\( y =x^{4} \)
\(y^{\prime} =4 x^{3} \)
\(y^{\prime \prime} =12 x^{2} \)
\(y^{\prime \prime}=0 \Rightarrow 12 x^{2}=0 \)
x  =0
Also y" >0 for x < 0 and x > 0
Hence the curve is concave upward. y'' does not change sign as the curve passes through x = 0. Thus the curve does not admit any point of inflection. The curve is concave in (\(-\infty\),0) and (0, \(\infty\)) .

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

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
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,

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

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

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

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

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

Since f(x) = x² - 2x - 3 , \(f'(x)=2x-2>0 \forall x\in (2, \infty)\). Hence f (x) is strictly increasing in \((2, \infty)\)

The given function is y mx +\(\frac{7}{m}\), where m is an arbitrary constant ....(1)
Differentiating both sides of equation (1) with respect to x, we get y' = m.
Substituting the values of y' and y in the given differential equation
we get xy'\(\frac{1}{y'}\)-y = xm +\(\frac{7}{m}\)- mx -\(\frac{7}{m}\) = 0
Therefore, the given function is a solution of the differential equation xy' + 7\(\frac{1}{y'}\) - y = 0

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.

Given = e^mx is the solution of
y' + 2y = 0 ...(1)
y = e^mx  ...... (2) 
\(\frac{dy}{dx} = e^{mx}. m\)
\(\frac{dy}{dx} = ym\)
\(\frac{dy}{dx} - my=0\)
⇒ y' - my = 0 ...(3)
Comparing equation (1) & (3),
we get m = -2

To prove  \(\neg(p \rightarrow q) \equiv p \wedge \neg q\)
Truth table for \(\neg(p \rightarrow q)\)

Truth table for \(p \wedge \neg q\)

The entries in the column \(\neg(p \rightarrow q) \text { and } p \wedge \neg q\) are identical and they are equivalent.

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.

Let X denote the winning amount. The possible events of selection are 
(i) both balls are black, or
(ii) one white and one black or
(iii) both are white 
Therefore X is a random variable that can be defined as
X (both are black balls) = Rs. 2(20) = Rs. 40
X (one black and one white ball) = Rs. 20 − Rs. 10 = Rs. 10
X (both are white balls) = (Rs. 20) = - Rs. 20
Therefore X takes on the values 40,10 and −20
Total number of balls n = 12
Total number of ways of selecting 2 balls = \(\left( \begin{matrix} 12 \\ 2 \end{matrix} \right) =\frac { 12\times 11 }{ 1\times 2 } =66\)
Number of ways of selecting 2 black balls =  \(\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) =6\) 
Number of ways of selecting one black ball and one white ball = \(\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) =32\) 
Number of ways of selecting 2 white balls = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Probability mass function is

Mean :
\(E(X)\Sigma xf(x)=40.\left( \frac { 6 }{ 66 } \right) +10.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) .\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
That is expected winning amount is 0
 Variance :
\(\Sigma x^{ 2 }=\Sigma { x }^{ 2 }f(x)=40^{ 2 }.\left( \frac { 6 }{ 66 } \right) +10^{ 2 }.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) ^{ 2 }.\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
 (E(X )² = 0² = 0
This gives  \(V(X)=E({ X }^{ 2 })-\left( E(X))^{ 2 } \right) =\frac { 4000 }{ 11 } -0=\frac { 4000 }{ 11 } \)
Therefore E(X ) = 0 and \(V(x)=\frac { 4000 }{ 11 } \)

First, we get the points of intersection of the parabolas. For this, we solve y² x = 4 and x² y = 4 simultaneously Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4. Then the points of intersection are (0, 0) and (4, 4). The required region is sketched.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2\(\sqrt x\) for 0\(\le x \le\) 4 and the equation of the lower boundary is \(y =\frac {x^2}{4}\)for \(0 \leq x \leq 4\). So, the required area \(\Delta\) is
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

