Mean of the numbers: (a+b+8+5+10)/5=6
a+b=7 ......... (i)
variance = 6.8 = {(a-6)²+(b-6)²+(8-6)²+(5-6)²+(10-6)²}
a²+b²+93-12(a+b)=34 ........... (ii)
By solving (i) and (ii)
a=3 , b=4

a, b and c are coplanar, so
a.(b x c) = 0
So, (αi + 2j + βk).(i - j + k) = 0
=> α + β = 2
a bisect the angle between b and c
So a = (λ/) (b + c)
=> a = (λ/) (i+2j+k)
=> (αi + 2j + βk) = (λ/) (i+2j+k)
=> So, λ =
=> So, α=1 , β=1

The vector and are parallel and are in opposite direction.
So angle between them is 180° = π

Let A = (5, 1, a)
Let B = (3, b, 1)
Let C = ( 0 , 17/2 , -13/2 )
Point A , B and C are in straight line
So,

5 1 a 3 b 1 0 17/2 -13/2

= 0

=> 51a - 65b = 46
So, a = 6, b = 4

As the two given lines are intersecting, so

2-1 3-2 1-3 k 2 3 3 k 2

= 0

So

1 1 -2 k 2 3 3 k 2

= 0

=> 1(4-3k)-1(2k-9)-2(k²-6)=0
=> 2k² + 5k - 25 = 0
=> k=-5 OR k=5/2

The conjugate of a complex number is given by changing the sign of the imaginary part.
For example: conjugate of a complex number a+ib is a-ib
z = 1/(i-1) = (i+1)/((i-1)(i+1)) = (i+1)/(-2) = -1/2-i/2
So conjugate complex number = -1/2+i/2 = (i-1)/2
So conjugate complex number = ((i-1)(i+1))/(2(i+1)) = -2/2((i+1)) = -1/(i+1)

In equivalence relation a set satisfies following three conditions:
a. Every member of the set is related to itself.
b. Whenever a is related to b then b is also related to a.
c. If a is related to b and b is related to c, then a is related to c.

y = 4x + 3
=> x = (y-3)/4
=> f(y) = (y-3)/4

p : x is an irrational number
q : y is a transcendental number
r : x is a rational number iff y is a transcendental number
=> r : ~ p ↔ q
s₁ : q or p
s₂ : ~ (p↔~q)

p	q	~p	~q	r:~p↔q	s1:q or p	p↔~q	s2:~(p↔~q)
T	T	F	F	F	T	F	T
T	F	F	T	T	T	T	F
F	T	T	F	T	T	T	F
F	F	T	T	F	F	F	T

Based upon above chart
s₁ and r are not equivalent
s₂ and r are not equivalent

x - cy - bz = 0
cx - y - az = 0
bx + ay - z = 0
So,

1 -c -b c -1 -a b a -1

= 0

(1 + a²) + c(-c - ab) - b(ca + b) = 0
a² + b² + c² + 2abc = 0

As det A = ±1
So inverse of matrix A exists
As all entries of matrix A are integers and Adjoint A is matrix of transpose of co-factors of matrix A.
So, Entries of Adjoint A is also integers.

Let α and β are roots of: x² - 6x + a = 0
Let α and γ are roots of: x² - cx + 6 = 0
So, α + β = 6 ........ (i)
α.β = a ........ (ii)

α + γ = c ........ (iii)
α.γ = 6 ........ (iv)

β/γ = 4/3 ........ (v)

Now using (ii) and (iii)
α.β/α.γ = a/6
=> β/γ = a/6
=> using (v), 4/3 = a/6
=> So, a = 8

Now , x² - 6x + a = 0
x² - 6x + 8 = 0
x = 2 OR 6
As α.β = 8
So α = 2

Count of M = 1
Count of I = 4
Count of S = 4
Count of P = 2
Number of ways MISSISSIPPI can be jumbled without using S = (7!)/(4! x 2!)
Now number od ways four S can be placed in MIIIPPI = ⁸C₄
So total number of ways = (7!)/(4! x 2!) x ⁸C₄
So total number of ways = 7 . ⁶C₄ x ⁸C₄

sinx	< 1 when x ∈ (0, 1) => So,	sin	<
x

So,

cosx	<	1	when x ∈ (0, 1)

So,

x + 2y² = 0 .............(i)
x + 3y² = 1 .............(ii)
By solving (i) and (ii)
x = -2 and y = ±1

p	q	p V q	q→q	p→(q→q)	p→(p V q)
F	F	F	T	T	T
F	T	T	F	T	T
T	F	T	T	T	T
T	T	T	T	T	T

cot(cosec^-15/3 + tan^-12/3) = cot(tan^-13/4 + tan^-12/3)
= cot(tan^-1(3/4 + 2/3)/(1-(3/4)x(2/3))) = cot(tan^-117/6) = 6/17

A = {4, 5, 6}
B = {1, 2, 3, 4}
A ∪ B = {1, 2, 3, 4, 5, 6}
P(A ∪ B) = 1

Let the equation of perpendicular bisector is y = mx + c
c = -4
So, y = mx - 4 .............(i)
Center point between P(1, 4) and Q(k, 3) will be ((1+k)/2, 7/2)
In equation (i) passes thru ((1+k)/2, 7/2)
So, 7/2 = m((1+k)/2) - 4
7 = m + mk - 8
m + mk - 15 = 0 .................(ii)
m = gradient of the perpendiculat line
m = - (gradient of PQ) = -(1-k)/(4-3) = k - 1
Replace value of m in equation (ii)
k - 1 + k(k - 1) - 15 = 0
k² = 16
k = ±4

a + ar = 12
a(1 + r) = 12
ar² + ar³ = 48
ar² (1 + r) = 48
r² = 4
r = ±2
=> r = -2
=> a = -12

f(x) = x³ - px + q
So, f'(x) = 3x² - p = 0
=> x = ±
=> f"(x) = 6x
=> f"(x) at x = is positive
=> So minima is at x =

=> f"(x) at x = is negative
=> So maxima is at x =