Explore topic-wise InterviewSolutions in Current Affairs.

Solution:
n³ – n = n (n² – 1) = n (n – 1) (n + 1)
Therefore, n³ – n is product of three consecutive positive integers, where n is any positive integer.
Since one out of every two consecutive integers is divisible by 2.

Therefore, The product n³ – n is divisible by 2.
Since one out of every three consecutive integers is divisible by 3.
Therefore, The product n³ – n is divisible by 3.
Any number which is divisible by 2 and 3 is also divisible by 6.
Hence, The product n³ – n is divisible by 6.

Solution:
By Euclid’s division lemma, we have
a = bq + r;  0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r, …(i)
where q is an integer
and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
n = 3q
So, n is divisible by 3.
n + 2 = 3q + 2
so, n + 2 is not divisible by 3.
n + 4 = 3q + 4
so, n + 4 is not divisible by 3.
Putting r = 1 in (i), we get
n = 3q + 1
so, n is not divisible by 3.
n + 2 = 3q + 3 = 3(q + 1)
so, n + 2 is divisible by 3.
n + 4 = 3q + 5

so, n + 4 is not divisible by 3.
Putting r = 2 in (i), we get
n = 3q + 2
so, n is not divisible by 3.
n + 2 = 3q + 4
so, n + 2 is not divisible by 3.
n + 4 = 3q + 6 = 3(q + 2)
so, n + 4 is divisible by 3
Thus for each value of r such that 0 ≤ r < 3  only one out of n, n + 2 and n + 4 is divisible by 3.

Solution:
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r …(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
n = 3q
So, n is divisible by 3.
n + 1 = 3q + 1
so, n + 1 is not divisible by 3.
n + 2 = 3q + 2
so, n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
n = 3q + 1
so, n is not divisible by 3.
n + 1 = 3q + 2
so, n + 1 is not divisible by 3.
n + 2 = 3q + 3 = 3(q + 1)
so, n + 2 is divisible by 3.
Putting r = 2 in (i), we get
n = 3q + 2
so, n is not divisible by 3.
n + 1 = 3q + 3 = 3(q + 1)
so, n + 1 is divisible by 3.
n + 2 = 3q + 4
so, n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.

let the 4 terms be a - 3d, a-d, a+d and a+3d

Sum = 4a = 36, so a = 9

Product of 2nd and 4th = (9-d)(9+3d)=105

81+27d-9d-3d² = 105

3d²-18d+24=0

d²-6d+8=0

(d-2)(d-4)=0

so d=2, or 4

numbers are

3, 7, 11, 15

or

-3, 5, 13, 21

A. into the page

Bayer is the C.G.S unit of pressure and is equal to 1-dyne/cm² . 

Weight is the gravitation force by which a body attracted to the earth. Gravitational unit of force in M.K.S system is kilogram weight or 9.81 Newton. Weight is the force with which 1-kilogram mass is attracted by the earth towards its center. 

One Newton is that amount of force which acting on one-kilogram mass for one second gives an acceleration 1 meter/sec/sec.

It is the unit of charge. One (1) coulomb is the quantity of electricity, which is circulated by a current of one (1) ampere in one second. The letter Q denotes it. So that 1 coulomb = 1 amp * 1 second. 

Farad is the unit of capacitance and the letter F denotes it. A condenser has a capacitance of one (1) farad, if it is capable to maintain a charge of one coulomb under a potential difference of one volt between its plates.

1 farad = 1 coulomb / 1 volt. = Q/V. 

Force is that which charge or tends to change a body state of rest or uniform motion through a straight line. The unit of force is Newton. 

It is the unit of inductance and the letter H denotes it. A circuit has inductance of one henry, if an electro-motive force of one volt if induced in that circuit, when the current in that circuit changes at the rate of one ampere per second. 1 henry = 1 volt sec / ampere. 

A conducting element used for converging (centering) current to and from a medium is called electrode. There are two types of electrode. A positive and other is negative. 

The least count of out-side micrometer is 0.01mm. 

A substance, which will not allow the flow of electric current to pass through it is called the insulator. The conductance and conductivity is zero in insulators. Insulators are used to isolate the electric current from neighbouring parts. Insulators will not allow the leakage of current, short-circuiting current, shock to the operator and isolates the electric current safely with out any diversion to any other place. 

Qualities of good insulator 

a. It should be flexible 

b. It should have good mechanical strength 

c. It should easily moulded into any shape 

d. It should not be effected by acid 

e. It should be non-inflammable 

f. It should have very high specific resistance to prevent leakage current 

g. It should be withstand high temperature. Because insulators posses negative temperature coefficient of resistance. That is resistance decreases with increasing temperature 

h. It should have high dielectric strength 

Conductance is the property of the conductor, which allows the flow of electric current through it. Conductance is denoted by the letter G and is reciprocal of resistance. The unit of conductance is mho. A substance, which posses conductance as its major property can be called as a good conductor.  

