Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

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

\(P(X=3)=\frac { 160 }{ 729 } \) 

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

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 \(f(x)=x-2logx, x\in [2,7]\)
(i) f(x) is continuous in [2, 7]
(ii) f(x) is differentiable in (2, 7)
f(2) = 2 - 2 log 2
= 2 - log 2² = 2 - log 4
f (7) = 7 - 2 log 7
= 7 - log 7² = 7 - log 49
Since f(2) ≠ f (7), Rolle's theorem is not applicable.

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +4=0\)
Given differential equation is
\({ \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }-3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) +5\frac { dy }{ dx } +4=0\)
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ \frac { 2 }{ 3 } }=3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4\)
Taking power 3 both sides,
\(\Rightarrow { \left( \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \right) }^{ 2 }={ \left( 3\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) -5\left( \frac { dy }{ dx } \right) -4 \right) }^{ 3 }\)
The highest derivative is 3 and its power is 2.
∴ Order 3, degree 2.

Given f(x) = 4k (x-1)², 1 ≤ x ≤ 3.
Since f(x) is a p.d.f\(\int_{1}^{3} f(x) d x=1\)
\( \Rightarrow \int_{1}^{3} 4 k(x-1)^{3} d x =1 \)
\(\Rightarrow 4 k\left[\frac{(x-1)^{4}}{4}\right]_{1}^{3} =1 \)
\(\Rightarrow \ k\left[2^{4}-0^{4}\right] =1 \)
\(\Rightarrow \ 16 k =1 \)
\(\Rightarrow \ k =\frac{1}{16} \)
\(\therefore \mathrm{P}(1 \leq \mathrm{X} \leq 2) =\int_{+1}^{2} f(x) d x \)
\(\Rightarrow \ =\int_{1}^{2} \frac{1}{4}(x-1)^{3} d x \)
\(\Rightarrow 1{ }^{2} =\frac{1}{4}\left[\frac{(x-1)^{4}}{4}\right]_{1}^{2} \)
\( =\frac{1}{16}\left(1^{4}-0^{4}\right)=\frac{1}{16} \)

\(\left( \frac { dy }{ dx } \right) ^{ 2 }\left( { x }^{ 2 }-2{ y }^{ 2 } \right) -4\left( \frac { dy }{ dx } \right) xy-{ x }^{ 2 }=0\)

Clearly 1 is the identity element of (G₁)
Inverse of 1 is 1 [∴ (1)(1) = 1]
Inverse of i is -i [∴ (i) (-i) = -i² = 1]
Inverse of -1 is -1 [∴ (-1)(-1) = 1]
Inverse of is i [∴ (-i)(i) = -i² = 1]

(i) Differentiating F(x) with respect to x at continuity points of f(x), we get
\(f(x)={ F }^{ 1 }(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & x\ge 1 \end{matrix} \end{cases}\) 
The pdf f(x) is not continuous at x = 0, or at x = 1. We can define f(0) and f(1) in any manner. Choosing f(0) = 1, and f(1) = 0 .
Therefore the probability density function f(x) is
\(f(x)=\begin{cases} \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(ii) P(0.2 ≤ X ≤ 0.7) = F(0.7) − F(0.2)
= 0.7-0.2 = 0.5
\(P(0.2\le X\le 0.7)=\int _{ 0.2 }^{ 0.7 }{ f(x) } dx=\int _{ 0.2 }^{ 0.7 }{ 1dx } =0.5\)

\(Let\quad t=\frac { x }{ 4 } \Rightarrow dt=\frac { dx }{ 4 } \Rightarrow dx=4dt\)
Here n = 7, which is odd
\(\therefore I=4\int _{ 0 }^{ \pi /2 }{ { sin }^{ 7 }tdt } \)

x	0	2\(\pi\)
t	0	\(\frac{\pi}{2}\)

\(=4\times \frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1\)
\(I=\frac { 64 }{ 35 } \)

