Explore topic-wise InterviewSolutions in Current Affairs.

Answer is (a) vector(|a x b|)

Answer is (b) sin²α - sin²β + sin²γ = 1

Answer is (c) -2 cot 2x + c

Concept:

The distance of the origin (0, 0, 0) from the plane ax + by + cz + d = 0 is given by \(\rm \left|\frac {(a)(0) +(b)(0) +(c)(0) + d }{\sqrt {a^2+b^2+c^2}}\right|\)

Calculation:

We know that the distance of the origin (0, 0, 0) from the plane ax + by + cz + d = 0 is given by \(\rm \left|\frac {(a)(0) +(b)(0) +(c)(0) + d }{\sqrt {a^2+b^2+c^2}}\right|\)

⇒ The distance of the origin from the plane 2x + 6y - 3z + 7 = 0

\(=\rm |\frac {(2)(0) +(6)(0) +(-3)(0) + 7 }{\sqrt {2^2+6^2+(-3)^2}}|\)

\(=\rm |\frac { 7 }{\sqrt {49}}|\)

= 1

Hence, the distance of the origin from the plane 2x + 6y - 3z + 7 = 0 is 1.

Concept:

If planes are parallel, no point in common.

If planes are perpendicular, only one point in common.

If plane coincides then infinite points in common

Calculations:

Given, the equation of two planes ax + by + cz + d = 0 and ax + by + cz + d1 = 0, where d ≠ d1.

Taking the ratio of direction ratios of the plane, we have

⇒ \(\rm \dfrac a a = \dfrac b b = \dfrac c c \)

⇒ 1 = 1= 1.

⇒The ratio of direction ratios of the plane ax + by + cz + d = 0 and ax + by + cz + d1 = 0 is same

⇒ The plane ax + by + cz + d = 0 and ax + by + cz + d1 = 0 are parallel.

Hence, The two planes ax + by + cz + d = 0 and ax + by + cz + d1 = 0, where d ≠ d1, have no points in common.

\(\frac1x - \frac1{x -2} = 3, \,x\ne0,2 \)

⇒ \(\frac{x -2-x}{x(x -2)}=3\)

⇒ \(-2 = 3x^2 - 6x\)

⇒ \(3x^2 - 6x + 2 = 0\)

⇒ \(x = \frac{6\pm\sqrt{36-24}}{6}\)

\(= \frac{6 \pm2\sqrt3}{6}\)

\(= \frac{3\pm\sqrt3}{3}\)

\(= 1\pm \frac1{\sqrt3}\)

Hence, solutions are \( 1+ \frac1{\sqrt3}\) or \( 1- \frac1{\sqrt3}\).

\(\lim\limits_{x\to \frac\pi4} \frac{2 - cosec^2x}{1 - cotx}\)     \((\frac 00 - case)\)

\(= \lim\limits_{x\to \frac\pi4} \frac{-2\,cosecx(-cosecx \;cotx)}{-(-cosec^2x)}\)    (By D.L.H. Rule)

\(=\lim\limits_{x\to \frac\pi4} \frac{2cosec^2x \,cotx}{cosec^2 x}\)

\(= \lim\limits_{x\to \frac\pi4 } 2\,cot x \)

\(= 2\,cot\frac\pi4\)

\( = 2\)

a) Equation of required circle is 

(x - 4)² + ( y + 3)² = 72

⇒ x² - 8x + 16 + y² + 6y + 9 = 49

⇒ x² + y² - 8x + 6y = 24    ....(i)

b) Put x = -5 & y = 2 in the L.H.S. of equation (i), we get

(-5)² + 2² - 8(-5) + 6(2)

= 25  + 4 + 40 + 12

= 81 > 24

∴ Point P(-5, 2) lie outside of circle (a).

