Verified Solution • 2025–26 Session moderate

QJEEM20Mad7f0: JEE Main 2020 Mathematics - Class 12 JEEM20Mad7f0 Solution

lim⁡x→1(∫0(x−1)2tcos⁡(t2)dt(x−1)sin⁡(x−1))\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)

Multiple Choice Options

  • A is equal to 0
  • B is equal to 12{1 \over 2}
  • C does not exist
  • D is equal to −12- {1 \over 2}

Step-by-Step Verified Solution

lim⁡x→1(∫0(x−1)2tcos⁡(t2)dt(x−1)sin⁡(x−1))\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right) (00)\left( {{0 \over 0}} \right)

Apply L Hospital Rule

= lim⁡x→12(x−1).(x−1)2cos⁡(x−1)4−0(x−1).cos⁡(x−1)+sin⁡(x−1)\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1} \right).{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4} - 0} \over {\left( {x - 1} \right).\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}} (00)\left( {{0 \over 0}} \right)

= lim⁡x→12(x−1)3cos⁡(x−1)4(x−1).[cos⁡(x−1)+sin⁡(x−1)(x−1)]\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^3}\cos {{\left( {x - 1} \right)}^4}} \over {\left( {x - 1} \right).\left[ {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right]}}

= lim⁡x→12(x−1)2cos⁡(x−1)4cos⁡(x−1)+sin⁡(x−1)(x−1)\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4}} \over {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}}}

on taking limit

= 01+1{0 \over {1 + 1}} = 0

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