We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.

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

Given g(x, y) = \(\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}g(x,y)=\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(=\frac { { 3(1) }^{ 2 }-1(2) }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+3 } =\frac { 3-2 }{ 8 } =\frac { 1 }{ 8 } \)

The region is sketched. It lies above the x − axis. Hence, the required area is given by 

\(A=\int _{ -1 }^{ 3 }{ ydx } =\int _{ -1 }^{ 3 }{ \left( \frac { 30-6x }{ 5 } \right) dx={ \left( \frac { 30x-3{ x }^{ 2 } }{ 5 } \right) }_{ -1 }^{ 3 } } \)
\(=\left( \frac { 90-27 }{ 5 } \right) -\left( \frac { -30-3 }{ 5 } \right) =\frac { 96 }{ 5 } \)

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
Let X be a random variable whose values x are the number of fours.
The sample space S is given in the table.
It can also be written as
S = {(i, j)} , where i = 1, 2, 3, 6 and j = 1, 2, 3, 6
Therefore X takes on the values of 0, 1 and 2.
We observe that
(i) X = 0, if (i,  j) for i ≠ 4, j≠ 4,
(ii) X = 1, if (1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(iii) X = 2, if (4, 4) ,
Therefore,

Values of the Random Variable X	0	1	2	Toatal
Number of elements in inverse images	25	10	1	36

The probabilities are
\(f(0)=P(X=0)\cfrac { 25 }{ 36 } \) 
\(f(1)=P(X=1)=\cfrac { 10 }{ 36 } \) 
and \(f(20=P(X=2)=\cfrac { 1 }{ 36 } \)
Clearly the function f(x) satisfies the conditions
(i) f (x) ≥ 0, for x = 0, 1, 2 and
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=-2 }{ f(x) } =f(0)+f(1)+f(2)=1\) 
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=1\)
The probability mass function is presented as

x	0	1	2
f(x)	\(\frac { 25 }{ 36 } \)	\(\frac { 10 }{ 36 } \)	\(\frac { 1 }{ 36 } \)

(or) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

Then the four types of conditional statements corresponding to p and q are respectively listed below.
(i) p →q : (conditional statement) “If the number of primes is infinite then Ooty is in Kerala”.
(ii) q → p : (converse statement) “If Ooty is in Kerala then the number of primes is infinite”
(iii) ¬p → ¬q (inverse statement) “If the number of primes is not infinite then Ooty is not in Kerala”.
(iv) ¬q → ¬p (contrapositive statement) “If Ooty is not in Kerala then the number of primes is not infinite”.

Given a*b = a.|b| on R
a,b ∈ R \(\Rightarrow\) a.|b| ∈R as a ∈ R and |b|∈R.
Hence * is a binary operation on R

Given u(x, y) = x² + 3xy + y²
u(x_o, y_o) = u(2,1)
= 2² + 3(2)(1) + 1²
= 4 + 6 + 1 = 11
\(\frac { \partial u }{ \partial x } \) = 2x+ 3y
\({ \left( \frac { \partial u }{ \partial x } \right) }_{ (2,1) }\)= 2 + 3 = 5
\(\frac { \partial u }{ \partial y } \) = 3x+ 2y
\({ \left( \frac { \partial u }{ \partial y } \right) }_{ (2,1) }\) = 6 + 2 = 8
Linear approximation 
L(x,y) = U(x_o, y_o) + \({ \left( \frac { \partial u }{ \partial x } \right) }_{ ({ x }_{ 0 },{ y }_{ 0 }) }\) (x - x_o) + \({ \left( \frac { \partial u }{ \partial y} \right) }_{ ({ x }_{ 0 }{ ,y }_{ 0 }) }\)(y - y_o)
L (x,y) = 11 + 5 (x - 2) + 8 (y - 1)
= 11 + 5x - 10 + 8y - 8
L(x,y) = 5x + 8y - 7

12π

Let a(t) be the distance of car A north of P at time t, and b (t) the distance of car B east of P at time t, and let c(t) be the distance from car A to car B at time t. By the Pythagorean Theorem, c(t)² = a(t)² + b(t)²

Taking derivatives, we get 2c(t)c'(t) = 2a(t)a'(t) + 2b(t)b'(t).
So, c′ = \(\frac { { aa }^{ ' }+{ bb }^{ ' } }{ c } =\frac { { aa }^{ ' }+{ bb }^{ ' } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
Substituting known values, we get
\(c' =\frac { (10\times 80)+(15\times 100) }{ \sqrt { { 10 }^{ 2 }+{ 15 }^{ 2 } } } =\frac { 460 }{ \sqrt { 13 } } \) ≈ 127.6 km/hr at the time of intersect