 \(\lim\limits_{x \to \frac \pi2} (cosec \, x)^{tan \, x}\)     \((1^\infty\, type)\) 

\(= Exp \left\{\lim\limits_{x\to \frac\pi2} (cosec\, x-1) tan\,x\right\}\)

\(= Exp \left\{\lim\limits_{x\to \frac \pi2} \frac {1 - sin \,x}{cos \,x}\right\}\)      \(\left(\frac 00 - case\right)\) 

\(= Exp \left\{\lim\limits_{x\to \frac \pi2} \frac {-cos \,x}{-sin\,x}\right\}\)      (By using D.L.H. Rule)

\(= Exp \left\{\lim\limits_{x\to \frac \pi2} cot \, x\right\}\)

\(= e^0 = 1\)                       \((\because cot \frac \pi 2 = 0)\) 

Hence, \(\lim\limits_{x\to \frac \pi 2} (cosec \, x)^{tan\, x} = 1\).

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) =  \(x = {-24 \pm \sqrt{576-64} \over 32}\) = \(x = {-3 + 2\sqrt{2} \over 4}\) OR \(x = {-3 - 2\sqrt{2} \over 4}\)

4x² - 13x + k = 0    .....(i)

Let roots are α & 12α.

∴ Sum of roots = \(\frac{-(-13)}{4}\)

⇒ \(\alpha + 12\alpha = \frac {13}4\)

⇒ \(13\alpha = \frac{13}4\)

⇒ \(\alpha = \frac 14\) and  \(12\alpha = \frac{12}{4} = 3\)

Hence, \(\frac14\) & 3 are roots of equation (i).

∴ 4(3)² - 13 x 3 + k = 0

⇒ k = 39 - 36 = 3

Correct option is (a) 5

\(f(x) = x^2 - 3x\)

\(\therefore f' (x) = 2x - 3\)

\(f(x) = f'(x)\)

⇒ \(x^2 - 3x = 2x - 3\)

⇒ \(x^2 - 5x + 3 = 0\)     ......(1)

Given that α & ß are roots of the equation f(x) = f'(x), it means α & ß are roots of equation( 1).

\(\therefore\) Sum of roots = α + ß = \(\frac{-(-5)}1\) = 5.

A is non singular matrix

∴ |A| \(\ne\) 0 and A^-1 will exist.

Given that 

\((5A)^{-1} = \frac{1}{n^2 - n - 7}A^{-1}\)

⇒ \(\frac{A^{-1}}{5} = \frac{A^{-1}}{n^2 - n - 7} \)       \(\left(\because (kA)^{-1} = \frac{A^{-1}}{k}\right) \)

⇒ \(n^2 - n - 7 = 5\)

⇒ \(n^2 - n-12 = 0\)

∴ Sum of all possible vowels of \(n=\frac{-b}a= \frac{-(-1)}{1}= 1\).

\(A = \begin{bmatrix}1 &1&1\\1&2&-3\\2&-1&3\end{bmatrix}\)

∵ c_ij = (-1)^ij is cofactor of element a_ij of matrix A where m_ij is minor of element a_ij of matrix A which equals to the determinant of matrix obtained by hiding i^th row and j^th column of matrix A

∴ Cofactor matrix

\(C = \begin{bmatrix} C_{11}& C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33} \end{bmatrix}= \begin{bmatrix}3&-9&-5\\-4&1&3\\-5&4&1 \end{bmatrix}\)

\(Adj(A) = C^T= \begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end{bmatrix}\)

\(|A| = 1(6 -3) - 1(3 + 6) + 1( -1 -4)\)

= 3 - 9 - 5 

= -11

\(\therefore A^{-1} = \frac{AdjA}{|A|} = \frac{-1}{11} \begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end {bmatrix}\)

Matrix form of given system of equations is A^Tx = B

where

\(A^T = \begin{bmatrix}1&1&2\\1&2&-1\\1&-3&3\end{bmatrix}\)

\(x= \begin{bmatrix}x\\y\\z\end{bmatrix}\)

\(B = \begin {bmatrix}0\\9\\-14 \end{bmatrix}\)

∴ \(x = (A^T)^{-1}B\)

\(= (A^{-1})^TB\)

\(= \frac{-1}{11} \begin{bmatrix}3 &-9&-5\\-4&1&3\\-5&4&1 \end{bmatrix} \begin{bmatrix}0\\9\\-14\end{bmatrix} \)

\(= \frac{-1}{11} \begin{bmatrix}-11\\-33\\22\end{bmatrix}\)

\(\therefore x = \begin{bmatrix}1\\3\\-2\end{bmatrix}\)

Hence, 

\(x= \begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}1\\3\\-2\end{bmatrix}\)

∴ x = 1, y = 3 & z = -2 is solution of given system.

|A| = 3

|3A| = 3^m|A|

On the bases of kinetic energy of gases, the mean kinetic energy of one mole of gas per degree of freedom is 1/2RT
where R is the universal gas constant.
On the basis kinetic energy of a molecule of gas per degree of freedom is 1/2kT
where k is the Boltzmann constant

Answer: Market Supply.