\(\int _{ 0 }^{ 1 }{ \frac { 2x+7 }{ { 5x }^{ 2 }+9 } dx } =\int _{ 0 }^{ 1 }{ \frac { 2x }{ { 5x }^{ 2 }+9 } } +7\int _{ 0 }^{ 1 }{ \frac { dx }{ (5{ x }^{ 2 })+{ 3 }^{ 2 } } =\frac { 1 }{ 5 } } log{ [{ 5x }^{ 2 }+9] }_{ 0 }^{ 1 }+\frac { 7 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }{ \left( \frac { 3 }{ \sqrt { 5 } } \right) }^{ 2 } } } \)
\(=\frac { 1 }{ 5 } [log14-log9]+\frac { 7 }{ 5 } \times \frac { \sqrt { 5 } }{ 3 } { \left[ { tan }^{ -1 }\frac { x }{ \left[ \frac { 3 }{ \sqrt { 5 } } \right] } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 5 } log\frac { 14 }{ 9 } +\frac { 7 }{ 3\sqrt { 5 } } { tan }^{ -1 }\frac { \sqrt { 5 } }{ 3 } \)

Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

This is an indeterminate form \(\frac{\infty}{\infty}\) and hence we use the l’Hôpital’s Rule to evaluate.
\(\underset{x\rightarrow 1^{-}}{lim}(\frac{log(1-x)}{cot(\pi x)})=\underset{x\rightarrow1^{-}}{lim}(\frac{-\frac{1}{1-x}}{-\pi cosec^{2}(\pi x)})\) \((\frac{\infty}{\infty})\) 
On Simplication,
\(=\underset{x\rightarrow-1}{lim}(\frac{sin^{2}(\pi x)}{\pi (1-x)})\) \((\frac{0}{0})\)
again applying the l’Hôpital Rule 
\(= \underset{x\rightarrow 1^{-}}{lim}(\frac{2\pi sin(\pi x). cos(\pi x)}{-\pi})\)
=\(\underset{x\rightarrow -1}{lim}(-2 sin(\pi x).cos (\pi x))\)
= 0.

Equating of the given curve is y = x⁴+ 2e^x
\(\frac { dy }{ dx } \) =  x⁴+ 2e^x
m = \(\frac { dy }{ dx } \)_{(0, 2)}
= 4(0) + 2e⁰ = 2(1) = 2
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - 2 = 2 (x - 0)
⇒ y-2 =2x
⇒ 2x - y = -2
Equating of the normal is y -y₁ = \(\frac{-1}{m}\) (x-x₁)
⇒ y - 2 = -\(\frac12\) (x - 0)
⇒ 2y - 4 = -x
⇒ x + 2y - 4 = 0

The equation of the family of  parabolas with latus rectum (4a) and whose axes are parallel to the x-axis is shown in sketch
Let vertex 'V' be (h, k) and focus at 'F" and let
L-L' be latus rectum = 4a
Hence, FL = 2a and VF = a.
Thus 'F' is at (h+a, k).

Equation of parabola with vertex at (h, k) and focal length 'a', latus rectum 4a' is
(y - k)² = 4a(x - h) .......(1)
Differentiating with respect to 'x' we got
2(y - k) \(\frac{dy}{dx}\) = 4a 
⇒ (y - k). \(\frac{dy}{dx}\) = 2a ....... (2)
Again differentiating with respect to x,
(y-k) y''+y'\(\times\) y' = 0  ............(3)
From(2),(y-k) = From(2),(y-k) = \(\frac{2a}{y'}\)
Putting in (3), we get
\((\frac{2a}{y'})y''+(y')^2=0 (or) 2ay'' +(y')^3 = 0\)
This is the required differential equation.

\({ tan }^{ -1 }y=\frac { \pi }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 2 } \left( 1+logx \right) ^{ 2 }\)

It can be obtained by using examples 12.15 and 12.16 that
p↔q ≡ (¬ p ∨ q)  ∧ (¬q ∨ p) ... (1)
≡ (¬p∨q)  ∧ ( p∨ ¬q) (by Commutative Law) ... (2)
≡ (¬p ∧ ( p ∨ ¬q)) ∨ (q ∧ ( p ∨ ¬q)) (by Distributive Law)
≡ (¬p ∧ p) ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ (q ∧ ¬q) (by Distributive Law)
≡ F ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ F; (by Complement Law)
≡ (¬p ∧ ¬q) ∨ (q ∧ p) ; (by Identity Law)
≡ ( p ∧ q) ∨ (¬p ∧ ¬q) ; (by Commutative Law)
Finally (1) becomes p ↔️ q ≡ ( p ∧ q) v (ㄱp ∧ ㄱq)

The table for the operation x₁₁ is as follows.

x11	1	3	4	5	9
1	1	3	4	5	9
3	3	9	1	4	5
4	4	1	5	9	3
5	5	4	9	3	1
9	9	5	3	1	4

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

If the region in the first quadrant bounded by the circle x² + y² = r², the x-axis, the lines x = r − h and x = r is revolved about the x-axis, then the solid generated is a spherical cap of height h cut of from a sphere of radius r. Hence, the required volume is given by

