Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

(1) Mount Everest is the highest mountain of the world.
(2) 3+ 4 = 8 .
(3) 7 + 5 >10 .
(4) Give me that book.
(5) (10 − x) = 7.
(6) How beautiful this flower is!
(7) Where are you going?
(8) Wish you all success.
(9) This is the beginning of the end.
The truth value of the sentences (1) and (3) are T, while that of (2) is F. Hence they are statements.
The sentence (5) is true for x = 3 and false for x ≠ 3 and hence it may be true or false but not both. So it is also a statement.
The sentences (4), (6), (7), (8) are not statements, because (4) is a command, (6) is an exclamatory, (7) is a question while (8) is a sentence expressing one’s wishes and (9) is a paradox.

\(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
Given \(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
\(h(\lambda x,\lambda y)=\frac { 6{ \lambda }^{ 2 }{ x }^{ 2 }{ \lambda }^{ 3 }{ y }^{ 3 }-\pi { \lambda }^{ 5 }{ y }^{ 5 }+9{ \lambda }^{ 4 }{ x }^{ 4 }\lambda y }{ 2020{ \lambda }^{ 2 }{ x }^{ 2 }+2019{ \lambda }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { { \lambda }^{ 5 }(6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y) }{ { \lambda }^{ 2 }(2020{ x }^{ 2 }+2019{ y }^{ 2 }) } \)
\( =\lambda^{3} \mathrm{~h}(\mathrm{x}, \mathrm{y}) \)
Thus f is homogeneous with degree 3.

\(\therefore Mean=E(x)=\Sigma xf(x)=0\left( \frac { 1 }{ 5 } \right) +1\left( \frac { 1 }{ 5 } \right) +2\left( \frac { 1 }{ 10 } \right) +3\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 1 }{ 5 } \right) +5\left( \frac { 1 }{ 10 } \right) \) 
\(\frac { 1 }{ 5 } +\frac { 1 }{ 5 } +\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { 1 }{ 2 } \) 
= \(\frac { 2+2+6+8+5+ }{ 10 } =\frac { 23 }{ 10 } =2.3\) 
= \(f({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\)
= \({ 0 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 1 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +2^{ 2 }\left( \frac { 1 }{ 10 } \right) +{ 3 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 4 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 5 }^{ 2 }\left( \frac { 1 }{ 10 } \right) \) 
= \(0+\frac { 1 }{ 5 } +\frac { 4 }{ 10 } +\frac { 9 }{ 5 } +\frac { 16 }{ 5 } +\frac { 25 }{ 10 } \) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\)
Variance = E(X²) - [E(X)]²
= 8.1- (2.3)2
= 8.1- 5.29
= 2.81

Given r = 2 mm
dr = (2.1 - 2) = 0.1 mm
Area = πr²
Approximate area dA = 2πr dr
= 2π (2) (0.1)
= 4 π (0.1) = 0.4 π mm²

We know that \(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x) } dx\quad if(2a-x)=f(x)\)
Take 2a = \(\pi\) and f(x) = g(sinx)
Then, f(2a-x) = g(sin(\(\pi\)-x)) = g(sinx) = f(x).
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)
\(\int _{ 0 }^{ \pi }{ g(sinx)dx } =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ g(sinx)dx } \)

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
\(\underset{x\rightarrow 0}{lim}(\frac{sin \ mx}{x})\)=\(\underset{x\rightarrow 0}{lim}(\frac{m\times cos \ mx}{1})\)
= m
The next example tells that the limit does not exist.

