0

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved

\(I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { dx }{ 1+\sqrt { \cot { x } } } } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { x } } }{ \sqrt { \sin { x } } +\sqrt { \cos { x } } } } dx\)
Apply property \(\int _{ a }^{ b }{ f\left( x \right) } =\int _{ a }^{ b }{ f\left( a+b-x \right) } \)
\(=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } }{ \sqrt { \sin { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } +\sqrt { \cos { \frac { \pi }{ 3 } +\frac { \pi }{ 6 } -x } } } } dx\)
\(\therefore I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { \cos { x } } }{ \sqrt { \cos { x } } +\sqrt { \sin { x } } } } dx\)
Adding, we get \(2I=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ dx } =\left[ x \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\)

Let the events defined are :
E₁: Person chosen is a cyclist
E_2: Person chosen is a scooter driver
E₃: Person chosen is a motorbike driver
A:Person meets with an accident
P(E₁_= 1/6, P(E₂) = 1/3, P(E₃) = 1/2
P(A/E₁) = 0.01, P(A/E₂) = 0.03,
P(A/E₃) = 0.15
\(P{ (E }_{ 2 }/A)=\frac { P({ E }_{ 2 }).P(A/{ E }_{ 2 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3) } } \)
\(=\frac { \frac { 1 }{ 3 } \times 0.03 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.03+\frac { 1 }{ 2 } \times 0.15 } \)
\(=\frac { 0.01 }{ 0.086 } \)
= 0.11627
Suggestion: Cycle should be promoted as it is:
(i) Good for health
(ii) Pollution free
(iii) Saves energy (no petrol).  

Let P(x, y) be the nearest point
\(\therefore \ D=\sqrt{(x-11a)^2+y^2}\)
\(S=(x-11a)^2+y^2=(x-11a)^2+4ax\)
\(\frac{dS}{dx}=2(x-11a)+4a\)
\(\frac{dS}{dx}=0\Rightarrow x=9a\)
\(\therefore\ y=pm6a\)
\(\Rightarrow\ \frac{d^2S}{dx^2}=2>0\)
For minimum distance, coordinates are \(p(9a,\pm6a)\)

Given that sin[ cot ^-1(x + 1)] = cos(tan^-1x)....(i)
We know that,
\({ cot }^{ -1 }(A)={ sin }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x + 1
Applying this identity in equation (i), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos({ tan }^{ -1 }x)\) ...(ii)
Also, we know that
\({ tan }^{ -1 }A={ cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x
Applying this identity in equation (ii), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos\left( { cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
\(\Rightarrow \frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } =\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)
\(\left[ \because { sin }^{ -1 }(sin\theta )=\theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
and cos^-1 (cos \(\theta \)) = \(\theta \) \(\forall \theta \) [0, \(\pi \)]]
Squaring and Reciprocating both side, we have
1+(1 + x)² = 1 + x²
\(\Rightarrow\) 1 + 1 + x² + 2x = 1 + x²
\(\Rightarrow\) 1 + 2x = 0
\(\Rightarrow\)  \(x=-\frac { 1 }{ 2 } \quad \)

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

Let y=\(log({ x }^{ x }+{ cosec }^{ 2 }x)\)
\(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ x }+{ cosec }^{ 2 }x } \frac { d }{ dx } ({ x }^{ x }+{ cosec }^{ 2 }x)\)
\(=\frac { 1 }{ { x }^{ x }+{ cosec }^{ 2 }x } \left[ { x }^{ x }(1+log\quad x)+2cosec\quad x(-cosec\quad x\quad cot\quad x) \right] \)
\(=\frac { 1 }{ { x }^{ x }+{ cosec }^{ 2 }x } \left[ { x }^{ x }(1+log\quad x)-2cot\quad x{ cosec }^{ 2 }x \right] \)

\({\left[\begin{array}{ll} x-2 & -3 \end{array}\right]\left[\begin{array}{l} x \\ 3 \end{array}\right]=0} \)
\(\Rightarrow\left[x^{2}-2 x-9\right]=[0] \Rightarrow x^{2}-2 x-9=0 \)
\(x=\frac { 2\pm \sqrt { 4+36 } }{ 2 } =1\pm \sqrt { 10 } \)

