Given n = 4
Getting  \(p=\frac { 1 }{ 2 } \ and\ q=\frac { 1 }{ 2 } \)

Mean = \(\sum { xp(x)=\frac { 32 }{ 16 } =2 } \)
\(\\ Variance=\sum { { x }^{ 2 }p(x)-\{ \sum { P(x){ \} }^{ 2 } } } \)
\(=\frac { 80 }{ 16 } -{ (2) }^{ 2 }=5-4=1\)

3e^x tan y dx + (2 - e^x)sec² y dy = 0
\(\Rightarrow\) (2 - e^x)sec² y dy = - 3e^x tan y dx
\(\frac { \sec ^{ 2 }{ y } }{ \tan { y } } dy=\frac { -3{ e }^{ x } }{ 2-{ e }^{ x } } dx\)
On integrating both sides
\(\int { \frac { \sec ^{ 2 }{ ydy } }{ \tan { y } } } =\int { \frac { -3{ e }^{ x }dx }{ 2-{ e }^{ x } } } \)
\(\Rightarrow \log { \left| \tan { y } \right| } =3\log { \left| 2-{ e }^{ x } \right| } +\log { C } \)
\(\Rightarrow \log { \left| \tan { y } \right| } =\log { \left| C\left( 2-{ e }^{ x } \right) ^{ 3 } \right| } \)
\(\Rightarrow \tan { y } =C\left( 2-{ e }^{ x } \right) ^{ 3 }\)
Putting x = 0, \(y=\frac { \pi }{ 4 } \), we get
\(\Rightarrow \tan { \frac { \pi }{ 4 } } =C\left( 2-{ e }^{ 0 } \right) ^{ 3 }\)
\(\Rightarrow C=1\)
Therefore particular solution is
\(\tan { y } =\left( 2-{ e }^{ x } \right) ^{ 3 }\)

\(I=\int { \frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } } } dx\)
\(\int { \frac { \frac { 5 }{ 2 } (2x+4)-7 }{ \sqrt { { x }^{ 2 }+4x+10 } } } dx\)
\(=\frac { 5 }{ 2 } \int { \frac { (2x+4)-7 }{ \sqrt { { x }^{ 2 }+4x+10 } } } dx-7\int { \frac { 1 }{ \sqrt { (x+2)^{ 2 }+(\sqrt { 6 } )^{ 2 } } } } dx\)
\(=5.\sqrt { { x }^{ 2 }+4x+10a } -7log\left| x+2+\sqrt { { x }^{ 2 }+4x+10 } \right| +C\)

We have:
\(f(x)=x^{ 2 }+2\) and \(g(x)=\frac { x }{ x-1 } \)
\(\therefore \) \(fog(x)=f(g(x))=(g(x))^{ 2 }+2\)
\(=\left( \frac { x }{ x-1 } \right) ^{ 2 }+2=\frac { x^{ 2 }+2(x-1)^{ 2 } }{ (x-1)^{ 2 } } \)
\(=\frac { x^{ 2 }+2(x^{ 2 }-2x+1) }{ (x-1)^{ 2 } } =\frac { 3x^{ 2 }-4x+2 }{ (x-1)^{ 2 } } \)
Hence, fog (2)\(=\frac { 3(4)-4(2)+2 }{ (2-1)^{ 2 } } =\frac { 12-8+2 }{ (1)^{ 2 } } =\frac { 6 }{ 1 } =6\)
And \(gof(x)=g(f(x))=\frac { f(x) }{ f(x)-1 } \)
\(=\frac { x^{ 2 }+2 }{ (x^{ 2 }+2)-1 } =\frac { x^{ 2 }+2 }{ x^{ 2 }+1 } \)
Hence, \(gof(-3)=\frac { (-3)^{ 2 }+2 }{ (-3)^{ 2 }+1 } =\frac { 9+2 }{ 9+1 } =\frac { 11 }{ 10 } \) .

\(Here\quad y={ tan }^{ -1 }\frac { 3x-{ x }^{ 3 } }{ 1-3{ x }^{ 2 } } \)
\(={ tan }^{ -1 }\frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \)
\(={ tan }^{ -1 }(tan3\theta )=3\theta =3{ tan }^{ -1 }x\)
\(Hence,\quad \frac { dy }{ dx } =\frac { 3 }{ 1+{ x }^{ 2 } } \)

The given equation are:
5x-y+4z=5
2x+3y+5z=2
5x-2y+6z=-1
Here A= \(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
Now |A|=\(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
=5(18+10)+1.(12-25)+4(-4-15)
=140-13-76=\(51\neq0\)
Hence, the given system of equations is consistent.