Given f(x) = tan x, x ∈ [0, π]
Rolle's theorem is not applicable since tan x is not continuous at x = \(\frac{\pi}{2}\) [∵ tan \(\frac{\pi}{2}\) = ∞]

2x = 3y =  −z \(\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{-z}{6}\) (Dividing by 6 all)
 \(\Rightarrow \frac{x-0}{3}=\frac{y-0}{2}= \frac{z-0}{-6}\)....(1)
6x = -y = -4z \(\Rightarrow \frac{x}{2}=\frac{-y}{12}=\frac{-z}{3} \) (Dividing by 6 all)
\(\Rightarrow \frac{x-0}{2}=\frac{y-0}{-12}=\frac{z-0}{-3}\) ....(2)
From (1) & (2), we get
\(\vec b = 3\vec i+2\vec j- 6\vec k\)and \( \vec d = 2\vec i-12\vec j- 3\vec k\)
Angle between lines (1) and (2) = Angle between \(\vec b\ and\ \vec d\)
Acute angle between lines cos 0 = \(\frac{|\vec b . \vec d|}{|\vec b||\vec d|} \)
\(\vec b . \vec d \)= (\(\vec b = 3\vec i+2\vec j- 6\vec k\)). (\( 2\vec i-12\vec j- 3\vec k\))
6-24+18 = 0
\( \cos \theta=0 \)
\(\theta=\frac{\pi}{2} \text { or } 90^{\circ}\)

  \(\vec { a } \)= \(a\hat { i } +a\hat { j } +c\hat { k } , \vec{b}=\hat { i } +\hat { k } \), \(\vec { c } \)= \(c\hat { i } +c\hat { j } +b\hat { k } \)
Given \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 0
⇒ \(\left| \begin{matrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{matrix} \right| \) = 0
⇒ \(a\left| \begin{matrix} 0 & 1 \\ c & b \end{matrix} \right| -a\left| \begin{matrix} 1 & 1 \\ c & b \end{matrix} \right| +c\left| \begin{matrix} 1 & 0 \\ c & c \end{matrix} \right| \) = 0
⇒ a(0-c)-a(b-c)+c(c-0) = 0
\(\Rightarrow-\not a c-a b+\not a c+c^{2}=0\)
⇒ c² = ab ⇒ c =\(\sqrt { ab } \).
Hence c is the geometric mean of a and b.

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.

The entries in column (3) and column (4) are not identical.

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

Given equation of the curves are y = tan x, y = cot x.
The intersection of y = tan x and y = cot x are
tan x = cot x \(\Rightarrow\) x = \(\frac{\pi}{2}\)
\(\therefore\) Required area \(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (tan\ x-cot\ x)dx } \)
\(={ [-log\ sin\ x+log\ sec\ x] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=log{ \left[ \frac { sec\ x }{ sin\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }==log{ \left[ \frac { 1 }{ sin\ x\ cos\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log{ (sin\quad x\ cos\ x) }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log\left( sin\frac { \pi }{ 4 } .cos\frac { \pi }{ 4 } \right) +log(sin0\quad cos0)\)
\(=-log\left( \frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } \right) +0\)
\(=-log\left( \frac { 1 }{ 2 } \right) =-(log1-log2)=log2\)

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 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

The equation of the latus-rectum is x = a. It intersects the parabola at the points L(a, 2a) and L₁ (a, −2). The required area is sketched. By symmetry, the required area A is twice the area bounded by the portion of the parabola

y = 2\(\sqrt a \sqrt x\), x -axis, x = 0 and x = a.
Hence, by taking vertical strips, we get
\(A=2\int _{ 0 }^{ a }{ ydx=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx=4 } \sqrt { a } } { \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }\)
\(=4\sqrt { a } \times \frac { 2 }{ 3 } { a }^{ \frac { 3 }{ 2 } }=\frac { 8{ a }^{ 2 } }{ 3 } \)

g(x, y) = xe^y + 3x²y
g_x = e^y + 6xy
gy = xey + 3x²
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })={ e }^{ y }+6x\)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ xy })=0+6y=6y\)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })={ xe }^{ y }\)
\({ g }_{ xy }=\frac { \partial }{ \partial y } ({ g }_{ x })\) = e^y + 6

\(put\sqrt { t } =z\Rightarrow \frac { 1 }{ 2\sqrt { t } } dt=dz\)
\(\Rightarrow \frac { 1 }{ 2z } dt=dz\Rightarrow dt=2zdz\)

