Explore topic-wise InterviewSolutions in Current Affairs.

Fe, due to large number of unpaired d -electrons/more interatomic attraction.

Generally +2 oxidation state is obtained with electronic configuration 5d¹⁰ 6s², So that two electrons lossed. Thus +4 oxidation state is shows by that elements. Which has 4f and 4f⁷ electronic configuration after losing 4 electron. So silicon have only +4 oxidation state and tin shows +2 and +4 both oxidation state.

(3) BaSO₄ + H₂O₂

BaO₂.8H₂O(s) + H₂SO₄(aq) BaSO₄(s) + H₂O₂(aq) + 8H₂O(l)

CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄

AgNO₃ + 2NH₃ → Ag(NH₃)₂ + NO₃

Volume strain is defined as ratio of change in the volume to the original volume  

Correct option is C) with

Correct option is B) to

Correct option is A) at

Longitudinal strain= ∆L / L where ∆L -change in the length, L-original length 

Volume strain= ∆V/ V 

where 

∆V -change in the volume, 

V-original volume

Here Z (12, 10) = 5(12) + 4(10)= 100 and Z (14, 5) = 5(14) + 4(5) = 90. (14,5) minimises the objective function.

Correct option:

(A) (1, 2) 

Option : (a) π/8

Correct option:

(D) - 1/3

Option : (a) 4

Option : (c) x + 5y = 2

Correct option:

(D) (0, 2)

Option : (d) k ≠ 5

Option : (d) (-∞,0) ∪ (2,∞)

Option : (b) {-2,-1,0,1,2}

• In a square matrix, the number of rows and columns are equal.
• A square matrix is a diagonal matrix in which non-diagonal elements are zeroes.
• A square matrix is a scalar matrix in which the non-diagonal elements are zeroes, and diagonal elements are the same.
• A matrix in which all non-diagonal elements are zeroes and diagonal elements are 1s is said to be an identity matrix.
• A matrix is a symmetric matrix if A=A^T
• A matrix is skew-symmetric if A=–A^T
• A matrix is an orthogonal matrix; if A×A^T=I
• A matrix is said to be a singular matrix, if |A|=0
• A matrix is said to be idempotent, if A²=A.
• A matrix is said to be involutory, if A²=I

<sub> : defines subscripted text 

<sup> : defines superscripted text 

Example 

To write a² : a² 

To write CO₂ : CO₂

Option : (B) K - \(\frac{1}{x-2}\) 

Option : (D) \(\frac{x^{19}}{19}\) + K

Option : (B) \(\frac{1}{2}\)sin2x

Option : (B) 1

Option : (B) \(\frac{1}{2}\)sec²\(\frac{x}{2}\)

Option : (A) \(\frac{a}{x}\) + logb

Option : (A) \(-\frac{1}{2x^5}\) 

Option : (C) \(\frac{\sqrt 5}{2\sqrt x}sec^2\sqrt{5x}\) 

Option : (C) \(\frac{1}{2}x^{\frac{-2}{3}}\) . \(e^{3\sqrt x}\) 

Option : (C) \(\frac{-1}{2\sqrt x}\) sin (√x+ 5)

Option : (B) \(\frac{\pi}{180}\)cos(\(\frac{\pi x}{180}\))

Given differential equation is

\(\frac{d^2y}{dx^2}+4y=cos\,x\)

It's auxiliary equation is 

m² + 4 = 0

⇒ m = \(\pm2i\)

\(\therefore\) C.F. = C₁ cos2x + C₂ sin 2x

Hence, cos2x and sin 2x are solution of given differential  equation 

Let = y₁ = cos2x ⇒ y'₁ = -2 sin 2x

y₂ = sin 2x ⇒ y'2 = 2 cos 2x

W(y_1, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}cos2x&sin2x\\-2sin2x&2cos2x\end{vmatrix}\)

 = 2cos²2x + 2sin²2x

 = 2(cos²2x + sin²2x)

