\( y=x^{\log x}+(\log x)^x \)
Take log on both sides
\( \log y=\log x^{\log x}+\log (\log x)^x \)
\(\log y=\log x(\log x)+x \log (\log x) \)
\(\frac{1}{y} \cdot \frac{d y}{d x}=\log x\left(\frac{1}{x}\right)+\log (x) \cdot \frac{1}{x}+\log (\log x) \) \(+x \frac{1}{\log x} \cdot \frac{1}{x}\)
\( \frac{1}{y d x} =2 \log x\left(\frac{1}{x}\right) \div \log (\log x) \div \frac{1}{\log x}\)
\( \frac{d y}{d x} =y\left[\frac{2}{x} \log x+\log (\log x) \div \frac{1}{\log x}\right] \)
\(=\left[x^{\log x} \div(\log x)^z\right] \) \( {\left[\frac{2}{x} \log x \div \log (\log x) \div \frac{1}{\log x}\right] } \)

\(y=e^{\sqrt{x}}\)
Take \(u=\sqrt{x}=x^{1 / 2}\)
\(\frac{d u}{d x}=1 / 2 x^{-1 / 2}=\frac{1}{2 \sqrt{x}}\)
\(y=e^2\)
\(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=e^u\left(\frac{1}{2 \sqrt{x}}\right)\)
\(=e^{\sqrt{x}}\left(\frac{1}{2 \sqrt{x}}\right)\)

Given \(\overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k}\), \(\overrightarrow{b}=3\hat{i}+5\hat{j}-2\hat{k}\)
\(\overrightarrow{a}\times \overrightarrow{b}\)= \(\left| \begin{matrix} i & j & k \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{matrix} \right| \)
Expanding along R₁ we get,
= \(\hat{i}\)(-2-15)-\(\hat{j}\)(-4-9)+ \(\hat{k}\)(10-3)
\(\overrightarrow{a}\times \overrightarrow{b}\) = -17\(\hat{i}\)+13\(\hat{j}\)+7\(\hat{k}\)
\(\left| \vec { a } \times \vec { b } \right| =\sqrt { ({ -17) }^{ 2 }+{ 13 }^{ 2 }+{ 7 }^{ 2 } } \) = \(\sqrt { 289+169+49 } =\sqrt { 507 } \)

Given \(\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}\)
\(\overrightarrow{b}=3\hat{i}-4\hat{j}-2\hat{k}\)|
\(\overrightarrow{a} . \overrightarrow{b}=(\hat{i}-2\hat{j}+\hat{k}).(3\hat{i}-4\hat{j}-2\hat{k})\)
= 1(3) - 2(-4) + 1(-2)
= 3 + 8 - 2 = 9
\(\therefore \overrightarrow{a} . \overrightarrow{b}=9\)

We know that a unit vector along the direction of the vector \(\overrightarrow{a}\) is given by \({\overrightarrow{a}\over|\overrightarrow{a}|}\). 
So a unit vector along the direction of 5 \(\hat{i}\) - 3 \(\hat{j}\) + 4 \(\hat{k}\) is given by \({5\hat{i}-3\hat{j}+4\hat{k}\over |5\hat{i}-3\hat{j}+4\hat{k}|}={5\hat{i}-3\hat{j}+4\hat{k}\over\sqrt{5^2+3^2+4^2}}={5\hat{i}-3\hat{j}+4\hat{k}\over \sqrt{50}}\).

\(\begin{vmatrix} cos \theta & sin \theta \\ -sin \theta & cos \theta \end{vmatrix}\) = (cos\(\theta\)cos\(\theta\)) - (-sin\(\theta\)sin\(\theta\)) = cos² \(\theta\) + sin²\(\theta\) = 1.