\(We\quad have:\quad x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]\)
\(\Rightarrow \left[ \begin{matrix} 2x \\ 3x \end{matrix} \right] +\left[ \begin{matrix} -y \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]\)
\(\Rightarrow \left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] .\)
Equating corresponding elements:
2x - y = 10...(1)
and 3x + y = 5....(2)
Adding (1) and (2), 5x = 15 \(\Rightarrow \) x = 3.
Putting in (1), 2, (3) - y = 10 \(\Rightarrow\) y = 6-10 = -4.
Hence,  x = 3 and y = -4.

First split \(\frac{2 \pi}{3} \text { as } \frac{(3 \pi-\pi)}{3} \text { or } \pi-\frac{\pi}{3}\)
After substuting in the given we get,
\(\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)=\frac{\pi}{3}\)

Let sin^-1 x = \(\theta\) then  sin^-1 x = \(\theta\). we have
sin^-1 \((2x\sqrt { 1-{ x }^{ 2 } } )\) = sin^-1\(2\sin { \theta } \sqrt { 1-{ sin }^{ 2 } } \)
= sin^–1 (2sinθ cosθ) = sin^–1 (sin2θ) = 2θ = 2 sin^–1 x

The given function is
y = a cos x + b sin x ... (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
\(\frac{d y}{d x}=-a \sin x+b \cos x \)
\(\frac{d^{2} y}{d x^{2}}=-a \cos x-b \sin x \)
Substituting the values of \(\frac{d^{2} y}{d x^{2}}\)and y in the given differential equation, we get
L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S..
Therefore, the given function is a solution of the given differential equation.

Given curve, \(y^{2}=4 x\)   ..(i)
is a parabola having vertex (0, 0) and open right sides and the curve x² = 4y  ...(ii)
is a parabola having vertex (0, 0) and open upward.

Also, draw the square bounded by the lines x = 0,x = 4
y = 4 and y = 0.To prove, Area A₁ = Area A₂ = Area A ₃
Now, the point of intersection of curves y² = 4x and
\(x^{2}=4 y \text { is given by }\left(\frac{y^{2}}{4}\right)^{2}=4 y\) 
\(\Rightarrow y^{4}=64 y \Rightarrow y\left(y^{3}-64\right)=0\) 
\(\Rightarrow y=0, y^{3}=64 \Rightarrow y=0, y=4\) 
When \(y=0, \text { then } 0^{2}=4 x \Rightarrow x=0\) 
When \(y=4, \text { then } 4^{2}=4 x \Rightarrow x=4\) 
So, the points of intersection are 0(0, 0) and B(4, 4). 
Now, the area of the region bounded by curves y² = 4x and x² = 4y is
\(A_{2}=\int_{0}^{4}[y\{\text { parabola }(\mathrm{i})\}-y\{\text { parabola }(\mathrm{ii})\}] d x\) 
\(=\int_{0}^{4}\left(\sqrt{4 x}-\frac{x^{2}}{4}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x\) 
\( =\left[2 \times \frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{12}\right]_{0}^{4}=\left[\frac{4}{3}(4)^{3 / 2}-\frac{4^{3}}{12}-(0-0)\right. \)
\(=\left[\frac{4}{3}(2)^{3}-\frac{64}{12}\right] \)
\(=\frac{32}{3}-\frac{16}{3}=\frac{16}{3} \mathrm{sq} \text { units } \) 
Now, the area of the region bounded by the curves 
\(x^{2}=4 y, x=4 \text { and } X \text { -axis is } A_{3}=\int_{0}^{4} y d x=\int_{0}^{4}\left(\frac{x^{2}}{4}\right) d x\) 
\( =\left[\frac{x^{3}}{12}\right]_{0}^{4}=\frac{(4)^{3}}{12}-0 \)
\(=\frac{16}{3} \text { sq units } \) 
Similarly, the area of the region bounded by the curve 
\(y^{2}=4 x, Y \text { -axis and } y=4 \text { is }\) 
\(A_{1}=\int_{0}^{4} x d y=\int_{0}^{4} \frac{y^{2}}{4} d y=\left[\frac{y^{3}}{12}\right]_{0}^{4}=\frac{1}{12}\left[4^{3}-0\right]\) 
\(=\frac{1}{12}(64)=\frac{16}{3} \text { sq units }\) 
Here, we see that
\(A_{1}=A_{2}=A_{3}=\frac{16}{3} \mathrm{sq} \text { units }\) 
Hence proved.