 = 2

u₁ = \(\int\frac{\begin{vmatrix}0&y_2\\R&y_2\end{vmatrix}}{W(y_1,y_2)}dx\) 

= \(\int\frac{\begin{vmatrix}0&sin2x\\cosx&2cos2x\end{vmatrix}}{2}dx\)

 = \(\int-\frac{cosx\,sin2x}2dx\)

 = \(-\frac12\int2sinx\,cos^2xdx\)

 = -1\(\int t^2(-dt)\) (\(\because\) By taking cos x = t ⇒ -sin x dx = dt)

 = t³/3 (\(\because\) t = cos x)

 = \(\frac{cos^3x}3\) 

u₂ = \(\int\frac{\begin{vmatrix}y_1&0\\y'_1&R\end{vmatrix}}{W(y_1,y_2)}dx\) = \(\int\frac{\begin{vmatrix}cos2x&0\\-2sin2x&cosx\end{vmatrix}}{2}dx\)

= \(\frac12\int\) cosx cos2x dx

 = \(\frac12\int\) cosx(1 - 2sin²x)dx

=\(\frac12\)\([\int cosxdx-2\int sin^2x\,cosxdx]\)

 = \(\frac12\)[sin x - \(\frac23\)sin³x]

 = \(\frac{sin x}2-\frac13sin^3x\)

 \(\therefore\) P.I. = u₁y₁ + u₂y₂

 = \(\frac{cos^3x}3cos2x+(\frac{sinx}2-\frac13sin^3x)sin2x\) 

 = \(\frac{cos^3x}3(2cos62x-1)+(\frac{sinx}2-\frac13sin^3x)2sinx cosx\)

 = \(\frac23\)cos⁵x - \(\frac13\)cos³x + sin²x sinx - \(\frac23\)sin⁴x cos x

 = \(\frac23\)cos⁵x -  \(\frac13\)cos³x + (1 - cos²x)cos x - \(\frac23\)(1 - cos²x)² cosx

= \(\frac23\)cos⁵x -  \(\frac13\)cos³x + cos x - cos³x - \(\frac23\) cos x (1 - 2cos²x + cos⁴x)

 =  \(\frac23\)cos⁵x -  \(\frac43\)cos³x + cos x - cos³x - \(\frac23\) cos x + \(\frac43\)cos³x - \(\frac23\)cos⁵x

 = cos x - \(\frac23\) cos x

= \(\frac13\) cos x

\(\therefore\) Complete solution of given differential equation is 

y = C.F. + P.I.

 = C₁ cos2x + C₂ sin 2x + \(\frac13\) cos x

Alternative:(Without variation of parameter)

P.I. = \(\frac{1}{D^2+4}cosx\)

 = \(\frac{cosx}{-1^2+4}\) = \(\frac{cosx}3\)

Given differential equation is

\(\frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y=0\)

its auxiliary equation is

m² - 4m + 3 = 0

⇒ (m - 3) (m - 1) = 0

⇒ m - 1 = 0 or m - 3 = 0

⇒ m = 1 or m = 3

\(\therefore\) y = C₁e^x + C₂e^3x is solution of given differential equation.

Option : (b)

|A| = 7, adj A \(=\begin{bmatrix}2&-1\\1&3\end{bmatrix}\)

\(\therefore14A^{-1}=14\frac{adj A}{|A|}=2\,adj A=\begin{bmatrix}4&-2\\2&6\end{bmatrix}\)

Option : (b) 

As every per-image x ∈ A has a unique image y ∈ B

⇒ f is injective function

Option : (c)

Since, principal value branch of \(tan^{-1}x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

\(\therefore -\frac{\pi}{2}<tan^{-1} x< \frac{\pi}{2}\)

\(\Rightarrow \frac{-\pi}{2}<y<\frac{\pi}{2}\)  \((\because tan^{-1}x =y)\)

\(3^{4x + 1} -2\times3^{2x+2}-81 =0\)

