Explore topic-wise InterviewSolutions in Current Affairs.

\(2x^3 + 3x^2 - 8x + 3 = 0\)

x = 1 satisfies the equation (1) as 2 + 3 - 8 + 3 = 0

\(\therefore (x - 1) (2x^2 + 5x -3) = 0\)

⇒ \((x - 1) (2x - 1) (x + 3) = 0\)

⇒ \(x = 1, \frac12,-3\)

\(x^3 - 6x^2 + 11x -6 = 0\)

Let roots are \(\alpha, \beta\) and \(\gamma\).

Then sum of roots = \(\alpha + \beta + \gamma = \frac{-b}a = \frac{-(-6)}1 = 6\)

Sum of products of two roots = \(\alpha\beta + \beta\gamma + \alpha\gamma = \frac ca = \frac{11}1 = 11\)

Product of roots = \(\alpha \beta\gamma = \frac{-d}a = \frac{-(-6)}1 = 6\).

Correct option is (C) \(2x^2 - \sqrt 3 x + 6 \)

(A) \(x + \frac 2x + 3 = \frac{x^2 + 2 + 3x}x\)    (Not a polynomial)

(B) \(3\sqrt x + \frac 2{\sqrt x} + 5 = \frac{3x + 2 + 5\sqrt x}{\sqrt x}\)    (Not a polynomial)

(C) \(2x^2 - \sqrt 3 x + 6 \)  is a polynomial in one variable (x).

(D) \(x^{10} + y^5 + 8\)  is a polynomial in two variables x & y.

\(P(x) = x^2 + 2x + 1\)

\(P(x) = 0\) gives 

\(x^2 + 2x + 1 = 0\)

⇒ \((x + 1)^2 = 0\)

⇒ \(x + 1 = 0 \,or\, x + 1 = 0\)

\(x = -1 \,or\,x = -1\).

Sum of zeros = \(-1 - 1 = -2 = \frac{-2}1 = \frac{-b}a = \frac{\text{-coefficient of }x}{\text{coefficient of }x^2}\)

Product of zeros = \((-1)(-1) = 1 = \frac11 = \frac ca = \frac{\text{constant term}}{\text{coefficient of }x^2}\) 

(b) Tharparkar

Correct option - (B) y=6, x=8, z=8

Explanation:-

y is assigned value of x that is 6, then x in incremented that is value of x=7, z is assigned value of x after incrementing that is z =8 so value of x =8. 

C function named ‘fiddle’ that takes two arguments, x and y and changes both values. x is an int while y is a pointer to int

#include<conio.h>

void main()

{

int x,z,*y,temp;

clrscr();

printf("\n Enter two numbers : \n");

scanf("%d %d",&x,&z);

y=&z;

printf("\n\n x = %d and y = %d",x,*y);

fiddle(&x,y);

printf("\n After changing their values ");

printf("\n x = %d and y = %d",x,*y);

getch();

}

fiddle(int *a,int *b)

{

int temp;

temp=*a;

*a=*b; 

*b=temp;

getch();

}  

\(\int\frac 1{(x -1 )^2 \sqrt{(1 - x)^2}}\,dx\)

\(= \int \frac{1}{(1 -x) (x - 1)^2}dx\)

\(= -\int \frac1{(x - 1)^3} \,dx\)

\(= -\int(x - 1)^{-3} \, dx\)

\(= - 1 \times\frac{(x - 1)^{-3 + 1}}{-3 + 1} + C\)

\(= \frac 12 (x -1)^{-2} + C\)

\(= \frac 1{2(x -1)^2} + C\)

√-a×√-b

=√-1×√a×√-1×√b

= i × √a × i × √b        (because √-1=i)

= i² × √ab

=-1√ab.        (because i² = -1)

=-√ab

Answer is √-ab

Interest = \(\frac {PRT}{100} = \frac {5000 \times 8.5 \times 5} {100}\)

\(= 5 \times 85 \times 5 \)

\(= 2125\)

Correct answer is (B) 12

Correct answer is (B) √3, -√3

Correct option is: (D) \(\frac{4}{9}\)π

Correct answer is (B) x = 2, y = 6

Correct option is: (A) 3

Correct answer is (B) 3

Correct option is: (C) third quadrant

Correct option is D) higher / four times that

P = Rs. 540000

R = 10%

T = 18 months = \(1 \frac{1}{2}\) = \(\frac{3}{2}\) years

S.I = \(\frac{P \times R \times T}{100}\)

= \(\frac{540000 \times 10 \times 3}{100 \times 2}\)

= Rs.5400 x 5 x 3

= Rs.81000

n(axb) = 20

20 = 5xn(B)

n(B) = 20/5

n(B) = 4 

n(A) = 5, n(A x B) = 20

⇒ n(A) n(B) = 20 (\(\because\) n(A x B) = n(A). n(B))

⇒ n(B) = \(\frac{20}{n(A)}=\frac{20}5=4\)

P = Rs. 240000

T = 3.5 years

S.I = Rs. 15000

R = ?

