Verified Solution • 2025–26 Session moderate

QJEEM18Ma7fe9: JEE Main 2018 Mathematics - Class 12 JEEM18Ma7fe9 Solution: If and then is equal to :

If x=2cos⁡ec−1x = \sqrt {{2^{\cos e{c^{ - 1}}}}} and y=2sec−1t  (∣t∣≥1),y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right), then dydx{{dy} \over {dx}} is equal to :

Multiple Choice Options

  • A yx{y \over x}
  • B xy{x \over y}
  • C −- yx{y \over x}
  • D −- xy{x \over y}

Step-by-Step Verified Solution

x = 2cos⁡ec−1t\sqrt {{2^{\cos e{c^{ - 1}}t}}}

∴    \therefore\,\,\,\, dxdt{{dx} \over {dt}} = 122cos⁡ec−1t{1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }} ×\times ( 2cos⁡ec−1t . log⁡2{2^{\cos e{c^{ - 1}}t}}\,.\,\log 2 ) ×\times −1tt2−1{{ - 1} \over {t\sqrt {{t^2} - 1} }}

dydt{{dy} \over {dt}} = 122sec⁡−1t{1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }} ×\times (2sec⁡−1tlog⁡2)\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right) ×\times 1tt2−1{1 \over {t\sqrt {{t^2} - 1} }}

∴    \therefore\,\,\,\, dydx{{dy} \over {dx}}

= dydtdxdt{{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}

= −2cos⁡ec−1t2cos⁡ec−1t{{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }} ×\times 2sec⁡−1t2cos⁡ec−1t{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}

= −2sec⁡−1t2cos⁡ec−1t- \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}}

= −- yx{y \over x}

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