Arrangement of the following areas between the curves is descending order:
$A : y^{2} = 4 x, x^{2} = 4 y$
$B . y = x, y = x^{3}$
$C . y^{2} = 8 x, y = 2 x$
$D . y = \sqrt{x}, y = x^{2}$

A,B,C,D

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A,C,B,D

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D,B,C,A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D,C,B,A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A A,C,B,D
A $\frac{16}{3} a^{2} = 16 / 3 s q u n i t s .$ $(a = 1)$
B $1 / 2 s q u n i t s .$
C $2 \int 0 (2 x - \sqrt{8 x}) d x$ . $4 - \frac{2}{3} \sqrt{8} 2 \sqrt{2}$
$∣ ∣ 4 - \frac{4}{3} \times 4 ∣ ∣ = 4 / 3 s q u n i t s .$
D $\frac{16}{3} (1 / 16) = 1 / 3 s q u n i t s .$ $(a = 1 / 4)$
So, descending order is A,C,B,D

Suggest Corrections

Similar questions

x^{2} = 8 y

y = 2 x + A, y = x + B, y = 3 x + C, y = - 4 x + D

A, B, C, D

A

x = 3, y^{2} = 3 x

B

y = 1 - | x |

C

y = x^{2}

y = \sqrt{3} x

A, B, C

\frac{x^{2} + 2 x + 3}{x^{3}} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} \Rightarrow A + B - C =

The normal to the curve x² = 4y passing (1, 2) is

(A) x + y = 3 (B) x − y = 3

Join BYJU'S Learning Program