Let the two curves intersect at a point (x₀ , y₀) This leads to (a-c)x₀² + (b-d)y₀² = 0
Let us now find the slope of the curves at the point of intersection (x₀, y₀). The slopes of the curves are as follows :
For the curve ax² + by² = 1,  \(\frac{dy}{dx}= -\frac{ax}{by}\)
For the curve cx² + dy² = 1, \(\frac{dy}{dx}= -\frac{cx}{by}\)
Now, two curves cut orthogonally, if the product of their slopes intersection (x₀, y₀) is −1. Hence, for the above two curves to cut orthogonally at (x₀, y₀) if
\((-\frac{ax_{0}}{by_{0}})\times(-\frac{cx_{0}}{dy_{0}})=-1\)
That is, acx₀² + bdy₀² = 0,
together with \((a-c)x^{2}_{0}+(b-d)y_{0}^{2}=0\)
gives, \(\frac{a-c}{ac}=\frac{b-d}{bd}\)
That is,  \(\frac{1}{c}-\frac{1}{a}=\frac{1}{d}-\frac{1}{b}\).
Hence,  \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}\).

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.

Given f(x) = 4x³ + 3x²- 6x + 1
f'(x) = 12x² + 6x - 6
f"(x) = 24x + 6
f'(x) = 0
⇒12x² + 6x - 6 = 0
⇒ 2x² + x - 1 = 0
⇒ (x + 1)(2x - 1) = 0
\(\Rightarrow x=-1,\frac { 1 }{ 2 } \)
The critical numbers are -1, \(\frac { 1 }{ 2 } \)
The possible intervals of monotonicity are
\(\left( -\infty ,-1 \right) \left( -1,\frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } ,\infty \right) \)

Interval	(∞,-1)	\(\left( -1,\frac { 1 }{ 2 } \right) \)	\(\left( \frac { 1 }{ 2 } ,\infty \right) \)
Sign of f'(x)	Say x = -2 12(-2)2 + 6 (-2)-6 = +ve	Say x = 0 = -6 -ve	Say x = 1 I2(1)2 + 6(1) - 6 = +ve
Monoto nicity	strictly increasmg	Strictly decreasing	Strictly increasing

∴ f(x) is strictly increasing in \(\left( -\infty ,-1 \right) \left( \frac { 1 }{ 2 } ,\infty \right) \) and strictly decreasing in \(\left( -1,\frac { 1 }{ 2 } \right) \)
f"(x) = 0
\(\Rightarrow 24x+6=0\Rightarrow 24x=-6\)
\(x=\frac { -6 }{ 24 } =\frac { -1 }{ 4 } \)
The possible intervals of concavity are \(\left( -\infty ,\frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } ,\infty \right) \) 

Interval	\(\left( -\infty ,\frac { -1 }{ 4 } \right) \)	\(\left( \frac { -1 }{ 4 } ,\infty \right) \)
Sign of f'(x)	Say x = -1 24(-1) + 6 = -ve	Say x = 0  +ve
Concavity	Concave down	Concave up

ஃf(x) concave down in \(\left( -\infty ,\frac { -1 }{ 4 } \right) \) and concave up in \(\left( \frac { -1 }{ 4 } ,\infty \right) \)
As f"(x) changes its sign when it passes through
\(x=\frac { -1 }{ 4 } \), the point of inflection is \(\left( -\frac { 1 }{ 4 } ,f\left( -\frac { 1 }{ 4 } \right) \right) \)
Now \(f\left( \frac { -1 }{ 4 } \right) =\left( -\frac { 1 }{ 4 } \right) ^{ 3 }+3\left( \frac { -1 }{ 4 } \right) ^{ 2 }-6\left( \frac { -1 }{ 4 } \right) +1\)
= \(4\left( \frac { -1 }{ 4 } \right) +\frac { 3 }{ 16 } +\frac { 6 }{ 4 } +1\)
= \(\frac { -1 }{ 16 } +\frac { 3 }{ 16 } +\frac { 3 }{ 22 } +1=\frac { 1 }{ 8 } +\frac { 3 }{ 2 } +1\)
= \(\frac { 1+12+8 }{ 8 } =\frac { 21 }{ 8 } \)
ஃ Point of  inflection. \(\left( \frac { -1 }{ 4 } ,\frac { 21 }{ 8 } \right) \)
Since f'(x) changes its sign from positive to negative at x = -1, it has a local maximum at x = -1.
ஃf (-1) = 4 (-1)³ + 3 (-1)² - 6(-1) + 1
Since f'(x) changes its sign from negative to positive at \(x=\frac { 1 }{ 2 } \) it has a local minimum at \(x=\frac { 1 }{ 2 } \)
\(\therefore f\left( \frac { 1 }{ 2 } \right) =4\left( \frac { 1 }{ 2 } \right) ^{ 3 }+3\left( \frac { 1 }{ 2 } \right) ^{ 2 }-6\left( \frac { 1 }{ 2 } \right) +1\)
= \(\frac { 4 }{ 8 } +\frac { 3 }{ 4 } -\frac { -3 }{ 2 } +1=\frac { 1 }{ 2 } +\frac { 3 }{ 4 } -\frac { 3 }{ 2 } +1\)
= \(\frac { 2+3-6+4 }{ 4 } =\frac { 3 }{ 4 } \)

