\(y=\frac { 2x-{ x }^{ 2 } }{ 2(1+sinx) } ,x\neq n\pi +(-1)^{ n }\frac { \pi }{ 2 } ,\)∀ n ε Z

\(y=-\frac { 1 }{ 2 } { cos }^{ 2 }x+\frac { { cos }^{ 5 }x }{ 5 } +{ xe }^{ x }-{ e }^{ x }+c\)

\(\frac { ab\left( \pi -2 \right) }{ 4 } \)

(i) not exist sq.units.
(ii) \(\frac { 32 }{ 15 } \)

100 m / s, t = 4 sec, 200 m / s, −100 m / s

Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

Equation of the given curves are y = sin x  ..(1)
Y = cos x ...(2)
from (1) and (2), sin x = cos x
y = sin x

y = cos x

\(\Rightarrow x=\frac { \pi }{ 4 } \)

\(\therefore \) Required area = \(2\int _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }{ (sinx-cosx)dx } \)
[\(\because\) the area is symmetrical about X - axis]
\(=2{ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }\)
\(=-2\left[ \left( cos\frac { 3\pi }{ 4 } +sin\frac { 3\pi }{ 4 } \right) -\left( cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } \right) \right] \)
= -2\(\left[ \left( -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) \right] \)
[cos 135^o = cos(180^o- 45) = -cos 45^o sin135^o = sin(180^o- 45) = -sin 45^o]
\(=-2\left[ \frac { -2 }{ \sqrt { 2 } } \right] =\frac { 4 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } \) 

X~ B(n, p)
Given mean np = 6
\(S.D=\sqrt { npq } =2\) 
\(\Rightarrow npq=4\) 
\( \rightarrow \frac { npq }{ np } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \) 
\(\Rightarrow q=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-P=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-\frac { 2 }{ 3 } =P\) 
\(\therefore P=\frac { 1 }{ 3 } \)  
\(n\times \frac { 1 }{ 3 } =6\Rightarrow n=18\)
(i) The probability mass function
P(X = x) nC_x p^x (1 - p )^n-x,
X = 0,1,2, ... , n
\(\therefore P(X=x)=\ ^{18}{ C }_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 18-x }\) 
x=0,1,2...,8
(ii) \(P(X=3)=\ ^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 18-3 }\) 
= \(^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 15 }\)
(iii) P(X ≥ 2)  
 P(X ≥ 2) 1 -P(X < 2)
= 1 - [P(X = 0) + P(X = 1)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\sqrt { tan\ x } +\sqrt { cot\ x } )dx } \) Then we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \sqrt { \frac { sinx }{ cosx } } +\sqrt { \frac { cosx }{ sinx } } \right) dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { sinxcosx } } dx } =\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { 2sinxcosx } } dx } \)
\(=\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{2 } }{ \frac { (sinx+cosx)dx }{ \sqrt { 1-sinx-cosx{ ) }^{ 2 } } } } \)
Put u = sin x − cos x 
Then, du = (cos x + sin x)dx .
When x = 0, u = −1
When x =\(\frac{\pi}{2}\), u = 1
\(\therefore I=\sqrt { 2 } \int _{ -1 }^{ 1 }{ \frac { du }{ \sqrt { 1-{ u }^{ 2 } } } =\sqrt { 2 } { [{ sin }^{ -1 }u] }_{ -1 }^{ 1 }=\sqrt { 2 } \left[ { sin }^{ -1 }(1)-{ sin }^{ -1 }(-1)) \right] } =\pi \sqrt { 2 } \)

The given function is defined and is differentiable at all \(x \in(0, \infty)\) 
f(x) = x log x + 3x.
Therefore f'(x) = log x+1+3 = 4 + log x.
The stationary points are given by 4 + log x = 0
That is x = e^-4
Hence the intervals of monotonicity are (0, e^-4) and \((e^{-4}, \infty)\) 
At \(x=e^{-5}\in(0,e^{-4})\), f'(e^-5) = -1<0 and hence in the interval (0, e^-4) -the function is strictly decreasing.
At \(x=e^{-3}\in(0,e^{-4})\), f'(e^-3) = 1>0 and hence strictly increasing in the interval \((e^{-4}, \infty)\). 
Since f′(x) changes from negative to positive when passing through x = e⁻⁴, the first derivative test tells us there is a local minimum at x = e^-4 and it is f(e-⁴) = -e^-4.

