(i) We have x² – 6x + 13 = x ²– 6x + 3² – 3² + 13 = (x – 3)² + 4
\(\text { So, } \ \int \frac{d x}{x^{2}-6 x+13}=\int \frac{1}{(x-3)^{2}+2^{2}} d x\)
Let x – 3 = t. Then dx = dt
\(\int \frac{d x}{x^{2}-6 x+13}=\int \frac{d t}{t^{2}+2^{2}}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)
\(=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+\mathrm{C}\)
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
\(3 x^{2}+13 x-10=3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)\)
\(=3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] \text { (completing the square) }\)
\(\text { Thus } \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(\text { Put } x+\frac{13}{6}=t \text { . Then } d x=d t . \)
\(\text { Therefore, } \quad \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(=\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C, \text { where } C=C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(\text { (iii) We have } \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}\)
\(=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} \text { (completing the square) }\)
\(\text { Put } x-\frac{1}{5}=t . \text { Then } d x=d t \text { . }\)
\(\text { Therefore, } \quad \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}\)
\(=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+\mathrm{C}\)
\(=\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+\mathrm{C}\)

(i) \({\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]} \)
\({\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]} \)
\(=\left[\begin{array}{cc} a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & -b(-b)+a(a) \end{array}\right] \)
\(=\left[\begin{array}{cc} a^{2}+b^{2} & -a b+a b \\ -a b+a b & b^{2}+a^{2} \end{array}\right]=\left[\begin{array}{cc} a^{2}+b^{2} & 0 \\ 0 & a^{2}+b^{2} \end{array}\right]\)
(ii) 
\({\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right][2,3,4]} \)
\(=\left[\begin{array}{lll} 1(2) & 1(3) & 1(4) \\ 2(3) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{array}\right]=\left[\begin{array}{llc} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{array}\right]\)
(iii)
\({\left[\begin{array}{cc} 1 & -2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]} \)
\(=\left[\begin{array}{lll} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\ 2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1) \end{array}\right] \)
\(=\left[\begin{array}{ccc} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{array}\right]=\left[\begin{array}{ccc} -3 & -4 & 1 \\ 8 & 13 & 9 \end{array}\right]\)
(iv)
\({\left[\begin{array}{lcc} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{ccc} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]} \)
\(=\left[\begin{array}{lll} 2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5) \end{array}\right] \)
\(=\left[\begin{array}{lccc} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{array}\right]=\left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right]\)
(v)
\({\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right]} \)
\(=\left[\begin{array}{ccc} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\ 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\ -1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1) \end{array}\right] \)
\(=\left[\begin{array}{ccc} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{array}\right]\)
(vi)
\({\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right]} \)
\(=\left[\begin{array}{cc} 3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\ -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{array}\right] \)
\(=\left[\begin{array}{cc} 6-1+9 & -9-0+3 \\ -2+0+6 & 3+0+2 \end{array}\right]=\left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right]\)

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

\(A=\left[\begin{array}{rrr} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]\)
|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13
Now,
\(A_{11}=14, A_{12}=11, A_{13}=-5 \)
\(A_{21}=11, A_{22}=4, A_{23}=-3 \)
\(A_{31}=-5, A_{32}=-3, A_{13}=-1\)
\(\operatorname{adj} A  =\left[\begin{array}{lll} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]\)
\(A^{-1}  =\frac{1}{|A|}(\operatorname{adj} A) \)
\(=-\frac{1}{13}\left[\begin{array}{lll} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{array}\right]=\frac{1}{13}\left[\begin{array}{lll} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{array}\right]\)
\(\mid \text { adj } A \mid =14(-4-9)-11(-11-15)-5(-33+20) \)
=14(-13)-11(-26)-5(-13)
=-182+286+65=169
We have,
\( \operatorname{adj}(\operatorname{adj} A)=\left[\begin{array}{ccc} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]\)
\( \therefore[\operatorname{adj} A]^{-1}=\frac{1}{|\operatorname{adj} A|}(\operatorname{adj}(\operatorname{adj} A))\)
\(=\frac{1}{169}\left[\begin{array}{lll} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]\)
\(=\frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]\)
Now, \(A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13}\end{array}\right]\)
\(\therefore \operatorname{adj}\left(A^{-1}\right) =\left[\begin{array}{lll} -\frac{4}{169}-\frac{9}{169} & -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{33}{169}+\frac{20}{169} \\ -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{14}{169}-\frac{25}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) \\ -\frac{33}{169}+\frac{20}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) & \frac{56}{169}-\frac{121}{169} \end{array}\right]\)
\( =\frac{1}{169}\left[\begin{array}{lll} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{array}\right]=\frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]\)
Hence, \([\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)\).
We have shown that:
\(A^{-1}=\frac{1}{13}\left[\begin{array}{lll} -14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1 \end{array}\right]\)
And, \(\operatorname{adj} A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]\)
Now,
\( \left|A^{-1}\right|=\left(\frac{1}{13}\right)^3[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=\left(\frac{1}{13}\right)^3 \times(-169)=-\frac{1}{13} \)
\(\therefore\left(A^{-1}\right)^{-1}=\frac{\operatorname{adj} A^{-1}}{\left|A^{-1}\right|}=\frac{1}{\left(-\frac{1}{13}\right)} \times \frac{1}{13}\left[\begin{array}{lll} -1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5 \end{array}\right]=\left[\begin{array}{lll} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{array}\right]=A\)
\(\left(A^{-1}\right)^{-1}=A\)