\(V=\pi \int _{ r-h }^{ r }{ { y }^{ 2 }dx=\pi \int _{ r-h }^{ r }{ \left( { r }^{ 2 }-{ x }^{ 2 } \right) dx } =\pi } { \left( { r }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ r-h }^{ r }\)
\(=\pi \left( { r }^{ 2 }(r-(r-h)-\frac { \left( { r }^{ 3 }-{ (r-h) }^{ 3 } \right) }{ 3 } \right) =\pi \left( { r }^{ 2 }h-\frac { \left( { r }^{ 3 }-\left( { r }^{ 3 }-3{ r }^{ 2 }h+3{ rh }^{ 2 }-{ h }^{ 3 } \right) \right) }{ 3 } \right) \)
\(\\ =\pi \left( \frac { { 3rh }^{ 2 }-{ h }^{ 3 } }{ 3 } \right) =\frac { 1 }{ 3 } \pi { h }^{ 2 }(3r-h)\)

Equation of the given curve is 2+x-x²+y = 0

x	0	2	-1
y	-2	0	0

\(\Rightarrow\)y = x²-x-2

\(\therefore\) Required area= \(\int _{ -3 }^{ -1 }{ ydx } +\int _{ -1 }^{ 2 }{ -y } dx+\int _{ 2 }^{ 3 }{ ydx } \)
\(\\ =\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-x-2 \right) } dx\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })dx } +\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }-x-2) } dx\)
\(={ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ -3 }^{ -1 }+{ \left( 2x+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 2 }+{ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ 2 }^{ 3 }\)
\(=\left( -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } +2 \right) -\left( -9-\frac { 9 }{ 2 } +6 \right) +\left( 4+2-\frac { 8 }{ 3 } \right) -\left( -2+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \right) +\left( 9-\frac { 9 }{ 2 } -6 \right) -\left( \frac { 8 }{ 3 } -2-4 \right) \)
\(\\ =\left( \frac { -2-3+6 }{ 6 } \right) -\left( \frac { -6-9 }{ 2 } \right) +\left( \frac { 18-8 }{ 3 } \right) -\left( \frac { -12+3+2 }{ 6 } \right) +\left( \frac { 6-9 }{ 2 } \right) -\left( \frac { 8-24 }{ 3 } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 15 }{ 2 } +\frac { 10 }{ 3 } +\frac { 7 }{ 6 } -\frac { 3 }{ 2 } +\frac { 16 }{ 3 } \)
\(=\frac { 1+45+20+7-9+32 }{ 6 } =\frac { 90 }{ 6 } =15 \)

Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 5 }{ kdx } +\int _{ 5 }^{ \infty }{ 0dx } =1\) 
\(0+k\left( x \right) _{ 1 }^{ 5 }+0=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 0 & Otherwise \end{matrix} \end{cases}\)
(i) Distribution function
The distribution function 
\(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1, \(F(x)=\int _{ -\infty }^{ x }{ f(u)du } =\int _{ -\infty }^{ x }{ oldu } =0\) 
When 1 ≤ x ≤ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u)du=\int _{ -\infty }^{ x }{ 0du } +\int _{ 1 }^{ x }{ odu } +\int _{ 1 }^{ x }{ \frac { 1 }{ 4 } du } =\frac { 1 }{ 4 } (x-1) } \) 
When x ≥ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du=\int _{ -\infty }^{ x }{ odu } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 5 }^{ 5 }{ odu } =1\) 
Thus \(F(x)=\begin{cases} \begin{matrix} 0 & x<1 \end{matrix} \\ \begin{matrix} \frac { x-1 }{ 1 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 1 & x>5 \end{matrix} \end{cases}\) 
(ii) P(X < 3) = P(X ≤ 3) = F(3) = \(\frac { 3-1 }{ 2 } =\frac { 1 }{ 2 } \) (Since F(x) is continuous)
(iii) P(2 < X < 4) = P(2 ≤ X ≤ 4) F(4) - F(2) = \(\frac { 3 }{ 4 } -\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \) 
(iv) P(3 ≤ X ) = P(X ≥ 3) = 1− P(X < 3) = 1 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Given \(f(x)=\begin{cases} \begin{matrix} 16{ xe }^{ -4x } & foex>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\) 
Mean :
= \(E(x)=\int _{ 0 }^{ \infty }{ x.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ x.16.xe^{ -4x }dx } \) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 2 } } .{ e }^{ -4x }dx=16\times \frac { 2! }{ { 4 }^{ 3 } } \)  \(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(16\times \frac { 2 }{ 64 } \)
\(=\frac { 1 }{ 2 } \) 
Variance :
\(E({ x }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.{ e }^{ -4x } } dx\) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 3 }{ e }^{ -4x }dx } \) 

