JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q27

← JEE Main 2022 · 26 Jul · Shift 1

MathematicsInteger+4 / -1Question 27 of 90

Let a curve $y=y(x)$ pass through the point $(3,3)$ and the area of the region under this curve, above the $x$ -axis and between the abscissae 3 and $x(>3)$ be $\left(\dfrac yx\right)^3$ . If this curve also passes through the point $\left(\alpha,6\sqrt{10}\right)$ in the first quadrant, then $\alpha$ is equal to ______.

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Answer and worked solution

Official answer: 6

Differentiate the area condition: $y=\dfrac d{dx}\left(\dfrac yx\right)^3$ . With $v=\frac yx$ : $vx=3v^2v'$ , so $v\,dv=\frac x3dx$ and $v^2=\frac{x^2}3+C$ ; $(3,3)$ gives $C=-2$ .

At α. $v=\frac{6\sqrt{10}}\alpha$ : $\frac{360}{\alpha^2}=\frac{\alpha^2}3-2$ , so $\alpha^4-6\alpha^2-1080=0$ , $\alpha^2=36$ , $\alpha=6$ .

Answer: 6

Topics:Formation and applications Fundamental theorem of calculus

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 27