Class 12th

Given equation of the curve is $\left|x\right|+\left|y\right|=1$ , which can be break down into each quadrant .

For Ist quadrant,  $\left|x\right|=x,\left|y\right|=y$

i.e.,  $x+y=1$ - (1)

Similarly for IInd, IIIRd nad IVth quadrant

$-x+y=1$ - (2)

$-x-y=1$ - (3)

$x-y=1$ - (4)

We draw the above focus lines on a graph and find the area enclosed which is a square.

$\therefore$ Required area $\left(?ABCD\right)=4*area\left(?AOB\right)$ .

$\begin{array}{l}=4*{\displaystyle \underset{0}{\overset{1}{\int }}y}dx\\ =4{\displaystyle \underset{0}{\overset{1}{\int }}\left(-x\right)}dx\\ =4{\left[x-\frac{x^2}{2}\right]}_0^1\\ =4\left[1-\frac{1}{2}\right]\\ =4*\frac{1}{2}\\ =2uni{t}^2\end{array}$

The given equation of the parabola is $x^2=y$ ---------(1)

and that the line is $y=x+2$ --------------(2)

Solving (1) and (2) for x and y

$\begin{array}{l}{x}^2=x+2\\ \to x^2-x-2=0\\ =x^2+x-2x-2=0\\ =x\left(x+1\right)-2\left(x+1\right)=0\\ =\left(x+1\right)\left(x-2\right)=0\\ =x=-1\&x=2\end{array}$

When $x=-1,y={\left(-1\right)}^2=1$

And $x=2,y=2^2=4$

$\therefore$ The point of intersection of the parabola and the lines $A(-1,1)\&B(2,4)$

Hence the required area enclosed region is $ea(AOBA)=area(DABCD)-area(DAOBCD)$

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{line}dx-}{\displaystyle \underset{-1}{\overset{2}{\int }}{y}_{curve}dx}\\ ={\displaystyle \underset{-1}{\overset{2}{\int }}\left(x+2\right)dx}-{\displaystyle \underset{-1}{\overset{2}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^2}{2}2x\right]}_{-1}^2-{\left[\frac{x^3}{3}\right]}_{-1}^2\end{array}$

$\begin{array}{l}=\left[\left(\frac{2^2}{2}2.2\right)-\left(\frac{{\left(-1\right)}^2}{2}+2\left(-1\right)\right)\right]-\left[\frac{2^3}{3}-\frac{{\left(-1\right)}^3}{3}\right]\\ =\left[2+4-\frac{1}{2}+2\right]-\left[\frac{8+1}{3}\right]=\frac{15}{2}-3=\frac{9}{2}uni{t}^2\\ \end{array}$

32.  (B) CuF₂

(A) Ag₂SO₄  (Ag⁺ )→ 5d¹⁰6s⁰

(B) CuF₂  (Cu²⁺ )→ 3d⁹4s⁰

(C) ZnF₂  (Zn²⁺)→ 3d¹⁰4s⁰

(D) Cu₂Cl₂  (Cu⁺)→ 3d¹⁰6s⁰

Unpaired electrons present in any compound impart colour to the salt of transition metal. Only CuF₂ has unpaired electrons in its 3d orbital that's why it is white coloured in its solid-state while rest of the salts are colourless.

The Given equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

And the equation of the line in $\frac{x}{a}+\frac{y}{b}=1$

With x and y intersept a and b

So, required area of the enclosed region is

$area(BCAB)=area(OBCAO)-area(?OBA)$

Given equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$ Which as major axis aling x- axis and that of the line is $\frac{x}{3}+\frac{y}{2}=1$ which has x and y intercepts at 3 and 2respectively.

Required area of enclosed region is area $area(BCAB)=area(OBAO)?area(ABOA)$

31.  (D) Cu

Density is mass by volume. As we move from left to right for a long period, the atomic radii decrease. Hence, volume decreases. Also, an increase in atomic masses is observed. 

So overall the density increases out of the above option from iron to copper, copper will have the highest density.

The given equation of parabola is $4y=3x^2\Rightarrow x^2=\frac{4}{3}y$ ------------(1)

And the line is $2y=3x+12\Rightarrow y=\frac{3}{2}x+6$ ----------------------(2)

Solving (1) and (2) for x and y,

$\begin{array}{l}{x}^2=\frac{4}{3}\left(\frac{3}{2}x+6\right)=2x+8\\ \Rightarrow x^2-2x-8=0\\ \Rightarrow x^2-4x+2x-8=0\\ \Rightarrow x\left(x-4\right)+2\left(x-4\right)=0\\ \Rightarrow \left(x-4\right)\left(x+2\right)=0\\ \Rightarrow x=4\&x=-2\end{array}$

At, $x=4,y=\frac{3}{2}*4+6=6+6=12$

And $x=-2,y=\frac{3}{2}*\left(-1\right)+6=-3+6=3$

Thus, the point of intersection of (1)&(2)are $A(4,12)\&B(-2,3)$

$\therefore$ Area of the enclosed region (BOAB)

=area (CBAD) – area (OADC)

$\begin{array}{l}={\displaystyle \underset{-2}{\overset{4}{\int }}{y}_{line}dx-}{\displaystyle \underset{-2}{\overset{4}{\int }}{y}_{curve}dx}\\ ={\displaystyle \underset{-2}{\overset{4}{\int }}\left(\frac{3}{2}x+6\right)dx-{\displaystyle \underset{-2}{\overset{4}{\int }}\frac{3x^2}{4}dx}}\\ ={\left[\frac{3}{2}\frac{x^2}{2}+6x\right]}_{-2}^4-{\left[\frac{3}{4}*\frac{x^3}{3}\right]}_{-2}^4\\ =\left[\left(\frac{3}{4}*4^2+6*4\right)-\left(\frac{3}{4}{\left(-2\right)}^2+6*\left(-2\right)\right)\right]-\frac{1}{4}\left[4^3-{\left(-2\right)}^3\right]\\ =\left[12+24-3+12\right]-\frac{1}{4}\left[64+8\right]\\ =45-18=27uni{t}^2\end{array}$

49. Option (i) A (3) B (4) C (2) D (1)

Sapphire is a gemstone containing Co.

Hence, option (A) from column I is matched with option (3) from column II. The Sphalerite single is ZnS.

Hence, option (B) from column I is matched with option (4) from column II. NaCN is also used as a depressant.

Hence, option (C) from column I is matched with option (2) from column II. Al2O3 is also called corundum.

Hence, option (D) from column I is matched with option (1) from column II.

30. (A) Cu (II) is more stable .

Electronic configuration of Cuis [Ar] 3d¹⁰ 4s¹

Cu (I) - [Ar] 3d¹⁰ 4s0

Cu (II)- [Ar] 3d⁹ 4s0

Despite the fact that Cu (I)  has fully filled 3d-orbital but Cu (II) is more stable than Cu (I) due to the greater effective nuclear charge of Cu (II) as nucleus has to hold 17 electrons rather than 18 electrons like in Cu (I).