Actual value = 12.5 cm,
Approximate value = 12.65 cm
Area of the circular plate = πr²
Percentage error = 0.024 x 100 = 2.4%
Volume of sphere = \(\frac43\)πr² 

Actual value = 12.5 cm,
Approximate value = 12.65 cm
Area of the circular plate = πr²
Relative error = \(\frac{160.225π-156.25π}{160.0225π}\)
= \(\frac{3.7725π}{160.0225π}\)= 0.024 cm

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

Let X be the random variable of number of black cards occur.
X = {0,1,2}
Sample space ⁵²C₂ = 1326
Let X denote the number of black cards drawn.
X = 0, X (both are red cards) = ²⁶C₂ = 325  
X = 1 (1 black card and 1 red card) = ²⁶C₁ \(\times\) ²⁶C₁ = 676  
X = 2 (both are black cards) = ²⁶C₂ = 325  
∴ X takes the values 0, 1, 2

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.

Given y = ae^-3x+ b   ........(1)
Differentiating cquation (1) w.r.t 'x', we get
\(\frac { d y }{ d x } =ae^{ -3x }(-3)+0\)
\(\frac { d y }{ d{ x } } =ae^{ -3x }(-3) \)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ ae^{-3 x } }(+9)\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-\frac{1}{3}\frac{dy}{dx}\times 9\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } = {-3}\frac{dy}{dx} \)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } +3\frac{dy}{dx} \) = 0 
Therefore, y = ae^-3x + b is a solution of the given differential equation.

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplifying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

We get AB = \(\left[ \begin{matrix} 0 & -3 \\ 1 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0+0 & 0+3 \\ -2+0 & -3-4 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 \\ -2 & -7 \end{matrix} \right] \)
(AB)^-1 = \(\frac { 1 }{ \left( 0+6 \right) } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \)...(1)
A^-1 = \(\frac { 1 }{ \left( 0+3 \right) } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] \)
B^-1 = \(\frac { 1 }{ \left( 2-0 \right) } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \)
B^-1A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \) ...(2).
As the matrices in (1) and (2) are same, (AB)⁻¹ = B⁻¹A⁻¹ is verified.

Given z₁ = 2 + 5i, z₂ = -3 - 4i and z₃ = 1 + i
Additive inverse of z₁ is
-z₁ = -(2 + 5i)
= -2 - 5i
Multiplicative inverse of z₁ is
\(\frac { 1 }{ { z }_{ 1 } } =\frac { 1 }{ 2+5i } \times \frac { 2-5i }{ 2-5i } \)
[Multiply and divide by the conjugate of denominator]
= \(\frac { 2-5i }{ { 2 }^{ 2 }-(5i)^{ 2 } } =\frac { 2-5i }{ 4-25^{ 2 } } =\frac { 2-5i }{ 4+25 } \)
(z₁)^-1 = \(\frac { 1 }{ 29 } \)(2- 5i)   [∴ i² = -1]
Additive inverse of z₂ is
-z₂ = -(3 - 4i)
= 3 + 4i
Multiplicative inverse of z₂ is
\(\frac { 1 }{ z_{ 2 } } =\frac { 1 }{ -3-4i } \times \frac { -3+4i }{ -3+4i } \)
= \(\frac { -3+4i }{ (-3)^{ 2 }-(4i)^{ 2 } } \)
= \(\frac { -3+4i }{ 9-16i^{ 2 } } =\frac { -3+4i }{ 9+16 } \)
(z₂)^-1 = \(\frac { 1 }{ 25 } \)(-3 + 4i)
Additive inverse of z₃ is
-z₃ = -(1 + i)
= -1- i
Multiplicative inverse of z₃ is
\(\frac { 1 }{ { z }_{ 3 } } =\frac { 1 }{ 1+i } \times \frac { 1-i }{ 1-i } =\frac { 1-i }{ { 1 }^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { 1-i }{ 1+i } \)
(z₃)^-1 \(=\frac { 1 }{ 2 } \)(1 - i)

