JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 4)
The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = ya is $${{364} \over 3}$$, is equal to :
3
5
7
9
Explanation
$$\mathrm{a}$$ is a odd natural number and
$$\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}$$
$$ \Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}$$
$$ \Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}$$
Solving with $$( - )$$ sign,
$${{{3^{a + 1}} - 1} \over {a + 1}} = {{364} \over 3} \Rightarrow (a = 5)$$
Solving with $$( + )$$ sign,
$${{{3^{a + 1}} - 1} \over {a + 1}} = {{ - 364} \over 3}$$, No a exist
$$\therefore$$ $$(a = 5)$$
Comments (0)