\(A=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\)
\(A^2=A \times A=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{ll} 1+0 & a+a \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 a \\ 0 & 1 \end{array}\right]\)
\(A^3:=A^2 \times A=\left[\begin{array}{cc} 1 & 2 a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{cc} 1+0 & a+2 a \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{cc} 1 & 3 a \\ 0 & 1 \end{array}\right]\)
\(A^4=A^3 \times A=\left[\begin{array}{cc} 1 & 3 a \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1+0 & a+3 a \\ 0+0 & 0+1 \end{array}\right]\)
\(=\left[\begin{array}{cc} 1 & 4 a \\ 0 & 1 \end{array}\right]\)
\(\therefore A^4=\left[\begin{array}{cc} 1 & 4 a \\ 0 & 1 \end{array}\right]\)

By the definition of sum of matrices, we have
\(A+B+C=\left[\begin{array}{cc} \sin ^2 \theta+\cos ^2 \theta+0 & 1+0-1 \\ \cot ^2 \theta-\operatorname{cosec}^2 \theta-1 & 0+1+0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right]\)

Given y = sin^-1x
y' = \(\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } =(1-x^{ 2 })^{ -\frac { 1 }{ 2 } }\)
 y' '= \(-\frac { 1 }{ 2 } (1-{ x }^{ 2 })^{ \frac { 1 }{ 2 } -1 }\)(-2x)
= x(1 - x²)^-3/2 = \(\frac { x }{ (1-x^{ 2 })^{ \frac { 3 }{ 2 } } } \)

\( \frac{x^2}{a^2}+\frac{y^2}{b^2} =1\)
\(\frac{2 x}{a^2}+\frac{2 y}{b^2} \cdot \frac{d y}{d x} =0 \)
\(\frac{2 y}{b^2} \frac{d y}{d x} =\frac{-2 x}{a^2}\)
\(\frac{d y}{d x}=\frac{-2 x}{a^2} \times \frac{b^2}{2 y} =\frac{-b^2 x}{a^2 y}\)

Take u = x^x,v = x log x
log u = x log x
\({1\over u}{du \over dx}=x.{1\over x}+1.log \ x=1+log \ x\)
\({du \over dx}=u({1}+log \ x)=x^x(1+log \ x)\)
\(\frac{d v}{d x}=1+\log x\)
\(\frac{d\left(x^x\right)}{d(x \log x)}=\frac{d u}{d V}=\frac{\frac{d u}{d x}}{\frac{d V}{d x}}=x^X\)

LHS =\(\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})+\overrightarrow{b}\times (\overrightarrow{c}+\overrightarrow{a})+\overrightarrow{c}\times (\overrightarrow{a}+\overrightarrow{b})\)
=\(\vec { a } \times \vec { b } +\vec { a } \times \vec { c } +\vec { b } \times \vec { c } +\vec { b } \times \vec { a } +\vec { c } \times \vec { a } +\vec { c } \times \vec { b } \) (By associative property)   
\(\left[ \therefore \vec { b } \times \vec { a } =-\vec { a } \times \vec { b } \vec { c } \times \vec { a } =-\vec { a } \times \vec { c } \vec { c } \times \vec { b } =-\vec { b } \times \vec { c } \right] \)
= \(\vec { a } \times \vec { b } +\vec { a } \times \vec { c } -\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { c } \times \vec { a } -\vec { c } \times \vec { b } \) = \(\vec{0}\) = RHS
Hence proved.

Let   \(2\hat{i}+3\hat{j}-6\hat{k}\) and \(6\hat{i}-3\hat{j}+2\hat{k}\)
Let \(\theta \) be the angle between the given vectors.
\(\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+3\hat{j}-6\hat{k}).(6\hat{i}-3\hat{j}+2\hat{k})\)
= 12 - 9 - 12 = -9
\(|\overrightarrow{a}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7\)
and \(|\overrightarrow{b}|=\sqrt{6^2+(-3)^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7\)
\(\therefore cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{a}|.|\overrightarrow{b}|}={-9\over 7(7)}={-9\over 49}\)
\(\Rightarrow \theta =cos^{-1}({-9\over 49})\)

|A| = [3(−1)(6) + 4(2)(5) + 1(0)(−2)] −[5(−1)(1) + (−2)(2)3 + 6(0)(4)]
= [−18 + 40 + 0]−[−5 −12 + 0] = 22 + 17 = 39.

A -B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 & 3 \\ -3 & 2 & 1 \\1 & 1 & 1 \end{bmatrix}\)
(A - B)^T =\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\) ..(1)
A^T - B^T =\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\).....(2)
From (1) and (2), (A - B)^T= A^T - B^T.

