QJEEM19Ma39ae: JEE Main 2019 Mathematics - Class 12 JEEM19Ma39ae Solution: Let S( ) = {(x, y) : y 2 x, 0 x } and A( ) is area of the…

Question

Verified Solution • 2025–26 Session moderate

Let S(

\alpha

) = {(x, y) : y²

\le

x, 0

\le

x

\le

\alpha

} and A(

\alpha

) is area of the region S(

\alpha

). If for a

\lambda

, 0 <

\lambda

< 4, A(

\lambda

) : A(4) = 2 : 5, then

\lambda

equals

Accepted Answer

A(

\lambda

) =

2\int\limits_0^\lambda {\sqrt x } dx

=

2\left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^\lambda

=

{4 \over 3}{\lambda ^{{3 \over 2}}}

\therefore

A(4) =

{4 \over 3}{\left( 4 \right)^{{3 \over 2}}}

Given,

{{A\left( \lambda \right)} \over {A\left( 4 \right)}} = {2 \over 5}

\Rightarrow

{{{4 \over 3}{{\left( \lambda \right)}^{{3 \over 2}}}} \over {{4 \over 3}{{\left( 4 \right)}^{{3 \over 2}}}}} = {2 \over 5}

\Rightarrow

{\lambda ^{{3 \over 2}}} = {2 \over 5} \times 8

\Rightarrow

\lambda

=

{\left( {{{16} \over 5}} \right)^{{2 \over 3}}}

\Rightarrow

\lambda

=

{\left( {{{256} \over {25}}} \right)^{{1 \over 3}}}

\Rightarrow

\lambda

=

{\left( {{{{4^3} \times 4} \over {25}}} \right)^{{1 \over 3}}}

=

4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}

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