Given, \(\left| a\right| =\left| b \right| \)
and a.b \(=\frac{1}{2}\)
Let \(\theta \) be thae angle between \(\overrightarrow a\) and \(\overrightarrow b\)
then, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

\(y={ tan }^{ -1 }\frac { 3x+2x }{ 1-3x2x } \)
= \({ tan }^{ -1 }3x+{ tan }^{ -1 }2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3 }{ 1+9{ x }^{ 3 } } +\frac { 2 }{ 1+4{ x }^{ 2 } } \)

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
Let \(\alpha \) makes with x-axis, y makes with z-axis and \(\beta \) makes with y-axis
\(\alpha =cos\quad 60^{ \circ },m=cos\quad \beta ,y=cos45^{ \circ }\)
\(\alpha ^{ 2 }+\beta ^{ 2 }+\gamma ^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }60^{ \circ }+cos^{ 2 }\beta +cos^{ 2 }45^{ \circ }\)
\(\Rightarrow \frac { 1 }{ 4 } +cos^{ 2 }\beta +\frac { 1 }{ 2 } =1\)
\(\Rightarrow cos^{ 2 }\beta =1-\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \)
\(=\frac { 4-1-2 }{ 4 } \)
\(=\frac { 1 }{ 4 } \)
\(\Rightarrow cos \beta =\pm \frac { 1 }{ 2 } \)
\(\beta =\frac { \pi }{ 3 } or\ \frac { 2\pi }{ 3 } \)
But angle between two lines in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
Hence required angles is \(\frac { \pi }{ 3 } \)

Given, \(P(E)=\frac { 1 }{ 4 } \)
\(P(F)=\frac { 1 }{ 2 } \)
and \(P(E\cap F) =\frac { 1 }{ 8 } \)
(a) P(E or F) = (E\(\cup\)F) = P(E) + P(F) - P(E\(\cap\)F)
\(=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } \)
\(=\frac { 2+4-1 }{ 8 } \)
\(=\frac { 5 }{ 8 } \)
(b) P(not E and not F) = \(P(\bar { E } \cap \bar { F } )=P\overline { (E\cup F) } \)
= 1 - P(E\(\cup\) F)
= 1 - \(\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)

\(y=\left( { 2x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow \frac { dy }{ dx } =2\left( 2x \right) +c{ e }^{ { -x }^{ 2 } }\times -2x\)
\(\Rightarrow \frac { dy }{ dx } =4x-2xc{ e }^{ { -x }^{ 2 } }\) ...(i)              
\(y=2\left( { x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow c{ e }^{ { -x }^{ 2 } }=y-{ 2x }^{ 2 }+2\)...(ii)
Putting the value of \(c{ e }^{ { -x }^{ 2 } }\) from eqn. (ii) to eqn.(i),
\(\frac { dy }{ dx } =4x-2x\left( y-{ 2x }^{ 2 }+2 \right) \)
\(\Rightarrow \frac { dy }{ dx } =4x-2xy+{ 4x }^{ 3 }-4x\)
\(\therefore \ \frac { dy }{ dx } +2xy+{ 4x }^{ 3 }=0\)

\(f(x)=x^{ 2 }+1,g(x)=2x+1\)
\(fog=f[(g(x)\} =f(2x+1)\)
\(=4x^{ 2 }+4x+1+1\)
\(=4x^{ 2 }+4x+2\)
\(gof=g(f(x)=g(x^{ 2 }+1)\)
\(=2(x^{ 2 }+1)+1\)
\(=2x^{ 2 }+2+1\)
\(=2x^{ 2 }+3\)

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

\(\int { \frac { a }{ b+ce^{ x } } } dx\)
Multiply by e^-x to numerator and denominator
\(\int { \frac { a{ e }^{ -x } }{ be^{ -x }+c } } dx\)
Put  \(be^{ -x }+c=t\)
\(\Rightarrow -be^{ -x }dx=dt\)
\(\Rightarrow { e }^{ -x }dx=\frac { dt }{ -b } \)
\(=-\frac { a }{ b } \int { \frac { dt }{ t } } \)
\(\Rightarrow -\frac { a }{ b } log\left| t \right| =-\frac { a }{ b } log\left| b{ e }^{ -x }+c \right| +K\)