From the figure, the area of the region ABA′B′A bounded by the ellipse
\(=4\left(\begin{array}{l} \text { area of the region AOBAin the first quadrantbounded } \\ \text { by thecurve, } x-\text { axis and the ordinates } x=0, x=a \end{array}\right)\)
(as the ellipse is symmetrical about both x-axis and y-axis)
\(=4 \int_{0}^{a} y d x \text { (taking verticalstrips) }\)
\(\text { Now } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { gives } y=\pm \frac{b}{a} \sqrt{a^{2}-x^{2}} \text { , }\)but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is
\(=4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2}-x^{2}} d x \)
\(=\frac{4 b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a} \)
\(=\frac{4 b}{a}\left[\left(\frac{a}{2} \times 0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-0\right] \)
\(=\frac{4 b}{a} \frac{a^{2}}{2} \frac{\pi}{2}=\pi a b \)

Alternatively, considering horizontal strips as shown in the Figure, the area of the ellipse is
\(=4 \int_0^b x d y=4 \frac{a}{b} \int_0^b \sqrt{b^2-y^2} d y\)
\(=\frac{4 a}{b}\left[\frac{y}{2} \sqrt{b^2-y^2}+\frac{b^2}{2} \sin ^{-1} \frac{y}{b}\right]_0^b\)
\( =\frac{4 a}{b}\left[\left(\frac{b}{2} \times 0+\frac{b^2}{2} \sin ^{-1} 1\right)-0\right] \\ =\frac{4 a}{b} \frac{b^2}{2} \frac{\pi}{2}=\pi a b \)

\(\text { Let } \mathrm{I}=\int_{-1}^{1} \sin ^{5} x \cos ^{4} x d x \text { . Let } f(x)=\sin ^{5} x \cos ^{4} x \text { . Then }\)
f (– x) = sin⁵ (– x) cos⁴ (– x) = – sin⁵ x cos⁴ x = – f (x), i.e., f is an odd function.
Therefore, by P₇ (ii), I = 0

Here the integrand \(\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-5x+6 } \)is not a proper rational function, so dividing \(({ x }^{ 2 }+1)by{ (x }^{ 2 }-5x+6)\), we devide \(x^2+1 \text { by } x^2-5 x+6\) and find that
\(\frac{x^{2}+1}{x^{2}-5 x+6} =1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)} \)
\(\frac{5 x-5}{(x-2)(x-3)} =\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\)
So that 5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10
\(\text {Thus,} \frac{x^{2}+1}{x^{2}-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}\)
\(\therefore \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=\int d x-5 \int \frac{1}{x-2} d x+10 \int \frac{d x}{x-3}\)
= x – 5 log | x – 2 | + 10 log | x – 3 | + C.

(i) We know that  \(\frac { d }{ dx } (\sin { 2x) } =2\cos { 2x } \)
\(\Rightarrow \cos { 2x } =\frac { 1 }{ 2 } \frac { d }{ dx } (\sin { 2x) } =\frac { d }{ dx } (\frac { 1 }{ 2 } \sin { 2x) } \)
Hence, antiderivative of \(\cos { 2x } =\frac { 1 }{ 2 } \sin { 2x } \)
(ii) We know that \(\frac { d }{ dx } ({ x }^{ 3 }+{ x }^{ 4 })={ 3x }^{ 2 }+{ 4x }^{ 3 }.\)
Hence, antiderivative of \({ 3x }^{ 2 }+{ 4x }^{ 3 }+{ x }^{ 3 }+{ x }^{ 4 }\)
(iii) We know that \(\frac { d }{ dx } (\log { x } )=\frac { 1 }{ x } ,x>0\) and \(\frac { d }{ dx } [\log { (-x) } ]=\frac { 1 }{ -x } (-1)=\frac { 1 }{ x } ,x>0.\)
combining, \(\frac { d }{ dx } (\log { 1\quad x1 } )=\frac { 1 }{ x } ,x\neq 0.\)
\(\Rightarrow \int { \frac { 1 }{ x } } dx=\log { 1\quad x1 } .\)
which is one of antiderivative of \(\frac { 1 }{ x } \).

