\({ e }^{ 2x }=1+\frac { 2x }{ 1! } +\frac { \left( 2x \right) ^{ 2 } }{ 2! } +\frac { \left( 3x \right) ^{ 3 } }{ 3! } +...\)

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

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
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as,
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{+}}{lim}(\frac{cos \ x}{2x})=\infty\)
\(\underset{x\rightarrow 0^{-}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{-}}{lim}(\frac{cos \ x}{2x})=\infty\)
As the left limit and the right limit are not the same we conclude that the limit does not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in \(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\) to conclude
\(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\)\(\underset{x\rightarrow 0^{+}}{lim} (\frac{-sin \ x}{2})\)=0
which is not true because it was not an indeterminate form.

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

Given \(f(x)=|\frac{1}{x}|, x\in [-1,1]\)
Rolle's theorem is not applicable since \(f(x)=(\frac{1}{x})\) is not continuous at x = 0 in [-1, 1] and not differentiable in (-1, 1)

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

By revolving the upper semicircular region enclosed between the circle x² + y² = a² and the x-axis, we get a sphere of radius a.

The boundaries of the region are y = \(\\ \\ \\ \\ \\ \\ \\ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \) x-axis, the lines x = −a and x = a. Hence, the volume of the sphere is given by
\(v=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx=\pi } \int _{ -a }^{ a }{ \left( { a }^{ 2 }-{ x }^{ 2 } \right) } dx\) 
\(=2\pi \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand (a²-x²) is an even function
\(=2\pi { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=2\pi \left( { a }^{ 3 }-\frac { { a }^{ 3 } }{ 3 } \right) =\frac { 4 }{ 3 } { \pi a }^{ 3 }\)

The sample space S = {H,T} \(\times\) {H,T}
That is S = {TT, TH, HT, HH}
Let X be the random variable denoting the number of heads.
Therefore
X (TT ) = 0 , X (TH ) = 1,
X (HT) = 1, and X (HH) = 2 .
Then the random variable X takes on the values 0, 1 and 2

Values of the Random Variable	0	1	2	Total
Number of elements in inverse images	1	2	1	4

The probabilities are given by

\(f(0)=P(X=0)=\cfrac { 1 }{ 4 } \) 
\(f(1)=P(X=1)=\cfrac { 1 }{ 2 } \) 
and \(f(2)=P(X=2)=\cfrac { 1 }{ 4 } \) 
The function f (x) satisfies the conditions
(i) f (x) ≥ 0 , for x = 0, 1, 2
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=2 }{ f(x) } =f(0)+f(1)+f(2)\) 
= \(\cfrac { 1 }{ 4 } +{ \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } =1 }\) 
Therefore f (x) is a probability mass function.
The probability mass function is given by

x	0	1	2
f(x)	\(\cfrac { 1 }{ 4 } \)	\(\cfrac { 1 }{ 2 } \)	\(\cfrac { 1 }{ 4 } \)

(or)
\(f(x)\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & forx=0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } & forx=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 4 } & forx=2 \end{matrix} \end{cases}\)

\(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\sqrt { tanx } } dx } \int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\frac { \sqrt { sin\quad x } }{ \sqrt { cos\quad x } } } dx } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx\quad \quad ..(1) } \)
By the property,\( \int _{ a }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(a+b-x)dx } } \)
we get \(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } }{ \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } +\sqrt { sin(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 2 } -x) } }{ \sqrt { cos(\frac { \pi }{ 2 } -x) } +\sqrt { sin(\frac { \pi }{ 2 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { cos\quad x } } ...(2) } \)
\((1)+(2)\rightarrow \)
\(2I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx+\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { sin\quad x } } } } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ 3\frac { \pi }{ 8 } }{ \frac { \sqrt { cos\quad x } +\sqrt { sin\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ dx={ [x] }_{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } } } } \)
\(2I=\frac { 3\pi }{ 8 } -\frac { \pi }{ 8 } =\frac { 2\pi }{ 8 } =\frac { \pi }{ 4 } \)
\(\therefore I=\frac { \pi }{ 8 } \)

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

\(\underset { x\rightarrow \infty }{ lim } { e }^{ -x }\sqrt { x } =\underset { x\rightarrow \infty }{ lim } \frac { { \sqrt { x } } }{ { e }^{ x } } =\frac { \infty }{ \infty } \)
Which is in indeterminate form. Applying L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { e }^{ -x } }{ \sqrt { x } } \)
\(\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } e^{ -x }\sqrt { \frac { 1 }{ x } } =\frac { 1 }{ 2 } { e }^{ -\infty }(0)=0\) [When x ➝ ∞, \(\frac1x\) ➝ 0 e^-∞ = 0]

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 that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

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.