We have, \(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(\left( X+Y \right) +\left( X-Y \right) =\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] +\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(2X=\left[ \begin{matrix} 8 & 8 \\ 12 & 4 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(\therefore \ X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
and \(Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -2 & -1 \\ -1 & -1 \end{matrix} \right] \)

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|

A square matrix \(A=[{ a }_{ ij }] \)is skew symmetric
if \({ a }_{ ij }=-{ a }_{ ji }\forall i,j \)
Let \(i=j\Rightarrow { a }_{ ii }=-{ a }_{ ii }\Rightarrow { 2a }_{ ii }=0\Rightarrow { a }_{ ii }=0\Rightarrow 0  \)
All the diagonal elements of a skew symmetric matrix are always zero.

\(\left( AB \right) ^{ -1 }=\left[ \begin{matrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix} \right] \)

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

\(\int _{ 1 }^{ 4 }{ \left( { x }^{ 2 }-x \right) dx } =\int _{ 1 }^{ 4 }{ { x }^{ 2 }dx } -\int _{ 1 }^{ 4 }{ { x }dx } ={ I }_{ 1 }-{ I }_{ 2 }\)
Let \({ I }_{ 1 }=\int _{ 1 }^{ 4 }{ { x }^{ 2 }dx } \)  [See Imp.keys and formulae]
Here  \(h-\frac { b-a }{ n } =\frac { 4-1 }{ n } =\frac { 3 }{ n } \)
\(=\left( 4-1 \right) \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ f\left( 1 \right) +f\left( 1+\frac { 3 }{ n } \right) +......+n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ { 1 }^{ 2 }+{ \left( 1+\frac { 3 }{ n } \right) }^{ 2 }+\left( 1+\frac { 6 }{ n } \right) ^{ 2 }+......+\ n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ \left( { 1 }^{ 2 }+{ 1 }^{ 2 }+....n\ terms \right) +\left\{ \left( \frac { 3 }{ n } \right) ^{ 2 }+{ \left( \frac { 6 }{ n } \right) }^{ 2 }+.....+\left( n-1 \right) \ terms \right\} +\left\{ 2\times \frac { 3 }{ n } +2\times \frac { 6 }{ n } +.....+\left( n-1 \right) terms \right\} \right] \)
\(=3\underset { n\rightarrow \infty}{ lim } \frac { 1 }{ n } \left[ n+\frac { 9 }{ { n }^{ 2 } } \frac { \left( n-1 \right) n\left( 2n-1 \right) }{ 6 } +\frac { 6 }{ n } \left( \frac { n\left( n+1 \right) }{ 2 } \right) \right] \)
\(=3\left[ 1+2\times \frac { 9 }{ 6 } +3 \right] =3\times 7=21\)
and \({ I }_{ 2 }=\int _{ 1 }^{ 4 }{ xdx } \)
\(=\left( 4-1 \right) \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ f\left( 1 \right) +f\left( 1+\frac { 3 }{ n } \right) +......+n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ { 1 }^{ 2 }+{ \left( 1+\frac { 3 }{ n } \right) }^{ 2 }+\left( 1+\frac { 6 }{ n } \right) ^{ 2 }+......+\ n\ terms \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \left[ n+\frac { { n }^{ 2 } }{ 2 } +\frac { 3n }{ 2 } \right] \)
\(= 3\underset { n\rightarrow \infty }{ lim } \left[ 1+\frac { 3 }{ 2 } \right] \)
\(= 3\left[ 1+\frac { 3 }{ 2 } \right] =\frac { 15 }{ 2 } \)
\(\therefore \int _{ 1 }^{ 4 }{ \left( { x }^{ 2 }-x \right) } .dx={ I }_{ 1 }-{ I }_{ 2 }=21-\frac { 15 }{ 2 } =\frac { 27 }{ 2 } .\)

E₁ = Event that 1 occur;
E₂ = Event that 1 does not occur;
A = Event that the man reports that 1 occur.
\(P({ E }_{ 1 })=\frac { 1 }{ 6 } ;P({ E }_{ 2 })=\frac { 5 }{ 6 } ;\)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 5 } ;P(A/{ E }_{ 2 })=\frac { 2 }{ 5 } ;\)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 6 } .\frac { 3 }{ 5 } }{ \frac { 1 }{ 6 } .\frac { 3 }{ 5 } +\frac { 5 }{ 6 } .\frac { 2 }{ 5 } } =\frac { 3 }{ 13 } \)

