Ncert Solutions Maths class 12th

(E) (i) a^ij = $\frac{1}{2}$ $|-3i+j|$ such that i = 1, 2, 3 and j = 1, 2, 3, 4 for 3 * 4 matrix

So, a₁₁= $\frac{1}{2}$ . $|-3.1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1.$

a₁₂ = $\frac{1}{2}|-3.1+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{13}=\frac{1}{2}|-3.1+3|=\frac{1}{2}*0=0$

a₁₄ = $\frac{1}{2}|-3.1+4|=\frac{1}{2}|1|=\frac{1}{2}$

a₂₁ = $\frac{1}{2}|-3.2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|5|=\frac{5}{2}$

a₂₂ = $\frac{1}{2}|-3.2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

a₂₃ = $\frac{1}{2}|-3.2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{24}=\frac{1}{2}|-3\cdot 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}+4|-2|=\frac{2}{2}=1$

$a_{31}=\frac{1}{2}|-3\cdot 3+1|=\frac{1}{2}|-0+1|=\frac{1}{2}|-8|=\frac{8}{2}=4.$

$a_{32}=\frac{1}{2}|-3\cdot 2+2|=\frac{1}{2}|-9+2|=?-\frac{7}{2}=\frac{7}{2}$

$a_{33}=\frac{1}{2}|-3\cdot 3+3|=\frac{1}{2}|-9+3|=\frac{+6?}{2}=\frac{6}{2}=3$

$a_{34}=\frac{1}{2}|-3\cdot 3+4|=\frac{1}{2}|-9+4|=?\frac{\left|5\right|}{2}=\frac{5}{2}.$

(E) (i) a^ij${\frac{\left(i+j\right)}{2}}^2$ such that i = 1, 2 and j = 1 * 2 for 2 * 2 matrix

Therefore a₁₁ = $\frac{{\left(1+1\right)}^2}{2}=\frac{2^2}{2}=2$ A _2*2 = $\left[\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right]$

a₁₂ = $\frac{{\left(1+2\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$

a₂₁ $\frac{{\left(2+1\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 8\hfill \end{array}\right]$

a₂₂ = ${\frac{\left(2+2\right)}{2}}^2=\frac{4^2}{2}=\frac{16}{2}=8$

As number of elements of matrix with order m * n

(E) Possible order of matrix with 18 elements are (1 * 18), (2 * 9), (3 * 6), (6 * 3), (9 * 2) and (18 * 1)

Similarly, possible order of matrix with 5 elements are (1 * 5) and (5 * 1)

As, number of elements of matrix having order m * n = m.n.

(b) So, (possible) order of matrix with 24 elements are (1 * 24), (2 * 12), (3 * 8), (4 * 6), (6 * 4), (8 * 3), (12 * 2), 24 * 1).

Similarly, possible order of matrix with 13 elements are (1 * 13) and (13 * 1)

Kindly go through the solution

Given curve is $y=cosx$

$y=sinx$ for $0\le x\le \frac{\pi }{2}$

And $y-axis$

We know that $sinx=cosx$ at $x=\frac{\pi }{4}and<\frac{\pi }{4}<\frac{\pi }{2}$ i.e,  $cos\frac{\pi }{4}=sin\frac{\pi }{4}=1/\surd 2$

So the point of intersection is at $x=\frac{\pi }{4}$

The given area of the circle is $x^2+y^2=16-------(1)$ is a circle with centre (0,0) and radius, $\pi =4$ and the parabola is $y^2=6x$ -------------(2)

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

$\begin{array}{l}{x}^2+6x=16\\ =x^2+6x-16=0\\ =x^2+8x-2x-16=0\\ =x\left(x+8\right)-2\left(x+8\right)=0\\ =\left(x+8\right)\left(x-2\right)=0\\ =x=-8\&x=2\end{array}$

For, $x=-8,y^2=6\left(-8\right)=-48$

Which is not possible.

For, $x=2,y^2=6\left(2\right)=12$

$y=\pm \mathrm{2\surd 3}$

$areaOACBO=2*\left\{area\left(OADO\right)+area\left(ACDA\right)\right\}$

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

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

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.

Given is $y=x^3$ and the ines $x=-2\&x=1$

For $y=x^3$

$\begin{array}{l}area\left(OAB\right)={\displaystyle \underset{0}{\overset{1}{\int }}ydx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}\\ area\left(ODC\right)={\displaystyle \underset{-2}{\overset{0}{\int }}ydx}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}{x}^{3}dx}\\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|=\left|\left[\frac{0^4}{4}-\frac{{\left(-2\right)}^4}{4}\right]\right|=4\end{array}$

$\therefore$ Total area of the bounded region $=\frac{1}{4}+4$

$=\frac{17}{4}uni{t}^2$

The given equation of curve $y^2\le 4x$ i.e,  $y^2=4x$ - (1) is a parabola and

$4x^2+4y^2\le 9\Rightarrow x^2+y^2\le \frac{9}{4}\Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2$ - (2) is a circle

With centre (0,0)and radius 

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

$\begin{array}{l}{x}^2+4x=\frac{9}{4}\\ =x^2+4x-\frac{9}{4}=0\\ =4x^2+16x-9=0\end{array}$