JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 12)
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
$$2{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$$
$$2{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}$$
$$4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$$
$$4{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}$$
Explanation
_8th_April_Evening_Slot_en_12_2.png)
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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