ஃ Var(X) = E(X²) - [E(x)]²
\( =\frac{3}{8}-\frac{1}{4} \)
\(=\frac{1}{8} \)
∴ Var(X) \(=\frac{1}{8} \)

Given w(x, y, z) = \(\frac { 1 }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } }\)
= (x² + y² + z²) -^\(\frac12\)
\(\frac { \partial w }{ \partial x } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2x)\)
= (-x² + y² + z²) -^\(\frac12\)
\(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial w }{ \partial x } \right) \)
= -[x\(\left( \frac { -3 }{ 2 } \right) \)( x² + y² +z²)^\(-\frac32\)
\((\not 2 x)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}\)
= (x² + y² + z²)^-\(\frac52\) [-3x² + x² +y + z²]
= - (x² + y² + z²)^-\(\frac52\) [y² + z² - 2x²] ....(1)
\(\frac { \partial w }{ \partial y } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2y)\)
= -y(x² +y² + z²)-^\(\frac32\) 
\(\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } =\frac { \partial }{ \partial y } \left( \frac { \partial w }{ \partial y } \right) \)
\(=-\left[y\left(\frac{-3}{\not 2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{\frac{5}{2}}(\not 2 y)+\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}(1)\right]\)
= -(x² + y² + z²)-^{\(\frac52 \)} 
= -(x² + y² + z²)-^{\(\frac52 \)} [3y² + x² + y² + z²]
= -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² + 2z²] ....(2)
Now \(\frac { \partial w }{ \partial z } =\frac { -1 }{ 2 } ({ x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ -\frac { 3 }{ 2 } }(2z)\)
= -z(x² + y² + z²)-^{\(\frac32 \) 
∴} \(\frac { { \partial }^{ 2 }w }{ \partial { z }^{ 2 } } \) = -(x² + y² + z²)-^{\(\frac52 \)} [x² + y² - 2z²] ...(3)
(1)+(2)+(3)⟶
^∴ \(\frac { { \partial }^{ 2 }w }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }w }{ \partial { y }^{ 2 } } \) ⁼ -(x² + y² + z²)-^{\(\frac52 \)} [y² + z² - 2x² + x² + z² - 2y + x²+ y-2z²]
= -(x² + y² + z²)-^{\(\frac52 \)}(0) = 0
Hence proved 

Diameter = 30 cm
Radius = 15 cm
Circumference (c) = 2πr
\(\frac{dc}{dr}\) = 2π(3) = 6πcm
dc = 2πdr
\(\frac{6}{2π}\) cm = dr
\(\frac{3}{π}\) cm = dr
Approximate growth of the diameter
= 2dr = 2 \(\times\) \(\frac{3}{π}\) cm = \(\frac{6}{2π}\)cm
(ii) A = πr²
dA = π ²r dr
\(d \mathrm{~A}=\not \pi 2(15) \frac{3}{\not \pi} \mathrm{cm}^{2}\)
dA = 90 cm²
Area = πr² = π \(\times\)15 \(\times\) 15 cm²
 

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.

Factorising the given function, we have
y = f (x) = (x − 3)(x + 2) .
(1) The domain of the given function f (x) is the entire real line.
(2) Putting y = 0 we get x = −2, 3. Therefore the x -intercepts are (−2, 0) and (3, 0) putting x = 0 we get y = −6. Therefore the y -intercept is (0, −6).
(3) f'(x) = 2x −1 and hence the critical point of the curve occurs at x = \(\frac{1}{2}\).
(4) f"(x) = 2 > 0,∀x . Therefore at xTherefore at x = \(\frac{1}{2}\) the curve has a local minimum which is \(f\left( \frac { 1 }{ 2 } \right) =-\frac { 25 }{ 4 } .\)
(5) The range of the function is \(y\ge-\frac{25}{4}\)
(6) Since f"(x) = 2 > 0,∀x the function is concave upward in the entire real line.
(7) Since f(x) = 2 ≠ 0, ∀x the curve has no points of inflection
(8) The curve has no asymptotes.
The rough sketch of the curve is shown on the right side.

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

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

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.