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.

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

Given
\(f(x)=\cfrac { 4-x }{ 6 } \) 
\(f(x)=\cfrac { 4-1 }{ 6 } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 } \) 
\(f(2)=\cfrac { 4-2 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \) 
\(f(3)=\cfrac { 4-3 }{ 6 } =\cfrac { 1 }{ 6 } \) 
ஃ The probability mass function is

Mean \(E(X)=\Sigma xf(x)\) 
= \(1\left( \cfrac { 1 }{ 2 } \right) +2\left( \cfrac { 1 }{ 3 } \right) +3\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 6 } =\cfrac { 3+4+6 }{ 6 } \) 
= \(\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } =1.67\) 
\(E({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \({ 1 }^{ 2 }\left( \cfrac { 1 }{ 2 } \right) +{ 2 }^{ 2 }\left( \cfrac { 1 }{ 3 } \right) +3^{ 2 }\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 4 }{ 9 } +\cfrac { 9 }{ 6 } =\cfrac { 3+8+9 }{ 6 } \) 
= \(\cfrac { 20 }{ 6 } =\cfrac { 10 }{ 3 } =3.33\)
var(X) E(X²) - [E(X)]²
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\) 

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 \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

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

The given differential equation is 
\({ \left( \frac { d^2y }{ dx^2 } \right) }^{ 3\times2 }= { 1+\left( \frac { dy }{ dx } \right) } \)
squaring both sides, we get
\({ \left( \frac { dy }{ dx } \right) }^{ 6 }=1+\left( \frac { dy }{ dx } \right) \)
In this equation, the highest order derivative is 2 and its power is 6.
∴ Order 2, degree 6.

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Let that number be x \( \\ \therefore \sqrt [ 3 ]{ x } +x=6\)
\(\Rightarrow \sqrt [ 3 ]{ x } =6-x\)
Taking power 3 both sides we get.
\({ \left( { x }^{ \frac { 1 }{ 3 } } \right) }^{ 3 }={ (6-x) }^{ 3 }\)
\(x={ 6 }^{ 3 }-3({ 6 }^{ 2 })x+3(6)({ x }^{ 2 })-{ x }^{ 3 }\)
\([\because { (a-b) }^{ 3 }={ a }^{ 3 }-3{ a }^{ 2 }b+{ 3ab }^{ 2 }-{ b }^{ 3 }]\)
\(\Rightarrow x=216-108x+18{ x }^{ 2 }-{ x }^{ 3 }\)
\(\Rightarrow { x }^{ 3 }+108x-18x-216+x=0\)
⇒ x³-18x²+109x -216 = 0. Which is the required mathematical problem.

Since the domain of y = sin^-1 is −[11], and 2\(\notin \)[-1, 1], sin⁻¹(2) does not exist.

Let A be a non-singular matrix of order 2m+1, where m = 0, 1, 2,... Then, we get |A| ≠ 0 and, by property (ii), we have |adj A| = |A|^(2m+1) − 1 = |A|^2m.
Since |A|^2m is always positive, we get that |adj A| is positive.

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx=\frac { (6-1) }{ (6+4) } .\frac { (6-3) }{ (6+4-2) } .\frac { (6-5) }{ (6+4-4) } .\frac { (4-1) }{ (4) } .\frac { (4-3) }{ (4-2) } .\frac { \pi }{ 2 } } \)
\(=\frac { (5) }{ (10) } \frac { (3) }{ (8) } \frac { (1) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)
Also, \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 4 }x{ cos }^{ 6 }xdx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }x{ cos }^{ 4 }xdx } =\frac { (3) }{ (10) } \frac { (1) }{ (8) } \frac { (5) }{ (6) } \frac { (3) }{ (4) } \frac { (1) }{ (2) } \frac { \pi }{ 2 } =\frac { 3\pi }{ 512 } \)

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 X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

Let f (t) be the distance travelled by the trucker in 't' hours. This is a continuous function in [0, 2] and differentiable in (0, 2). Now, f (0) = 0 and f (2) =164. By an application of the Mean Value Theorem, there exists a time c such that, \(f'(c)=\frac{164-0}{2-0}=82>80\)
Therefore at some point of time, during the travel in 2 hours the trucker must have travelled with a speed more than 80 km which justifies the issuance of a speed violation ticket.

The curve y = sin x intersects the positive x -axis. When y = 0 which gives, x = 
\( x=n\pi , n=1,2,3,...\)
Now, \(\frac{dy}{dx}=cos x\). The slpoe \(x=n\pi\)  are \(cos(n\pi)=(-1)^{n}\).   
Hence, the required angle of intersection is m₂ = 0
\(tan \theta = \frac{(-1)^n - 0}{1+((-1)^n(0)} = 1 ∀ n\)

Let A = \(\left[ \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 3. So ρ(A) ≤ min {3, 3} = 3. 
The highest order of minors of A is 3. There is only one third order minor of A.
It is \(\left| \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right| \) = 3(6 - 6) - 2(6 - 6) + 5(3 - 3) = 0. So, ρ(A) < 3.
Next consider the second - order minors of A.
We find that the second order minor \(\left| \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right| \) = 3 - 2 ≠ 0. So  ρ(A) = 2.

Given that equation of the circle is
x² + y² = 16
⇒ a² = 16
and equation of the tangent is
y = \( 2\sqrt { 2 } \)x + c
⇒ m = \( 2\sqrt { 2 } \)  and c = c
[∵ y = mx + c is the tangent]
The condition for the line y = mx + c is a tangent to the circle x² +y² = a² is
c² = a² (1 + m²)
⇒ c² = 16(1 + \(\left( { \left( 2\sqrt { 2 } \right) }^{ 2 } \right) \)
⇒ c² = 16(1 + 8)
⇒  c² = 16(9)
⇒ c = ± 4(3)
⇒ ±12

Given r = 5 cm

Since the circle touches the x axis, its centre is (0, ±5)
Equation of the circle is (x - h)² + (y - k)² = r²
⇒ (x - 0)² + (y ± 5)² = 25
\(\Rightarrow x^{2}+y^{2}+\not 25 \pm 10 y=\not 25\)
⇒ x²+y²+10y = 0

We consider \(\frac { 1+i }{ 1-i } =\frac { \left( 1+i \right) \left( 1+i \right) }{ \left( 1-i \right) \left( 1+i \right) } =\frac { 1+2i }{ 1+1 } =\frac { 2i }{ 2 } =i\)
and \(\frac { 1-i }{ 1+i } =\left( \frac { 1+{ i } }{ 1-i } \right) ^{ -1 }=\frac { 1 }{ i } =-i\)
Therefore,\(\left( \frac { 1+i }{ 1-i } \right) ^{ 3 }-\left( \frac { 1-i }{ 1-i } \right) ^{ 2 }\)= i³-(-i)³ = - i - i = -2i

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)

Given roots are 1, 2 and 3
Here a = 1, β = 2 and ૪ = 3
A cubic polynomial equation whose roots are α, β, ૪ is
x³-(α+β+૪)+x²(αβ+β૪+૪α)x-∝β૪ = 0
⇒ x³-(1+1+2)x²(2+6+3)x-6 = 0
⇒ x³-6x²+11x-6 = 0

(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Given z₁= 1-3i, z₂ - 4i and z₃ = 5
LHS = (z₁+ z₂) + z₃
= [1- 3i + (- 4i)] + 5
[1-7i] + 5
= 6 -7i
RHS = z₁+ (z₂ + z₃)
= 1- 3i + (-4i + 5)
= 6 - 7i
LHS = RHS
∴ (z₁+ z₂)+ z₃ = z₁+(z₂+ z₃)

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