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

Let C₁, C₂, C₃, C₄ be the events that the lost card is of heart, spades, diamond or club respectively.
Obviously P(C₁) = P(C₂) = P(C₃) = P(C₄)
\(={13\over 52}={1\over 4}\)
Let S be the event of drawing two cards of heart from the remaining 51 cards.We wish to find \(P\left(C_1\over S\right)\)
Now \(P\left(C_1\over S\right)\) is the probability of drawing two heart cards from 51 cards given that one heart card is lost
\(={^{12}C_2\over ^{51}C_2}={12\times11\over 1\times2}\times{1\times2\over 51\times50}={22\over 425}\)
\(P\left( S\over C_3\right)=P\left( S\over C_3\right)=P\left( S\over C_4\right)={^{13}C_2\over ^{51}C_2}\)
\(={13\times12\over 1\times2}\times{1\times2\over 51\times50}={26\over 425}\)
By Bayes' Theorem
\(P\left(C_1\over S\right)={P(C_1).P\left(S\over C_1\right)\over \sum P(C_1).P\left(S\over C_1\right)}\)
\(={{1\over4}\times{22\over 425}\over{1\over 4}\times{22\over 425}+{1\over 4}\times{26\over 425}+{1\over4}\times{26\over 425}}\)
\(={22\over 22+26+26+26}\)
\(={11\over 50}\)

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

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

y = e^ax
\(\frac { dy }{ dx } \)= e^ax(-sinbx)b+cosbx \(\times \) e^ax\(\times \) a
\(\frac { dy }{ dx } -be^{ ax }sinbx+ay\Rightarrow \left[ \frac { -1 }{ b } \left( \frac { dy }{ dx } -ay \right) \right] =e^{ ax }sinbx\)
\(\frac { dy }{ dx } =-b\left[ e^{ a^{ x }cosb }\times a \right] +\frac { ady }{ dx } \)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } -b\left[ by\frac { -a }{ -b } \left( \frac { dy }{ dx } -ay \right) \right] +\frac { ady }{ dx } \) from (1)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } -b^{ 2 }y+a\frac { dy }{ dx } -a^{ 2 }y+a\frac { dy }{ dx } \)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } +\left( a^{ 2 }+b^{ 2 } \right) y-2a\frac { dy }{ dx } =0\)
Hence proved

Let x, y and z be the values of adaptable new techniques, careful and alert in difficult situations and keeping calm in tense situations respectively.
Then the system of equations is
2x + 4y + 3z = 29,000 
5x + 2y + 3z = 30,500 
x + y + z = 9,500
Matrix equation is
\(\left[ \begin{matrix} 2 & 4 & 3 \\ 5 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 29000 \\ 30500 \\ 9500 \end{matrix} \right] \)
i.e, AX = B
\(|A|=\left| \begin{matrix} 2 & 4 & 3 \\ 5 & 2 & 3 \\ 1 & 1 & 1 \end{matrix} \right| \)
= 2(2-3)-495-3)+3(5-2)
= -1 \(\ne\) 0
\(\therefore\) A^-1 exists.
adj A = \({ \left[ \begin{matrix} -1 & -2 & 3 \\ -2 & -1 & 2 \\ 3 & 2 & -16 \end{matrix} \right] }^{ T }\)
\(=\left[ \begin{matrix} -1 & -1 & 6 \\ -2 & -1 & 9 \\ 3 & 2 & -16 \end{matrix} \right] \)
\(X={ A }^{ -1 }B\)
\(=-1\left[ \begin{matrix} -29,000-30,500+57,000 \\ -58,000-30,500+85,500 \\ 87,00+61,000-15,2,000 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2,500 \\ 3,000 \\ 4,000 \end{matrix} \right] \)
x = 2,500, y = 3,000, z = 4,000
i.e., Rs. 2,500 for adaptable n.ew technique Rs. 3,000 for careful and alert in difficult situations and Rs. 4,000 for keeping calm in tense situations,
Value: Award should'be given to a person who truly deserves it.

