mathrm{c}\Rightarrow \mathrm{c}=0$

$\Rightarrow \ln (x\ln y)=0\Rightarrow x\ln y=1\Rightarrow y=f(x)=e^{\frac{1}{x}}$

Now, ${\mathrm{Lim}}_{x\to \infty }x(f(x)-1)={\mathrm{Lim}}_{x\to \infty }\left(e^{\frac{1}{x}}-1\right)=1=e^0$

$\Rightarrow \lambda =0$ ]

(R) ${\int }_1^e\left(2^{\ln x}+e\right)dx+{\int }_{1+e}^{2+e}{e}^{{\log }_2(x-e)}dx$ $=\mathrm{e}(2+\mathrm{e})-(1+\mathrm{e})$

$\Rightarrow {\int }_1^{\mathrm{e}}{2}^{\ln \mathrm{x}}\mathrm{dx}+{\int }_{1+\mathrm{e}}^{2+\mathrm{e}}{\mathrm{e}}^{{\log }_2(\mathrm{x}-\mathrm{e})}\mathrm{dx}$ $=(2\mathrm{e}-1)$

$\therefore [\lambda ]=4$

(S) As two given lines are parallel, so we must have $(\alpha +1+2(\alpha -3)-1)(2(\alpha +1)+4(\alpha -3)-14)<0$

$\Rightarrow (3\alpha -6)(6\alpha -24)<0\Rightarrow 2<\alpha <4$

Hence only one integral value of $\alpha =3$ exist.

Note: $P(\alpha +1,\alpha -3)$ lies on line $x-y=4$