We have,
\(\frac{d}{dt}W(t)=-20\times(1-0.1t)\)
Therefore at t = 2, \(\frac{d}{dt}W(t)=-16\)
That is, the person forgets at the rate of 16 words after 2 days of studying.

Given the circles centre on r-axis & the circle is passing through the origin.
Let it be (r, 0) & its radius r.

Equation of the circle is
(x - a)² + (y - b)² = r²
(x - r)² + (y - 0)² = r²
⇒ x² - 2xr + r² + y² = r²
⇒ x² - 2xr + y² = 0 ...(1)
defferentiating equation (1) with respect to 'x' we get
⇒ 2x - 2r + 2y \(\frac { dy }{ dx } =0\)
⇒ 2x + 2y \(\frac { dy }{ dx } =2r\)
⇒ x + y \(\frac { dy }{ dx } =r\) ...(2)
Substituting r value in equation (1), we get
x² - 2x \(\left( x+y\frac { dy }{ dx } \right) +{ y }^{ 2 }=0\)
\(\Rightarrow \ { x }^{ 2 }-{ 2x }^{ 2 }-2xy\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \ { -x }^{ 2 }{ -2x }y\left( \frac { dy }{ dx } \right) { +y }^{ 2 }\)
Multiply by '-', we get
\(\Rightarrow \ { x }^{ 2 }{ +2x }y\left( \frac { dy }{ dx } \right) { -y }^{ 2 }\) which is the required differential equation.

p	q	r	~ q	~q V r	p➝(¬qVr)	~p	~pV(~qVr)
T	T	T	F	T	T	F	T
T	T	F	F	F	F	F	F
T	F	T	T	T	T	F	T
T	F	F	T	T	T	F	T
F	T	T	F	T	T	T	T
F	T	F	F	F	T	T	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴  p➝(¬q V ~r) ≡ ~pV(~q V r) 
Hence proved.

(i) The output m+ n - mn is clearly an integer and hence∗ is a binary operation on Z.
(ii) m*n = m+ n − mn = n + m − nm = n*m, ∀m,n∈Z. So ∗ has commutative property.
(iii) Consider (m*n)*p = (m+ n −m n)* p= (m+ n −mn) + p − (m+ n −m n) p
= m+ n + p −mn −m p − n p + m n p ... (1)
Similarly m*(n*p) = m*(n + p − n p) = m+ (n + p − n p) −m (n + p − n p)
= m+ n + p − n p −m n −mp + m n p ... (2)
From (1) and (2), we see that m*(n*p) = (m*n)*p. Hence * has associative property.
(iv) An integer e is to be found such that 
m*e = e*m = m, ∀m∈Z ⇒ m + e - m e = m
⇒e(1-m) = 0 ⇒ e = 0 or m = 1. But m is an arbitrary integer and hence need not be equal to 1. So the only possibility is e = 0. Also m*0 = 0*m = m, ∀m∈Z. Hence 0 is the identity element and hence the existence of identity is assured.
(v) An element m'∈ Z is to be found such that m*m' = m' * m = e = 0, ∀m∈Z.
m*m' = 0 ⇒ m+m'-m m'= 0⇒ m m' = 0 ⇒ \(\frac{m}{m-1}\). when m = 1, m' is not defined.
When m = 2, m' is an integer. But except m = 2, m′ need not be an integer for all values of m. Hence inverse does not exist in Z.