\(\therefore I=\int _{ 0 }^{ sin\quad x }{ 2z{ sin }^{ -1 }zdz } +\int _{ 0 }^{ cosx }{ 2z{ cos }^{ -1 }zdz } ...(1)\)
= I₁+ I₂ 
\(=2\int _{ 0 }^{ x }{ \theta sin\theta cos\theta d\theta +2\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta cos\theta (-sin\theta )d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta 2sin\theta cos\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta 1sin\theta cos\theta d\theta } } \)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta -\int _{ \frac { \pi }{ 2 } }^{ x }{ \theta sin2\theta d\theta } } \)
\([\because sin2\theta =sin\theta cos\theta ]\)
\(=\int _{ 0 }^{ x }{ \theta sin2\theta d\theta \int _{ x }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta sin2\theta d\theta } } \)
\(\left[ \because \int _{ a }^{ c }{ f(x)dx+\int _{ c }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(x)dx } } } \right] \)
\(={ \left[ -\frac { \theta cos2\theta }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos2\theta }{ 2 } d\theta } \)

\(={ \left[ -\frac { \theta cos2\theta }{ 2 } +\frac { sin2\theta }{ 4 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=\frac { -\frac { \pi }{ 2 } cos\pi }{ 2 } +\frac { sin\pi }{ 4 } -(0+0)\)
\(=\frac { -\frac { \pi }{ 2 } (-1) }{ 2 } +\frac { 0 }{ 4 } =\frac { \pi }{ 4 } \)

Let x be the length of the rectangular pasture and y be the breadth of the rectangular pasture.
Given xy = 1,80,000
\(\Rightarrow y=\frac { 1,80,000 }{ x } \) 
Since fencing is not needed along the river side, 
Perimeter = 2x + y
Letf(x) = 2x+y

= \(2x+\frac { 1,80,000 }{ x } \) 
\(f'(x)=2-\frac { 1,80,000 }{ { x }^{ 2 } } \) 
f'(x) = 0
\(\Rightarrow 2=\frac { 1,80,000 }{ { x }^{ 2 } } \) 
\(\Rightarrow { x }^{ 2 }=90,000\) 
\(\Rightarrow x=\pm 300\) 
ஃ The critical number is 300, -300
\(f''(x)=-1,80,000\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 360000 }{ { x }^{ 3 } } \) 
\(f''\left( 300 \right) =\frac { 360000 }{ \left( 300 \right) ^{ 3 } } >0\) 
From (1), when x = 300, \(y=\frac { 1,80,000 }{ 300 } =600\) 
ஃ Length of the minimum needed fencing material = 2x + y
= 2(300) + 600 = 600 + 600 = 1200 m

f (x) is defined and differentiable for all x∈(-∞, ∞)
\(f'\left( x \right) =\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \)
\(f''(x)=\frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) \)
f"(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) =0\Rightarrow { e }^{ x }-{ e }^{ -x }=0\) 
\(\Rightarrow { e }^{ x }={ e }^{ -x }\Rightarrow { e }^{ x }=\frac { 1 }{ { e }^{ x } } \) 
\(\Rightarrow { e }^{ 2x }=1\Rightarrow { e }^{ 2x }={ e }^{ 0 }\) 
\(\Rightarrow 2x=0\Rightarrow x=0\) 
The possible intervals are (-∞,0) and (0,∞)

ஃ f(x) is concave up in (0, ∞) and concave down in (∞, 0).
Since f"(x) changes its position from negative to positive, when it passes through x = 0 the points of inflection is (0,1(0))
\(f(0)=\frac { 1 }{ 2 } \left( { e }^{ o }-{ e }^{ o } \right) =\frac { 1 }{ 2 } \left( 1-1 \right) =0\) 
ஃ (0, 0) is the point of inflection.