Given (3 -i) x - (2 - i) y + 2i + 5
= 2x + (-1 + 2i) y + 3 + 2i
⇒ 3x - ix - 2y + iy + 2i + 5 = 2x - y + 2iy + 3 + 2i
choosing the real and imaginary parts
(3x-2y + 5) + i (-x + y + 2) = 2x - y + 3 + i (2y+ 2)
Equating the real and imaginary parts both sides, we get
3x- 2y+ 5 = 2x-y+3
⇒ 3x - 2y + 5 - 2x +y - 3 = 0
⇒ x-y = -2... (1)
-x+y+2 = 2y+2
⇒ -x+y+2-2y-2 = 0
⇒ -x-y = 0 ⇒ x+y = 0.. (2)
(1)-(2) we get,

y = 1
Substituting y = 1 in (2) we get.
x+1 = 0 ⇒ x = -1
∴ x = -1 and y = 1

\( \tan ^{-1} x+2 \cot ^{-1} x =\frac{2 \pi}{3} \)
\(\tan ^{-1} x+\cot ^{-1} x+\cot ^{1} x =\frac{2 \pi}{3} \)
\(\frac{\pi}{2}+\cot ^{-1} x =\frac{2 \pi}{3} \)
\(\cot ^{-1} x =\frac{2 \pi}{3}-\frac{\pi}{2} \)
\(\cot ^{-1} x =\frac{4 \pi-3 \pi}{6} \)
\(\cot ^{-1} x =\frac{\pi}{6} \)
\(x =\cot \frac{\pi}{6} \)
\(x =\sqrt{3} \in \mathrm{R} \)

Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.

I = \(\int ^\frac{\pi}{4}_{0} \frac{1}{sin x+cos x}\) dx = \(\int ^\frac{\pi}{4}_{0} \frac{1} {\sqrt 2( \frac {1}{\sqrt 2}sin x+ \frac {1}{\sqrt 2}cos x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {( cos \frac {\pi}{4}cos x+ sin \frac {\pi}{4}sin x)}\) dx = \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos (\frac{\pi}{4}- x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos x}\)dx since\(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\) f (a-x)dx
\(=\frac { 1 }{ \sqrt { 2 } } \int _{ 0 }^{ \frac { \pi }{ 4 } } secxdx=\frac { 1 }{ \sqrt { 2 } } { \left[ log(secx+tanx) \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
=  \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1) - log (1+0)]\)
= \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1)\).

Put u = tan \(\frac{x}{2}\) 
Then, sin x = \(\frac{2 tan \frac {x}{2}}{1 + tan^2 \frac{x}{2}}\) = \(\frac{2u}{1+u^2}\), du = \(\frac{1}{2}\) sec² \(\frac{x}{2}\) dx  \(\Rightarrow \) dx = \(\frac {2du}{1+u^2}\)
When x = 0,u = tan 0 = 0
When x = \(\frac{\pi}{2},u=tan\frac{\pi}{4}=1\)
∴ I =\(\int ^\frac{\pi}{2}_0\) \(\frac{dx}{4+5 sin x}\) =\(\int ^{1}_0\) \(\frac{\frac {2du}{1+u^2}}{4+5 (\frac{2u}{1+u^2})}\) =\(\int ^{1}_0\) \(\frac {du}{2u^2+5u +2}\)=\(\frac{1}{2}\)\(\int ^{1}_{0}\) \(\frac {du}{u^2+\frac{5}{2}u +1}\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { du }{ (u+\frac { 5 }{ 4 } )^2-(\frac { 3 }{ 4 } )^{ 2 } } { \left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2\times \left( \frac { 3 }{ 4 } \right) } log\left( \frac { (u+\frac { 5 }{ 4 } )-\frac { 3 }{ 4 } }{ (u+\frac { 5 }{ 4 } )+\frac { 3 }{ 4 } } \right) \right] }_{ 0 }^{ 1 } = \frac{1}{3} [log (\frac{u+\frac{1}{2}}{u+2})] =\frac { 1 }{ 3 } log2\)