The given ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both axes,
Therefore, area of the ellipse = 4 (Shaped area).

\(=4(area\ OAB)\)
\(But\ area\ OAB=\overset { 4 }{ \underset { 0 }{ \int { } } } ydx\quad [Taking\ vertical\ strips]\)
\(=\overset { 4 }{ \underset { 0 }{ \int { } } } \frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } dx\)
\([\because \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { y }^{ 2 } }{ 9 } =1-\frac { { x }^{ 2 } }{ 16 } \Rightarrow y=\frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } (\because y>0)]\)
\(=\frac { 3 }{ 4 } \left[ \frac { x\sqrt { 16-{ x }^{ 2 } } }{ 2 } +\frac { 16 }{ 2 } { sin }^{ -1 }\frac { x }{ 4 } \right] _{ 0 }^{ 4 }\)
\(=\frac { 3 }{ 4 } [[2(0)+8{ sin }^{ -1 }(1)]-[0-0]]\)
\(=\frac { 3 }{ 4 } \left[ 8\frac { \pi }{ 2 } \right] =3\pi \)
\(\therefore From\ (2),area\ of\ the\ ellipse =4(3\pi )=12\pi sq.units.\)

\(\text { The given function is } f(x)=x^{2}-x+1 \text { . }\)
\(\therefore f^{\prime}(x)=2 x-1 \)
\(\text {Now, } f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2} . \)
\(\text {The point } \frac{1}{2} \text { divides the interval ( }-1,1 \text { ) into two disjoint intervals i.e., }\left(-1, \frac{1}{2}\right) \text { and }\left(\frac{1}{2}, 1\right) \text { . }\)
\(\text {Now, in interval }\left(-1, \frac{1}{2}\right), f^{\prime}(x)=2 x-1<0 \text { . }\)
 Therefore, f is strictly decreasing in interval (−1,12) . " id="MathJax-Element-1011-Frame" role="presentation" style="position: relative;" tabindex="0">Therefore, f is strictly decreasing in interval \( \left(-1, \frac{1}{2}\right) \text { . }\)
\(\text {However, in interval }\left(\frac{1}{2}, 1\right), f^{\prime}(x)=2 x-1>0 \text { . }\)
 Therefore, f is strictly increasing in interval (12,1) . " id="MathJax-Element-1013-Frame" role="presentation" style="position: relative;" tabindex="0"> Therefore, f is strictly increasing in interval \( \left(\frac{1}{2}, 1\right) \text { . }\)
Hence, f is neither strictly increasing nor decreasing in interval  (-1,1) 

(i) Area of the triangle
\({1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}={1\over2}\begin{vmatrix}1&0&1\\6&0&1\\4&3&1 \end{vmatrix}\)
= \({1\over2}(-3)[1-6]\)
= \({1\over2}(15)=7{1\over7}\)
(ii) Area of the triangle
\({1\over2}\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}={1\over2}\begin{vmatrix} 2&7&1\\1&1&1\\10&8&1\end{vmatrix} \)
\({1\over2}[2(1-8)-7(1-10)+(8-10)]\)
\({1\over2}[-14+63-2]={1\over2}(47 )\)
\(23{1\over2}\) sq.units
(iii) Area of the triangle
\({1\over2}\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}={1\over2}\begin{vmatrix}-2&-3&1\\3&2&1\\-1&8&1 \end{vmatrix}\)
\({1\over2}[(-2)(2+8)+3(3+1)+1(-24+2)]\)
\({1\over2}[-20+12-22]=-{30\over2}=-15\)
Area if the triangle = 15 sq.units.

Since 6 lies in the second row and third column, its minor M₂₃ is given by
\(M_{23}=\left|\begin{array}{ll} 1 & 2 \\ 7 & 8 \end{array}\right|=8-14=-6\)(obtained by deleting R₂ and C₃ in Δ).

