Let, the given vectors be \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=p \hat{i}+\hat{j}-2 \hat{k}\).
Also, given projection of \(\vec{a} \text { on } \vec{b}=\frac{1}{3}\)
\(\therefore\) Projection of \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
\(\begin{array}{ll} \Rightarrow & \frac{1}{3}=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(p \hat{i}+\hat{j}-2 \hat{k})}{\sqrt{p^2+(1)^2+(-2)^2}} \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{1}{3}=\frac{p+1-2}{\sqrt{p^2+1+4}}=\frac{p-1}{\sqrt{p^2+5}} \end{array}\)
\(\Rightarrow \quad \sqrt{p^2+5}=3 p-3\)
On squaring both sides, we get
p² + 5 = 9p² + 9 - 18p
\(\begin{aligned} \Rightarrow 8 p^2-18 p+4=0 \Rightarrow 4 p^2-9 p+2=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow 4 p^2-8 p-p+2=0 \Rightarrow(4 p-1)(p-2)=0 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad p=\frac{1}{4} \text { or } p=2 \end{aligned}\)

\(e^{6 x}(3 \cos 3 x-4 \sin 3 x)\)

The given system of equation is
x + 3y = 5 and
2x + 6y = 8.
The given system of equations can be written in the form of  A X = B, where 
\(A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 8 \end{array}\right] \).
Now, |A| = 1(6) - 3(2) = 6 - 6 = 0
∴ A  is a singular matrix.
\(\text { Now, }(\operatorname{adj} A)=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\)
\((a d j A) B=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \end{array}\right]=\left[\begin{array}{c} 30-24 \\ -10+8 \end{array}\right]=\left[\begin{array}{l} 6 \\ -2 \end{array}\right] \neq O\)
Thus, the solution of the given system of equations does not exist. Hence, the system of equation is inconsistent 

\(\left| \begin{matrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ 0 & \lambda & 3 \end{matrix} \right| =0\)
\(\Rightarrow \lambda=7\)

\(X=\begin{bmatrix} 2 & 4 \\ 5 & 0 \end{bmatrix}-\begin{bmatrix} 2 & -1 \\ 3 & -1 \end{bmatrix} \)
\(=\begin{bmatrix} 2-2 & 4+1 \\ 5-3 & 0+1 \end{bmatrix}=\begin{bmatrix} 0 & 5 \\ 2 & 1 \end{bmatrix}. \)

\(a *(b * c)=a *\left(\frac{b+c}{2}\right)=\frac{a+\frac{b+c}{2}}{2}=\frac{2 a+b+c}{4} \ldots(1)\)
and \((a * b) * c=\left(\frac{a+b}{2}\right) * c=\frac{\frac{a+b}{2}+c}{2}=\frac{a+b+2 c}{4} \ldots(2)\)
From (1) and (2), we get
a * (b* c) ≠ (a * b) * c
Hence, not associative

\( f(x)=27 x^3 and g(x)=x^{1 / 3}\\ \operatorname{gof}(x)=g[f(x)]=g\left[27 x^3\right]=\left[27 x^3\right]^{1 / 3} \\ =\left[(3 x)^3\right]^{1 / 3}=3 x \therefore \operatorname{gof}(x)=3 x \)

\(\lambda= \pm 1\)

Given, \(R= \big\{(x,y):x+2y\}\) on a set of natural numbers,Consider, x + 2y = 8, which can be rewritten as \(y= \frac{8-x}{2} \)
Now, as x, y \(\in\) N, therefore substitute values of x from natural numbers such that y \(\in\) N.
On putting x = 2, we get \(y= \frac{8-2}{2} =3 \) 
On putting x = 4,we get \(y= \frac{8-4}{2} =2 \) 
On putting x = 6, we get \(y= \frac{8-6}{2} =1 \) 
Thus, R = {(2, 3), (4, 2), (6, 1)}
[\(\therefore\)there is no other value of x, for which y \(\in\) N]
\(\therefore\) Domain of R = {2, 4, 6}, codomain of R = N and range of R = {3, 2, 1}.