\(R = \frac{S.I \times 100}{P \times T}\)

= \(\frac{15000 \times 100}{240000 \times 3.5}\)

= \(\frac{150}{84}\)

= 1.8 % aprox.

1. Forecasting Demand for Products
Today, there are still many manufacturers who have difficulty forecasting future demand. The main problem is that they do not have advanced reporting tools that allow them to estimate how many items they should sell in the next few months or the following year. As a result, their products fail to meet the customer demand and they suffer lower sales.

2. Controlling Inventory
Inventory management is still one of the main challenges in the manufacturing industry, but thanks to the help of automated solutions, it has become much simpler. Nevertheless there are still many manufacturers, especially the small ones, who still manage their inventory manually.

Inventory tracking is a time-consuming process that can be streamlined with the help of software. Checking stock manually is very inefficient and prone to errors that can lead to inaccuracies, shortages and overstock, as well as unidentified damages.

3. Improving Efficiency at Manufacturing Plants
Up until now, manufacturers have been looking for effective ways to reduce costs and improve efficiency at their plants. Many of them choose to sacrifice the quality of their products to reduce their production costs, but this will only reduce their profitability, because dissatisfied customers will stop buying from them.

4. Increasing ROI
Any manufacturer would want to be able to increase their ROI. They would usually increase their sales or the price of their products. However, these aren’t effective ways, especially when economic conditions are being erratic, reducing consumer purchasing power.

5. Skilled Labor Shortage
Although automation and robotics can help fill the labor gap, human capabilities will still be needed to analyze and solve problems as well as manage outputs. With the baby boomer generation entering retirement, the manufacturing industry is facing a looming labor shortage. This is one of the biggest threats facing the manufacturing sector today.

6. Managing Sales Leads
Another challenge often faced by manufacturers is managing and prioritizing sales leads. Many of them treat their leads in the same way, but this isn’t the right method. Every sales lead must be treated specifically since they have different characters, preferences, and needs.

Manufacturers also often find it difficult to identify potential leads so they often focus on unpromising opportunities and forget to follow up with high potential leads.

Types of cybercrime

• Email and internet fraud.
• Identity fraud (where personal information is stolen and used).
• Theft of financial or card payment data.
• Theft and sale of corporate data.
• Cyberextortion (demanding money to prevent a threatened attack).
• Ransomware attacks (a type of cyberextortion).

The correct option is (D) data to information to knowledge

• The main transformation cycle for information is “data, information, knowledge”.

• This is often stated as the DIKAR model i.e. Data, Information, Knowledge, Action, and the Result. This model gives a solid hint as to the layers elaborate in aligning technology and structural strategies, and it is an essential moment in changing defiance to information management.  

B) always false

The correct option is (A) A family of concurrent lines.

Given 

ax² + bx + c = 0

One root is −2

−2 + α = −b/a ​−−−(1)

−2α = c/a ​−−−(2)

On solving eq (1) and (2)

c = 2(a + b)

Given eq 

x + by + c = 0

x + by + 2a + 2b = 0

Hence above equation represents family of lines.

Computer is an electronic device.which is used by human to reduce time.it is used to calculate millions of calculations in a second.

VR = 2ⁿ , where n is the number of movable pulleys. 

VR = 23 = 8 

Now, MA = η × VR 

= 0.8 × 8 

= 6.4

\(\frac WP\) = 6.4

P = \(\frac W{6.4}\) = \(\frac {6000}{6.4}\)

P = 937.5 N 

In the second case, 

Effort = 520 N

Efficiency η = 0.80 – n₁ × 0.05 

where n₁ = number of additional pulleys required and equal to (n – 3). 