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

Given \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =-\hat { i } +2\hat { j } -4\hat { k } ,\vec { c } =\hat { i } +\hat { j } +\hat { k } \)
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ 2 & -4 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ -1 & -4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ -1 & 2 \end{matrix} \right| \)
= \(\hat { i } (-12+2)-\hat { j } (-8-1)+\hat { k } (4+3)\)
= \(-10\hat { i } +9\hat { j } +7\hat { k } \)
\(\vec { a } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= \(\hat { i } (3+1)-\hat { j } (2+1)+\hat { k } (2-3)\)
= \(4\hat { i } -3\hat { j } -\hat { k } \)
∴ \((\vec { a } \times \vec { b } ).(\vec { a } \times \vec { c } )=(-10\hat { i } +9\hat { j } +7\hat { k } ).(4\hat { i } -3\hat { j } -\hat { k } )\)
= -40-27-7 = -74

Taking \(\vec { p } =(\vec { a } \times \vec { b } )\) as a single vector and using the vector triple product expansion, we get 
\((\vec { a } \times \vec { b } )\times(\vec{c}\times\vec{d})=\vec{p}\times(\vec{c}\times\vec{d})\)
= \((\vec { p } .\vec { d } )\vec { c } -(\vec { p } .\vec { c } )\vec { d } \)
= \(((\vec { a } \times \vec { b } ).\vec { d } )\vec { c } -((\vec { a } \times \vec { b } ).\vec { c } )\vec { d } =[\vec { a } ,\vec { b } ,\vec { d } ]\vec { c } -[\vec { a } \vec { b } \vec { c } ]\vec { d } \)
Similarly, taking \(\vec { q } \) = \(\vec { c } \times \vec { d } \)
\((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=(\vec { a } \times \vec { b } )\times \vec { q } \)
= \((\vec { a } .\vec { q } )\vec { b } -(\vec { b } .\vec { q } )\vec { a } \)
\(= [\vec { a } ,\vec { c }, \vec d ]\vec { b } -[\vec { b } ,\vec { c },\vec d ]\vec { a } \)

Truth Table for (~p ⟶ r) ∧ ( p ↔️ q)

Truth Table for ( p V q) ∧ ~q

Let us denote white and red balls as w₁, w₂, r₁, r₂ and r₃
The sample space consists of 5C₃ = 10 different samples of size 3.
That is S = \(\left\{w_{1} w_{2} r_{1}, w_{1} w_{2} r_{2}, w_{1} w_{2} r_{3}, w_{1} r_{1} r_{2}, w_{1} r_{2} r_{3}, w_{1} r_{1} r_{3}, w_{2} r_{1} r_{2}, w_{2} r_{2} r_{3}, w_{2} r_{1} r_{3}, r_{1} r_{2} r_{3}\right\} .\)
The random variable X takes on the values 1, 2, and 3.

Given (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

\(Let\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { sinx }{ 1+cosx } \right) dx } \)
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1+sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\left[ \because cosx=2{ cos }^{ 2 }x-1\Rightarrow 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } +\frac { sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } +\frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\\ [\because sin2x=2sinx\quad cosx]\)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } 2tan\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }(f'(x)+f(x)dx } \)
Where \(f(x)=2tan\frac { x }{ 2 } \)
\(=\frac { 1 }{ 2 } .{ e }^{ x }.f(x)={ \left[ \left( \frac { 1 }{ 2 } { e }^{ x }.2tan\frac { x }{ 2 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ \left[ { e }^{ x }tan\frac { x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ e }^{ \frac { \pi }{ 2 } }tan\frac { \pi }{ 4 } -{ e }^{ 0 }tan0={ e }^{ \frac { \pi }{ 2 } }(1)\)
\(\therefore I={ e }^{ \frac { \pi }{ 2 } }\)