Given inequations are
\(2 x+4 y \leq 8 \text { or } \ x+2 y \leq 4 \)
\(3 x+y \leq 6, x+y \leq 4, \ x \geq 0, \ y \geq 0 \)
Maximise Z = 2x + 5y on plotting the graph of the inequations we notice shaded portion as feasible solution

Possible points for maximum Z are A(2, 0),\(B\left(\frac{8}{5}, \frac{6}{5}\right)\) C (0,2)

Z is maximum at qo, 2), i.e. x = 0, y = 2, maximum value = 10

Taking L.H.S = \(\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 3 & 4+3p & 2+4p+3q \\ 4 & 7+4p & 2+7p+4q \end{matrix} \right| \)
Applying \(R_{ 2 }\rightarrow R_{ 2 }+R_{ 2 }+3R_{ 1 }\)
\(=\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 4 & 7+4p & 2+7p+4q \end{matrix} \right| \)
Applying \(R_{ 2 }\rightarrow R_{ 3 }+R_{ 3 }-4R_{ 1 }\)
\(=\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 3 & -2+3p \end{matrix} \right| \)
Applying \(R_{ 3 }\rightarrow R_{ 3 }-3R_{ 2 }\)
\(=\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 0 & 1 \end{matrix} \right| \)
Expanding along R₃
= 1 = R.H.S

\(\frac { dy }{ dx } ={ e }^{ x }\left[ 2cot2x+log\quad (sin2x) \right] \)

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

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.

\(\text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{B C} \times \overrightarrow{B A}| \)
\(\overrightarrow{B C}=(4 \hat{i}+5 \hat{j}-\hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k})=2 \hat{i}+6 \hat{j}-5 \hat{k} \)
\(\text { and } \overrightarrow{B A}=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k})=-\hat{i}+3 \hat{j}-\hat{k} \)
\(\overrightarrow{B C} \times \overrightarrow{B A}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & -5 \\ -1 & 3 & -1 \end{array}\right|=\hat{i}(9)-\hat{j}(-7)+\hat{k}(12) \)
\(=9 \hat{i}+7 \hat{j}+12 \hat{k} \)
\(\therefore \text { Area of triangle }=\frac{1}{2}|9 \hat{i}+7 \hat{j}+12 \hat{k}|\)
\(=\frac{1}{2} \sqrt{81+49+144}=\frac{1}{2} \sqrt{274} \text { sq units } \)

The given vectors \(3\overset { \wedge }{ i } -2\overset { \wedge }{ j } +9\overset { \wedge }{ k } \) and \(\overset { \wedge }{ i } -2p\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)  are parallel
if  \(\frac { 3 }{ 1 } =\frac { 2 }{ -2p } =\frac { 9 }{ 3 } 3=\frac { 1 }{ -p } =3\) 
if   \(p=-\frac { 1 }{ 3 } \) 

We know that two nonzero vectors are perpendicular if their scalar product is zero.
\(\text{Here} \ \vec{a}+\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k})=6 \hat{i}+2 \hat{j}-8 \hat{k} \)
\(\text{and} \ \vec{a}-\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k})=4 \hat{i}-4 \hat{j}+2 \hat{k} \)
\(\text{so} \ (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})=24-8-16=0 . \)
\(\text {Hence } \ \vec{a}+\vec{b} \text { and } \vec{a}-\vec{b} \) are perpendicular vectors.

\(Let\quad y={ e }^{ { sin }^{ -1 }x }\)
\(\frac { dy }{ dx } ={ e }^{ { sin }^{ -1 }x }\frac { d }{ dx } ({ sin }^{ -1 }x)\)
\(={ e }^{ { sin }^{ -1 }x }\left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\(=\frac { { e }^{ { sin }^{ -1 }x } }{ \sqrt { 1-{ x }^{ 2 } } } ,x\epsilon (-1,1)\)

Let x = a cos³ θ, y = a sin³ θ. Then
\( x^{\frac{2}{3}}+y^{\frac{2}{3}} =\left(a \cos ^3 \theta\right)^{\frac{2}{3}}+\left(a \sin ^3 \theta\right)^{\frac{2}{3}} \)
\( =a^{\frac{2}{3}}\left(\cos ^2 \theta+\left(\sin ^2 \theta\right)=a^{\frac{2}{3}}\right. \)
Hence, \(x=a \cos ^3 \theta, y=a \sin ^3 \theta\) is parametric equation of \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\)
\(Therefore\ \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}=-\tan \theta=-\sqrt[3]{\frac{y}{x}} \)
\(=\frac { -\sqrt [ 3 ]{ \frac { y }{ a } } }{ \sqrt [ 3 ]{ \frac { y }{ x } } } \)
\(=-\sqrt [ 3 ]{ \frac { y }{ x } } \)

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

Operating R₁ \(\rightarrow\) R₁ + R₂ + R₃, we get
\(\Delta =\left| \begin{matrix} x+y+z & z+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{matrix} \right| \) = 0
Since the elements of R₁ and R₃ are proportional, \(\Delta\) = 0.