Given v (x, y) = log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \)
Since log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \) is not homogeneous,
left f (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \)
⇒ v = log f
⇒ ev = f
Now, f(tx, ty) = \(\frac { { t }^{ 2 }{ x }^{ 2 }+{ t }^{ 2 }{ y }^{ 2 } }{ tx+ty } =\frac { { t }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ t(x+y) } \)
= t.f(x, y)
∴ f is a homogeneous function of degree 1.
∴ By Euler's theorem,
⇒ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 1.f
From (1), \(x.\frac { \partial }{ \partial x } \left( { e }^{ v } \right) +y.\frac { \partial }{ \partial y } \left( { e }^{ v } \right) ={ e }^{ v }\) [using (1)]
⇒ \(x.{ e }^{ v }.\frac { \partial }{ \partial x } +y.{ e }^{ v }.\frac { \partial }{ \partial y } ={ e }^{ v }\)
⇒  \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = 1 [Divided by e^v]

(i) Since X denotes the number of success, X can take the values 0,1, 2, ...10
The probability for success is \(p=\frac { 1 }{ 4 } \) and for failure \(q=1-p=\frac { 3 }{ 4 } \) and n = 10
Therefore X follows a binomial distribution denoted by \(X\sim B\left( 10,\frac { 1 }{ 4 } \right) \) 
This gives,\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ x }\left( \cfrac { 3 }{ 4 } \right) ^{ 10-x }\)  x = 0, 1, 2,..,10
(ii) Probability for seven correct answers is 
\(P(X=7)=f(7)=\left( \begin{matrix} 10 \\ 7 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 7 }\left( \cfrac { 3 }{ 4 } \right) ^{ 10-7 }=120\left( \cfrac { { 3 }^{ 2 } }{ { 4 }^{ 10 } } \right) \)
Probability that the student will get seven correct answers is \(120\left( \cfrac { { 3 }^{ 2 } }{ { 4 }^{ 10 } } \right) \) 
(iii) Probability for at least one correct answer is
P(X ≥1) = 1- P(X <1) = 1- P(X = 0)
= \(1-\left( \begin{matrix} 10 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 0 }\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }=1-\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }\)
Probability that the student will get for at least one correct answer is \(1-\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }\)  

Observe that the given distribution is continuous
By definition \(\mu =E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -2x } \right) dx } +\int _{ 0 }^{ \infty }{ x\left( { \lambda e }^{ -\lambda x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ x\left( { e }^{ -\lambda x } \right) dx } \) 
=  \(0+\lambda \left( \frac { 1 }{ { \lambda }^{ 2 } } \right) \) (using Gamma integral for positive integer n,\(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\cfrac { n }{ { a }^{ n+1 } } \))
= \(\frac { 1 }{ \lambda } \) 
Variance :
By definition,\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -\lambda x } \right) } dx+\int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \lambda { e }^{ -2x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( { e }^{ -2x } \right) dx } \)
 (using Gamma integral for positive integer)
Therefore Var(X ) = E(X² )- E(X )²
= \(\frac { 2 }{ { \lambda }^{ 2 } } -\left( \frac { 1 }{ \lambda } \right) ^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } } \) 
Hence the mean and variance are respectively \(\frac { 1 }{ \lambda } \)  and \(\frac { 1 }{ { \lambda }^{ 2 } } \)

The lines 5x − 2y = 15, x + y + 4 = 0 intersect at (1, −5). The line 5x − 2y =15 meets the x-axis at (3, 0). The line x + y + 4 = 0 meets the x-axis at (−4, 0). The required area is shaded. It lies below the x-axis. It can be computed either by considering vertical strips or horizontal strips.
When we do by vertical strips, the region has to be divided into two sub-regions by the line x = 1. Then, we get