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Let X be the random variablc denotes number of red balls.
Then X take the values 0, 1, 2
Sample space = ⁷C₂ = 21
Let X denote the drawing the red ball.
Then X take the values 0, 1, 2
P(X = 0), X^-1 (BB) = ³C₂ = 3
P(X = 1), X^-1 (BR) = ³C₁ x ⁴C₁ = 12
P(X = 2), X^-1 (BR) = ³C₂ = 6

Values of random variable	0	1	2	Total
Number of elements in inverseimage	3	12	6	21

The probability mass function is

x	0	1	2
f(x)	\(\cfrac { 1 }{ 7 } \)	\(\cfrac { 4 }{ 7 } \)	\(\cfrac { 2 }{ 7 } \)

Mean :
\(E(x)=\Sigma x.f\left( x \right) \) 
= \(0(\frac { 1 }{ 7 } )+1\left( \frac { 4 }{ 7 } \right) +2\left( \frac { 2 }{ 7 } \right) \) 
= \(\frac { 4 }{ 7 } +\frac { 4 }{ 7 } =\frac { 8 }{ 7 }\)

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)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

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 \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

Given equation of the parabola is (y-1) = (x-2)²
\(\Rightarrow\) y = (x - 2)² + 1
It vertex is (2, 1) which is the minimum point P. Let Q(x, y) be a point on the parabola given slope of PQ = 2
\(\Rightarrow \frac { y-1 }{ x-2 } =2\ \left[ \because slope=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\Rightarrow\) y-1 = 2(x-2) \(\Rightarrow\) y-1 = 2x-4 \(\Rightarrow\) y = 2x-4+1
\(\Rightarrow\) y = 2x + 3
From (1) and (2), (x-2)²+1 = 2x-3
\(\Rightarrow\) x²-4x + 4 + 1 = 2x - 3 \(\Rightarrow\) x²- 6x + 8 = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4

\(\therefore\) Required area \(=\int _{ 2 }^{ 4 }{ ({ y }_{ 1 }-{ y }_{ 2 }) } dx\)
\(=\int _{ 2 }^{ 4 }{ (2x-3)-{ (x-2) }^{ 2 }-1dx } \)
\(=\int _{ 2 }^{ 4 }{ (2x-3-{ x }^{ 2 }+4x-4-1)dx } \)
\(=\int _{ 2 }^{ 4 }{ (-{ x }^{ 2 }+6x-8)dx } \)
\({ \left[ \frac { -{ x }^{ 3 } }{ 3 } +3{ x }^{ 2 }-8x \right] }_{ 2 }^{ 4 }=\left( \frac { -64 }{ 3 } +48-32 \right) -\left( \frac { -8 }{ 3 } +12-16 \right) \)
\(=\left( \frac { -64 }{ 3 } +16 \right) -\left( -\frac { 8 }{ 3 } -4 \right) \)
\(=\left( \frac { -64+48 }{ 3 } \right) -\left( \frac { -8-12 }{ 3 } \right) \)
\(=\frac { 16 }{ 3 } \) sq.units

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.

