JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 5)
If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to :
$${y \over x}$$
$${x \over y}$$
$$-$$ $${y \over x}$$
$$-$$ $${x \over y}$$
Explanation
x = $$\sqrt {{2^{\cos e{c^{ - 1}}t}}} $$
$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$
$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$
$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$
= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$
= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$
= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$
= $$-$$ $${y \over x}$$
$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$
$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$
$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$
= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$
= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$
= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$
= $$-$$ $${y \over x}$$
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