Maths

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,                                 

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

The given equation of the lines are

$\begin{array}{l}2x+y=4------(1)\\ 3x-2y=6-----(2)\\ x-3y+5=0----(3)\end{array}$

$\therefore$ Area of $?ABC=area(PCBQ)-area(?APC)-area(?AQB)$

$\begin{array}{l}={\displaystyle \underset{1}{\overset{4}{\int }}{y}_{(3)}dx-}{\displaystyle \underset{1}{\overset{2}{\int }}{y}_{(1)}dx-{\displaystyle \underset{2}{\overset{4}{\int }}{y}_{(2)}dx-}}\\ ={\displaystyle \underset{1}{\overset{4}{\int }}\frac{\left(x+5\right)}{3}dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(4-2x\right)dx-{\displaystyle \underset{2}{\overset{4}{\int }}\left(\frac{3x-6}{2}dx\right)}}}\end{array}$

$\therefore$ The point of intersection of the circle and the parabola is . $A\left(\frac{1}{2},\right)\&C\left(\frac{1}{2},-\right)$

Taking in first quadrant

Area of $(OABO)=area(OADO)+area(BADB)$

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is $y-0=\frac{5-0}{4-2}\left(x-2\right)$

$=y=\frac{5}{2}\left(x-2\right)$

Similarly equation of BC is

$\begin{array}{l}y-5=\frac{3-5}{6-4}\left(x-4\right)\\ =y=5+\left(-x+4\right)\\ y=9-x\end{array}$

And equation of AC is $y-0=\frac{3-0}{6-2}\left(x-2\right)$

= $y=\frac{3}{4}\left(x-2\right)$

$\therefore$ Area of $?ABC$

= $area(?ABE)+area(BCDE)-area(?ACD)$

$\begin{array}{l}={\displaystyle \underset{2}{\overset{4}{\int }}{y}_{AB}dx+}{\displaystyle \underset{4}{\overset{6}{\int }}{y}_{BC}dx-{\displaystyle \underset{2}{\overset{6}{\int }}{y}_{BC}dx}}\\ ={\displaystyle \underset{2}{\overset{4}{\int }}\frac{5}{2}\left(x-2\right)dx+{\displaystyle \underset{4}{\overset{6}{\int }}\left(9-x\right)dx-{\displaystyle \underset{2}{\overset{6}{\int }}\frac{3}{4}}}}\left(x-2\right)dx\\ =\frac{5}{2}{\left[\frac{x^2}{2}-2x\right]}_2^4+{\left[9x-\frac{x^2}{2}\right]}_4^6-\frac{3}{4}{\left[\frac{x^2}{2}-2x\right]}_2^6\end{array}$

$\begin{array}{l}=\frac{5}{2}\left[\left(\frac{4^2}{2}-2.4\right)-\left(\frac{2^2}{2}-2.2\right)\right]+\left[\left(9.6-\frac{6^2}{2}\right)-\left(9.4-\frac{4^2}{2}\right)\right]-\frac{3}{4}\left[\left(\frac{6^2}{2}-2.6\right)-\left(\frac{2^2}{2}-2.2\right)\right]\\ =\frac{5}{2}\left[\left(8-8\right)-\left(2-4\right)\right]+\left[54-18-36+8\right]-\frac{3}{4}\left[18-12-2+4\right]\\ =5+8-6=7uni{t}^2\end{array}$