\(\int\limits_0^{\pi/2}\frac{d\theta}{\sqrt{tan\theta}}\) = \(\int\limits_0^{\pi/2}\cfrac{d\theta}{\sqrt{\frac{sin\theta}{cos\theta}}}\)

\(\int\limits_0^{\pi/2}\frac{\sqrt{cos\theta}}{\sqrt{sin\theta}}d\theta\) = \(\int\limits_0^{\pi/2}sin^{-1/2}\theta\,cos^{1/2}\theta d\theta\) 

 \(=\cfrac{\Gamma(\frac{-1/2+1}2)\Gamma(\frac{1/2+1}2)}{2\Gamma(\frac{-1/2+1/2+2}2)}\) 

\(\Big(\)\(\because\) \(\int\limits_0^{\pi/2}\)sin^m θ cosⁿ θ dθ = \(\cfrac{\Gamma(\frac{m+1}2)\Gamma(\frac{n+1}2)}{2\Gamma(\frac{m+n+2}2)}\)\(\Big)\) 

\(=\cfrac{\Gamma(1/4)\Gamma(3/4)}{2\Gamma(1)}\)

\(=\frac{\pi}{2sin(\pi/4)}\) (\(\because\) \(\Gamma(n)\Gamma(1-n)=\frac{\pi}{sin\,n\pi}\) and here n = 1/4 and \(\Gamma\)(1) = 0! = 1)

\(=\frac{\pi}{2\times1/\sqrt2}\) (\(\because\) sin \(\pi/4\) = 1/√2)

\(=\frac{\pi}{\sqrt2}\) or \(\frac{\pi/2}2\)

Hence proved

_-π/2∫^π/2 \(\frac1{1+e^{sin\,x}}dx\) = I (Let)  ...(1)

⇒ I =  _-π/2∫^π/2 \(\frac1{1+e^{sin\,x}}dx\) =  _-π/2∫^π/2 \(\frac{e^{sin\,x}}{1+e^{sin\,x}}dx\)  ...(2)

By adding equations (1) and (2), we get

2I = _-π/2∫^π/2 \(\frac{1+e^{sin\,x}}{1+e^{sin\,x}}dx\) \(=[x]^{\pi/2}_{-\pi/2}\) = π/2 + π/2 = π

⇒ I = π/2

Hence, _-π/2∫^π/2 \(\frac1{1+e^{sin\,x}}dx\) = π/2.

(3x+2y)¹⁸

General term in the expansion is

T_r+1 = ¹⁸c_r (3x)^8-r (2y)^r

∴ 4^th term = T₄ = ¹⁸ c₃ (3x)^18-3 (2y)³

= 18 c₃ 3¹⁵ 2³ x¹⁵ y³

7^th term = T₇ = ¹⁸c₆ (3x)¹² (2y)⁶

8^th term = T₈ = ¹⁸c₇ (3x)¹¹ (2y)⁷.

First 10 odd natural numbers are: 1,3,5,7,9,11,13,15,17,19

Mean = Sum / Number of observations

 = 1+3+5+7+9+11+13+15+17+19/10

 =10

Correct option is (d) none of the above

\(p = \frac4{12}= \frac13\)

\(q = 1 - p = 1 - \frac13 = \frac23\)

\(n= 6\)

\(P(X =r) =\, ^nC_r \,p^r\,q^{n -r}\)

\(= \,^6C_r\,p^r\,q^{6-r}\)     \((\because n = 6)\) 

\(E(X) = \displaystyle\sum^4_{i = 0} x_i(P(x_i))\)