MA = η × VR

\(\frac WP\) = η × VR

W = P × η × 2ⁿ 

= P(0.8 – n1 × 0.05) × 2ⁿ 

= P[0.8 – (n – 3) × 0.05] 2ⁿ

By going for a trial and error solution, starting with one additional pulley i.e., totally with four pulleys, 

W = 520 [0.8 – (4 – 3) × 0.05] 2⁴ = 6240 N 

i.e., if four pulleys are used, a load of 6240 N can be raised with the help of 520 N effort.

∴ Number of movable pulleys required = 4

 5. (a) D² + 2pD + (p² + q²)y = 0

It's auxiliary equation is

m² + 2pm + p² + q² = 0

⇒ (m + p)² + q² = 0

⇒ m + p = \(\pm q\)i

⇒ m = -p \(\pm q\)i

\(\therefore\) C.F. = e^-px (C₁ cos9x + C₂sin qx)

Hence, y = e^-px (C₁cos 9x + C₂ sin 9x) is solution of given differential equation.

(b) m² + 1 = 0

⇒ m = \(\pm i\)

C.F. = C₁ cos x + C₂ sin x

y₁ = cos x, y₂ = sin x

w(y₁, y₂) = \(\begin{vmatrix}cosx&sin x\\-sin x&cos x\end{vmatrix}\) = cos²x + sin²x = 1

u₁ = \(\int\frac{\begin{vmatrix}0&sin x\\cosxsin3x&cosx\end{vmatrix}}{W(y_1,y_2)}dx\) = \(\int-sin xcos xsin 3xdx\)

 = \(\frac{-1}2\int sin2xsin 3x dx\) 

 = \(\frac{-1}4\int 2sin2xsin 3x dx\) 

= \(\frac{-1}4\int4[cos(2x-3x)-cos(2x+3x))dx\)

 = \(\frac{-1}4[\int cos xdx-\int cos 5xdx)\) 

 = \(\frac{-1}4[sin x - \frac{sin 5x}5]\)

u₂ = \(\int\frac{\begin{vmatrix}cos x&0\\-sin x&cosxsin 3x\end{vmatrix}}{1}dx\) 

 = \(\int\)cos²x sin 3x dx

 = \(\int\) \(\frac{1+cos2x}2sin3xdx\)

 = \(\frac12\)\(\int\)sin 3x dx + \(\frac14\)\(\int\)2sin 3x cos 2x dx

 = -\(\frac12\) \(\frac{cos 3x}3+\frac14\int\)(sin(3x + 2x) - sin (3x - 2x)]dx

 = -\(\frac16\)cos3x + \(\frac14\int\) sin 5x dx - \(\frac14\int\) sin x dx

 = -\(\frac16\) cos 3x - \(\frac1{20}\) cos 5x + \(\frac14\) cos x

\(\therefore\) P.I. = u₁y₁ + u₂y₂ 

  = -\(\frac14\)sin x cos x + \(\frac1{20}\) cos x sin 5x - \(\frac1{6}\)sin x cos 3x

- \(\frac1{20}\) sin x cos 5x + \(\frac1{4}\) sin x cos x

 = \(\frac1{20}\) (cos x sin 5x - sin x cos 5x) - \(\frac1{6}\) sin x cos 3x

\(\therefore\) y = C.F. + P.I.

= C₁ cos x + C₂ sin x + \(\frac1{20}\) (cos x sin 5x - sin x cos 5x)

- \(\frac1{6}\)sin x cos 3x

(c) m³ + m² + m = 0

m(m² + m + 1) = 0

m = 0, \(\frac{-1\pm\sqrt3i}2\) 

C.F. = C₁ + e^-x/2(C₂ cos(\(\frac{\sqrt{3x}}{2}\)) + C₃ sin(\(\frac{\sqrt{3x}}{2}\)))

P. I. = \(\frac{1}{D^3+D^2+D}\) (e^2x + x² + x)

 = \(\frac{e^{2x}}{8+4+2}+\frac1D\)(1 + (D + D²))^-1 (x² + x)

 = \(\frac1{14}e^{2x}\)+ (\(\frac1D\) - 1 - D + D + 2D² - D²) (x² + x)

 = \(\frac1{14}e^{2x}\) + (\(\frac1D\) - 1 + D²) (x² + x)

 = \(\frac1{14}e^{2x}\) + \(\frac{x^3}3-\frac12x^2-x+2\)

y = C.F. + P.I.