We have: \(3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix},\)
Equating corresponding elements:
3x = x + y  \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8 \(\Rightarrow \) y = 4
3w = 2w + 3 \(\Rightarrow \) w = 3   and  
3z = z + w - 1\(\Rightarrow \) 2z = 3 - 1 = 2 \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

We have
\(\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}=(x)(x)-(x-1)(x-1)\)
\(=x^{2}-\left(x^{2}-1\right)=x^{2}-x^{2}+1=1\)

(i) We have : \({ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right| \)
\(\therefore \ { a }_{ 11 }=\frac { 1 }{ 2 } \left| -3+1 \right| =\frac { 2 }{ 2 } =1;\quad { a }_{ 12 }=\frac { 1 }{ 2 } \left| -3+2 \right| =\frac { 1 }{ 2 } =1;\)
\({ a }_{ 13 }=\frac { 1 }{ 2 } \left| -3+3 \right| =0;\quad { a }_{ 14 }=\frac { 1 }{ 2 } \left| -3+4 \right| =\frac { 1 }{ 2 } ;\)
\({ a }_{ 21 }=\frac { 1 }{ 2 } \left| -6+1 \right| =\frac { 5 }{ 2 } ;\quad { a }_{ 22 }=\frac { 1 }{ 2 } \left| -6+2 \right| =2;\)
\( { a }_{ 23 }=\frac { 1 }{ 2 } \left| -6+3 \right| =\frac { 3 }{ 2 } ;\quad { a }_{ 24 }=\frac { 1 }{ 2 } \left| -6+4 \right| =1;\)
\({ a }_{ 31 }=\frac { 1 }{ 2 } \left| -9+1 \right| =4;\quad { a }_{ 32 }=\frac { 1 }{ 2 } \left| -9+2 \right| =\frac { 7 }{ 2 } ;\)
\({ a }_{ 33 }=\frac { 1 }{ 2 } \left| -9+3 \right| =3;\quad { a }_{ 32 }=\frac { 1 }{ 2 } \left| -9+4 \right| =\frac { 5 }{ 2 } ;\)
Hence, \(A=\left[ \begin{matrix} 1 & \frac { 1 }{ 2 } & 0 & \frac { 1 }{ 2 } \\ \frac { 5 }{ 2 } & 2 & \frac { 3 }{ 2 } & 1 \\ 4 & \frac { 7 }{ 2 } & 3 & \frac { 5 }{ 2 } \end{matrix} \right] \)
(ii) We have: \({ a }_{ ij }=2i-j.\)
\(​​\therefore {a }_{ 11 }=2-1=1;\quad { a }_{ 12 }=2-2=0;\)
\({ a }_{ 13 }=2-3=-1;\quad { a }_{ 14 }=2-4=-2;\)
\({ a }_{ 21 }=4-1=3;\quad { a }_{ 22 }=4-2=2;\)
\({ a }_{ 23 }=4-3=1;\quad { a }_{ 24 }=4-4=0;\)
\({ a }_{ 31 }=6-1=5;\quad { a }_{ 32 }=6-2=4;\)
\({ a }_{ 33 }=6-3=3;\quad { a }_{ 32 }=6-4=2;\)
Hence, \(A=\left[ \begin{matrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{matrix} \right] .\)

Let \(\overset { \rightarrow }{ d } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \) ..(1)
Now  \(\overset { \rightarrow }{ d } \) is perp.to \(\overset { \rightarrow }{ a } \) 
\(\therefore\) \(\overset { \rightarrow }{ d } \).\(\overset { \rightarrow }{ a } \) = 0
\(\Rightarrow (x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } ).(\overset { \wedge }{ i } +4\overset { \wedge }{ j } +2\overset { \wedge }{ k } )=0\) ...(2)
\(\Rightarrow x+4y+2z=0\)
and \(\overset { \rightarrow }{ d } \) is prep.to \(\overset { \rightarrow }{ b } \)
\(\therefore\) \(\overset { \rightarrow }{ d } \).\(\overset { \rightarrow }{ b } \) = 0
\(\therefore(\hat{i} \hat{i}+\hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})=0\)
\(\Rightarrow\) 3x - 2y + 7z = 0..(3)
Also  \(\overset { \rightarrow }{ c } \).\(\overset { \rightarrow }{ d } \) = 15
\(\Rightarrow\) \((2\overset { \wedge }{ i } -\overset { \wedge }{ j } +4\overset { \wedge }{ k } ).(x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } )=15\) 
2x - y + 4z = 15 ......(4)
(3)-3(2) gives: -14y + z = 0        ..(5)
(4)-2(2) gives :  -9y = 15        ....(6)
from (6). y = -\(\frac { 5 }{ 3 } \) 
putting in (5), -14(\(\frac {- 5 }{ 3 } \)) + z = 0
\(\Rightarrow\) z = -\(\frac {- 70 }{ 3 } \)
Putting in (2), \(x -\frac {- 20 }{ 3 } -\frac {140}{3}=0 \rightarrow x =\frac {160}{3}\) 
putting in (1)
\(\vec{d} =\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{8} \hat{k} \)
\(=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \)