\(= 0.\left(\frac23\right)^6 + 1. 2\left(\frac23\right)^5 + 2. \left(\frac53\right)\left(\frac23\right)^4+ 3.\left(\frac{20}{27}\right)\left(\frac23\right)^3 + 4 . \frac5{27}.\left(\frac23\right)^2\)

\(= \frac{64}{243}+ \frac{160}{243}+ \frac{160}{243}+\frac{80}{243}\)

\(= \frac{464}{243}\)

\(= 1.9\)

Given curve is :

y = \(\frac{3}{2}x^3+4x\) ....(1)

∴ Slope of tangent,

\(\frac{dy}{dx}\) = \(\frac{9}{2}x^2+4\)

∴ Slope of tangent to given curve at x = 2 is :

\((\frac{dy}{dx})\)_{(x = 2)}

= \(\frac{9}{2}\) x 4 + 4

= 18 + 4

= 22.

Put x = 2 in equation (1), we get

y = \(\frac{3}{2}\) x 8 + 8

= 12 + 8 

= 20.

Hence,

(2,20) is point on the given curve when x = 2.

Now,

Equation of tangent to curve at x = 2 is :

y - 20 =  \((\frac{dy}{dx})\)_{(x = 2)} (x-2)

⇒ y - 20 = 22(x-2)

⇒ y - 20 = 22(x-2)

⇒ y = 22x - 44 + 20

⇒ y = 22x - 24.

Slope of normal to given curve at x = 2 is :

\(\frac{-1}{(\frac{dy}{dx})_{(x=2)}}\) = \(\frac{-1}{22}\).

∴ Equation of normal to curve at (2,20) is :

y - 20 = ( slope of normal)(x - 2)

⇒ y - 20 = \(\frac{-1}{22}(x-2)\)

⇒ 22y - 440 = - (x - 2)

⇒ x - 2 + 22y - 440 = 0

⇒ x + 22y - 442 = 0

Hence,

Equation of normal to the curve at x = 2 is x + 22y- 442 = 0.

Given curve is :

y = \(\frac{3}{2}x^3+4x\) ....(1)

∴ Slope of tangent,

\(\frac{dy}{dx}\) = \(\frac{9}{2}x^2+4\)

∴ Slope of tangent to given curve at x = 2 is :

\((\frac{dy}{dx})\)_{(x = 2)}

= \(\frac{9}{2}\) x 4 + 4

= 18 + 4

= 22.

Put x = 2 in equation (1), we get

y = \(\frac{3}{2}\) x 8 + 8

= 12 + 8 

= 20.

Hence,

(2,20) is point on the given curve when x = 2.

Now,

Equation of tangent to curve at x = 2 is :

y - 20 =  \((\frac{dy}{dx})\)_{(x = 2)} (x-2)

⇒ y - 20 = 22(x-2)

⇒ y - 20 = 22(x-2)

⇒ y = 22x - 44 + 20

⇒ y = 22x - 24.

Slope of normal to given curve at x = 2 is :

\(\frac{-1}{(\frac{dy}{dx})_{(x=2)}}\) = \(\frac{-1}{22}\).

∴ Equation of normal to curve at (2,20) is :

y - 20 = ( slope of normal)(x - 2)

⇒ y - 20 = \(\frac{-1}{22}(x-2)\)

⇒ 22y - 440 = - (x - 2)

⇒ x - 2 + 22y - 440 = 0

⇒ x + 22y - 442 = 0

Hence,

Equation of normal to the curve at x = 2 is x + 22y- 442 = 0.

Let 9/((x - 1)(x+ 2)²) = A/(x - 1) + B/(x + 2) + C/(x + 2)²

Multiplying with (x –1) (x + 2)²

9 = A (x + 2)² + B(x – 1) (x + 2) + C(x – 1) x = 1

⇒ 9 = 9A ⇒ A = 1

x = –2 ⇒ 9 = –3C ⇒ C = –3

Equating the coefficients of x²

A + B = 0 ⇒ B = – A = –1

∴ 9/((x - 1)(x + 2)²) = 1/(x - 1) - 1/(x + 2) - 3/(x + 2)²

(3x + 5)/((x + 2)(x - 1)²) = A/(x + 2) + B/(x - 1) + C/(x - 1)²

3x + 5 = A(x – 1)² + (x + 2)(x – 1) + C(x + 2)