\(A=\left| \int _{ -4 }^{ 1 }{ ydx } \right| +\left| \int _{ 1 }^{ 3 }{ ydx } \right| \)
\(=\left| \int _{ -4 }^{ 1 }{ (-4-x)dx } \right| +\left| \int _{ 1 }^{ 3 }{ \left( \frac { 5x-15 }{ 2 } \right) dx } \right| \)
\(=\left| { \left( -4x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -4 }^{ 1 } \right| +\left| { \left( \frac { { 5x }^{ 2 } }{ 4 } -\frac { 15x }{ 2 } \right) }_{ 1 }^{ 3 } \right| \)
\(=\left| \left( -\frac { 9 }{ 2 } \right) -\left( 8 \right) \right| +\left| \left( -\frac { 45 }{ 4 } \right) -\left( -\frac { 25 }{ 4 } \right) \right| \)
\(=\frac { 25 }{ 2 } +5\)
\(=\frac { 35 }{ 2 } \)
When we do by horizontal strips, there is no need to subdivide the region. In this case, the area is bounded on the right by the line 5x − 2y = 15 and on the left by x + y + 4 = 0. So, we get
\(A=\int _{ -5 }^{ 0 }{ [{ x }_{ R }-{ x }_{ L }]dy } =\int _{ -5 }^{ 0 }{ \left[ \frac { 15+2y }{ 5 } -(-4-y) \right] dy } \)
\(=\int _{ -5 }^{ 0 }{ \left[ 7+\frac { 7y }{ 5 } \right] dy={ \left[ 7y+\frac { { 7y }^{ 2 } }{ 10 } \right] }_{ -5 }^{ 0 } } \)
\(\\ =0-\left[ -35+\frac { 35 }{ 2 } \right] =\frac { 35 }{ 2 } \)

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(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

Let us put I = \(\int ^\frac{\pi}{4}_{0}\) log(1 + tan x) dx
Applying the property \(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\)f(a-x) dx in equation (1), we get
I = \(\int ^\frac{\pi}{4}_{0}\) log \([1+tan (\frac{\pi}{4}-x)]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([1 + \frac{tan \frac {\pi}{4}- tan x}{1 + tan {\frac {\pi}{4} tan x}}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([1 + \frac {1 - tan x}{1 + tan x}]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([\frac {1+tan x +1 - tan x}{1 + tan x}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([\frac{2}{1+tanx}]\) dx = \(\int ^\frac{\pi}{4}_{0}\) [log 2 - log (1+tan x)] dx
= log 2  \(\int ^\frac{\pi}{4}_{0}\) dx - \(\int ^\frac{\pi}{4}_{0}\)log (1+tan x)] dx
= \(\frac{\pi}{4}\)log 2 - I
So, we get 2I = \(\frac{\pi}{4}\)log 2.
Hence, we get I = \(\frac{\pi}{8}\)log 2.

Given probability mass function is

x	0	1	2	3	4
f(x)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

Let A be the mass of the isotope remaining after t years, and let −k be the constant of proportionality, where k > 0. Then the rate of decomposition is modeled by \(\frac{da}{dt}=-kA,\) where the minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables,we get\(\frac{da}{dt}=-kdt\).
Integrating on both sides, we get log |A| = −kt + log |C| or A = Ce−^kt.
Given that the initial mass is 200mg. That is, A = 200 when t = 0 and thus, C = 200.
Thus, we get A = − 200e^-kt.
Also, A =150when t = 2 and therefore, k = \(\frac{1}{2}log(\frac{4}{3})\)
Hence, A(t) = 200e^{\(\frac{1}{2}log(\frac{4}{3})\)} is the mass of isotope remaining after t years.
The half-life t_h is the time corresponding to A = 100 mg
Thus, \({ t }_{ k }=\frac { 2log\left( \frac { 1 }{ 2 } \right) }{ log\left( \frac { 3 }{ 4 } \right) } \).

Given AC = b and CD = a
Let hand r be the height and radius of the cylinder
ஃ AB = AC - BC = b - h
In the diagram

