8. The area between x = y² and x = 4 is divided into two equal parts by the line x = a find the value of a

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Vishal Baghel | Contributor-Level 10

8 months ago

Kindly go through the solution

\begin{array}{l} \Rightarrow {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{a} = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{a}^{} \\ \Rightarrow \frac{2}{3} [a^{\frac{3}{2}} - 0] = \frac{2}{3} [4^{\frac{3}{2}} - a^{\frac{3}{2}}] \\ \Rightarrow a^{\frac{3}{2}} = 4^{\frac{3}{2}} - a^{\frac{3}{2}} \\ \Rightarrow 2 a^{\frac{3}{2}} = 4^{\frac{3}{2}} \end{array}

$\Rightarrow 4 a^{3} = 4^{3}$ (Squaring both sides)

$\Rightarrow a^{3} = 4^{2}$

$\Rightarrow a = 4^{\frac{2}{3}}$ (Taking cube on both sides)

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Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

Raj Pandey

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) Þ c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Raj Pandey

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$