 = C₁ + e^-x/2(C₂cos(\(\frac{\sqrt{3x}}{2}\)) + C₃sin(\(\frac{\sqrt{3x}}{2}\)))

 + \(\frac1{14}e^{2x}\) + \(\frac{x^3}3-\frac{x^2}2-x+2\) 

6. (a) Let x = e^z

Then xd = \(\frac d{dz}=D'\)

x²D² = D' (D' - 1)

Given D.E. converts to

(D' (D' - 1) - 4D' + 6)y = e^yz

⇒ ((D')² - 5D' + 6)y = e^yz

m² - 5m + 6 = 0 by taking D' = m

⇒ m = 2, 3

C.F. = C₁e^2z + C₂e^3z

P.I. = \(\frac{1}{(D')^2 - 5D' + 6}e^{yz}\)

 = \(\frac{e^{4z}}{4^2-5\times4+6}\) = \(\frac{e^{4z}}2\)

\(\therefore\) y = C.F. + P. I.

 = C₁(e^z)² + C₂(e^z)³ + \(\frac12\)(e^z)⁴

 = C₁x² + C₂x³ + \(\frac{x^4}2\) (\(\because\) e^z = x)

(b) (x²D² + xD + 1)y = log x + cos (log x)

Let x = e^z

Then  given D.E. converts to

(D'(D'- 1) + D' + 1)y = a + cos z

⇒ ((D')² + 1)y = z + cos z

\(\therefore\) m² + 1 = 0

m = \(\pm i\)

C.F. = C₁cos z + C₂ sin z

P.I = \(\frac1{(D')^2+1}\)(z + cos z)

 = (1 - (D')² + (D')⁴+....) z + \(\frac{z}2\)sin z

 = z + \(\frac{z}2\) sin z

\(\therefore\) y = C.F. + P. I.

= C₁cos z + C₂ sin z + z + \(\frac{z}2\) sin z

= C₁cos(log x) + C₂ sin (log x) + log x + \(\frac{log x}2\) sin (log x)

(c) (x + 1)²\(\frac{d^2y}{dx^2}\) - 3(x + 1) \(\frac{dy}{dx}\) + 4y = x

Let x + 1 = e^z

Then D.E. converts to

(D'(D'-1) - 3 D' + 4)y = e^z - 1

⇒ ((D')² - 4D' + 4)y = e^z - 1

⇒ (D' - 2)²y = e^z - 1

\(\therefore\) (m - 2)² = 0

m = 2, 2

C.F. = (C₁ + C₂z)e²z

P.I. = \(\frac{1}{(D'-2)^2}(e^z-e^{0z})\)

 = \(\frac{e^z}{(1-2)^2}-\frac{e^{0z}}{(0-z)^2}\)

 = \(\frac{e^z}1-\frac14\)

\(\therefore\) y = C.F. + P. I.

= (C₁ + C₂z)e²z + e^z - \(\frac14\)

 = (C₁ + C₂ log x) x² + x - 1/4 is solution of given differential equation.

The differential equation is 

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

Highest order derivative is \(\frac{d^3y}{dx^3}\)

But differential equation is not a polynomial equation in derivative of y.

∴ Degree is not defined

Order = Order of highest order derivative=3

Hence, order of differential equation is 3 but degree is not defined.

Given differential equation is

\(\frac{d^2x}{dt^2}-4\frac{dx}{dt}+13x=0\)

It's auxiliary equation is

m² - 4m + 13 = 0

⇒ m = \(\frac{4\pm\sqrt{16-52}}2\) = \(\frac{4\pm 6i}2\) = 2\(\pm\)3i

∴ x = e^2t(C₁ cos 3t + C₂ sin 3t)

\(\frac{dx}{dt} \) = e^2t(2C₁ cos 3t + 2C₂ sin 3t - 3C₁ sin 3t + 3C₂ cos 3t)

 = e^2t((2C₁ + 3C₂) cos 3t + (2C₂ - 3C₁) sin 3t)

Given that x(0) = 0

∴ C₁ = 0

also given that \(\frac{dx}{dt}(0)=2\)

∴ 2C₂ = 2

⇒ C₂ = 2/3

∴ x = e^2t(2/3 sin 3t)

⇒ 2/3 e^2t sin 3t which is solution of given differential equation.

It's auxiliary equation is 

m³ + 2m² + m = 0

⇒ m(m² + 2m + 1) = 0

⇒ m(m + 1)² = 0

⇒ m = 0, -1, -1

∴ y = C.F. = C₁e^0x + (C₂ + C₃x)e^-x

⇒ y = C₁ + (C₂ + C₃x)e^-x which is solution of given differential equation.