The common conductors in sequence with high conductivity.

a. Silver 

b. Silver copper alloy 

c. Copper (Hard down and Annealed) 

d. Gold 

e. Zinc 

f. Platinum 

g. Tin 

h. Aluminum 

i. Iron 

j. Brass 

k. Phosphorous bronze 

l. Nickel 

m. Lead 

n. Germanium silver 

o. Antimony 

p. Platinoid 

q. Mercury 

r. Bismuth 

s. Platinum iridium 

The correct option (A) 5

Explanation.

t₈ = t₃ + 25 

a + 7d = a + 2d + 25 

5d = 25 

d = 5

The correct option (B) – 2

Explanation.

If – 2 is a root of this equation then 

(– 2)² + (– 2) + k = 0 

4 – 2 + k = 0 

k = – 2

Correct option (C) 30 – 40   

Explanation:

Modal class = highest frequency

∴ 30 – 40

Correct option (C)  1/3

Explanation:

D = [1, 2, 3, 4, 5, 6] 

Multiple of 3 = {3, 6}

P(multiple of 3) = 2/6 = 1/3

The correct option (A) x² – 2x – 3 = 0

Explanation.

equation is 

x ² – x (sum of roots) + (product of roots) = 0 

x ² – x (2) + (– 3) = 0 

x ² – 2x – 3 = 0

Both assertion and reason are correct but, Reason is not the correct explanation of the assertion.

Correct option:

(B) B'A' 

Correct option:

(C) (–1,1)

Correct option:

(D) m × n

Correct option:

(C) e^x/x

Margaret Mitchell

i. 3,  ii. 2,  iii. 4,  iv. 2

Answer is (3) (I), (II) and (III) only

(I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin complex.

(II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of Ni(II) are low spin complexes.

(III) With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron = 1)

While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron = 5)

(IV) Aqueous solution of Mn(II) ions is pink in colour.

Atom, smallest unit into which matter can be divided without the release of electrically charged particles. It also is the smallest unit of matter that has the characteristic properties of a chemical element. As such, the atom is the basic building block of chemistry.

|x - 5| < 3

⇒ -3 < x - 5 < 3

⇒ -3 + 5 < x < 3 + 5

⇒ 2 < x < 8

Therefore,

Required set = {3, 4, 5, 6, 7}

Given curve is 3x² - y³ + 6x + 2y = 0.....(i)

Let the chord be y = mx + c 

⇒ \(\frac{y + mx}c=1\)

Given that chord are substanded a right angle at the origin.

By using homogenisation

Since, 3x² - y² + (3x + y) X 1 = 0

⇒ 3x² - y² + 2(3x + y)\((\frac{y + mx}c)\) = 0

\((\because \frac{y-mx}c=1)\)

⇒ 3x²c - cy² + 6xy - 6mx² + 2y² - 2mxy = 0

⇒ (3c - 6m)x² + (6 - 2m)xy + (2 - c)y² = 0......(ii)

Since, Curve (ii) is subtended a right angle at the origin.

Therefore, Coefficient of x² + coefficient of y² = 0

⇒ 3c - 6m + 2 - c = 0

⇒ 2c - 6m + 2 = 0

⇒ 2 = 6m - 2c........(iii)

Given chord is y = mx + c

Therefore for fixed point,

\(\frac{2}y=\frac6x=\frac{-2}1\)

⇒ x = \(\frac{-6}2\) = -3

and \(\frac2y\) = -2

⇒ y = \(\frac2{-2}\) = -1

Hence, fixed point is (-3, -1) ≡ (α, β) (given)

⇒ α = -3, β = -1

⇒ |α| = 3, |β| = 1

⇒ |α| > |β|

option (B) is correct.

If |sin x| = sin x

Then |sin x| - sin x = 0 but \(\sqrt{|sin\,\text x|-sin \text x}\neq0\)

Therefore f(x) is not designed at points where 

|sin x| = sin x

Therefore |sin x| = - sin x

Also |sin x| - sin x > 0

⇒ -2 sin x > 0   (\(\because\) |sin x| = - sin x)

⇒ sin x < 0 (Multiplying both sides by negative number -2)

⇒ x ∈ (-π, 0) ⋃ (π, 2π).........(i)

Also 4 - x² ≥ 0 (\(\because\) domain of √x is [0, 0])

⇒ x² ≤ 4

⇒ x ∈ [-2, 2]......(ii)

Since, π = 3.14 >2 and  - π < -2

Therefore from equation(i) and (ii), we observed that domain of function f(x) is

x ∈ [-2, 2] ∩ (-π, 0)  (\(\because\) π = 3.14 >2)

⇒ x ∈ [-2, 0)

Hence, domain of given function f(x) is [-2, 0).