JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 16)
Consider a region R = {(x, y) $$ \in $$ R : x2 $$ \le $$ y $$ \le $$ 2x}.
if a line y = $$\alpha $$ divides the area of region R into
two equal parts, then which of the following is
true?
3$$\alpha $$2 - 8$$\alpha $$ + 8 = 0
$$\alpha $$3 - 6$$\alpha $$3/2 - 16 = 0
3$$\alpha $$2 - 8$$\alpha $$3/2 + 8 = 0
$$\alpha $$3 - 6$$\alpha $$2 + 16 = 0
Explanation
_2nd_September_Evening_Slot_en_16_2.png)
y $$ \ge $$ x2 $$ \Rightarrow $$ upper region of y = x2
y $$ \le $$ 2x $$ \Rightarrow $$ lower region of y = 2x
According to question, area of OABC = 2 $$ \times $$ area of OAC
$$ \Rightarrow $$ $$\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$$ = 2$$\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$$
$$\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha $$
$$ \Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$$
$$ \Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$$
$$ \Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}$$
$$ \Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}$$
$$ \Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0$$
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