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 

Let I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 x}{1+ a^x} dx\)-- (1) 
Using \(\int ^{b}_{a}\) f(x) dx =\(\int ^{b}_{a}\) f(a+b -x)dx  we get,
I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 (\pi -\pi - x)}{1+ a^{\pi -\pi - x}} dx\)
= \(\int ^{\pi}_{-\pi} \frac{cos ^2 (-x)}{1+ a^{-x}} dx\)
= \(\int ^{\pi}_{-\pi} a^x (\frac{cos ^2 x }{a^x+ 1} )dx\) --- (2)
Adding (1) and (2) we get
2I =  \(\int ^{\pi}_{-\pi} \frac{cos ^2 x }{a^x+ 1}(a^x+ 1) dx\) =  \(\int ^{\pi}_{-\pi} cos ^2 x dx\)
= 2 \(x =\int ^{\pi}_{-\pi} cos ^2 x dx\)  (since cos² x is an even function)
Hence, I =  \(\int ^{\pi}_{0} (\frac{1 + cos2x }{2} )dx\) 
=  \(\frac {1}{2} [ x + \frac {sin 2x}{2}]^{x}_{0}\) 
= \(\frac {1}{2} [\pi]\)
= \(\frac {\pi}{2}\)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

This is an indeterminate of the form \(\infty^{0}\).
Let \(g(x)=(1+2x)^{\frac{1}{2log \ x}}\)
Taking the logarithm, we get
log g(x) = \(\frac{log(1+2x)}{2log \ x}\)
\(\underset{x\rightarrow \infty}{lim}log \ g(x)=\underset{x\rightarrow \infty}{lim}(\frac{log(1+2x)}{2log \ x})\) \((\frac{\infty}{\infty})\) 
\(=\underset{x\rightarrow \infty}{lim}(\frac{\frac{2}{1+2x}}{\frac{2}{x}})\) (by l’Hôpital Rule)
= \(\underset{x\rightarrow \infty}{lim}(\frac{x}{1+2x})\) \((\frac{\infty}{\infty})\)
= \(\underset{x\rightarrow \infty}{lim}(\frac{1}{2})=\frac{1}{2}\) but,
\(\underset{x\rightarrow \infty}{lim}log \ g(x)=log (\underset{x\rightarrow \infty}{lim}g(x))\).
Hence by exponentiating, we get the required limit as  \(\sqrt{e}\)

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.

Since the open box has square area, let the length, breadth and height of the box be I, l and b cm respectively.
ஃ Surface area = l² + 41b = 108
\(\Rightarrow l+4b=\frac { 108 }{ l } \) 
\(\Rightarrow 4b=\frac { 108 }{ l } -1\) 
\(\Rightarrow b=\frac { 108 }{ 4l } -\frac { l }{ 4 } \) 
\(\Rightarrow b=\frac { 27 }{ l } -\frac { l }{ 4 } \) 
Let f(1) = Volume of the box = 1 \(\times\) 1\(\times\) b = l² b
=  \({ l }^{ 2 }\left( \frac { 27 }{ l } -\frac { l }{ 4 } \right) \)
\(f(1)=27l-\frac { { l }^{ 3 } }{ 4 } \) 
\(f'(l)=27-\frac { { 3l }^{ 2 } }{ 4 } \) 
f'(1) = 0
\(\Rightarrow 27-\frac { 3{ l }^{ 2 } }{ 4 } =0\) 
\(\Rightarrow \frac { 3l^{ 2 } }{ 4 } =27\)
\(\Rightarrow l^{2}=\not 27 \times \frac{4}{\not 3}=36\) 
\(\Rightarrow 1=\pm 6\) 
\(\Rightarrow 1=6\) 
ஃThe critical number is 6
\(f''(l)=\frac { 6l }{ 4 } =\frac { 3l }{ 2 } \) 
\(\therefore f''(6)=-\frac { 3(6) }{ 2 } <0\)
ஃ f"(1) is maximum when l = 6
When l = 6,\(b=\frac { 27 }{ 6 } -\frac { 6 }{ 4 } \) 
= \(\frac { 9 }{ 2 } -\frac { 3 }{ 2 } =\frac { 6 }{ 2 } =3cm\)
Hence the dimensions of the required box are 6 cm, 6 cm and 3 cm respectively.