Let T be the temperature of water at any time t.
Then, by Newton's law of cooling,
\(\frac { dT }{ dt } \infty (T-{ 25 }^{ o })\)
\(\Rightarrow \frac { dT }{ dt } =-\lambda (T-25)\)
\(\Rightarrow \int { \frac { dT }{ T-25 } } =-\lambda \int { dt } \)
\(\Rightarrow log(T-25)=-\lambda t+C\)  ...(1)
At t = 0, T = 100⁰e in (1) we get
\(\therefore(1)\Rightarrow\)log75 = 0+C
\(\Rightarrow\)C = log75
\(\therefore\)(1) becomes, log (T-25) = -\(\lambda\)t + log 75
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =-\lambda t...(2)\)
When t = 10, T = 80^oC
\(\therefore log\left( \frac { 80-25 }{ 75 } \right) =-10|\)
\(\Rightarrow log\left( \frac { 11 }{ 15 } \right) =-10\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \)
Substituting \(\lambda\) in (2) we get
\(log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t ...(3)\)
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \times 20\)
\(={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow \frac { T-25 }{ 75 } ={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow T-25=\frac { 121 }{ 225 } \times 75=\frac { 121 }{ 3 } =40.33\)
\(\Rightarrow T=40.33+25\)
\(\\ =65{ .33 }^{ 0 }C\)
So the temperature of water after 20 minutes is 65.33°C
(ii) putting T = 40^oC in (3) we get
\(log\left( \frac { 40-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(\Rightarrow log\left( \frac { 1 }{ 5 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(t=\frac { 10log\left( \frac { 1 }{ 5 } \right) }{ log\left( \frac { 11 }{ 15 } \right) } =\frac { -10log5 }{ log\left( \frac { 11 }{ 15 } \right) } \)
\(=\frac { -10\times 1.6094 }{ -0.3101 } \)
\(\therefore\) t = 53.46 minutes

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.

\( y e^{\left(\frac{1}{r}\right)} \cdot d x   =\left(x e^{\frac{1}{y}}+y\right) d y \)
\(\frac{d x}{d y}   =\frac{x \cdot e^{\left(\frac{6}{y}\right)}+y}{y e^{\left(\frac{x}{y}\right)}} \)
\(\frac{d x}{d y}   =\left(\frac{x}{y}\right)+\frac{1}{e^{\left(\frac{6}{y}\right)}}\)
Put x = \( \mathrm{vy}\) \( \Rightarrow\left(\frac{x}{y}\right)=\mathrm{v}\) and \(\frac{d x}{d y}=\cdot v+y \cdot \frac{d v}{d y} \)
\((1) \Rightarrow \ v+y \cdot \frac{d v}{d y}=v+\frac{1}{e^y} \)
\(\mathrm{e}^v \cdot \mathrm{dv}=\frac{d y}{y}\)
Integrating on both sides,
ie) \(\int e^v \cdot d v   =\int \frac{d y}{y} \)
\(e^v   =\log |y|+\log |c|  \) 
\(e^{\left(\frac{x}{y}\right)}   =\log |c y|\)

Given that s(t) = t³ − 6t² + 9t + 1. On differentiating, we get v(t) = 3t² -12t + 9 and a(t) = 6t −12.
(i) The particle is at rest when v(t) = 0 . Therefore, v(t) = 3(t −1)(t − 3) = 0 gives t = 1 and t = 3.
(ii) The particle changes direction when v (t) changes its sign. Now.
if 0 ≤ t < 1 then both (t −1) and (t − 3) < 0 and hence, v(t) > 0.
If 1< t < 3 then (t −1) > 0 and (t − 3) < 0 and hence, v(t) < 0.
If t > 3 then both (t −1) and (t − 3) > 0 and hence, v(t) > 0.
Therefore, the particle changes direction when t = 1 and t = 3.
(iii) The total distance travelled by the particle from time t = 0 to t = 2 is given by,
|s(0) − s(1)| + |s(1) − s(2)| = |1− 5 | + | 5 − 3| = 6 metres.

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

Completing the square on x and y of 4x²+36y²+40x−288y+532 = 0,
4(x² + 10x + 25 − 25) + 36(y² − 8y + 16 − 16) + 532 = 0 , gives
4(x² + 10x + 25) + 36(y² − 8y + 16) = −532 + 100 + 576
4(x + 5)² + 36(y − 4)² = 144.
Dividing both sides by 144, the equation reduces to \(\frac { { \left( x+5 \right) }^{ 2 } }{ 36 } \frac { { \left( y-4 \right) }^{ 2 } }{ 4 } =1\)
This is an ellipse with centre (-5, 4), major axis is parallel to x-axis, length of major axis is 12 and length of minor axis is 4. Vertices are (1, 4) and (-11, 4).
Now, c² = a²−b² = 36 − 4 = 32
and c = ±4 \(\sqrt { 2 } \)
Then the foci are (−5 − 4\(\sqrt { 2 } \), 4) and (−5 + 4\(\sqrt { 2 } \) , 4).
Length of the major axis = 2a = 12 units and
the length of the minor axis = 2b = 4 units.