ΔABE - ΔACD
\(\therefore \frac { AB }{ BE } =\frac { AC }{ CD } \) 
\(\Rightarrow \frac { b-h }{ r } =\frac { b }{ a } \)
 \(\Rightarrow r=\frac { a\left( b-h \right) }{ b } \)
Volume of the cylinder = \({ \pi r }^{ 2 }h\)
= \(\pi \left( \frac { a }{ b } \left( b-b \right) ^{ 2 }h \right) \) 
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\)
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }+{ h }^{ 2 }-2bh \right) h\)
 \(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }{ h }^{ 2 }+{ h }^{ 3 } \right) \)
Differentiating with respect to 'h',
\(\frac { dV }{ dh } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( { b }^{ 2 }+{ 3h }^{ 2 }-4bh \right) \)
\(\frac { dV }{ dh } =0\Rightarrow { b }^{ 2 }-4bh+{ 3j }^{ 2 }=0\)
\(\Rightarrow (b-3h)+(b-h)=0\Rightarrow b-3h=0\quad \)or b = h
\(\Rightarrow h=\frac { b }{ 3 } asb\neq h\) 
\(\frac { dV }{ { dh }^{ 2 } } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
\(\left( \frac { { d }^{ 2 }V }{ { dh }^{ 2 } } \right) _{ at\ h=\frac { b }{ 3 } }=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
= \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi (-2b)<0\) 
ஃ V is maximum at \(h=\frac { b }{ 3 } \)
ஃ Maximum volume of the cylinder
\(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\) 
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-\frac { b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \left[ \therefore h=\frac { b }{ 3 } \right] \)
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( \frac { 2b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \)
\(=\frac{\pi a^{2}}{\not b^{2}}\left(\frac{4\not b^{2}}{9}\right)\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^{2} b\right)\)
= \(\frac { 4 }{ 9 } \)(Volume of the cone)
Hence, volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 } \) times volume of the cone.

The given differential equation may be written
\(\Rightarrow \frac { dy }{ dx } =\frac { -3{ e }^{ \frac { y }{ x } }\left( 1-\frac { y }{ x } \right) }{ 1+3{ e }^{ \frac { y }{ x } } } ...(1)\)
This is a homogeneous differential equation
\(\therefore put\quad y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(v+x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } -v\)

\(=\frac { -3{ e }^{ v }-v }{ 1+3{ e }^{ v } } =-\left( \frac { -3{ e }^{ v }+v }{ 1+3{ e }^{ v } } \right) \)
Separating the variables we get,
\(\frac { 1+3{ e }^{ v } }{ 3{ e }^{ v }+v } dv=-\frac { dx }{ x } \)
\(\Rightarrow log(3{ e }^{ v }+v)=-log\quad x+log\quad c\)
\(\Rightarrow log(3{ e }^{ v }+v)=log\left( \frac { c }{ x } \right) \Rightarrow { 3e }^{ v }+v=\frac { c }{ x } \)
\({ 3e }^{ \frac { y }{ x } }+\frac { y }{ x } =\frac { c }{ x } \)
\(\Rightarrow \frac { 3x{ e }^{ \frac { y }{ x } }+y }{ x } =\frac { c }{ x } \)
\(\Rightarrow y+3x{ e }^{ \frac { y }{ x } }=c\)
Given that y = 0 when x = 1
3(1)e⁰+ 0 = c \(\Rightarrow\) c = 3
\(\therefore\) (2) becomes, 3x\({ e }^{ \frac { y }{ x } }\)+ y = 3

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 h = 12 m
r = 5 m
∴ tan α = \(\frac { OA }{ VO } =\frac { 5 }{ 12 } =\frac { r }{ h } \)
⇒ 12r = 5h 
⇒ r = \(\frac{5h}{12}\) ...(1)
Given rate of changeof volume = \(\frac{dv}{dt}=10\)
Volume of cone V = \(\frac13\) πr²h
= \(\frac { 1 }{ 3 } \)π\({ \left( \frac { 5h }{ 12 } \right) }^{ 2 }.h\) [From (1)]
\(v=\frac { 1 }{ 3 } \pi { \left( \frac { 25h }{ 144 } \right) }^{ 2 }=\frac { { 25\pi h }^{ 3 } }{ 3\times 144 } \)
Differentiating with respect to 't' we get,
\(\frac { dv }{ dt } =\frac { 25\pi }{ 3\times 144 } 3{ h }^{ 2 }\frac { dh }{ dt } \)
= \(\frac { 25\pi }{ 144 } { h }^{ 2 }\frac { dh }{ dt } \)
[∵ Given h = 8, \(\frac { dh }{ dt } \) = 10m³/min]
\(\Rightarrow \frac { 10\times 144 }{ 25\pi \times 64 } =\frac { dh }{ dt } \)
\(\Rightarrow \frac { dh }{ dt } =\frac { 9 }{ 10\pi } \)
The depth of the water increasing at the rate of \(=\frac { 9 }{ 10\pi } \)m / min.