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QJEEM18Maed59: JEE Main 2018 Mathematics - Class 12 JEEM18Maed59 Solution: If the area of the region bounded by the curves, and the lines…

If the area of the region bounded by the curves, y=x2,y=1xy = {x^2},y = {1 \over x} and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :

Multiple Choice Options

  • A e32{e^{{3 \over 2}}}
  • B 43{4 \over 3}
  • C 32{3 \over 2}
  • D e23{e^{{2 \over 3}}}

Step-by-Step Verified Solution

JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Area Under The Curves Question 124 English Explanation

Point of intersection of y = x2 and y = 1x{1 \over x} .

put y = 1x{1 \over x} in y = x2, then we get,

1x=x2{1 \over x} = {x^2}

⇒\Rightarrow    \,\,\, x3 −- 1 = 0

⇒\Rightarrow    \,\,\, x = 1

∴   \therefore\,\,\, y = 1

∴   \therefore\,\,\, point B = (1, 1)

Area of region ABCDA

= ∫01x2\int\limits_0^1 {{x^2}} dx + ∫1t1x\int\limits_1^t {{1 \over x}} dx

== [x33]01\left[ {{{{x^3}} \over 3}} \right]_0^1 + [ℓn x]1t\left[ {\ell n\,x} \right]_1^t

= 13{1 \over 3} + ℓn t\ell n\,t −- ℓn\ell n 1

= 13{1 \over 3} + ℓn t\ell n\,t     [ as    ℓn\ell n 1 = 0]

given this Area = 1 sq unit.

∴   \therefore\,\,\, 13{1 \over 3} + ℓn t\ell n\,t = 1

⇒\Rightarrow    ℓn t\ell n\,t = 23{2 \over 3}

⇒\Rightarrow    t = e 23{^{{2 \over 3}}}

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