Given \(a*b=\frac { a+b }{ 2 } \), where a.b ∈Q Let a,b ∈Q
An element e has to found out such that 
a*e = e*a = a
Let a = 5, Then 5*e = 5
\(\Rightarrow \frac { 5+e }{ 2 } =\)5 ⇒ 5 + e = 10
Let a = \(\frac{2}{3}\). Then \(\frac{2}{3}\)*e = \(\frac{2}{3}\)
\(\Rightarrow \frac { \frac { 2 }{ 3 } +e }{ 2 } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2 }{ 3 } +e=\frac { 4 }{ 3 } \)
\(\Rightarrow e=\frac { 4 }{ 3 } -\frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
It is seen that for the binary operation * defined on Q, identity element e is not unique. Hence identity element not defined for the binary operation * on Q. 
The identity does not exist. Hence inverse also does not exist for the operation * on Q.

~(p V q) V (~p ∧ q) and ~p

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

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

P(x) = \(x^{4}-6x^{3}-11x^{2}+24x+28\), \(\alpha\) = 2, \(\beta\) = 7 
and observing \(\frac{P'(x)}{2}=2x^{3}-9x^{2}-11x+12=Qx\), (say).
This implies that there is a zero of the polynomial Q(x) in the interval (2, 7)
For verification,
Q(2) = 16 - 36 - 22 +12 = 28 - 58 = -30 < 0
Q(7) = 686 - 441 - 77 +12 = 698 - 518 = 180 > 0
From this we may see that there is a zero of the polynomial Q(x) in the interval (2, 7)

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.

We have,
\(f'(x)=2x>0, \forall x\in(2,7)\) and
\(f'(x)=2x>0, \forall x\in(-2,0)\)
and hence the proof is completed.

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

Given P (hitting the target) = \(\frac { 1 }{ 4 } \Rightarrow P=\frac { 1 }{ 4 } \) 
n = 10,
(i) P(X = 4)
\(P(X+4)=\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }(1-p)^{ n-x },x\) 
= 0,1,2,...n
(i) Probability of hitting the target exactly 4 times
P(X = 4) = \(^{10}{ C }_{ 4 } \times\left( \begin{matrix} 1 \\ 4 \end{matrix} \right) ^{ 4 }\times \left( \cfrac { 3 }{ 4 } \right) ^{ 6 }\)
\( =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2} \times \frac{1}{4^{4}} \times \frac{3^{6}}{4^{6}} \\ =210 \times \frac{3^{6}}{4^{10}} \) 
(ii) Probability of hitting atleast one time
= P(X≥1) = 1-P(x<1)
= 1-P(X = 0)
\( =1-{ }^{10} \mathrm{C}_{0} \times\left(\frac{1}{4}\right)^{0} \times\left(\frac{3}{4}\right)^{10} \\ =1-\frac{3^{10}}{4^{10}} \)

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(i) The sample space
S = {1, 2, 3, 4, 5, 6}\(\times\){1, 2, 3, 4, 5, 6}
consists of 36 ordered pairs (α, β) where α and β can take any integer value between 1 and 6 as shown. X is assigned to each point (α, β) the sum of the numbers on the dice .
That is X (α, β) = α + β
Therefore
X (1,1) = 1+1 = 2
X (1, 2) = X (2,1) = 3
X (1,3) = X (2,2) = X (3,1)= 4
X (1, 4) = X (2,3) = X (3, 2)  X (4,1) = 5
X (1,5) = X (2,4) = X (3,3) = X (4, 2) = X (5,1) = 6
X (1,6) = X (2,5) = X (3, 4) = X (4,3 = X (5, 2) X (6,1) = 7
X (2,6) = X (3,5) = X (4,4) = X (5,3) = X (6,2) = 8
X (3,6) = X (4,5) = X (5,4)  X (6,3) = 9
X (4,6) = X (5,5) X (6,4) = 10
X (5,6) = (6,5) = 11
X (6,6) = 12
(ii) Then the random variable X takes on the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
(iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}.
(iv) The number of inverse images are given below

Values of the random variable	2	3	4	5	6	7	8	9	10	11	12	Total
Number of elements in inverse image	1	2	3	4	5	6	5	4	3	2	1	36

(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
\(P(0.2\le X\le 0.7)=\int _{ 0.2 }^{ 0.7 }{ f(x) } dx=\int _{ 0.2 }^{ 0.7 }{ 1dx } =0.5\)

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
\(=2\pi { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=2\pi \left( { a }^{ 3 }-\frac { { a }^{ 3 } }{ 3 } \right) =\frac { 4 }{ 3 } { \pi a }^{ 3 }\)

\(\sqrt { 2 } -1\)