A² = AA =\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)=\(\begin{bmatrix} C_{11} &C_{12} &C_{13} \\ C_{21} & C_{22} &C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}\)

We have x = a(t - sin t), y = a(1 - cos t).
Now \({dx\over dt}=a(1-cos \ t);{dy\over dt}=a \ sin \ t\)
Therefore, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \sin t}{a(1-\cos t)}=\frac{\sin t}{(1-\cos t)}\)

\(y=\tan (\cos x)\)
Take \(u=\cos x \Rightarrow \frac{d u}{d x}=-\sin x\)
\(y =\tan u\)
\(\frac{d y}{d x} =\frac{d y}{d u} \times \frac{d u}{d x}=\sec ^2 u(-\sin x\)
\(=\sec ^2(\cos x)(-\sin x)\)
\(=-\sin x \sec ^2(\cos x)\)

Given \(\vec{a}, \vec{b}\) and \(\vec{c}\)are unit vectors.
\(\Rightarrow \left| \vec { a } \right| =\left| \vec { b } \right| =\left| \vec { c } \right| =1\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0, and angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0 \(\Rightarrow \vec{a}\) is \({ \bot }^{ r }\) to both\(\vec{b}\) and \(\vec{c}\)
\(\vec{a}\) is \({ \bot }^{ r }\) to \(\vec { b } \times \vec { c } \Rightarrow \vec { a } =\lambda (\vec { b } \times \vec { c } )\) for some scalar \(\lambda\).
\(\therefore { \left| \vec { a } \right| }^{ 2 }={ \lambda }^{ 2 }{ \left| \vec { b } \times \vec { c } \right| }^{ 2 }\)
\(\Rightarrow 1={ \lambda }^{ 2 }[{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }-{ (\vec { b } \times \vec { c } ) }^{ 2 }]\left[ \therefore \left| \vec { a } \right| =1{ and\left| \vec { a } \times \vec { b } \right| }^{ 2 } \right] ={ \left| \vec { a } \times \vec { b } \right| }^{ 2 }={ \left| \vec { a } \right| }^{ 2 }{ \left| \vec { b } \right| }^{ 2 }-{ (\vec { a } .\vec { b } ) }^{ 2 }]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[(1)(1)-{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }{ cos }^{ 2 }\frac { \pi }{ 3 } ]\) [\(\therefore\)angle between \(\vec{a}\)and\(\vec{b}\) is\(\frac{\pi}{3}\) ]
\(\Rightarrow 1={ \lambda }^{ 2 }[1-{ cos }^{ 2 }\frac { \pi }{ 3 } ][\because \left| \vec { b } \right| \left| \vec { c } \right| =1]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1-\frac { 1 }{ 4 } ]\Rightarrow 1={ \lambda }^{ 2 }\left( \frac { 3 }{ 4 } \right) \)
\(\Rightarrow 1={ \lambda }^{ 2 }\frac { 4 }{ 3 } \Rightarrow 1={ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \)
Substituting \(\)\({ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \) in (1) we get,
\(\vec { a } =\pm \frac { 2 }{ \sqrt { 3 } } (\vec { b } \times \vec { c } )\)

