Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

We compute
F(\(\lambda \)x, \(\lambda \)y) = \(\frac { { \left( { \lambda x } \right) }^{ 2 }+5\left( \lambda x \right) \left( \lambda y \right) -10(\lambda y)^2 }{ 3\lambda x+7\lambda y } =\frac { { \lambda }^{ 2 } }{ \lambda } \left( \frac { { x }^{ 2 }+5xy-10{ y }^{ 2 } }{ 3x+7y } \right) =\lambda F(x,y)\)
for all \(\lambda \) ∈ R. So F is a homogeneous function of degree 1.

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }x } =\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2\)
\(Let\quad { I }_{ 10 }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 10 }xdx=\frac { 9 }{ 10 } { I }_{ 8 } } \)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times { I }_{ 6 }=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times { I }_{ 4 }\)
\(\\ =\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } { I }_{ 2 }\)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } \)
\(=\frac { 315 }{ 1280 } \times \frac { \pi }{ 2 } =\frac { 63\pi }{ 256(2) } =\frac { 63\pi }{ 512 } \)

\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi +\frac { \pi }{ 4 } \right) \right) \) \(\because \frac { 5\pi }{ 4 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
= \({ sin }^{ -1 }\left( sin\left( -\frac { \pi }{ 4 } \right) \right) \)
= \( \frac {- \pi }{ 4 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) 

Comparing the given equation of the plane with \(\vec { r } .\vec { n } \) = p, we have \(\vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \).
We know that the perpendicular distance from the given point with position vector u to the plane \(\vec { r } .\vec { n } \)= p is given by \(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } \). Therefore, substi \(\vec { u } \)= (2, 5, -3) = \(2\hat { i } +5\hat { j } -3\hat { k } \) and \(\ \vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \) in the formula, we get
\(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } =\frac { |(2\hat { i } +5\hat { j } -3\hat { k } ).(6\hat { i } -3\hat { j } +2\hat { k } )-5| }{ |6\hat { i } -3\hat { j } +2\hat { k } | } \) = 2 unit.

\(Re\left( \frac { 1 }{ z } \right) \)
= \(Re\left( \frac { 1 }{ x+iy } \right) =Re\left( \frac { 1 }{ x+iy } \times \frac { x-iy }{ x-iy } \right) \)
= \(Re(\left( \frac { x-iy }{ { x }^{ 2 }({ i }^{ 2 }{ y }^{ 2 }) } \right) =Re\left( \frac { x-iy }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
∴ Real part is \(\frac { x }{ { x }^{ 2 }+{ y }^{ 2 } } \)

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Equation of the circle in standard form is (xr − h)² + (y − k)² = r²
\( \Rightarrow (x-(-3))^{2}+(y-(-4))^{2} =3^{2} \)
\( \Rightarrow (x+3)^{2}+(y+4)^{2} =3^{2} \)
\( \Rightarrow x^{2}+y^{2}+6 x+8 y+16 =0 .\)

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

~( p ∧ q) ≡ ~p V ~q

p	q	p ∧ q	~(p ∧ q)	~p	~p	~p V ~q
T	T	T	F	F	F	F
T	F	F	T	F	T	T
F	T	F	T	T	F	T
F	F	F	T	T	T	T

From column (4) and column (7), the entries are Identical.
ஃ ~( p ∧ q) ≡ ~p V ~q are identical and hence they are equivalent.

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

By the mean value theorem we have, there exists 'c'∈(2, 7) such that,
\(\frac { f(7)-f(2) }{ 7-2 } \) = f'(c) ≤ 29
Hence, f(7) ≤ 5× 29 +17 = 162
Therefore, the maximum value of f (7) is 162.

Given y = ax + \(\frac { b }{ x } \) .......(1)
Differentiating with respect to x
y' = ax - \(\frac { b }{ x ^2} \) ......(2)
Differentiating again with respect to x
\(y'' = \frac{-b(-2)}{x^3}= \frac{2b}{x^3} \)
\(Now, x^2y'' + xy'-y \)
\( = x^2 \times \frac{2b}{x^3}+x(a- \frac{b}{x^2})-(ax+\frac{b}{x}) \)
\(= 2\times (\frac{b}{x})+ax-(\frac{b}{x})-ax-(\frac{b}{x})\)
= 0
Hence, y = ax + b is the solution of the differential equation x²y"+xy'-y = 0.

