(A)   $\because {\mathrm{A}}^{\mathrm{T}}=\left[\begin{array}{cc}1& -\mathrm{tanx}\\ \mathrm{tanx}& 1\end{array}\right]\mathrm{and}{\mathrm{A}}^{-1}=\frac{1}{{\sec }^2\mathrm{x}}\left[\begin{array}{cc}1& -\mathrm{tanx}\\ \mathrm{tanx}& 1\end{array}\right]$

        $\therefore {\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}=\frac{1}{{\sec }^2\mathrm{x}}\left[\begin{array}{cc}1-{\tan }^2\mathrm{x}& -2\mathrm{tanx}\\ 2\mathrm{tanx}& 1-{\tan }^2\mathrm{x}\end{array}\right]$

            | A^TA^–1 | = $\frac{1}{{\sec }^4\mathrm{x}}$ (1 + tan2x)2 = 1.

(B)        Q f(x) = x³ – 3ax² + 3 (a² – 1) x + 1

            ⇒ f’(x) = 3x² – 6ax + 3 (a² – 1)

            = 3 (x² – 2ax + a² – 1) = 3 ((x – a)² – 1)

            ⇒ 3(x – a + 1) (x – a – 1)

            $\therefore$  a – 1 > – 2  ⇒ a > – 1

            and a + 1 < 4 ⇒ a < 3

            ⇒ a $\in$ (– 1, 3)

          $\therefore$ a₂ – a₁ = 4.

(C)        Image (h, k) of focus (– 1, – 1) in tangent 3x – y – 8 = 0 will lie on directrix.

             $\frac{\mathrm{h}+1}{3}=\frac{\mathrm{k}+1}{-1}=\frac{-2\left(3(-1)-1(-1)-8\right)}{3^2+1^2}=2$

          $\therefore$ Image (5, – 8) i.e. on directrix

          $\therefore$ Slope of PM = $\frac{13+3}{7-5}=8$

          $\therefore$ Equation of directrix will be

                        y + 3 = $\frac{-1}{8}$ (x – 5)                                 

           ⇒x + 8y + 19 = 0

         $\therefore$ Length of LR = 2 × $\left(\frac{-1-8+19}{\sqrt{65}}\right)=\frac{20}{\sqrt{65}}=\mathrm{\ell }$ 

          $\therefore \left[\ell \right]=2.$         

(D)        Distance between centre and focus = ae = 10

            Distance between directrices = $\frac{2\mathrm{a}}{\mathrm{e}}$ = 4

           $\therefore \frac{\mathrm{ae}}{\frac{2\mathrm{a}}{\mathrm{e}}}=\frac{10}{4}\Rightarrow {\mathrm{e}}^2=5$

           $\Rightarrow \frac{4{\mathrm{e}}^2}{5}=4$