Let f (x) = tan x, then the Mclaurin series of f (x) is 
\(f(x)=\sum^{n=\infty}_{n=0} a_{n}x^{n}\), where , \(a_{n}=\frac{f^{(n)}(0)}{n!}\).
Various derivative’s of the function f (x) evaluated at x = 0 is given below :
Now,
\(f'(x)=\frac{d}{dx}(tanx)=sec^{2}(x)\)
\(f''(x)=\frac{d}{dx}sec^{2}(x))=2sec x.sec x.tan x= 2sec^{2}x.tanx\)
\(f'''(x)=\frac{d}{dx}(2sec^{2}(x).tan x)= 2sec^{2}(x).sec^{2}x+tan x.4secx.secx.tanx\)
=\(2sec^{4}x+4sec^{2}x.tan^{2}x\)
\(f^{(iv)}(x)=8sec^{3}(x).secx+tan x+4sec^{2}x.2tanx sec^{2}x+8secx.secx.tanx.tan^{2}x\)
=\(16sec^{4}xtanx+8sec^{2}x.tan^{3}x\)
\(f^{(v)}(x)=16sec^{4}x.sec^{2}x+64sec^{3}x.sec x.tan x.tan x+8 sec^{2}x.3tan^{2}x.sec^{2}x+16secx.secx.tanx.tan^{3}x\)
=\(16sec^{6}x+88sec^{4}x.tan^{2}x+16sec^{2}x.tan^{4}x\).

Substituting the values and on simplification we get the required expansion of the function as
\(tan x=x+\frac{1}{3}x^{3}+\frac{2}{15}x^{5}+...; -\frac{\pi}{2}

Given f(x) = x (x - 4)³
f'(x) = x .3(x - 4)² + (x - 4)³(1)
= 3x (x - 4)² + (x - 4)³
= (x - 4)² + (3x +x- 4)³
= (x - 4)² + (4x - 4)
= 4 (x-1) (x-4)²
f"(x) = 4 [(x - 1)² (x - 4) +(x - 4)² (1)]
= 4[2(x-4)(x-1)+(x - 4)²]
= 4(x- 4)[2x - 2 +x - 4)]
= 4(x - 4)(3x - 6)
= 12 (x - 4)(x- 2)
f"(x) = 0
⇒ 12 (x - 4)(x- 2) = 0
⇒ x = 2, 4
The possible intervals are (-∞, 2) (2, 4) and (4, ∞)

ஃThe curve is concave upward on (-∞, 2) (4, ∞) and concave downward on (2, 4).
As f"(x) changes its sign when it passes through x = 2 and x = 4, the points of inflection are (2, f(2)) and (4,f(4)). 
f(2) = 2(2 - 4)³
= 2(-2)³ = 2(-8) = -16
f(4) = 4 (4 - 4)³ = 0
\(\therefore\) (2, -16) and (4, 0) are the points of inflection

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)

Let h and r be the height and the base radius. Therefore h = 2r. Let V be the volume of the salt cone.

\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{12}\pi h^{3}; \frac{dV}{dt}=30\) mtr³ / min.
Hence,  \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\)
Therefore,  \(\frac{dh}{dt}=4 \frac{dV}{dt}.\frac{1}{\pi h^{2}}\)
That is,  \(\frac{dh}{dt}=4\times30\times \frac{1}{100 \pi}\)
=\(\frac{6}{5\pi}\) mtr / min. 

Let (S, *) be an algebraic structure and a ∈ S.  Assume that the inverse of a exists in S. It is to be proved that the inverse of a is unique. The existence of inverse in S ensures the existence of the identity element e in S. 
Let a ∈ S. It is to be proved that the inverse a (if exists) is unique.
Suppose that a has two inverses, say a₁, a₂
Treating a₁ as an inverse of a gives \(a * a_{1}=a_{1} * a=e\)  ........(1)
Next treating a₂ as the inverse of a gives \(a * a_{2}=a_{2} * a=e\)  ........(2)
\(a_{1}=a_{1} * e=a_{1} *\left(a * a_{2}\right)=\left(a_{1} * a\right) * a_{2}=e * a_{2}=a_{2}(\text { by }(1) \text { and }(2))\)
So, a₁= a₂. Hence the inverse of a is unique which completes the proof.