\(I=\int { \frac { { x }^{ 2 }+1 }{ (x-1)^{ 2 }(x+3) } } dx\)
Let  \(\int { \frac { { x }^{ 2 }+1 }{ (x-1)^{ 2 }(x+3) } } =\frac { A }{ x-1 } +\frac { B }{ (x-1)^{ 2 } } +\frac { C }{ x+3 } \)
\(\Rightarrow { x }^{ 2 }+1=A(x-1)(x+3)+B(x+3)+C(x-1)^{ 2 }\)
Getting the values \(A=\frac { 3 }{ 8 } , B=\frac { 1 }{ 2 } , C=\frac { 5 }{ 8 } \)
\(\therefore \ I=\int { \frac { { x }^{ 2 }+1 }{ (x-1)^{ 2 }(x+3) } } dx\)
\(=\frac { 3 }{ 8 } \int { \frac { dx }{ x-1 } +\frac { 5 }{ 8 } \int { \frac { dx }{ x+3 } +\frac { 1 }{ 2 } \int { \frac { 1 }{ (x-1)^{ 2 } } } } } dx\)
\(I=\frac { 3 }{ 8 } log\left[ x-1 \right] +\frac { 5 }{ 8 } log\left[ x+3 \right] -\frac { 1 }{ 2 } \frac { 1 }{ (x-1) } +C\)

Let a, b \(\in A\), a " b = a + b + ab
Commutatively : for all a, b \(\in A\)
.a * b = a + b + ab
= b + a + ba
= b * a
Associatively: Let a, b, c \(\in A\)
(a * b) " c = (a + b + ab) " c
= (a + b + ab) + c + (a + b + ab)c
(a * b) " c = a + b + c + ab + bc + ac + abc
a * (b * c) = a * (b + c + bc)
= a + b + c + ab + bc + ac + abc
Clearly
(a * b) * c = a * (b "c)  \(\forall \) a, b, c  \(\in A\)}
 * is associative.
Identity: Let e \(\in A\)}such that
a * c = a
c + a + ea = a
c = 0
Identity element of A is e = O.
Inverse: Let b \(\in A\)such that
a*b = b*a = e
\(\Rightarrow \) a + b + ab = 0 and b + a + ba = 0
\(\Rightarrow \)  a = -b-ab
\(\Rightarrow \)  a = -b(l + a)
\(\Rightarrow \) b = \(\frac { -a }{ 1+a } \) \([\because \quad a\in A\therefore a\neq -1]\)
Invertible element of A is \(\frac { -a }{ 1+a } \) for all \(a\in A\)

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

Let Rs. 30,000 be divided into two parts:
Rs. x invested in 1st type and Rs. (30,000-x) in 2^nd type.
The values of the bonds are represented by \(1\times 2\) row matrix as: \(A=\left[ \begin{matrix} x & 30,000-x \end{matrix} \right] \)
The amount received as interest per Rs. annually are represented by \(2\times 1\) column matrix as: \(B=\left[ \begin{matrix} \frac { 5 }{ 100 } \\ \frac { 7 }{ 100 } \end{matrix} \right] .\)
\(\therefore \) The interest to be obtained is a \(1\times 1\) matrix, which is represented by the product matrix \(1\times 1\).
\(AB=\left[ \begin{matrix} x & 30,000-x \end{matrix} \right] \left[ \begin{matrix} \frac { 5 }{ 100 } \\ \frac { 7 }{ 100 } \end{matrix} \right] \)
\(=\left[ \frac { 5x }{ 100 } +\left( 30,000-x \right) \frac { 7 }{ 100 } \right] \)
\(=\left[ 2100-\frac { 2x }{ 100 } \right] .\)
(a) By the question,
\(\left[ 2100-\frac { 2x }{ 100 } \right] =\left[ 1800 \right] \)
\(\Rightarrow 2100-\frac { 2x }{ 100 } =1800\)
\(\Rightarrow \frac { 2x }{ 100 } =300\)
\(\Rightarrow x=15000\)
Hence, the reqd.amounts are Rs.15,000 and Rs. (30,000 - 15,000) i.e., Rs. 15,000 and Rs. 15,000.
(b) By the question,
\(\left[ 2100-\frac { 2x }{ 100 } \right] =\left[ 2000 \right] \)
\(\Rightarrow 2100-\frac { 2x }{ 100 } =2000\)
\(\Rightarrow \frac { 2x }{ 100 } =100\)
\(\Rightarrow x=5000\)
Hence, the reqd .amounts are Rs. 5,000 and Rs. (30,000 - 5,000)
i.e., Rs. 5,000 and Rs. 25,000.          

