Class 12th

Given equation of lines is $y=|x+3|$ -------(1)

The point (x,y)satisfying (1)are

Hence plotting the above in graph we get     

Now, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}$

We know that, $\begin{array}{l}{\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}|x+3|dx+{\displaystyle \underset{-3}{\overset{0}{\int }}|x+3|dx}}\\ y=|x+3|=\left\{\begin{array}{l}x+3,if,x+3\ge 0\Rightarrow x\ge -3\\ -\left(x+3\right),if,x+3\angle 0\Rightarrow x\angle -3\end{array}\right\}\end{array}$

So, ${\displaystyle \underset{-6}{\overset{0}{\int }}|x+3|dx=}{\displaystyle \underset{-6}{\overset{-3}{\int }}-\left(x+3\right)dx+}{\displaystyle \underset{-3}{\overset{0}{\int }}\left(x+3\right)dx}$

$\begin{array}{l}=-{\left[\frac{x^2}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{x^2}{2}+3x\right]}_{-3}^0\\ =-\left\{\left[\frac{{\left(-3\right)}^2}{2}+3\left(-3\right)\right]-\left[\frac{{\left(-6\right)}^2}{2}+3\left(-6\right)\right]\right\}+\left\{\left[\frac{0^2}{2}+3*0\right]-\left[\frac{{\left(-3\right)}^2}{2}+3*\left(-3\right)\right]\right\}\\ =-\left\{\frac{9}{2}-9-\frac{36}{2}+18\right\}+\left\{-\frac{9}{2}+9\right\}\\ =-\frac{9}{2}+9+18-18-\frac{9}{2}+9\\ =9uni{t}^2\end{array}$

Given curve is $y=4x^2\Rightarrow x^2=\frac{1}{4}y$ - (1)

$x=0$  i.e, y-axis and y=4 and y=1

Hence, the required area in Ist quadrant i.e, area ABCD = ${\displaystyle \underset{y=1}{\overset{y=4}{\int }}xdy}$

48. Correct code (i) A (4) B (2) C (3) D (1)

The cyanide process is used in the extraction of Au.

Hence, option (A) from column I is matched with option (4) from column II. Froth floatation process is used in the dressing of ZnS.

Hence, option (B) from column I is matched with option (2) from column II. Electrolytic reduction is used in the extraction of AI.

Hence, option (C) from column I is matched with option (3) from column II. Zone refining is used to get ultrapure Ge.

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

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

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

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

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

Where, $x=0,y=0^2=0$

And when $x=1,y=1^2=1$

$\therefore$ The point of intersection of the parabola $y=x^2$ and the line $y=x$

Is O(0,0) and B (1,1)

Hence, area between the curve and the line is

$area\left(DCAO\right)=area\left(?OAB\right)-area\left(OABO\right)$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}{y}_{line}}dx-{\displaystyle \underset{0}{\overset{1}{\int }}{y}_{curve}}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}xdx-{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}}\\ ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\end{array}$

$\begin{array}{l}=\frac{1}{2}-\frac{1}{3}\\ =\frac{3-2}{6}=\frac{1}{6}uni{t}^2\end{array}$

(i) Given of curve is $y=x^2\Rightarrow x^2=y$ and the equation are $x=1\&x=2.$

$\therefore$ Area enclosed

$\begin{array}{l}={\displaystyle \underset{x=1}{\overset{x=2}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{2}{\int }}{x}^2}dx\\ ={\left[\frac{x^3}{3}\right]}_1^2=\frac{2^3}{3}-\frac{1^3}{3}\\ =\frac{8-1}{3}=\frac{7}{3}uni{t}^2\end{array}$

(ii) Given equation of curve is $y=x^4$ and the lines are $x=1\&x=5$

So, area enclosed

$\begin{array}{l}={\displaystyle \underset{1}{\overset{5}{\int }}y}dx\\ ={\displaystyle \underset{1}{\overset{5}{\int }}{x}^4}dx\\ ={\left[\frac{x^5}{5}\right]}_1^5=\left(\frac{5^5-1^5}{5}\right)\\ =\frac{\left(3125-1\right)}{5}\\ =\frac{3124}{5}=624.8uni{t}^2\end{array}$

47. Option (ii) A (4) B (3) C (1) D (2)

Coloured bands are found in chromatography.

Hence, option (A) from column I is matched with option (4) from column II. Impure metals are converted to volatile complexes in Mond's process.

Hence, option (B) from column I is matched with option (3) from column II. Purification of Ge and silicon is done using zone refining.

Hence, option (C) from column I is matched with option (1) from column II. Purification of mercury is done using fractional distillation.

Hence, option (D) from column I is matched with option (2) from column II.

The given equation of the curve is $y^2=4x$ - (1) and

the line is $y=2x$ - (2)

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

$\begin{array}{l}{\left(2x\right)}^2=4x\\ =4x^2=4x\\ =x^2-x=0\\ =x\left(x-1\right)=0\end{array}$

So,  $x=0\&x=1$

for $x=0$ we get $y=2*0=0$

for $x=1$ , we get $y=2*1=2$

so, the point of intersection are (0,0)and (1,2)

area (DCAO)=area (DCABO)-area ( $?OAB$ )

46. Option (ii) A (2) B (4) C (5) D (3)

Pendulum is always made of nickel steel.

Hence, option (A) from column I is matched with option (2) from column II. Malachite is the ore of copper.

Hence, option (B) from column I is matched with option (4) from column II. Calamine is the ore of zinc.

Hence, option (C) from column I is matched with option (5) from column II. Cryolite is an ore of aluminum.

Hence, option (D) from column I is matched with option (3) from column II.

28. Ionisation enthalpies are the main factor that influence the reactivity of transition elements. Higher the ionisation enthalpy, lesser is the reacting of the transition element. 

When we move along the period from Sc to Cu, a regular increase in the ionisation enthalpy is observed which results in the almost regular decrease in the reactivity of elements.

The equation of circle is $x^2+y^2=4$ which has centre at (0,0) & radius,

$\pi =2$

And the line $x+y=2=y=2-x$

The smaller area of circle is given by

Area (ABCA) area (BOAB) – area (BOA)