The given equation is:
\(\left( x^{ 2 }-1 \right) +2xy=\frac { 2 }{ { x }^{ 2 }-1 } \)
\(\Rightarrow \)  \(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }-1 } y=\frac { 2 }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 } } \)  ..(1)
| Linear Equation 
Comparing with \(\frac { dy }{ dx } +Py=Q,\)we have: 
\('P'=\frac { 2x }{ { x }^{ 2 }-1 } \) 
\('Q'=\frac { 2 }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 }. } \)
\(\therefore \)  \(\int { Pdx } =\int { \frac { 2x }{ { x }^{ 2 }-1 } } dx=log|{ x }^{ 2 }-1|\)
\(\therefore \)  I.F. \(={ e }^{ \int { Pdx } }={ e }^{ log|{ x }^{ 2 }-1| }={ x }^{ 2 }-1.\)
Multiplying (1) by \(\left( { x }^{ 2 }-1 \right) ,\) we get:
\(\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } +2xy=\frac { 2 }{ { x }^{ 2 }-1 } \)
\(\Rightarrow \) \(\frac { d }{ dx } \left( y\left( { x }^{ 2 }-1 \right) \right) =\frac { 2 }{ { x }^{ 2 }-1 } \)
Integrating,  \(y\left( { x }^{ 2 }-1 \right) =2\int { \frac { 1 }{ { x }^{ 2 }-1 } } dx+c\)
\(\Rightarrow \) \(y\left( { x }^{ 2 }-1 \right) =\frac { 2 }{ 2\left( 1 \right) } log|\frac { x-1 }{ x+1 } |+c,\)
\(\Rightarrow \) \(y\left( { x }^{ 2 }-1 \right) =log|\frac { x-1 }{ x+1 } |+c,\)
which is the required solution

Maximize P = 3x + 4y + 5z subject to:
\(​​​​​​​2x+3y \le 8,2y+5z\le 10,3x+2y+4z\le 15,x\ge 0,y\ge 0,z\ge 0.\)

\(Let\quad y=sin({ x }^{ 2 }+5)\)
\(Put\quad { x }^{ 2 }+5=t\)
\(y=sin\quad t,\quad where\quad t={ x }^{ 2 }+5\)
\( \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(=cos\quad t.\quad (2x+0)=2xcos({ x }^{ 2 }+5)\)

The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

\(\text { Area of } \mathrm{ABCD} =\int^{1} y d x \)
\(=\int^{4} \sqrt{x} d x \)
\(=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4} \)
\(=\frac{2}{3}\left[(4)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right] \)
\(=\frac{2}{3}[8-1] \)
\(=\frac{14}{3} \text { units } \)

The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1). If we add any one, say (2, 3) to R₁ , then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1, 3) and (3, 1). Thus, the only equivalence relation bigger than R1 is the universal relation. This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two

\({ A }^{ 2 }=AA=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\)
\(=\begin{bmatrix} 3\times 3+(-2)\times 4 & 3\times (-2)+(-2)\times (-2) \\ 4\times 3+(-2)\times 4 & 4\times (-2)+(-2)\times (-2) \end{bmatrix}\)
\(=\begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix}=\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}.\)
\(Now { A }^{ 2 }=kA-2I\)
\(\Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}=k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}-2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}=\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}+\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
\(=\begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}.\)
Equating corresponding elements:
1 = 3k-2 
\(\Rightarrow \)3k = 3\(\Rightarrow \) k = 1
-2 = -2k \(\Rightarrow \) k = 1
4 = 4k \(\Rightarrow \) k = 1
-4 = -2k - 2 \(\Rightarrow \) -2k = 4 - 2 = 2 \(\Rightarrow \) k = 1.
Hence, k = 1.

|A| = \(\left| \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right| \)
= \((1)\left| \begin{matrix} 1 & -3 \\ 4 & -9 \end{matrix} \right| -1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix}-2\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}\)
= 1(-9 + 12) - (-18 + 15) -2 (8 - 5)
= 3 + 3 - 6 = 0
= 6 - 6 = 0
= 0

Let \({ cosec }^{ -1 }\left( -\sqrt { 2 } \right) =y\), where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] -\left\{ 0 \right\} \)
\(\Rightarrow cosecy=-\sqrt { 2 } =-cosec\frac { \pi }{ 4 } \)
\(=cosec\left( -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

\({1\over4}{[x^2\sqrt{x^4-1}-log |x^2+\sqrt{x^4-1}|]+c}\)

Curve is y² + 1 = x, curve is symmetrical about the x-axis.

\(\text { Area } =2 \int_{1}^{2} \sqrt{x-1} d x=\frac{4}{3}\left[(x-1)^{3 / 2}\right]_{1}^{2} \)
\(=\frac{4}{3}\left[(2-1)^{3 / 2}-(1-1)^{3 / 2}\right]=\frac{4}{3} \text { sq units }\)

Since ABCD is a parallelogram and diagonal are bisect each other.
Let A(3, - 1, - 1), B(5, - 4, 0), C(2, 3, - 2), \(D(x_{ 2 },y_{ 2 },z_{ 2 })\)
Midpoint of AC \(\left( \frac { 2+3 }{ 2 } ,\frac { 3-1 }{ 2 } .\frac { -2-1 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,\frac { 2 }{ 2 } ,\frac { -3 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,1,\frac { -3 }{ 2 } \right) \)
Mid-point of BD is same let P be the mid-point of ACand BD
\(\frac { 5 }{ 2 } =\frac { 5+x_{ 2 } }{ 2 } ,1=\frac { -4+y_{ 2 } }{ 2 } \)
\(\frac { -3 }{ 2 } =\frac { 0+z_{ 2 } }{ 2 } \)
\(\Rightarrow x_{ 2 }=0,y_{ 2 }=6,z_{ 2 }=-3\)
D(0,6,-3)