The vertices of △ABC are given as A(1, 2, 3), B(−1, 0, 0), and C(0, 1, 2) .
Also, it is aiven that \( \square \mathrm{ABC} \text { is the anale between the vectors } \overrightarrow{\mathrm{BA}} \text { and } \overrightarrow{\mathrm{BC}}\)
\(\overrightarrow{\mathrm{BA}}=\{1-(-1)\} \hat{i}+(2-0) \hat{j}+(3-0) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k} \)
\(\overrightarrow{\mathrm{BC}}=\{0-(-1)\} \hat{i}+(1-0) \hat{j}+(2-0) \hat{k}=\hat{i}+\hat{j}+2 \hat{k} \)
\(\therefore \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})=2 \times 1+2 \times 1+3 \times 2=2+2+6=10 \)
\(|\overrightarrow{\mathrm{BA}}|=\sqrt{2^{2}+2^{2}+3^{2}}=\sqrt{4+4+9}=\sqrt{17} \)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1+1+2^{2}}=\sqrt{6} \)
Now, it is known that:
\(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}} \| \overrightarrow{\mathrm{BC}}| \cos (\angle \mathrm{ABC}) \)
\(\therefore 10=\sqrt{17} \times \sqrt{6} \cos (\angle \mathrm{ABC}) \)
\(\Rightarrow \cos (\angle \mathrm{ABC})=\frac{10}{\sqrt{17} \times \sqrt{6}} \)
\(\Rightarrow \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right) \)

We have
\(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \)
\(\therefore \frac{d y}{d x} =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \)
\(=\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \)
\(\text { Now, } \frac{d y}{d x}=0 .\)
\(\Rightarrow \frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}=1\)
\(\Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta \)
\(\Rightarrow \cos ^{2} \theta-4 \cos \theta=0\)
\(\Rightarrow \cos \theta(\cos \theta-4)=0 \)
\(\Rightarrow \cos \theta=0 \text { or } \cos \theta=4\)
\(\text { Since } \cos \theta \neq 4, \cos \theta=0\)
\(\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
\(\text { Now, }\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}}=\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}\)
\(\text { In interval }\left(0, \frac{\pi}{2}\right), \text { we have } \cos \theta>0 . \text { Also, } 4>\cos \theta \Rightarrow 4-\cos \theta>0 \text { . }\)
\(\therefore \cos \theta(4-\cos \theta)>0 \text { and also }(2+\cos \theta)^{2}>0 \)
\(\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 \)
\(\Rightarrow \frac{d y}{d x}>0 \)
 Therefore, y is strictly increasing in interval \( \left(0, \frac{\pi}{2}\right) \text { . }\)
Also, the given function is continuous at  \( x=0 \text { and } x=\frac{\pi}{2} \text { . }\)
Hence, y is increasing in interval  \( \left[0, \frac{\pi}{2}\right] \text { . }\)

Radius (say r) of sphere 
\(=\frac { 1 }{ 2 } (diameter)\)
\(=\frac { 1 }{ 2 } ,\frac { 3 }{ 2 } (2x+3)\)
\(=\frac { 3 }{ 4 } (2x+3)\)
Let V be the volume of the sphere 
Then\(V=\frac { 4 }{ 3 } \pi r^{ 3 }=\frac { 4 }{ 3 } \pi \left( \frac { 3 }{ 4 } (2x+3 \right) ^{ 3 }\)
\(=\frac { 9 }{ 16 } \pi (2x+3)^{ 3 }\)
\(\therefore \) Rate of change of volume w,r.t.x
\(=\frac { dv }{ dx } =\frac { 9 }{ 16 } \pi .3(2x+3)^{ 2 }.2\)
\(=\frac { 27 }{ 8 } \pi (2x+3)^{ 2 }\)

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

\(cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta & cos\theta sin\theta \\ -cos\theta sin\theta & { cos }^{ 2 }\theta \end{bmatrix}+\begin{bmatrix} { sin }^{ 2 }\theta & -sin\theta cos\theta \\ sin\theta cos\theta & { sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta & cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta & { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).