A = -1/9, C = 8/3, 0 = A + B, B = 1/9

Substitution of A,B,C

(x - 1)/(x(x + 2)(x + 4) = A/x + B/(x + 2) + c/(x + 4)

Getting Getting A = - 1/8 B = 3/4 C = -5/8 

\(x = \frac{32 \times 50}{75}\)

= \(\frac{64}{3}\)

Correct option is (2) 150π

\(\sum^{300}_{i =1} cos^{-1}x_i = 0\)    (Given)

\(\therefore\sum^{300}_{i =1} \left(\frac{\pi}{2} - sin^{-1}x_i\right)= 0\)           \(\left(\because cos^{-1}x = \frac{\pi}{2} - sin^{-1}x\right)\)

⇒ \(\frac{\pi}{2}\times 300 - \sum^{300}_{i =1}sin^{-1}x_i = 0\)    \(\left(\because \sum^{300}_{i = 1} = 300\right)\) 

⇒ \(\sum^{300}_{i = 1 }sin^{-1}x_i = 150 \pi\)

D) ANGULAR MOMENTUM

1. Heat is represented by the symbol ‘q’. The ‘q’ is positive, when heat is transferred from the surroundings to the system and ‘q’ is negative when heat is transferred from system to the surroundings.

Work is represented by the symbol ‘w’. The ‘w’ is positive when work is done on the system and ‘w’ is negative when work is done by the system.

2. Every substance is associated with a definite amount of energy due to its physical and chemical constitution. This is called internal energy. It is the sum of the different types of energies such as chemical, electrical, mechanical etc.

1. iii) formal

( Letter to the Editor is formal )

2 iii) photo

( Not essential)

3. D) all of these

( Who to whom address is required)

4. iii) yours faithfully

( Being respectful in formal tone)

(3) Scattering

If size of particle is very small then wave length of light then scattering will happen.

A i. Missing person

( notification od missing boy)

B ii. is hereby informed

( to draw the attention of the people)

C ii. responds

( acts to the sound)

D i. abouts the whereabout

( details of he missing person)

A  iii. vehicle for sale

(Model no of car is given)

B. iv. Swift available for sale

( The model of the car other options are different)

C iii. 12000 km average self-driven

(driven by the owner)

D. iii. contact Amar / Amira 18/57 Rajnagar, Srinagar mobile 99999 24245

( contact details written)

Correct option is (B) 40 N 

For equilibrium m₂g = m₁g sinθ

Sinθ = m₂ / m₁ = 3/5

Cosθ = 4/5

Normal force on m₁ = 5g cosθ

= 5 x 10 x 4/5 = 40N

10Ω resistance and 8Ω resistance are connected in series.

R = R₁ + R₂

R = 10 + 8

R = 18Ω

(i) Current through circuit

V = IR

I = \(\frac{V}R\)

I = 6/18

I = \(\frac13\) ampeare

(ii) Potential across the heater

V = IR

V = 1/3 x 10

V = 10/3 v

(iii) Potential across the wire

V = IR

V = 1/3 x 8

V = 8/3 v

Human development refers to the psychosocial, physical, and cognitive development of humans. It extends to the entire lifespan of humans. Physical development of humans involves growth and changes in the brain and the body including sense organs, motor skills, health, etc. Psychosocial development involves changes in personality, social relationships, and emotions. Cognitive development involves changes in memory, language, reasoning, creativity, etc.

• Development in humans is both qualitative and quantitative.
  • Qualitative change occurs when individual progress and become different from the way they were earlier like the way an individual perceives the world, behave and think.
  • Quantitative change occurs when individuals acquire novel information and experiences like, the individuals become stronger, bigger, gain more weight or height.
  • It is concerned with the overall and progressive changes taking place in an individual including both quantitative as well as qualitative aspects.
  • It is a wide and complex process, thus there are some principles that need to be followed for a better understanding.
  • Sudden changes in development are actually the result of both qualitative and quantitative development.

  Hence, it can be concluded that human development is both quantitative and qualitative.

