Match the following. List – I List – II (A) Let y(x) = (3 ^ - 1 x),...

\left(P\right)y = \cos \left(3\cos^{- 1} x\right)$<br />${y}^{′} = \dfrac{3\sin \left(3\cos^{- 1} x\right)}{\sqrt{1 - {x}^{2}}}$<br />$\sqrt{1 - {x}^{2}}{y}^{′} = 3\sin \left(3\cos^{- 1} x\right)$<br />$⇒ \dfrac{- x}{\sqrt{1 - {x}^{2}}}{y}^{′} + \sqrt{1 - {x}^{2}}y " = 3\cos ( 3\cos^{- 1}x ) . \dfrac{- 3}{\sqrt{1 - {x}^{2}}}$<br />$⇒ - x{y}^{′} + \left(1 - {x}^{2}\right)y " = - 9y$<br />$⇒ \dfrac{1}{y} [ \left({x}^{2} - 1\right)y " + xy′ ] = 9$<br />\left(Q\right)\left({a}_{k} \times {a}_{k + 1}\right) = {r}^{2}sin \dfrac{2\pi}{n}$
${a}_{k} . {a}_{k + 1} = {r}^{2}cos \dfrac{2\pi}{n}$
$⇒ \left|{∑}_{k = 1}^{n - 1}\overset{\to}{{a}_{k}} \times \overset{\to}{{a}_{k + 1}}\right| = \left|\mathop{∑}_{k = 1}^{n - 1}{a}_{k} . {a}_{k + 1}\right|$
$⇒ {r}^{2}\left(n - 1\right)sin \dfrac{2\pi}{n} = {r}^{2}\left(n - 1\right)cos \dfrac{2\pi}{n}$
$tan \dfrac{2\pi}{n} = 1 ⇒ n = \dfrac{8}{4k + 1}$
$⇒ n = 8$
\left(R\right)\dfrac{{h}^{2}}{6} + \dfrac{{1}^{2}}{3} = 1, h = \pm 2$<br />Tangent at (2, 1) is$\dfrac{2x}{6} + \dfrac{y}{3} = 1x + y = 3$<br />\left(S\right)\tan^{- 1} \left(\dfrac{1}{2x + 1}\right) + \tan^{- 1} \dfrac{1}{4x + 1} = \tan^{- 1} \dfrac{2}{{x}^{2}}$
$\tan^{- 1} \left(\dfrac{3x + 1}{4{x}^{2} + 3x}\right) = \tan^{- 1} \dfrac{2}{{x}^{2}}$
$⇒ 3{x}^{2} - 7x - 6 = 0$
$x = - \dfrac{2}{3}, 3$