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 \(a\in A\)

A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \);  A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
L.H.S = A³ - 6A² + 7A + 2I
\(=\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) -\left( \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right) +\left( \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right) +\left( \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
= RHS

(1-sin x)² y"=cos x

\(X=\begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}\)

Let x be the random variable denoting no. of white balls selected.
No. of balls can be 0, 1, 2, 3 or 4.
\(P_{ (0) }=\frac { { 6 }_{ C_{ 4 } }\times 4_{ C_{ 0 } } }{ { 10 }_{ { C }_{ 4 } } } =\frac { 15 }{ 210 } =\frac { 1 }{ 14 } \)
\(P_{ (1) }=\frac { { 6 }_{ { C }_{ 3 } }\times { 4 }_{ { C }_{ 1 } } }{ { 10 }_{ { C }_{ 4 } } } =\frac { 80 }{ 210 } =\frac { 8 }{ 21 } \)
\({ p }_{ (2) }=\frac { 6{ c }_{ 2 }\times { 4 }_{ C_{ 2 } } }{ 10_{ { C }_{ 4 } } } =\frac { 90 }{ 210 } =\frac { 3 }{ 7 } \)
\({ P }_{ (3) }=\frac { { 6 }_{ { C }_{ 1 } }\times { 4 }_{ { C }_{ 3 } } }{ { 10 }_{ { C }_{ 3 } } } =\frac { 24 }{ 210 } =\frac { 4 }{ 35 } \)
\(P_{ (4) }=\frac { { 6 }_{ { C }_{ 0 } }\times { 4 }_{ { C }_{ 4 } } }{ { 10 }_{ { C }_{ 4 } } } =\frac { 1 }{ 210 } \)

We have,
\(I=\int { \frac { 1 }{ sinx-sin2x } } dx\)
\(\Rightarrow I=\int { \frac { dx }{ sinx-2sinxcosx } } \)
\(\Rightarrow I=\int { \frac { dx }{ sinx(1-2cosx) } dx } \)
\(\Rightarrow I=\int { \frac { sinx }{ sin^{ 2 }x(1-2cosx) } dx } \)
\(\Rightarrow I=\int { \frac { sinxdx }{ (1-cos^{ 2 }x)(1-2cosx) } } \)
Putting, \(cosx=t\Rightarrow -sinxdx=dt\) or sinxdx=-dt, we get
\(I=\int { \frac { -dt }{ (1-t^{ 2 })(1-2t) } } \)
\(\Rightarrow I=\int { \frac { -1 }{ (1+t)(1-t)(1-2t) } dt } \)
Let \(\int { \frac { 1 }{ (1+t)(1-t)(1-2t) } } \)
\(=\frac { A }{ (1+t) } +\frac { B }{ (1-t) } +\frac { C }{ 1-2t } \)
\(A=\left[ \frac { 1 }{ (1-t)(1-2t) } \right] _{ t=-1 }=\frac { 1 }{ 6 } \)
\(B=\left[ \frac { 1 }{ (1+t)(1-2t) } \right] _{ t=1 }=-\frac { 1 }{ 2 } \)
\(C=\left[ \frac { 1 }{ (1+t)(1-t) } \right] _{ t=\frac { 1 }{ 2 } }=\frac { 4 }{ 3 } \)
\(\therefore I=-\frac { 1 }{ 6 } \int { \frac { dt }{ 1+t } +\frac { 1 }{ 2 } \int { \frac { dt }{ 1-t } -\frac { 4 }{ 3 } \int { \frac { dt }{ 1-2t } } } } \)
\(\Rightarrow I=-\frac { 1 }{ 6 } log\left| 1+t \right| -\frac { 1 }{ 2 } log\left| 1-t \right| -\frac { 4 }{ 3 } \times \left( -\frac { 1 }{ 2 } \right) log\left| 1-2t \right| +C\)
\(\Rightarrow -\frac { 1 }{ 6 } log\left| 1+cosx \right| -\frac { 1 }{ 2 } log\left| 1-cosx \right| +\frac { 2 }{ 3 } log\left| 1-2cosx \right| +C\)