\(\frac { dx }{ d\theta } =\theta cos\theta +sin\theta \)
\(\frac { dy }{ d\theta } =-\theta sin\theta +cos\theta \)
\(\frac { dy }{ dx } =\frac { cos\theta -\theta sin\theta }{ \theta cos\theta +sin\theta } \)
dy/dx at \(\theta\)= \(\pi/4\)
\(=\frac { cos\frac { \pi }{ 4 } -\frac { \pi }{ 4 } sin\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } } \)
\(=\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } }{ \frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } } \)
=\(\frac { 1-\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } +1 } \)
\(\Rightarrow\)\(\frac { dy }{ dx } =\frac { 4-\pi }{ 4+\pi } \)

\(P(E\cap F)=P(E)\times P(F)\)
\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
\(\Rightarrow P(E\cup F)=P(E)+P(F)-P(E)\times P(F)\)
\(\Rightarrow 0.85=P(E)+0.35-P(E)\times 0.35\)
\(\Rightarrow 0.85=P(E)(1-0.35)+0.35\)
\(\Rightarrow 0.85-0.35=P(E)(0.65)\)
\(\Rightarrow \frac { 0.52 }{ 0.65 } =P(E)\)
\(\Rightarrow P(E)=\frac { 50 }{ 65 } =\frac { 10 }{ 13 } \)
\(\therefore P(F)=1-\frac { 10 }{ 13 } =\frac { 3 }{ 13 } \)

\(\overset\rightarrow {BA}\)= Position vector of A- Position vector of B
\(\overset\rightarrow{BA}=(\overset\wedge i+2\overset\wedge j-3\overset\wedge k)-(-\overset\wedge i-2\overset\wedge j+\overset\wedge k)\)
\(\overset\rightarrow {BA}=2\overset\wedge i+4\overset\wedge j-\overset\wedge k \)
\(\left| \overset\rightarrow {BA}\right| =\sqrt{2^{2}+4^{2}+(-4)^{2}}\)
\(=\sqrt{4+16+16}\)
\(=\sqrt{36}=6 \)
Direction cosines of the vector \(\overset\rightarrow{BA}\) are \((\frac{2}{6},\frac{4}{6},-\frac{4}{6})\)
\(=(\frac{1}{3},\frac{2}{3},-\frac{2}{3})\)

\(\frac { dy }{ dx } ={ e }^{ 3x-4y }\)
\(\frac { dy }{ dx } ={ e }^{ 3x }.{ e }^{ -4y }\)
\(\Rightarrow \frac { dy }{ { e }^{ -4y } } ={ e }^{ 3x }dx\)
integrating both the sides
\(\int { { e }^{ 4y }dy } =\int { { e }^{ 3x }dx } \)
\(\Rightarrow \frac { 1 }{ 4 } { e }^{ 4y }=\frac { 1 }{ 3 } { e }^{ 3y }+C\)

Let x = cos 2\(\theta \)
\(=sin\left[ 2t{ an }^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \right] \)
\(=sin\left[ 2tan^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \right] \)
\(\left[ \because cos2\theta =1-2{ sin }^{ 2 }\theta \ and\ cos\ 2\theta =2{ cos }^{ 2 }\theta -1 \right] \)
\(=sin\left[ 2{ tan }^{ -1 }\left( tan\quad \theta \right) \right] \)
\(=sin(2\theta )=\sqrt { 1-{ cos }^{ 2 }2\theta } \)
\(=sin\quad 2\theta =\sqrt { 1-{ x }^{ 2 } } \)

\(\int { tan^{ -1 } } \sqrt { \frac { 1-cos2x }{ 1+cos2x } } dx=\int { tan^{ -1 } } \sqrt { \frac { 2sin^{ 2 }x }{ 2cos^{ 2 }x } } \)
\(=\int { { tan }^{ -1 }(tanx) } dx\)
\(=\int { x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } +C\)

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

We have, \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)
\(\Rightarrow\)  x² - 3x = - 2 and y² - 6y = - 9
\(\Rightarrow\)  x² - 3x + 2 = 0 and y² - 6y + 9 = 0
\(\Rightarrow\)  x² - 2x - x + 2 = 0 and y² - 3y - 3y + 9 = 0
\(\Rightarrow\)  x(x - 2) - 1(x - 2) = 0 and y(y - 3) - 3(y - 3) = 0
\(\Rightarrow\)  (x - 2)(x - 1) = 0 and (y - 3)(y - 3) = 0
\(\therefore\)   x = 1, 2 and y = 3, 3

x + y + z = 0
Alternate method :
\(\left[ \begin{matrix} xy & 4 \\ z+6 & x+y \end{matrix} \right] =\left[ \begin{matrix} 8 & w \\ 0 & 6 \end{matrix} \right] \)
By equating z + 6 = 0 and x + y = 6
⇒  z = - 6, x + y = 6
x + y + z = 6 - 6
⇒  x + y + z = 0

T₁ and T₃ are similar as their sides are proportional.