False

a = 2b {a=4, b=2}

Given,
\(\overrightarrow a\times \overrightarrow b=\overrightarrow a\times \overrightarrow b \) and \(\overrightarrow a\times \overrightarrow c=\overrightarrow b\times \overrightarrow d\)
\(\Rightarrow \overrightarrow a\times \overrightarrow b-\overrightarrow a\times \overrightarrow c=\overrightarrow c\times \overrightarrow d-\overrightarrow b\times \overrightarrow d\)
\(\Rightarrow \overrightarrow a\times \overrightarrow d-\overrightarrow a\times \overrightarrow c+\overrightarrow b\times \overrightarrow b-\overrightarrow c\times \overrightarrow d=\overrightarrow 0\)
\(\Rightarrow \overrightarrow a\times(\overrightarrow b-\overrightarrow c)+(\overrightarrow b-\overrightarrow c)\times \overrightarrow d=\overrightarrow 0\)[By left and right distributive law]
\(\Rightarrow \overrightarrow a\times(\overrightarrow b-\overrightarrow c)-\overrightarrow b\times (\overrightarrow b-\overrightarrow c)=\overrightarrow 0\)\([\because \overrightarrow a\times \overrightarrow b=-\overrightarrow b\times \overrightarrow a]\)
\(\Rightarrow (\overrightarrow a-\overrightarrow d)\times (\overrightarrow b-\overrightarrow c)=\overrightarrow 0\)
\(\Rightarrow(\overrightarrow a-\overrightarrow d)\parallel(\overrightarrow b-\overrightarrow c)\)

\(\Rightarrow I=\int _{ 0 }^{ 1 }{ x{ e }^{ x } } dx\)
\(\Rightarrow I=\left[ \left[ x\int { { e }^{ x }dx } -\int { \left[ \frac { d }{ dx } x\int { { e }^{ x }dx } \right] dx } \right] \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ x{ e }^{ x }-\int { 1.{ e }^{ x }.dx } \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ xe^{ x }-e^{ x } \right] ^{ 1 }_{ 0 }\)
\(\Rightarrow I=\left[ (1e^{ 2 }-e^{ 1 })-({ 0e }^{ 0 }-{ e }^{ 0 } \right] \)
\(\Rightarrow I=(0)-(0-1)=1\)

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

Let A be the point (5, 4, - 6)
Image A' be the point (5, 4, - 6)
\(\therefore\) A'(5, 4, - 6)
Distance between AA'
\(=\sqrt { (5-5)^{ 2 }+(4-4)^{ 2 }+(6+6)^{ 2 } } \)
\(=\sqrt { 0+0+12^{ 2 } } \)
=123 units

\(P(\bar { A } /\bar { B } )=\frac { P(\bar { A } \cap \bar { B } ) }{ P(\bar { B } ) } \)
\(=\frac { 1-P(A\cup B) }{ 1-P(B) } \)
\(=\frac { 1-[P(A)+P(B)-P(A\cap B)] }{ 1-P(B) } \)
\(=\frac { 7 }{ 10 } \)

\(\frac { dy }{ dx } =xy+2y\)
\(\Rightarrow \frac { dy }{ dx } =y\left( x+2 \right) \)
Integrating both the sides,
\(\int { \frac { dy }{ y } } =\int { \left( x+2 \right) dx } \)
\(\Rightarrow \log { y } =\frac { { x^{ 2 } } }{ 2 } +2x+c\)

we have f = {(a, 3) (b, 2) (e, 1)}, f is a one-one function.
\(f^{ -1 },f^{ 0 }=[(3,a),(2,b)(1,c)]\)
(ii) f = {(a, 2), (b, 1)(e, 1)}. f is not one-one, f is not onto because 3 \(\epsilon\) B as it has no Pre-image.

\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }x\)             [Given]
Put, x = tan \(\theta \)        \(\Rightarrow\) tan^-1 x=\(\theta \)
\(\Rightarrow \quad { tan }^{ -1 }\left[ \frac { 1-tan\theta }{ 1+tan\theta } \right] =\frac { 1 }{ 2 } \theta \)
\(\Rightarrow \quad { tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\theta \right) \right] =\frac { \theta }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 4 } -\theta =\frac { \theta }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 4 } =\frac { \theta }{ 2 } +\theta \)
\(\Rightarrow \quad \frac { \pi }{ 4 } =\frac { 3\theta }{ 2 } \Rightarrow \theta =\frac { \pi }{ 6 } \)
\(\Rightarrow \quad { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow \quad x=tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)         

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|

Here (1,2) e R, (2,1) € R, if transitive (1,1) should belong to R.