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

We get \([\vec { a } +\vec { c } ,\vec { a } +\vec { b } ,\vec { a } +\vec { b } +\vec { c } ]\) = \(\left| \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right| \)\([\vec { a } ,\vec { b } ,\vec { c } ]\) = \([\vec { a } ,\vec { b } ,\vec { c } ]\)

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

Given \(\frac { \pi }{ 2 } <{ cos }^{ -1 }(3x-1)<\pi \)
\((cos2\frac { \pi }{ 2 } <3x-1\))
\(\Rightarrow 0<3x-1<-1\)
\(0+1<3x<-1+1\)
\(1<3x<0\)
This inequality is true, only when 0 < x < \(\frac { 1 }{ 3 }\)

Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

Let X denote the winning amount. The possible events of selection are 
(i) both balls are black, or
(ii) one white and one black or
(iii) both are white 
Therefore X is a random variable that can be defined as
X (both are black balls) = Rs. 2(20) = Rs. 40
X (one black and one white ball) = Rs. 20 − Rs. 10 = Rs. 10
X (both are white balls) = (Rs. 20) = - Rs. 20
Therefore X takes on the values 40,10 and −20
Total number of balls n = 12
Total number of ways of selecting 2 balls = \(\left( \begin{matrix} 12 \\ 2 \end{matrix} \right) =\frac { 12\times 11 }{ 1\times 2 } =66\)
Number of ways of selecting 2 black balls =  \(\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) =6\) 
Number of ways of selecting one black ball and one white ball = \(\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) =32\) 
Number of ways of selecting 2 white balls = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Values of Random Variable X	40	10	-20	Total
Number of elements in inverse images	6	32	28	66

Probability mass function is

X	40	10	-20	Total
f (x)	\(\cfrac { 6 }{ 66 } \)	\(\cfrac { 32 }{ 66 } \)	\(\cfrac { 28 }{ 66 } \)	1

Mean :
\(E(X)\Sigma xf(x)=40.\left( \frac { 6 }{ 66 } \right) +10.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) .\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
That is expected winning amount is 0
 Variance :
\(\Sigma x^{ 2 }=\Sigma { x }^{ 2 }f(x)=40^{ 2 }.\left( \frac { 6 }{ 66 } \right) +10^{ 2 }.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) ^{ 2 }.\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
 (E(X )² = 0² = 0
This gives  \(V(X)=E({ X }^{ 2 })-\left( E(X))^{ 2 } \right) =\frac { 4000 }{ 11 } -0=\frac { 4000 }{ 11 } \)
Therefore E(X ) = 0 and \(V(x)=\frac { 4000 }{ 11 } \)

The ellipse is symmetric about both major and minor axes. It is sketched. So, viewing in the positive direction of y -axis, the required area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant 
\(\left( y=\frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } ,0

x -axis, x = 0 and x = a .
Hence, by taking vertical strips, we get \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

Let T be the temperature of the body at any time t and with time 0 taken to be 8 p.m. By Newton’s law of cooling \(\frac { dT }{ dt } =k(T-50)or\frac { dT }{ T-50 } =dt\).
Integrating on both sides, we get log |50 −T| = kt + logC or 50 −T = Ce^kt.
When t = 0, T = 70, and so C = −20
When t = 2,T = 60, we have −10 = −20 e^k2.
Thus, \(k=\frac { 1 }{ 2 } log\left( \frac { 1 }{ 2 } \right) \)
Hence, the solution is 50-T = -20e^{\(\frac{1}{2}\)tlog\((\frac{1}{2})\)} or T = 50 + 20\((\frac{1}{2})^\frac{t}{2}\)
Now, we would like to find the value of t, for which T(t) = 98.6 , and t = 2\(\left( \frac { log\left( \frac { 48.6 }{ 20 } \right) }{ log\left( \frac { 1 }{ 2 } \right) } \right) \approx -2.56\)
It appears that the person was murdered at about 5.30 p.m.