Let O be the origin and let \(\overrightarrow{OA}\), \(​​\overrightarrow{OB}\), and\(​​\overrightarrow{OC}\) be the vectors 2\(\hat{i}\) + 3\(\hat{j}\) − 5\(\hat{k}\), 3\(\hat{i}\) + \(\hat{j}\) − 2\(\hat{k}\) and, 6\(\hat{i}\) − 5\(\hat{j}\) + 7\(\hat{k}\) respectively. Then
\(\overrightarrow{AB}=\hat{i}-2\hat{j}+3\hat{k} \ and \ \overrightarrow{AC}=4\hat{i}-8\hat{j}+12\hat{k}\).
Thus \(\overrightarrow{AC}=4\overrightarrow{AB}\) and hence \(​​\overrightarrow{AB}\) and\(​​\overrightarrow{AC}\) are parallel. They have a common point namely A. Thus, the three points are collinear.
Alternative method
Let O be the point of reference.
Let \(\overrightarrow {OA} = 2\hat i+3\hat j-5\hat k, \) \(\overrightarrow {OB} = 3 \hat j+\hat j-2\hat k\ and\ \overrightarrow {OC} = 6\hat i-5\hat j+7\hat k \)
\(\overrightarrow {AB} = \hat i- 2\hat j+3\hat k; \overrightarrow {BC} = 3\hat i-6\hat j+9\hat k; \overrightarrow {CA} = -4\hat i+8\hat j-12 \hat k\\ |\overrightarrow {AB}| = \sqrt 14; |\overrightarrow {BC}|= \sqrt 126 = 3 \sqrt 14; |\overrightarrow {CA}|= \sqrt 224 = 4 \sqrt 4\)
Thus, AC = AB + BC.
Hence A, B, C are lying on the same line. That is, they are collinear.

Let the position vector of the vertices of the \(\triangle\) ABC be \(\overrightarrow{a},\overrightarrow{b}\) and \(\overrightarrow{c}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}.\)
Since G is the centroid of \(\triangle\) ABC, we have
\(\Rightarrow \overrightarrow{OG}={\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\over3}\) \(\Rightarrow 3\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)
Now,LHS \(=\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\)
\(=\overrightarrow{OA}-\overrightarrow{OG}+\overrightarrow{OB}-\overrightarrow{OG}+\overrightarrow{OC}-\overrightarrow{OG}=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\)\(-3\overrightarrow{OG}\)
\(=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})-(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}=RHS\)
Hence proved.

\(\begin{vmatrix} log_364 &log_43 \\ log_38 & log_49 \end{vmatrix}\times \begin{vmatrix} log_23 & log_83 \\ log_34 & log_34 \end{vmatrix}\) = \(\left| \begin{matrix} { log }_{ 3 }64.{ { log }_{ 2 }3+log }_{ 4 }3.{ log }_{ 3 }4 & { log }_{ 3 }64.{ log }_{ 8 }3+{ log }_{ 4 }3.{ log }_{ 3 }4 \\ { log }_{ 3 }8.{ log }_{ 2 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 & { log }_{ 3 }8.{ log }_{ 8 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 \end{matrix} \right| \)
= \(\left| \begin{matrix} { log }_{ 2 }64+1 & { log }_{ 8 }64+1 \\ { log }_{ 2 }8+{ log }_{ 3 }9 & 1+{ log }_{ 3 }9 \end{matrix} \right| [\therefore { log }_{ y }x.{ log }_{ x }y=1]\)
= \(\left| \begin{matrix} { log }_{ 2 }{ 2 }^{ 6 }+1 & { log }_{ 8 }{ 8 }^{ 2 }+1 \\ { log }_{ 2 }{ 2 }^{ 3 }+{ log }_{ 3 }{ 3 }^{ 2 } & 1+{ log }_{ 3 }{ 3 }^{ 2 } \end{matrix} \right| [\therefore { log }_{ x }x=1]\)
= \(\left| \begin{matrix} 6+1 & 2+1 \\ 3+2 & 1+2 \end{matrix} \right| =\left| \begin{matrix} 7 & 3 \\ 5 & 3 \end{matrix} \right| \) = 21 - 15 = 6

LHS = \(\begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}^2=\begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}\times \begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}\)
\(=\begin{vmatrix} 0+c^2+b^2 & 0+0+ab &0+ac+0 \\ 0+0+ab & c^2+0+a^2 &bc+0+0 \\ 0+ac+0 & bc+0+0 & b^2+a^2+0 \end{vmatrix}\)
\(=\begin{vmatrix} c^2+b^2 & ab & ac \\ ab & c^2+a^2 & bc \\ ab & bc & b^2+a^2 \end{vmatrix}\) = \(\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ a b & c^{2}+a^{2} & b c \\ a c & b c & a^{2}+b^{2} \end{array}\right|\) = RHS.