We have
ho(gof) (x) = h(gof (x)) = h(g(f (x))) = h(g(2x))
= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4) \(\forall x\in N\).
Also, ((hog)of ) (x) = (hog) ( f (x)) = (hog) (2x) = h(g(2x))
= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4),\(\forall x\in N\)
This shows that ho(gof) = (hog) o f.
This result is true in general situation as well.

\(A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-\left|\begin{array}{ll} h & g \\ f & c \end{array}\right|\)

 We have, \(\left[\begin{array}{cc}1 & 2 \\ -2 & -b\end{array}\right]+\left[\begin{array}{ll}a & 4 \\ 3 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 0\end{array}\right]\)
= \(\left[\begin{array}{cc}a+1 & 6 \\ 1 & 2-b\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 0\end{array}\right]\)
\(\Rightarrow\)a + 1 = 5, 2 - b = 0
\(\Rightarrow\)a = 4, b = 2
\(\Rightarrow\)\(a^{2}+b^{2}=20\)

As \({ cos }^{ 2 }\alpha +{ cos }^{ 2 }\beta +{ cos }^{ 2 }\gamma =1\)
\(\Rightarrow \frac { 1+cos2\alpha }{ 2 } +\frac { 1+cos2\beta }{ 2 } +\frac { 1+cos2\gamma }{ 2 } =1\)
= cos 2α + cos 2β + cos 2γ = 1

As (gof) (x) = g[f(x)] = g(e^x)
= loge e^x = x
∴ (gof)' (x) = 1

Δ = 6(-1)- 1(1) = -7

Total injective mappings/functions
= ⁴ P₃ = 4! = 24.

(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.

Given, \(\vec{F}_1=6 \hat{i}+0 \hat{j} k N\)
\(\begin{aligned} \therefore \quad\left|\vec{F}_1\right|=\sqrt{6^2+0^2}=6 \mathrm{kN} \\ \end{aligned}\)
\(\vec{F}_2=-4 \hat{i}+4 \hat{j} k N\)
\(\therefore \quad\left|\vec{F}_2\right|=\sqrt{(-4)^2+4^2}=4 \sqrt{2} k N\)
\(\vec{F}_3=-3 \hat{i}-3 \hat{j} k N\)
\(\Rightarrow\left|\vec{F}_3\right|=\sqrt{(-3)^2+(-3)^2}=3 \sqrt{2} \mathrm{kN}\)
(i) Magnitude of force of team \(A=\left|\vec{F}_1\right|=6 \mathrm{kN}\)
(ii) Since, magnitude of force of team A is greater than other teams, therefore team A will win the game.
(iii) Resultant force,
\( \vec{F}=\vec{F}_1+\vec{F}_2+\vec{F}_3\)
\(=(6 \hat{i}+0 \hat{j})+(-4 \hat{i}+4 \hat{j})+(-3 \hat{i}-3 \hat{j}) \)
\(\vec{F}=(-\hat{i}+\hat{j}) k N \)
\(\Rightarrow |\vec{F}|=\sqrt{(-1)^2+(1)^2}=\sqrt{2} k N \)
Or
Resultant force \(\vec{F}=-\hat{i}+\hat{j}\)
Let \(\vec{F}\) makes an angle \(\theta\) with the X-axis,then its direction cosine along X-axis is cos \(\theta\).
\(\therefore \cos \theta=\frac{f_x}{\sqrt{f_x^2+f_y^2}} \text { (where } f_x \text { and } f_y\) are direction ratios along X-axis and Y-axis, respectively.)
\(\begin{aligned} \cos \theta & =\frac{-1}{\sqrt{(-1)^2+(1)^2}}=\frac{-1}{\sqrt{2}} \end{aligned}\)
\(\begin{aligned} \Rightarrow & & \theta & =\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \end{aligned}\)
\(\Rightarrow \theta =\pi-\frac{\pi}{4}=\frac{3 \pi}{4}\)
where '\(\theta\)' is the angle made by the resultant force with the '+' ve direction of the X-axis.