(i) Since f (x) is a probability mass function, f (x) ≥ 0 for all x , and d \(\sum_{x} f(x)=1\)
Thus, \(\sum_{x} f(x)=1\)
\(c^{2}+2 c^{2}+3 c^{2}+4 c^{2}+c+2 c=0\)
\(c=\frac{1}{5} \text { or }-\frac{1}{2}\)
Since f x( ) ≥ 0 for all x , the possible value of c is \(\frac{1}{5}\)
Hence, the probability mass function is

 (ii) To find mean and variance, let us use the following table

Mean :  \(E(X)=\Sigma xf(x)=\frac { 115 }{ 25 } =4.6\) 
Variance : \(V(x)=E\left( x \right) ^{ 2 }=\Sigma { x }^{ 2 }f(x)-\left( \Sigma xf(x) \right) ^{ 2 }\) 
= \(\frac { 585 }{ 25 } -\left( \frac { 115 }{ 25 } \right) ^{ 2 }=23.40-21.16=2.24\)
Therefore the mean and variance are 4.6 and 2.24 respectively.

Given g(x, y) = \(\frac { { e }^{ y }sinx }{ x } \) for x ≠ 0 and g(0, 0) = 1
g(0, 0) = 1
The function g is defined for all (x, y) ∈ R²
To check if g has a limit L at (0,0) and if L= g(0,0) = 1
Consider \(\left| g(x,y)-g(0,y) \right| =\left| \frac { { e }^{ y }sinx }{ x } -0 \right| \)
= \(\left| \frac { { e }^{ y }sinx }{ x } \right| =\left| \frac { \left| { e }^{ y } \right| \left| sinx \right| }{ x } \right| =\left| { e }^{ x } \right| \left| \frac { sinx }{ x } \right| =1\)
\(\left[ \because (x,y)\longrightarrow (0,0)\Rightarrow \left| { e }^{ y } \right| =1and\left| \frac { sinx }{ x } \right| =1 \right] \)
\(\because \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\frac { { e }^{ y }sinx }{ x } =1=g(0,0)\) Which proves that is continuous at (0, 0)
∴ g(x, y) is continuous at (0, 0)

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 = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Let \(\sqrt{x}\) = u
Then x = u², and so dx = 2u du
When x = 0, u = 0 
When x = 9, u = 3
\(\therefore \int _{ 0 }^{ 9 }{ \frac { 1 }{ x+\sqrt { x } } dx=\int _{ 0 }^{ 3 }{ \frac { 1 }{ { u }^{ 2 }+u } (2u)du=2\int _{ 0 }^{ 3 }{ \frac { 1 }{ 1+u } du=2{ \left[ log|1+u \right] }_{ 0 }^{ 3 }=2[log4-0]=log16 } } } \)

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

n (S) = 36
X = {4,5,6,7,8}
From the sample space

Observe that, f(0) = 0 = f (1), is continuous in the interval [0,1] and is differentiable in (0,1). Now,
 \(f'{x}=2x(1-x)(1-2x)\).
Therefore,  \(f'(c)=0 \) gives c = 0, 1 and \(\frac{1}{2}\)
which \(\Rightarrow c= \frac{1}{2}\in (0,1)\).

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

The slope of the line y = x is 1. The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also 1. Hence,
\(\frac{dy}{dx}=3x^{2}-6x+1=1\)
which gives \( 3x^{2}-6x=0\)
Hence, x = 0 and x = 2.
Therefore, at (0, –2) and (2, –4) the tangent is parallel to the line y = x.

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec  
Instantaneous Velocity at t = 6
= V(6) = 27m/sec