\(|A|=\left|\begin{array}{lll} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b \end{array}\right|\)
Put a = 0
\(|A|=\left|\begin{array}{ccc} b+c & -c & -b \\ b-c & c & b \\ c-b & c & b \end{array}\right|=0 \quad\left(C_2 \cong C_3\right)\)
\(\therefore\) (a - 0) is a factor. (i.c) a is a factor.
Since |A| is in cyclic symmetric form in a, b, c and hence b, c also factors.
The degree of the product of the factor a, b, c is 3. The delerminant is a culbic polynomial.
The olher faclor must be a constant k.
\(\left|\begin{array}{ccc} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c+a & a+b \end{array}\right|=k(a b c)\)
Put a = 1, b = 1, c = 1
\(\left|\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right|=k(1)(1)(1)\)
8 = k
\((1) \Rightarrow \quad|A|=8 a b c\)
Hence proved.

\(f(x)=|x-1|+|x-3|+\sin x\)
\(\text { Since } \sin x \text { is differentiable everywhere }\)

At x = 1 and x = 3. The graph admit cups.
\(\therefore\) The derivative is not exist.
\(\therefore\) The number of points in R is 2.

\(f^{\prime}\left(2^{-}\right)=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x+1-(2+1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{-}} \frac{x+1-3}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x-2}{x-2}=1\)
\(f^{\prime}\left(2^{+}\right)=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{(2 x-1)-(4-1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2 x-1-3}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{2 x-4}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{(x-2)}=2\)
\(f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)\)
\(\therefore f^{\prime}(2) \text { does not exist. }\)

\(\text { Given } f(x)=x+2\)
\(f^{\prime}(x) =1 \)
\(f^{\prime}(f(x)) =f^{\prime}(x+2)=1 \)

\(y =(a x-5) e^{3 x} \)
\(\frac{d y}{d x} =(a x-5) e^{3 x}(3)+e^{3 x}(a) \)
\(-13 =(-5)(3)+a \quad(\text { At } x=0) \)
\(-13 =-15+a \)
\(a =-13+15=2 \)
\(\therefore a =2 \)

\(y =\cos \left(\sin x^{2}\right) \)
\(\frac{d y}{d x} =-\sin \left(\sin x^{2}\right) \cos \left(x^{2}\right)(2 x) \)
\(\text { At } x =\sqrt{\pi / 2}, \frac{d y}{d x}=-\sin \sin \left(\frac{\pi}{z}\right) \cos (\pi / 2) 2(\sqrt{\pi / 2}) \)
\(=(\sin 1)(0) 2\left(\frac{\sqrt{\pi}}{2}\right)=0 \quad[\because \cos \pi / 2=0] \)

\(u =\frac{2}{3} x^{3}+5 \)
\(\frac{d u}{d x} =\frac{2}{3}\left(3 x^{2}\right)=2 x^{2} \)
\(y =\frac{1}{4} u^{4} \)
\(\frac{d y}{d x} =\frac{1}{4}\left(4 u^{3}\right) \frac{d u}{d x}=u^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2 x^{3}+15}{3}\right)^{3} \times 2 x^{2}=\frac{\left(2 x^{3}+15\right)^{3}}{27}\left(2 x^{2}\right) \)

\(\overrightarrow{O A}=10 \hat{i}+3 \hat{j}, \overrightarrow{O B}=12 \hat{i}-5 \hat{j}, \overrightarrow{O C}=a \hat{i}+11 \hat{j} \)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=2 \hat{i}-8 \hat{j} \)
\(\overrightarrow{B C} =\overrightarrow{O C}-\overrightarrow{O B} \)
\(=(a-12) \hat{i}+16 \hat{j} \)
Condition:
\(\overrightarrow{B C} =-2 \overrightarrow{A B} \)
\((a-12) \hat{i}+16 \hat{j} =2(2 \hat{i}-8 \hat{j}) \)
\((a-12) \hat{i}+16 \hat{j} =-4 \hat{i}+16 \hat{j} \)
\(a-12 =-4 \)
\(a =-4+12=8 \)
\(a =8 \)

\(\text { W.K.T }|\vec{a} \cdot \vec{b}|^{2}+|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\)
\(|\vec{a} \times \vec{b}|^{2} =13^{2} \cdot 5^{2}-|\vec{a} \cdot \vec{b}|^{2} \)
\(=(169)(25)-3600=4225-3600 =625 \)
\(|\vec{a} \times \vec{b}| =25 \)

