Verified Solution • 2025–26 Session moderate

QJEEM19Ma95ff: JEE Main 2019 Mathematics - Class 12 JEEM19Ma95ff Solution: If   x 3 tan t and y 3 sec t, then the value of…

If   x == 3 tan t and y == 3 sec t, then the value of d2ydx2{{{d^2}y} \over {d{x^2}}} at t =π4,= {\pi \over 4}, is :

Multiple Choice Options

  • A 132{1 \over {3\sqrt 2 }}
  • B 162{1 \over {6\sqrt 2 }}
  • C 322{3 \over {2\sqrt 2 }}
  • D 16{1 \over 6}

Step-by-Step Verified Solution

x = 3 tan t and y = 3 sec t

So that dxdt{{dx} \over {dt}} = 3sec2t and dydt{{dy} \over {dt}} = 3 sec t tan t

dydx{{dy} \over {dx}} = dy/dtdx/dt{{dy/dt} \over {dx/dt}} = sin t

d2ydx2{{{d^2}y} \over {d{x^2}}} = (cos t) .dtdx.{{dt} \over {dx}}

d2ydx2=(cos⁡t).13sec⁡2t{{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}

d2ydx2{{{d^2}y} \over {d{x^2}}} = 13{1 \over 3} (cos3 t)

(d2ydx2)t=π/4{\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}} = 13×(12)3=162{1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}

Related Learning Resources & Exam Preparation