Truth Table for (~p ⟶ r) ∧ ( p ↔️ q)

\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } } \)
\(=\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+1+4 } } =\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| \right] \)
\(={ \left[ \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { x+1 }{ 2 } \right) \right] }_{ -1 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }(1)-{ tan }^{ -1 }(0) \right] \)
\(\\ =\frac { 1 }{ 2 } \left[ \frac { \pi }{ 4 } \right] =\frac { \pi }{ 8 } \)

\(\underset { x\rightarrow \infty }{ lim } \frac { x }{ logx } =\frac { \infty }{ \infty } \)
Which is in indeterminate form
∴ By L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1 } }{ x } =\underset { x\rightarrow \infty }{ lim } x=\infty \)

Given f(x) = x² − x, x ∈ [0, 1] 
(i) f(x) is continuous in [0, 1]
(ii) f(x) is differentiable in (0, 1)
(iii) f(0) = 0² - 0 = 0
f(1) = 12-1 = 1-1 = 0
∴ f(0) = f(1)
By Rolle's theorem, there exists C ∈ [0, 1] such that
f'(c) = 0
⇒ 2c - 1 = 0
⇒ 2c = 1
⇒ c = \(\frac12\) ∈ [0, 1]

\({ sec }^{ -1 }\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) \)
Note that \(\frac{5\pi}{3}\) is not in [0, \(\pi\)]\{\(\frac{\pi}{2}\)}, the principal range of sec^-1 x.
we write \(\frac { 5\pi }{ 3 } =2\pi -\frac { \pi }{ 3 } \).
Now, sec\(\left( \frac { 5\pi }{ 3 } \right) =sec\left( 2\pi -\frac { \pi }{ 3 } \right) =sec\left( \frac { \pi }{ 3 } \right) and\frac { \pi }{ 3 } \in [0,\pi ]\)\{\(\frac{\pi}{2}\)}
Hence, sec^-1\(\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) ={ sec }^{ -1 }\left( sec\left( \frac { \pi }{ 3 } \right) \right) =\frac { \pi }{ 3 } \)

The distance between origin to z = i, −2 + i, and 3 are
|z| = |i| = 1
|z| = |−2+i| = \(\sqrt { \left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
|z| = |3| = 3
Since \(1<\sqrt { 5 } <3\), the farthest point from the origin is 3 .

The roots are rational if Δ = b²−4ac = (−2p)²−4(p²−q²+2qr−r²).
But this expression reduces to 4(q²−2qr+r²) or 4(q−r)² which is a perfect square.
Hence the roots are rational.

\(2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and\({ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=cos\frac { \pi }{ 3 } \) \(\left[ \therefore \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) 
[\(\therefore\) Principal domain of sin is \(\therefore \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and principal domain of cos is\(\left[ 0,\pi \right] \)
\(siny=\frac { 1 }{ 2 } \)
\(siny=sin\frac { \pi }{ 6 } \)  \(\left[ \because \frac { \pi }{ 6 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) and \(y=\frac { \pi }{ 6 } \)
\(\therefore \quad 2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
= \(2\left( \frac { \pi }{ 3 } \right) +\frac { \pi }{ 6 } =\frac { 2\pi }{ 3 } +\frac { \pi }{ 6 } \)
= \(\frac { 4\pi +\pi }{ 6 } =\frac { 5\pi }{ 6 } \)

(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
Given z₁= 1-3i, z₂ - 4i and z₃ = 5
LHS = (z₁+ z₂) + z₃
= [1- i + (- 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)

|xy| = ce^x

\(\frac { 1 }{ 6 } \left[ log\left( \frac { 35 }{ 8 } \right) \right] \)

concave upward everywhere

f(x) is strictly decreasing on [0, 1]

c = 3

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.

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

Given y = e^−x + mx + n .... (1) 
Derentiating cquation (1) wr.t 'x', we get
\(\frac { dx }{ dx } =-e^{ -x }(-1)+m\)
\(\frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-e^{ -x }+m\)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ e^{ x } }+0\)
\(\Rightarrow \left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) =e^x\)
Substituting the value of \(\frac{d^2y}{dx^2}\) in the given differential equation, we get
\(\Rightarrow e^{ x }\left( \frac { d^{ 2 }y }{ d{ x }^{ 2 } } \right) -1=e^x(e^x)-1\\= e^{x-x}-1\\ e^0-1= 1-1=0\)
Thus the solution of the given differential equation e^x \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) -1 = 0

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
The given differential equation is 
\(\sqrt { \frac { dy }{ dx } } =4\frac { dy }{ dx } +7x\)
Squaring both sides,
\(\frac { dy }{ dx } =\quad { \left( 4\frac { dy }{ dx } +7x \right) }^{ 2 }\)
\(16{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ 49 }x^{ 2 }+56x{ \left( \frac { dy }{ dx } \right) }\)
The highest derivative is 1 and its maximum power is 2.
∴ Order 1, degree 2.