\(\text { W.K.T }|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=2\left[|\vec{a}|^{2}+|\vec{b}|^{2}\right]\)
\(60^{2}+40^{2} =2\left(|\vec{a}|^{2}+46^{2}\right) \)
\(3600+1600 =2\left(|\vec{a}|^{2}+2116\right) \)
\(\frac{5200}{2} =|\vec{a}|^{2}+2116 \)
\(2600-2116 =|\vec{a}|^{2} \)
\(|\vec{a}|^{2} =484\)
\(|\vec{a}| =22 \)

\(2 \vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}+\vec{a}=\vec{b}+\vec{c} \Rightarrow \vec{a}-\vec{b}=\vec{c}-\vec{a} \)
\(\overrightarrow{O A}-\overrightarrow{O B}=\overrightarrow{O C}-\overrightarrow{O A} \Rightarrow \overrightarrow{B A}=\overrightarrow{A C} \)
\(\Rightarrow \vec{a}, \vec{b}, \vec{c} \text { are collinear }\)

\(\overrightarrow{A B}=-\overrightarrow{C D}\)
\(\overrightarrow{A D}=\overrightarrow{B C}=-\overrightarrow{C B}\)
\(\therefore \overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}=-\overrightarrow{C D}-\overrightarrow{C B}+\overrightarrow{C B}+\overrightarrow{C D}=\overrightarrow{0}\)

\(\overrightarrow{B A}=3 \hat{i}+2 \hat{j}+\hat{k} \)
\(\overrightarrow{O A}-\overrightarrow{O B}=3 \hat{i}+2 \hat{j}+\hat{k} \)
\(\overrightarrow{O A}=3 \hat{i}+2 \hat{j}+\hat{k}+\overrightarrow{O B}=3 \hat{i}+2 \hat{j}+\hat{k}+\hat{i}+3 \hat{j}-\hat{k} \)
\(=4 \hat{i}+5 \hat{j} \)

\(\underbrace{\overrightarrow{A B}}+ \overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A C}+\overrightarrow{C D}}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A D}+\overrightarrow{D A}} \)
\(=\overrightarrow{A A}=\overrightarrow{0} . \)

\(A \& B \text { are symmetric }\)
\(A^{T}=A \text {, and } B^{T}=B\)
\(\text { 1) }(A+B)^{T}=B^{T}+A^{T}=B+A=A+B \Rightarrow A+B\text { is symmetric }\)
\(\text { 2) }(A B)^{T}=B^{T} A^{T}=B A \neq A B\)
\(\therefore A B \text { is not symmetric }\)
\(\text { 3) } A B=A^{T} B^{T}=(B A)^{T}\)
\(\text { 4) } A^{T} B=A B=A B^{T}\)

\(\left|\begin{array}{ccc} a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b \end{array}\right| =0 \)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{lll} a & 2 b & 2 c \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{2} \rightarrow 2 R_{2} \)
\(\frac{1}{2}\left|\begin{array}{ccc} a-6 & 0 & 0 \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{1} \rightarrow R_{1}-R_{2} \)
\(\Rightarrow \frac{1}{2}\left[(a-6)\left(2 b^{2}-2 a c\right)\right] =0 \)
\((a-6)\left(2 b^{2}-2 a c\right) =0 \)
\(a=6,2 b^{2} =2 a c \)
\(b^{2} =a c \quad \therefore b^{3}=a b c \)

\(A A^{T}=9 I \)
\({\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]} \)
\({\left[\begin{array}{ccc} 1+4+4 & 2+2-4 & a+4+2 b \\ 2+2-4 & 4+1+4 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^{2}+4+b^{2} \end{array}\right]} \)
\(=\left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \)
\(a+4+2 b =0 \)
\(a+2 b =-4 \)
\(2 a-2 b+2 =0 \)
\(2 a-2 b =-2 \)
\(\text { Solve (1) } \&(2) \text { we get } a=-2 b=-1\)
\(\therefore(a, b)=(-2,1)\)