Let the point of intersection of the two curves be (a,b) . Hence,
 \(a^{2}+4b^{2}=8\) and  \(a^{2}-2b^{2}=4\) ...(4)
It is enough if we show that the product of the slopes of the two curves evaluated at (a, b) is −1.
Differentiation of \(x^{2}+4y^{2}=8\)  with respect x, gives
\(2x+8y=\frac{dy}{dx}=0\).
Therefore \(\frac{dy}{dx}= -\frac{-x}{4y}\),
\((\frac{dy}{dx})_{(a,b)}=m_{1}= -\frac{a}{4b}\)
Differentiation of x²-2y² = 4 with respect to x, gives
\(2x-4y\frac{dy}{dx}=0\)
Therefore, \(\frac{dy}{dx}=\frac{x}{2y}\), 
at \((a,b)(\frac{dy}{dx})=m_{2}= \frac{a}{4b}\).
Therefore, \(m_{1}\times m_{2}=(-\frac{a}{4b})\times (\frac{a}{2b})= -\frac{a^{2}}{8b^{2}}\)  ...(5)
Applying the ratio of proportions in (4), we get
\(\frac{a^{2}}{-16-16}=\frac{b^{2}}{-8+4}=\frac{1}{-2-4}\)
Therefore, \(\frac{a^{2}}{b^{2}}=\frac{32}{4}=8\) Substituting in (5), we get \(m_{1}\times m_{2}=-1\) Hence, the curves cut orthogonally.

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

Let the parabolic bridge to a open downward parabola.
Then its equation is x² = -4ay ......(1)
Since 10 m is the height in the centre (0, 0) and AD = 30 m.

∴ A is the point (15, -10) which is in the IV quadrant A(15, -10) lies in (1)
15² = -4a(-10)
⇒ \(\frac { 225 }{ 10 } =4a\Rightarrow \frac { 45 }{ 2 } \)
(1) becomes x² = \(\frac { -45 }{ 2 } \)y .....(2)
To find the height ofthe arch at x = 6 m, the point B(6, -y₁) lies in (2)
6² = \(\frac { -45 }{ 2 } \)(-y₁)
⇒ \(36=\frac { -45{ y }_{ 1 } }{ 2 } \Rightarrow \frac { 36\times 2 }{ 45 } ={ y }_{ 1 }\)
⇒ \({ y }_{ 1 }=\frac { { 8 } }{ 5 } =1.6\)
∴ Height of the arch on either sides
= 10-1.6 = 8.4m

Giveny = ax² + bx + c .............(1)
(-6, 8) lies on (1)
⇒ 8 = a(-6)²+b(-6)+c
⇒ 8 = 36z-6b+c ..........(2)
(-2,12) lies on (1)
⇒ -12 = a(-2)²+b(-2)+c
⇒ -12 = 4a-2b+c ...........(3)
Also (3, 8) lies on (1)
⇒ 8 = a(3)²+b(3)+c
⇒ 8 = 9a+3b+c ............(4)
Reducing the augment matrix to an equivalent row-echelon form by using elementary. row operations, we get,
\(\left[ \begin{matrix} 36 & -6 & 1 \\ 4 & -2 & 1 \\ 0 & 3 & 1 \end{matrix}|\begin{matrix} 8 \\ -12 \\ 8 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { 9R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow 4{ R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -12 & 8 \\ 0 & 18 & 3 \end{matrix}|\begin{matrix} 8 \\ -116 \\ 24 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 4\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 3 }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -50 \end{matrix} \right] \)
Writing the equivalent equation from the row echelon matrix, we get
36a - 6b + c = 8 ........(1)
-3b+2c = -29 ....(2)
5c = -50
⇒ c = \(\frac{-50}{5}\) = -10
Substituting c = -10 in (2) we get,
-3b+2(-10)= -29
⇒ -3b+2(-10) = -29
⇒ -3b-20 = -29
⇒ -3b = -9
⇒ b = \(\frac{-9}{-3}\) = 3
Substituting b = 3 and c = -10 in (1) we get,
36a-6(3)-10 = 8
⇒ 36a-18-10 = 8
⇒ 36a-28 = 8
⇒ 6a = 8+28 = 36
⇒ a = \(\frac{36}{36}\) = 1
∴ a = 1, b = 3, c = -10
Hence the path of the boy is
y = 1(x²)+3(x)-10
⇒ y = x²+3x-10
Since his friend is at P(7, 60),
60 = (7)²+3(7)-10
⇒ 60 = 49+21-10
⇒ 60 = 70-10 = 60
⇒ 60 = 60
Since (7, 60) satisfies his path, he can meet his friend who is at P(7, 60)

Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

This equation is Type- 2 even degree reciprocal equation. Hence, it can be rewritten as
\(6({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } )-5(x+\frac { 1 }{ x } )-38=0...(1)\)
Put \(x+\frac { 1 }{ x } =y\Rightarrow \left( x+\frac { 1 }{ x } \right) ^{ 2 }={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\Rightarrow { x }^{ 2 }+\frac { 1 }{ x } ={ y }^{ 2 }-2\)
∴ (1)becomes as,

\(6{ (y }^{ 2 }-2)-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-12-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-5y-50=0\)
\(\Rightarrow (3y-10)(2y+5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { -5 }{ 2 } \)
Case (i) when \(y=\frac { +10 }{ 3 } \)
\(x+\frac { 1 }{ x } =\frac { +10 }{ 3 } \)
\(\frac { { x }^{ 2 }+1 }{ x } =\frac { +10 }{ 3 } \Rightarrow { 3x }^{ 2 }+3=10x\)
\({ 3x }^{ 2 }-10x+3=0\)
\((x-3)(x-\frac { 1 }{ 3 } )=0\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=-\frac { 5 }{ 2 } \Rightarrow x+\frac { 1 }{ x } =-\frac { 5 }{ 2 } \)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { -5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2+5x=0\)
\(\Rightarrow { 2x }^{ 2 }+5x+2=0\)
\(\Rightarrow (x+2)(2x+1)=0\)
\(\Rightarrow x=-2,\frac { -1 }{ 2 } \)
\(\therefore\) The roots are \(3,\frac { 1 }{ 3 } ,-2,\frac { -1 }{ 2 } \)

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α, β respectively with positive x-axis Draw AL and BM 丄 to axis
Then \(|\vec { OL } |=|\vec { OA } |\) = cos α = cos α
\(\\ |\vec { LA } |=|\vec { OB } |\) sin α = sin α
[∵ ΔOAL, sin α = \(\frac { opp }{ hyp } cos\alpha =\frac { adj }{ hyp } \)]
∴ \(\vec { OL } =|\vec { OL } |\hat { i } =cos\alpha \hat { i } \)
\(\vec { LA } =|\vec { LA } |\hat { j } =sin\ \alpha \hat { j } \)
∴ \(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } \) [Using Δ law of addition]
\(\hat { a } =cos\alpha \hat { i } +sin\alpha \hat { j } \)
Similarly \(\\ \hat { b } =cos\alpha \hat { i } +sin\alpha \hat { j } \) ...(2)
The angle between \(\hat { a } \) and \(\hat { b } \) is (α-β)
∴ \(\hat { a } .\hat { b } =|\vec { a } ||\vec { b } |\)cos(α-β) = cos(α-β)..(3)
[∵ \(|\vec { a } |=|b|\) = 1]
Also \(\vec { a }. \vec { b } =(cos\alpha \hat { i } +sin\alpha \hat { j } )(cos\beta \hat { i } +sin\beta \hat { j } )\)
\(\hat { a } \hat { b } \) = cosα cosβ + sinα sinβ....(4)
From (3) and (4), we get
cos(α-β) = cos α cos β + sin α sin β

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)  ...(2)
∴ From (1) and (2), f_xy = f_yz

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)