Whine-List I-List IINWhline P-Area bounded - by - y - x - x - - x-axis- - 1 - -frac103 1- x -1 - and - x -18 - ll lhlineQ- Area - bounded - by - x - y -2- - 2 - - frac 643 1- x - sqrt y and - x -0 - 11 hline R-Area enclosed by - y - 2- x - 3 - -frac23 1- and y-x- - lllhline S-Area bounded - by - y - 2- x - 4 - -frac16 1- y -4 - and - x -0 - W hineend array A- P-3- Q-1- R-2- S-4B- P-1- Q-3- R-2- S-4C- P-1- Q-2- R-4- S-3D- P-3- Q-1- R-4- S-2

The correct option is D P 3; Q 1; R 4; S 2(P) A=2∫^1_0 x^2 dx = 2/3 (Q) A∫^1_0 (x+2 x^2)dx=10/3 (R) A∫^1_0 (√(x) x)dx = 1/6 (S) A=64 ∫^16_0 (√(x))dx = 64/3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Square Root of a Complex Number

Whine&List I&...

Question

$\begin{matrix} L i s t I & L i s t I I P . & A r e a b o u n d e d b y y = x | x |, x - a x i s, & 1. & \frac{10}{3} x = 1 a n d x = - 1 Q . & A r e a b o u n d e d b y x - y + 2, & 2. & \frac{64}{3} x = \sqrt{y} a n d x = 0 R . & A r e a e n c l o s e d b y y^{2} = x & 3. & \frac{2}{3} a n d y = | x | S . & A r e a b o u n d e d b y y^{2} = x & 4. & \frac{1}{6} y = 4 a n d x = 0 \end{matrix}$

P-3; Q-1; R-2; S-4

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

P-3; Q-1; R-4; S-2

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

P-1; Q-3; R-2; S-4

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

P-1; Q-2; R-4; S-3

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

Open in App

Solution

The correct option is B P-3; Q-1; R-4; S-2
(P) $A = 2 \int_{0}^{1} x^{2} d x = \frac{2}{3}$
(Q) $A \int_{0}^{1} (x + 2 - x^{2}) d x = \frac{10}{3}$
(R) $A \int_{0}^{1} (\sqrt{x} - x) d x = \frac{1}{6}$
(S) $A = 64 - \int_{0}^{16} (\sqrt{x}) d x = \frac{64}{3}$

Suggest Corrections

Similar questions

Matching type questions:

$\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & N o . o f p o i n t s o f i n t e r s e c t i o n o f c u r v e s | y | = l n | x | a n d (x - 1)^{2} + y^{2} - 4 = 0 & (1) & 1 (Q) & S_{n} = \sum_{r = 1}^{n} (n > 6) t h e n S_{n} - 7 [\frac{S_{n}}{7}] i s (w h e r e [x] d e n o t e s i n t e g e r l e s s t h a n o r e q u a l t o x) & (2) & 2 (R) & E q u a t i o n o f t a n g e n t a t z_{0} t o t h e c i r c l e | z | = r i s R e (\frac{z}{z_{0}}) = λ, t h e n l i s & (3) & 3 (S) & N o r m a l d r a w n a t a n y p o i n t o f a n e l l i p s e \frac{x^{2}}{25} + \frac{y^{2}}{9}, i s t a n g e n t t o t h e & (4) & 5 c i r c l e x^{2} + y^{2} = r^{2} t h e n m a x i m u m v a l u e o f r i s (5) & 5 \end{matrix}$

Q. Match the following:

\begin{matrix} C o l u m n - I & C o l u m n - I I (P) & M i n i m u m o f \frac{x^{4}}{y^{4}} + \frac{y^{4}}{x^{4}}; & (1) & \frac{1}{4} (x, y \neq 0; x, y > 0) i s (Q) & x ϵ (0, 1), t h e f u n c t i o n x^{25} (1 - x)^{75} & (2) & 2 t a k e s t h e m a x i m u m v a l u e a t x (R) & T h e m i n i m u m v a l u e o f & (3) & 8^{2} \times 16^{4} 2^{(x^{2} - 3)^{3} + 27} i s (S) & I f x + y = 24, (x > 0, y > 0), t h e n & (4) & 1 t h e m a x i m u m v a l u e o f x^{2} y^{4} i s \end{matrix}

Match the column

$\begin{matrix} Equation & Name of the curve 1) x^{2} - 2 x - y - 3 = 0 & P) Circle 2) x^{2} + 3 x y + 2 y^{2} - x - 4 y - 6 = 0 & Q) Parabola 3) x^{2} + y^{2} - 20 = 0 & R) Ellipse 4) 7 x^{2} + 7 y^{2} + 2 x y + 10 x - 10 y + 7 = 0 & S) Hyperbola 5) 6 x^{2} - x y - y^{2} - 23 x + 4 y + 15 = 0 & T) Pair of straight lines \end{matrix}$

Consider the lines
$L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2},$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4.$ Let $a x + b y + c z = d$ the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ and perpendicular to planes $P_{1}$ and $P_{2}$ .
Match List I with List II and select the correct answer using the code given below the lists.

$\begin{matrix} L i s t I & L i s t I I P & a = & 1 & 13 Q & b = & 2 & - 3 R & c = & 3 & 1 S & d = & 4 & - 2 \end{matrix}$

To plot the graphs of the following equations:
Equation 1: $3 x + 2 y = 1$
Equation 2: $y - 2 x = 2$
Fill in the values (p, q, r, s) in the given tables.
Equation 1:
$\begin{matrix} x & 0 & 5 y & p & q \end{matrix}$
Equation 2:
$\begin{matrix} x & r & s y & 0 & 5 \end{matrix}$

Join BYJU'S Learning Program

List IList IIP.Area bounded by y=x|x|, x−axis,1.103x=1 and x=−1Q.Area bounded by x−y+2,2.643x=√y and x=0R.Area enclosed by y2=x3.23andy=|x|S.Area bounded by y2=x4.16y=4 and x=0

The correct option is B P-3; Q-1; R-4; S-2(P) A=2∫10x2dx=23 (Q) A∫10(x+2−x2)dx=103 (R) A∫10(√x−x)dx=16 (S) A=64−∫160(√x)dx=643

The correct option is B P-3; Q-1; R-4; S-2
(P) $A = 2 \int_{0}^{1} x^{2} d x = \frac{2}{3}$
(Q) $A \int_{0}^{1} (x + 2 - x^{2}) d x = \frac{10}{3}$
(R) $A \int_{0}^{1} (\sqrt{x} - x) d x = \frac{1}{6}$
(S) $A = 64 - \int_{0}^{16} (\sqrt{x}) d x = \frac{64}{3}$