Let the price of a pen, bag and instrument box are ₹ x, ₹y and ₹z respectively, then
\(5 x+3 y+z=160\)
(for Gautam)
\(2 x+y+3 z=190\)
(for Vikram)
\(\text { and } x+2 y+4 z=250\)
(for Ankur)
(i) This system of equation can be written as AX=B, where
\(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\)
\(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 160 \\ 190 \\ 250 \end{array}\right]\)
(ii) |A|  =5(4-6)-3(8-3)+1(4-1)
\(=5(-2)-3(5)+1(3) =-10-15+3=-22 \neq 0\)
(iii) Cofactors of |A| are
\(A_{11}=(-1)^{1+1}\left|\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right|=(4-6)=-2\)
\(A_{12}=(-1)^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right|=-(8-3)=-5\)
\(A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=4-1=3\)
\(A_{21}=(-1)^{2+1}\left|\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right|=-(12-2)=-10 \)
\( A_{22}=(-1)^{2+2}\left|\begin{array}{ll} 5 & 1 \\ 1 & 4 \end{array}\right|=20-1=19\)
\(A_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=-(10-3)=-7\)
\( A_{31}=(-1)^{3+1}\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|=(9-1)=8 \)
\(A_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 1 \\ 2 & 3 \end{array}\right|=-(15-2)=-13\)
\(A_{33}=(-1)^{3+3}\left|\begin{array}{ll} 5 & 3 \\ 2 & 1 \end{array}\right|^2=(5-6)=-1 \)
\(\therefore \operatorname{adj} A =\left[\begin{array}{lll} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}\right]^\tau\)
\(=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^T\)
\( =\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]^2\)
\( \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=-\frac{1}{22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \)
Or 
(iii) P =A²-5 A
\(=\left[\begin{array}{ccc} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]-5\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right]\)
\( =\left[\begin{array}{ccc} 25+6+1 & 15+3+2 & 5+9+4 \\ 10+2+3 & 6+1+6 & 2+3+12 \\ 5+4+4 & 3+2+8 & 1+6+16 \end{array}\right] \)
\(=\left[\begin{array}{lll} 32 & 20 & 18 \\ 15 & 13 & 17 \\ 13 & 13 & 23 \end{array}\right]-\left[\begin{array}{ccc} 25 & 15 & 5 \\ 10 & 5 & 15 \\ 5 & 10 & 20 \end{array}\right] \)
\(=\left[\begin{array}{lll} 7 & 5 & 13 \\ 5 & 8 & 2 \\ 8 & 3 & 3 \end{array}\right]\)

We have, R = {(l₁,l₂) :l₁ is parallel to l₂}
(i) If l₁ is parallel to l₂, then l₂ is parallel to l_1.
So, if (l₁, l₂) \(\in R,\) then (l₂, l₁) \(\in R\)
\(\therefore\) R is symmetric.
(ii) If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃.
So, if \(\left(l_1, l_2\right) \in R,\left(l_2, l_3\right) \in R\), then (l₁,l₃)\(\in R\)
\(\therefore\) R is transitive.
(iii) Let equation of line parallel to y = 3x + 2 be y = mx + c, where m is the slope of line. Since, y = 3x + 2 and y = mx + c are parallel. Slope of (y =3x + 2) = Slope of(y = mx + c)
\(\Rightarrow\) 3 = m i.e. m = 3
Hence, the required line is
y = 3x + c, where c \(\in R\)
Or
We have, S = {(I₁, I₂) : l₁ is perpendicular to l₂}
For Symmetric If I₁, is perpendicular to I₂, then l₂ is perpendicular to l₁.
So, if (l₁,l₂) \(\in S\), then (l₂, l₁) \(\in S\)
\(\therefore\) S is symmetric.
For Transitive If I₁, is perpendicular to l₂, and l₂, is perpendicular to l₃, then I₁, is not perpendicular to l₃, it is parallel to I₃.
So, if \(\left(l_1, l_2\right) \in S,\left(l_2, l_3\right) \in S \text {, then }\left(l_1, l_3\right) \notin S \text {. }\)
\(\therefore\) S is not transitive.