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\theta \Rightarrow cos\theta =\frac { 1 }{ 2 } \)
\(\Rightarrow sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 4 } } =\sqrt { \frac { 3 }{ 4 } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) =\frac { \sqrt { 3 } }{ 2 } \)

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

The matrix form of the system is
\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where
A=\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
|A| =\(\left| \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right| =3\left| \begin{matrix} 2 & 12 \\ 1 & 7 \end{matrix} \right| -1\left| \begin{matrix} 3 & 12 \\ 2 & 7 \end{matrix} \right| +9\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| \)
= 3 (14 - 12) - 1 (21 -24) + 9 (3 -4)
= 3 (2) -1 (-3) + 9 (-1)
= 6 + 3 - 9 = 9 - 9 = 0
Since |A| = 0, the homogeneous system of equations have non-trivial solutions also.

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0

Equation of the circle is x² + y² - x + 2y - 3 = 0
Here 2g = -1 ⇒ g = \(\frac { -1 }{ 2 } \)
2f = 2 ⇒ f = 1 and c = -3
Centre is (-g, -f) = \(\left( \frac { 1 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
= \(\sqrt { \frac { 1 }{ 4 } +1+3 } \)
= \(\sqrt { \frac { 1 }{ 4 } +4 } =\sqrt { \frac { 1+16 }{ 2 } } \)
r = \(\sqrt { \frac { 17 }{ 2 } } \) units.

\(\sqrt { 3 } -i\)

r = 2 and \(-\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =-\alpha =-\frac { \pi }{ 6 } \)
Therefore, the modulus and principal argument of
\(-\sqrt { 3 } -i\) are 2 and \(\frac { \pi }{ 6 } \)

tan(tan^-1(0.2021))
= -0.2021[\(\because \) tan (tan^-1x) = x for any real number]

2i(3−4i)(4−3i)
Let z = 2i(3−4i)(4−3i).
∴ |z| = |2i(3-4i)(4-3i)|
= |2i| |3-4i| |4-3i|
= \(\sqrt { { 2 }^{ 2 } } \sqrt { { 3 }^{ 2 }+(-4)^{ 2 } } \sqrt { { 4 }^{ 2 }+(-3)^{ 2 } } \)
= \(\\ 2.\sqrt { 9+16 } \sqrt { 16+9 } =2.\sqrt { 25 } .\sqrt { 25 } \)
= 2(5)(5) = 50

(1-i)¹⁰
Let z = (1- i)¹⁰
|z| = |1- i|¹⁰ = \(\left[ \sqrt { { 1 }^{ 2 }+(-1)^{ 2 } } \right] ^{ 10 }\)
= \(\left[ \sqrt { 2 } \right] ^{ 10 }\) = 2^{1/2 x 10} = 2⁵ = 32

z w
= (5-2i)(-1+3i)
= -5+15i+2i-6i²
= -5+17i-6(-1)
= -5+17i+6
= 1+17i

The angle between a line \(\vec { r } =\vec { a } +t\vec { b } \) and a plane \(\vec { r } .\vec { n } \) = p with normal \(\vec { n } \) is θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad \)
Here, \(\vec { b } =\hat { i } -\hat { j } +\hat { k } \) and \(\vec { n } =2\hat { i } \hat { j } +\hat { k } \)
So, we get θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad =sin^{ -1 }\left( \frac { |(\hat { i } -\hat { j } +\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )| }{ |\hat { i } -\hat { j } +\hat { k } ||2\hat { i } -\hat { j } +\hat { k } | } \right) =sin^{ -1 }\left( \frac { 2\sqrt { 3 } }{ 3 } \right) \).

We observe that the straight line \(\frac { x-1 }{ 4 } =\frac { 2-y }{ 6 } =\frac { z-4 }{ 12 } \) is parallel to the vector \(4\hat { i } -6\hat { j } +12\hat { k } \) and the straight line \(\frac { x-3 }{ -2 } =\frac { y-3 }{ 3 } =\frac { 5-z }{ 6 } \) is parallel to the vector \(2\hat { i } +3\hat { j } -6\hat { k } \)
Since \(4\hat { i } -6\hat { j } +12\hat { k } =-2(-2\hat { i } +3\hat { j } -6\hat { k } )\), two vectors are parallel, and hence the two straight lines are parallel.

ang \(z=\frac { -2 }{ 1+i\sqrt { 3 } } \)
= arg(-2)-arg \(\left( 1+i\sqrt { 3 } \right) \) \(\left( \because arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg{ z }_{ 1 }=g_{ 2 } \right) \)
= \(\left( \pi -{ tan }^{ -1 }\left( \frac { 0 }{ 2 } \right) \right) -tan^{ -1 }\left( \frac { \sqrt { 3 } }{ 1 } \right) \)
= \(\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \)
This implies that one of the values of arg z is \(\frac { 2\pi }{ 3 } \)
Since \(\frac { 2\pi }{ 3 } \) lies between \(-\pi \), the principal argument Argz is \(\frac { 2\pi }{ 3 } \)

A = 11, C = -25, D = -44, E = - 50, and F = -256
Here A ≠ C and A and C are of opposite signs. Hence, the given equation represents a hyperbola.