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 − x − y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B.
Similarly, (3000 − y) L and 3500 − (7000 − x − y) = (x + y − 3500) L will be transported from depot B to petrol pump E and F respectively.
The given problem can be represented diagrammatically as follows.
Minimize: \(C=\frac { 3x }{ 10 } +\frac { y }{ 10 } +3950\)
Subject to:   \(4500-x\ge 0\Leftrightarrow x\le 4500\) ...(1)
\(3000-y\ge 0\Leftrightarrow y\le 3000\)....(2)
\(x+y-3500\ge 0\Leftrightarrow x+y\ge 3500\) ...(3)
\(7000-(x+y)\ge 0\Leftrightarrow x+y\le 7000\) ..(4)
and  \(x,y\ge 0\) ....(5)
Cost of transporting \(10 \mathrm{~L} \text { of petrol }=\operatorname{Re} 1\)
Cost of transporting \( 1 \mathrm{~L} \text { of petrol }=\mathrm{Rs} \frac{1}{10}\)
Therefore, total transportation cost is given by,
\(z =\frac{7}{10} \times x+\frac{6}{10} y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)\)
\(=0.3 x+0.1 y+3950 \)
The problem can be formulated as follows.
Minimize z = 0.3x + 0.1y + 3950 … (1)
subject to the constraints,
\(x+y \leq 7000 \)
\(x \leq 4500 \)
\(y \leq 3000 \)
\(x+y \geq 3500 \)
\(x, y \geq 0 \)
The feasible region determined by the constraints is as follows.


The corner points of the feasible region are A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000), and E (500, 3000).
The values of z at these corner points are as follows.

The minimum value of z is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.
The minimum transportation cost is Rs. 4400.

(i) Commutativity.
(a,b) * (c,d) = (a+c, b+d)
= (c+a, d+b)
[∵ Addition is commutative in N]
= (c,d) * (a,b).
Hence ' * ' is commutative
(ii) Associativity.
[(a*b)*(c,d)]*(e,f)] = (a+c, b+d)*(e,f)
= [(a+c)+e,(b+d)+f)
= (a+(c+e),b+(d+f)
[∵ Addition is associative in N]
= (a,b)*(c+e,d+f)
= (a,b)*[(c,d)*(e,f)]
Hence, '*' is associative.
(iii) Let (x, y) be the identity element.
Then (a,b)*(x,y) = (a,b)
\(\Rightarrow \) (a+x, b+y) = (a,b)
\(\Rightarrow \) a+x = a, b+y = b
\(\Rightarrow \) x = 0, y = 0.
But \(0\notin N\).
Hence, identity element does not exist. 

\(Let\ y=sin\ (log\ x)\)
\(\frac { dy }{ dx } =cos(log\ x).\frac { 1 }{ x } \)
\(and\ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { x.\frac { d }{ dx } (cos(log\ x))-cos(log\ x).1 }{ { x }^{ 2 } } \)
\(=\frac { x.(-sin(log\ x))\frac { 1 }{ x } -cos(log\ x) }{ { x }^{ 2 } } \)
\(=-\frac { sin(log\ x)+cos(log\ x) }{ { x }^{ 2 } } \)

We have: \(\frac { dy }{ dx } +\sqrt { \frac { 1-{ y }^{ 2 } }{ 1-{ x }^{ 2 } } } =0.\)
\(\Rightarrow \) \(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } +\frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } =0\)
| Variables Separable
Integrating, \(\int { \frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } } +\int { \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } } =C\)
\(\Rightarrow \) \({ sin }^{ -1 }\quad y+{ sin }^{ -1 }x=C,\)
Which is the reqd. solution.

The given circle is 4x² + 4y² = 9
i.e., x² + y² = \(\frac{9}{4}\)
which has centre (0,0) and radius \(\frac{3}{2}\)
The given parabola is x² = 4y which is an upward parabola with vertex (0,0).
The reqd. area is shown shaded in the figure. 
Solving (1) and (2):