The term ‘development’ refers to qualitative changes in an individual such as a change in personality or other mental and emotional aspects. However, very often growth and development are used interchangeably. The process of development continues even after the individual has attained physical maturity (growth). The individual is continuously changing as he/she interacts with the environment.

Cognitive development:- Cognitive development refers to thinking, understanding, and concept formation.

• It involves cognitive processes such as knowing, thinking, remembering, recognizing, categorizing, imagining, reasoning, decision-making, and so forth.
• According to Piaget, children’s understanding of the world expands as they experience new ideas and challenges. Children construct their own knowledge through interaction with their surroundings. 

​​

• Piaget’s theory of cognitive development is a comprehensive theory about the nature and development of human intelligence. 
• Piaget’s idea is primarily known as a developmental stage theory. The theory deals with the nature of knowledge itself and how humans gradually come to acquire, construct, and use it. To Piaget, cognitive development was a progressive reorganization of mental processes resulting from biological maturation and environmental experience. It provides a rich and varied environment. 
• He believed that children construct an understanding of the world around them, experience discrepancies between what they already know and what they discover in their environment, then adjust their ideas accordingly.
• Moreover, Piaget claimed that cognitive development is at the centre of the human organism, and language is contingent on knowledge and understanding acquired through cognitive development.
• Piaget’s earlier work received the greatest attention. Many parents have been encouraged to provide a rich, supportive environment for their child’s natural propensity to grow and learn.

 Thus, it is concluded that cognitive development is supported by providing a rich and varied environment.

Jean Piaget, a Swiss psychologist proposed widely known perspectives about cognitive development in the stages from birth through the end of adolescence. Jean Piaget's theory focuses on the mental development of a child that moves through four different stages of intellectual development. In Piaget's view, early cognitive development involves processes based upon actions and later progresses to changes in mental operations.

Piaget’s Four Stages of Cognitive Development:

Hence, it can be concluded that the Sensory Motor stage is associated with when infants "think" with their eyes, ear, and hands.

5 tan^-1x + 3 cot^-1x = \(\frac{7\pi}4\)

⇒ 2 tan^-1x + 3(tan^-1x + cot^-1x) = \(\frac{7\pi}4\)

⇒ 2 tan^-1x + \(\frac{3\pi}2\) = \(\frac{7\pi}4\)

⇒ 2 tan^-1x = \(\frac{7\pi}4\) - \(\frac{3\pi}2\) = \(\frac{7\pi-5\pi}4=\frac{\pi}4\)

⇒ tan^-1x = \(\frac{\pi}8\)

⇒ x = tan\((\frac{\pi}8)\) ≅ 0.414
 

y = sin(cos x)

\(\frac{dy}{dx}\) = cos (cos x) \(\times\)  - sin x

 = -sin x cos (cos x)

r = cos x cot x

\(\frac{dr}{dx}\) = cos x \(\frac{d}{dx}\) cot x + (\(\frac{d}{dx}\)cos x) cot x

 = cos x (- cosec²x) - sin x cot x

 = - cos x. cosec² x - sin x\(\frac{cosx}{sin x}\) 

 = - cos x (cosec²x + 1)

Given:

Ankita is two years younger than Anu.

After four years Anu's age will be two times of Ankita's age three years ago

Calculation:

Let the age of Ankita be x

⇒ age of Anu will be = x+2

After four years Anu's age will be two times of Ankita's age three years ago.

⇒ (x + 2) + 4 = 2 (x – 3)

⇒ x + 6 = 2x – 6

⇒ x = 12

⇒ x + 2 = 12 + 2 = 14

∴ Age of Ankita is 12 years and age of Anu is 14 years.

(c) 760 W/m² 

Here G = 1000 W/m² 

E_b = 400 W/m² (Assume)

Radiosity, J = E + pG

p = Reflectivity

as surface is opaque.

So, t = 0 (Transmitivity)

So, p = 1 - \(\alpha\) = 1 - \(\varepsilon\) = 1 - 0.4 = 0.6

(by Krischoff’s law)

So, J = \(\varepsilon\)E_b + pG

= 0.4 x 400 + 0.6 x 100

= 760 W/m²

Note : In this question emissive power = 400 W/m² is considered for black surface.

Otherwise, for opaque surface emissive power = 400 W/m² , surface turned out to be black.