\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
A is a matrix of order 2 \(\times\) 2
∴ \(\rho \)(A) ≤ min(2,2) = 2
The highest order of minor of A is 2
it is \(\left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| \)= 4 - 4 = 0
So, \(\rho \)(A)<2
Next consider the minor of order 1 |2| = 2 ≠ 0
∴ \(\rho \)(A) = 1

\(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right| \)= 6 - 4 = 2 ≠ 0
Since A is non singular, A^-1 exists
A^-1 = \(\frac { 1 }{ |A| } \)
Now, adj A = \(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \)
[Inter change the entries in leading diagonal and change the sign of elements in the off diagonal]
∴ A^-1 = \(\frac{1}{2}\)\(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \).

Given U(x, y, z) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy } +3{ z }^{ 2 }y\)
\(\frac { \partial U }{ \partial x } =\frac { xy(2x)-({ x }^{ 2 }+{ y }^{ 2 })(y) }{ { x }^{ 2 }{ y }^{ 2 } } +0\)
\(=\frac { 2{ x }^{ 2 }y-{ x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { y({ x }^{ 2 }-{ y }^{ 2 }) }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }y } \)
\(\frac { \partial U }{ \partial y } =\frac { xy(2y)-({ x }^{ 2 }+{ y }^{ 2 })(x) }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
= \(\frac { { 2xy }^{ 2 }-{ x }^{ 3 }-{ xy }^{ 2 } }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { xy }^{ 2 }-{ x }^{ 3} }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { y }^{ 2 }-{ x }^{ 2 } }{ x{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(\frac { \partial U }{ \partial z } =0+3y(2z)=6yz\)

(i) Probability mass function
For a discrete random variable we have
f(x) = p(X = x)
\(\therefore f(0)=F(0)=\frac { 1 }{ 2 } \) 
f(1) = F(1) - F(0)
= \(\frac { 3 }{ 5 } -\frac { 1 }{ 2 } =\frac { 6-5 }{ 10 } =\frac { 1 }{ 10 } \) 
f(2) = F(2)-F(1)
= \(\frac { 4 }{ 5 } -\frac { 3 }{ 5 } =\frac { 1 }{ 5 } \) 
f(3) = F(3) - F(2)
\(\frac { 9 }{ 10 } -\frac { 4 }{ 5 } =\frac { 9-8 }{ 10 } =\frac { 1 }{ 10 } \)
f(4) = F(4)-F(3)
= \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \) 
ஃThe probability mass function is

(ii) p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)
= \(\frac { 1 }{ 2 } +\frac { 1 }{ 10 } +\frac { 1 }{ 5 } =\frac { 5+1+2 }{ 10 } =\frac { 8 }{ 10 } \) 
= \(\frac { 4 }{ 5 } \)
(iii) p(x≥2) = p(x = 2) + p(x = 3) + p(x = 4)
= \(\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2+1+1 }{ 10 } =\frac { 4 }{ 10 } \)
= \(\frac { 2 }{ 5 } \)

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are 2 and 1 respectively. Hence it has at most two positive roots and at most one negative root. Since the difference between number of sign changes in coefficients of P(−x) and the number of negative roots is even, we cannot have zero negative roots. So the number of negative roots is 1. Since the difference between number of sign changes in coefficient of P(x) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, 1 is a root. Thus we must have two and only two positive roots Obviously the other two roots are imaginary numbers.

\({ tan }^{ -1 }x+{ tan }^{ -1 }y+{ tan }^{ -1 }z={ tan }^{ -1 }\left( \frac { x+y+z-xyz }{ 1-xy-yz-zx } \right) \)
Given \({ tan }^{ -1 }x+{ tan }^{ -1 }x+{ tan }^{ -o }y+{ tan }^{ -1 }z=\pi \)
\(\Rightarrow \pi ={ tan }^{ -1 }\left( \frac { x+y+zxyz }{ 1-xy-yz-zx } \right) \)
\(\Rightarrow tan\pi =\frac { x+y+z-xyz }{ 1-xy-yz-zx } \)
\(\Rightarrow 0=\frac { x+y+z-xyz }{ 1-xy-yz-zx } \quad [\therefore tan\pi =0]\)
\(\Rightarrow x+y+z-xyz=0\)
\(\Rightarrow x+y+z=xyz\)