Putting the value of x² from (2) in (1), we get:
\(4y+{ y }^{ 2 }=\frac { 9 }{ 4 } \Rightarrow { 4y }^{ 2 }+16y-9=0\)
\(\Rightarrow (2y+9)(2y-1)=0\Rightarrow y=-\frac { 9 }{ 2 } ,\frac { 1 }{ 2 }\)
\(Thus\ y=\frac { 1 }{ 2 } \left[ \because y\neq -\frac { 9 }{ 2 } \right] \)
\(\therefore Reqd.area=ar(OABC)\)
\(=ar(OAC)+ar(ACB)\)
\(=2\left[ \left( arOAD \right) +ar\left( DAB \right) \right] \)
\(=2\left[ 2\overset { 1/2 }{ \underset { 0 }{ \int { } } } \sqrt { y } dx+\overset { 3/2 }{ \underset { 1/2 }{ \int { } } } \frac { \sqrt { { 9-4y }^{ 2 } } }{ 2 } dy\left[ \because { x }^{ 2 }+{ y }^{ 2 }=\frac { 9 }{ 4 } \Rightarrow x=\sqrt { \frac { { 9-4y }^{ 2 } }{ 2 } } \right] \right] \)
\(=4\overset { 1/2 }{ \underset { 0 }{ \int { } } } { y }^{ 1/2 }dx+2\overset { 3/2 }{ \underset { 1/2 }{ \int { } } } \sqrt { \frac { 9 }{ 4 } -{ y }^{ 2 } } dy\)
\(=4\left[ \frac { { y }^{ 3/2 } }{ 3/2 } \right] _{ 0 }^{ 3/2 }+2\left[ \frac { y }{ 2 } \sqrt { \frac { 9 }{ 4 } -{ y }^{ 2 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2y }{ 3 } \right) \right] _{ 1/2 }^{ 3/2 }\)
\(=\frac { 8 }{ 3 } \left( \frac { 1 }{ 2 } \right) ^{ 3/2 }+2\left[ \frac { y }{ 2 } \sqrt { \frac { 9 }{ 4 } -{ y }^{ 2 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2y }{ 3 } \right) \right] _{ 1/2 }^{ 3/2 }\)
\(=\frac { 8 }{ 3 } \times \frac { 1 }{ 2\sqrt { 2 } } +2\left[ \left( 0+\frac { 9 }{ 8 } { sin }^{ -1 }(1) \right) -\left( \frac { 1 }{ 4 } \sqrt { 2 } +\frac { 9 }{ 8 } { sin }^{ -1 }\frac { 1 }{ 3 } \right) \right] \)
\(=\frac { 2\sqrt { 2 } }{ 3 } -\frac { \sqrt { 2 } }{ 2 } +\frac { 9 }{ 4 } \left( { sin }^{ -1 }1-{ sin }^{ -1 }\frac { 1 }{ 3 } \right) \)
\(=\frac { \sqrt { 2 } }{ 6 } +\frac { 9 }{ 4 } { sin }^{ -1 }\left( (1)\sqrt { 1-\frac { 1 }{ 9 } } -\frac { 1 }{ 3 } \sqrt { 1-1 } \right) \)
\(=\frac { \sqrt { 2 } }{ 6 } +\frac { 9 }{ 4 } { sin }^{ -1 }\left( \frac { 2\sqrt { 2 } }{ 3 } -0 \right) \)
\(=\left( \frac { \sqrt { 2 } }{ 6 } +\frac { 9 }{ 4 } { sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } \right) sq.units.\)

\(\Delta=\begin{vmatrix}x+4&x+4&x+4\\1&x&1\\3&2&x \end{vmatrix}\)
\((x+4)\begin{vmatrix} 1&1&1\\1&x&1\\3&2&x\end{vmatrix}\)
\(=(x+4)\begin{vmatrix}1&0&0\\1&x-1&0\\3&-1&x-3 \end{vmatrix}\)
= (x+4) (1)[(x-1) (x-3)+0]
= (x+4) (x-1) (x-3)
Thus \(\Delta=0\Rightarrow x=-4,1,3.\)
Hence the other two roots are 1 and 3.

\(We\quad have:\quad 2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
\(\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\( \Rightarrow \begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} y+2 & 6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
Equating corresponding elements:
2x + 2 = 8 and y + 2 = 5
\(\Rightarrow \) 2x = 6 and y = 3.
Hence, x = 3 and y = 3.

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 = \(-\frac { \pi }{ 4 } \)

If perpendicular then aa' + 1 + cc' = 0

\(\int e^{y} d y=\int\left(e^{x}+x^{3}\right) d x\)
\(\Rightarrow { e }^{ y }={ e }^{ x }+\frac { { x }^{ 4 } }{ 4 } +c\)