JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 20)
If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = $${3 \over 2}$$ and the curve y = 1 + 4x $$-$$ x2, then 12 m is equal to _____________.
Answer
26
Explanation
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According to the question,
$${1 \over 2}\int_0^{3/2} {(1 + 4x - {x^2})dx = \int_0^{3/2} {mx\,dx} } $$
$$ \Rightarrow {1 \over 2}\left[ {\left( {x + 2{x^2} - {{{x^3}} \over 3}} \right)} \right]_0^{3/2} = {m \over 2}[x]_0^{3/2} \Rightarrow {3 \over 2} + {9 \over 2} - {9 \over 8} = {{9m} \over 4}$$
$$\Rightarrow$$ m = 39/18 $$\Rightarrow$$ 12m = 26
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