Let T be the triangle with vertices $(0, 0), (0, c^{2})$ and $(c, c^{2})$ and let R be the region between $y = c x$ and $y = x^{2}$

Area

(R) = \frac{c^{3}}{6}

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Area of

R = \frac{c^{3}}{3}

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lim c \to 0^{+} \frac{A r e a (T)}{A r e a (R)} = 3

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lim c \to 0^{+} \frac{A r e a (T)}{A r e a (R)} = \frac{3}{2}

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Solution

The correct options are
B Area $(R) = \frac{c^{3}}{6}$
D $lim c \to 0^{+} \frac{A r e a (T)}{A r e a (R)} = 3$
Area(T) $== \frac{c . c^{2}}{2} = \frac{c^{3}}{2}$
Area(R) $= \frac{c^{3}}{2} - \int_{0}^{c} x^{2} d x = \frac{c^{3}}{2} - \frac{c^{3}}{3} = \frac{c^{3}}{6}$
$∴ lim c \to 0^{+} \frac{A r e a (T)}{A r e a (R)} = lim c \to 0^{+} \frac{\frac{c^{3}}{2}}{\frac{c^{3}}{6}} = 3$

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The correct options are B Area (R)=c36 D limc→0+Area(T)Area(R)=3Area(T) ==c.c22=c32Area(R) =c32−∫c0x2dx=c32−c33=c36∴limc→0+Area(T)Area(R)=limc→0+c32c36=3

Let T be the triangle with vertices $(0, 0), (0, c^{2})$ and $(c, c^{2})$ and let R be the region between $y = c x$ and $y = x^{2}$