By the definition of a conic  \(\frac{SP}{PM}\)= e or SP² = e²PM²
Then, (x−2)² + (y−3)² = \(\frac { 1 }{ 4 } \) (x-7)² 
3x²+ 4y²−2x − 24y + 3 = 0
\({ 3\left( x-\frac { 1 }{ 3 } \right) }^{ 2 }+4(y-3)^{ 2 }=3\left( \frac { 1 }{ 9 } \right) +4\times 9-3=\frac { 100 }{ 3 } \)
\(\frac { { \left( x-\frac { 1 }{ 3 } \right) }^{ 2 } }{ \frac { 100 }{ 9 } } +\frac { (y-3{ ) }^{ 2 } }{ \frac { 100 }{ 12 } } \) = 1 which is in the standard form.
Therefore, the length of major axis = 2a = 2\(\sqrt { \frac { 100 }{ 9 } = } \frac { 20 }{ 3 } \) and
the length of minor axis = 2b = 2\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).

 f(x) = sin^-1 \((\frac{|x|-2}{3})+cos^-1(\frac{1-|x|}{4})\)
From the definition of sin^-1
\(-1\le \frac { \left| x \right| -2 }{ 3 } \le 1\)
\(\Rightarrow -3\le \left| x \right| -2\le 3\)
\(\Rightarrow -3+2\le \left| x \right| \le \left| x \right| \le 3+2\)
\(\Rightarrow -1\le \left| x \right| \le 5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(\Rightarrow 0\le \left| x \right| and\left| x \right| \le 5\)
\(\Rightarrow \left| x \right| \ge 0and-5\le x\le 5\)
From the definition of cos^-1x.
\(-1\le \frac { 1-\left| x \right| }{ 4 } \le 1\)
\(\Rightarrow -4\le 1-\left| x \right| \le 4\)
\(\Rightarrow -4-1\le \left| x \right| \le 4-1\)
\(\Rightarrow -5\le -\left| x \right| \le 3\)
\(\Rightarrow -3\le \left| x \right| >5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(-5\le |x|\le 5\)
From (1) & (2),
Domain is [-5, 5]

Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)

Given p, q are the roots of lx² + nx + n = 0 
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S 
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α and β, respectively, with positive x-axis, where A and B are as in the diagram.
Draw AL and BM perpendicular to the x-axis. Then \(\left| \vec { OL } \right| =\left| \vec { OA } \right| \) cos α = cos α, \(\left| \vec { LA } \right| =\left| \vec { OA } \right| \) sin α =  sin α
So, \(\vec { OL } =\left| \vec { OL } \right| \)\(\hat { i } \) =  cos,α \(\hat { i } \), \(\overrightarrow { LA } \) = sin α (-\(\hat { j } \))
Therefore, \(\hat { a } =\overrightarrow { OA} = \overrightarrow { OL } +\overrightarrow { LA } \) = cos α \(\hat { i } \) - sin α \(\hat { j } \) ..(1)
Similarly \(\hat { b } \) = cos β \(\hat { i } \)+ sin β \(\hat { j } \) ....(2)
The angle between \(\hat { a } \) and \(\hat{b}\) is α + β and so,
\(\hat { a } .\hat { b } =\left| \hat { a } \right| \left| \hat { b } \right| \) cos (α + β) = cos (α + β) ... (3)

On the other hand, from (1) and (2)
\(\hat { a } .\hat { b } =(cos\alpha \hat { i } -sina\hat { j } )(cos\beta \hat { i } -sin\beta \hat { j } )\) = cos α cos β - sin α sin β....(4)
From (3) and (4), we get cos(α + β) = cos α cos β - sin α sin β

(i) At the maximum height, the velocity v(t) of the particle is zero.
Now, we find the velocity of the particle at time t.
\(v(t)=\frac{ds}{dt}=128-32t\)
\(v(t)=0 \Rightarrow 128-32t=0 \Rightarrow t=4.\)
After 4 seconds, the particle reaches the maximum height.
The height at t = 4 is s(4) = 128(4) - 16(4)² = 256 ft.
(ii) When the particle hits the ground then s = 0 .
s = 0 ⇒ 128t −16t² = 0
⇒ t = 0, 8 seconds.
The particle hits the ground at t = 8 seconds. The velocity when it hits the ground v(8) = –128 ft /s.