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

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

Given f (x) = (x − 2)(x − 7), x ∈ [3,11]
a) f(x) is continuous in [3, 11]
b) f(x) is differentiable in (3, 11)
c) f(11) = (11-2)(11-7)
= (9) (4) = 36
f(3) = (3 - 2)(3 - 7)
= (1)(-4) = - 4
∴ By Lagrange's mean value theorem, there exists c ∈ [3,11] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
\(\left[ \begin{matrix} f(x)\begin{matrix} = & (x \end{matrix}- & 2) & \begin{matrix} (x & - \end{matrix}7) \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 7x & -2x\begin{matrix} + & 14 \end{matrix} \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 9x & +\begin{matrix} 14 & \end{matrix} \end{matrix} \right] \)
⇒ 2c - 9 = \(\frac{36+4}{11-3}\)
⇒ 3c - 9 = \(\frac{40}{8}\) = 5
⇒ 2c = 14
⇒ c = 7 ∈ [3 , 11]

Let (x₁, y₁, z₁) is (6, 7, 4) and (x₂, y₂, z₂) (8, 4, 9)
The cartesian equation of a straight line passing through two points is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
⇒ \(\frac { x-6 }{ 8-6 } =\frac { y-7 }{ 4-7 } =\frac { z-4 }{ 9-4 } \)
⇒ \(\frac { x-6 }{ 2 } =\frac { y-7 }{ -3 } =\frac { z-4 }{ 5 } \) = s...(1)
⇒ x-6 = 2s 
⇒ y-7 = -3s 
⇒ y = -3s+7
\(\frac { y-7 }{ -3 } \)=s ⇒ y-7 = -3s ⇒ -3s + 7
\(\frac { z-4 }{ 5 } \) = s 
⇒ z-4 = 5s 
⇒ z = 5s + 4
∴ Point on the line is (2s + 6, -3s + 7, 5x + 4)...(2)
To find the point of intersection of (1) and x z plane, put y = 0, in (2)
∴ -3s+7 = 0
⇒ -s = -7 
⇒ s = \(\frac{7}{3}\)
Put s = \(\frac{7}{3}\) in (2) we get, the point of intersection
as \(\left( 2\left( \frac { 7 }{ 3 } \right) +6,0,5\left( \frac { 7 }{ 3 } \right) +4 \right) \)
⇒ \(\left( \frac { 14 }{ 3 } +6,0,\frac { 35 }{ 3 } +4 \right) \)
⇒ \(\left( \frac { 14+18 }{ 3 } ,0,\frac { 35+12 }{ 3 } \right) \Rightarrow \left( \frac { 32 }{ 3 } ,0,\frac { 47 }{ 3 } \right) \)
To find the point of intersection of (1) and yz plane, put x = 0 in  (2)
∴ 2s + 6 = 0
⇒ 2s = -6 
⇒ s = -3
∴ (2) ⟶ (2(-3) + 6, -3(-3) + 7, 5(-3)+4)
= (0, 16, -11)

We know that the line \(\vec { r } =(\hat { i } +2\hat { j } +4\hat { k } )+t(2\hat { i } +2\hat { j } +\hat { k } )\) is parallel to the vector \(2\hat { i } +2\hat { j } +\hat { k } \).
Direction ratios of the straight line joining the two given points (5, 1, 4) and (9, 2, 12) are 4,1,8 and hence this line is parallel to the vector \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, the angle between the given two straight lines is
\(\theta ={ cos }^{ -1 }\left( \frac { \left| \vec { b } .\vec { d } \right| }{ \left| \vec { b } \right| \left| \vec { d } \right| } \right) \), where \(​​\vec { b } \) = \(2\hat { i } +2\hat { j } +\hat { k } \) and \(\vec { d } \) = \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, \(\theta ={ cos }^{ -1 }\left( \frac { \left| (2\hat { i } +2\hat { j } +\hat { k } ).(4\hat { i } +\hat { j } +8\hat { k } ) \right| }{ \left| 2\hat { i } +2\hat { j } +\hat { k } \right| \left| 4\hat { i } +\hat { j } +8\hat { k } \right| } \right) ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)