⇒ \(3\times3^{4x}-2\times3^2\times3^{2x}-81=0\) ⇒ \(3\times3^{4x}-18\times3^{2x}-81=0\)

⇒ 3a² - 18a - 81 = 0 ⇒ (Put 3^2x = a) ⇒ a² - 6a - 27 = 0

⇒ a² - 9a + 3a - 27 = 0 ⇒ a(a - 9) + 3(a - 9) = 0

⇒(a - 9)(a - 3) = 0 ⇒ a - 9 = 0 or a + 3 = 0

⇒ a = 9 or a = -3 ⇒ 3^2x = 9 or 3^2x = -3

⇒3^2x = 3² (∵ 3^2x ≠ -3) ⇒ 2x = 2 ⇒ x = 1.

(i) Notes:
1. Types of cinema
1.1. art cinema
1.2. commercial cinema.

2. The success of commercial films
2.1. spicy elements
2.2. song
2.3. dance

3. Art films
3.1. low success
3.2. concern for social issues

4. Purpose of cinema
4.1. entertainment
4.2. education.
4.3. social awareness

5. Difference between art cinema and commercial cinema

(ii) Summary
Cinema can be put into two categories, i.e., commercial cinema, and art cinema. Commercial films are more popular among masses. These films present spicy elements, i.e., songs, dance, and comedy, etc. They provide the perfect entertainment to the masses. On the other hand, art cinema is not so popular among the masses. It has a limited audience because they don’t present spicy elements. These films are mainly based on social issues and reality, hence not easy to understand for ordinary audience.
Title – Cinema: Art versus Commercial

(i) Notes:
1. Problems before our educationists:
1.1.religious and moral instructions because we have children
1.1.1 born in different faiths
1.1.2. living in diverse ways of life
1.1.3. following different religions
1.1.4. following different forms of worship
1.1.5. speaking a large variety of languages

2. The children should be taught to
2.1. love one another
2.2. be kind and helpful to all
2.3. be tender to animals
2.4. observe and think right

3. Primary Aim
3.1. personality development in the right way
3.2. develop suitable techniques and methods for
3.2.1. serving spiritual need
3.2.2. promoting an atmosphere of mutual respect
3.2.3. being helpful
3.2.4. speaking and understanding more than one language
3.2.5. respecting and appreciating different religions

4. Wrong and unwise to
4.1.use any kind of pressure to accept
4.2. impose a single way of life and form of worship

(ii) Summary
In a land of many faiths, there are two big problems before our educationists. One is the problem of religious and moral instruction and the other problem arises out of a large variety of languages. We should teach our children to love one another, to be kind and helpful and to be tender to animals. They should not only be provided formal education but also mould their personality in the right way. We should promote mutual respect, fuller understanding, and cooperation among different communities. We should train our children to speak and understand more languages and to appreciate and respect other religions. Imposition will be futile. No single way of life or worship should be promoted.
Title – Human Values and Education.

Given:

A's profit, while selling to B = 15%

B's loss, while selling to C = 15%

C's profit, while selling to D = 15%

Formula Used:

While selling an article at a profit, we have:

SP = [(100 + %profit)/100] × CP

Similarly, while selling an article at a loss, we have:

SP = [(100 – %profit)/100] × CP

Here SP = Selling Price, and CP = Cost Price.

Calculation:

Let the CP for A be Rs.x

After successive selling, the price became Rs.8993,

So we get:

[(100 + 15)/100] of [(100 – 15)/100] of [(100 + 15)/100] of x = 8993

⇒ 115/100 × 85/100 × 115/100 × x = 8993

⇒ x = 20 × 20 × 20 = Rs.8000

∴ CP for A is Rs.8000

Given:

4, 9, 17, 35, 69, ?

Calculation:

⇒ 4 × 2 + 1 = 9

⇒ 9 × 2 - 1 = 17

⇒ 17 × 2 + 1 = 35

⇒ 35 × 2 - 1 = 69

⇒ 69 × 2 + 1 = 139

∴ The value of ? is 139.