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

(x²+y²)dy = xy dx
\(\frac { dy }{ dx } =\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } ...(1)\)
\(\therefore put=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dv }{ dx } =\frac { xvx }{ { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } \)
\(=\frac { { x }^{ 2 }v }{ { x }^{ 2 }(1+{ v }^{ 2 }) } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } v=\frac { v-v-{ v }^{ 3 } }{ 1+{ v }^{ 2 } } =\frac { -{ v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
Separating the variables we get,
\(\frac { 1+{ v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \frac { 1 }{ { v }^{ 3 } } +\frac { { v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \int { { v }^{ -3 }dv } +\int { \frac { dv }{ v } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { { v }^{ -2 } }{ -2 } +log\ v=-logx+logc\)
\(\Rightarrow -\frac { 1 }{ 2{ v }^{ 2 } } +log\ v=-log\ x+log\quad c\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } -log\ v=logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =log\left( \frac { vx }{ c } \right) \)
\(\Rightarrow \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =log\left( \frac { y }{ c } \right) \Rightarrow { e }^{ \frac { { x }^{ 2 } }{ { e }^{ 2{ y }^{ 2 } } } }=\frac { y }{ c } \)
\(\Rightarrow y={ ce }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } } ...(2)\)
Given y(1) = 1
\(1={ ce }^{ \frac { 1 }{ 2 } }\Rightarrow 1=c\sqrt { e } \)
\(\Rightarrow c=\frac { 1 }{ \sqrt { e } } \)
\(\therefore\)(2) becomes,
\(y=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } }\)
\(Also\ y({ x }_{ 0 })=e\Rightarrow e=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow e\sqrt { e } ={ e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =log\quad e\sqrt { e } =log{ e }^{ \frac { 3 }{ 2 } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =\frac { 3 }{ 2 } { log }_{ e }^{ e }=\frac { 3 }{ 2 } (1)\)
\(\left[ \because { log }_{ e }^{ e }=1 \right] \)
\(\Rightarrow { x }_{ 0 }^{ 2 }=\frac { 3 }{ 2 } (2{ e }^{ 2 })={ 3e }^{ 2 }\)
\(\Rightarrow { x }_{ 0 }=\pm \sqrt { 3{ e }^{ 2 } } =\pm \sqrt { 3 } .e\)
\(\therefore { x }_{ 0 }=\pm \sqrt { 3 } .e\)

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

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

Augmented matrix [A|B] is \(\left[ \begin{matrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)
[A|B]\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ -2 & 1 & 1 \\ 1 & 1 & -2 \end{matrix}|\begin{matrix} b \\ a \\ c \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 3 & -3 \end{matrix}|\begin{matrix} b \\ a+2b \\ c-b \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} b \\ c+2b \\ a+b+c \end{matrix} \right] \)
Here \(\rho\) (A) = 2
The given system is consistent only when \(\rho\)([A|B]) = 2\(\rho\)([A|B]) = 2 only if a + b + c = 0 Hence proved.

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

Now, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } =tan^{ -1 }{ a }_{ 2 }-tan^{ -1 }{ a }_{ 1 }\)
Similarly, \(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) =tan^{ -1 }{ a }_{ 3 }-tan^{ -1 }{ a }_{ 2 }\)
Continuing inductively, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ n-1 } }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) =tan^{ -1 }{ a }_{ n }-tan^{ -1 }{ a }_{ n-1 }\)
Adding vertically, we get
\(tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +....+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) tan[tan^{ -1 }{ a }_{ n }-{ tan }^{ -1 }{ a }_{ 1 }]\\ \)
\(tan\left[ tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 1 }{ a }_{ 2 } } \right) +tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ 2 }{ a }_{ 3 } } \right) +...+tan^{ -1 }\left( \frac { d }{ 1+{ a }_{ n }{ a }_{ n-1 } } \right) \right] =tan\left[ tan^{ -1 }{ a }_{ n }-tan^{ -1 }a_{ 1 } \right] \)\(=\left[ tan^{ -1 }\left( \frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \right) \right] =\frac { { a }_{ n }-{ a }_{ 1 } }{ 1+{ a }_{ 1 }{ a }_{ n } } \)

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\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).

Observe that the given curve is neither a circle nor an ellipse. For your reference the curve is shown in Figure.

Now,  \(\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} } \)
= -\(\frac{6 cos2t}{6sin3t} = -\frac{cos2t}{sin3t} \).
Therefore, the tangent at any point is
\(y-3sin2t= -\frac{cos2t}{sin3t}(x-2cos3t)\)
That is, x cos 2t + y sin 3t = 3sin 2t sin 3t + 2cos 2t cos 3t.
The slope of the normal is the negative of the reciprocal of the tangent which in this case is \(\frac{sin3t}{cos2t}\). Hence, the equation of the normal is \(y-3sin2t=\frac{sin3t}{cos2t}(x-2cos3t)\).
That is, x sin 3t - y cos 2t = 2sin 3t cos3t 3sin 2t cos 2t = sin 6t - \(\frac{3}{2}\) sin 4t.

\(f(x)=\frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } \) 
\(\underset { x\rightarrow \infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } \) 
= \(\frac { 3 }{ \sqrt { 1+0 } } =3\) 
ஃy = 3 is the horizontal asymptote.
Also \(\underset { x\rightarrow -\infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow { 0 }^{ - } }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } =-3\) 
ஃ y = -3 is the horizontal asymptote