Explore topic-wise InterviewSolutions in .

(b) LibreOffice Calc

Answer is (d) one-time password

(c) TO the mobile, the number is given to the .. for configuring the account online

Answer is c and d

Answer is a and d

Set - I

(a) What things are posted on the media is removed, to prevent others see the matter

Set - II 

(a) Posting of vulgar messages and bullying are legally criminal 

Set - III

(c) Hide your personal details and don’t share to others 

Set - IV

(d) Delete your friendship with those people who misuse the internet

(b) One time Password

(a) World web-wide

Set -1 

(a) We can find out the details of our bank account form any place at any time

Set-II

(b) We can examine our recent bank transaction details form any place at any time

Set - III

(c) WE can transfer money to another account at any place at any time.

Set – IV

(d) We can pay for things or services that we have bought any time at any place.

Answer is (b) OTP is send to the mobile for that transition and is carried out by that password for safety

Answer is a and c

(c) e-banking

(b) Prepared in HTML Language

(a) Uniform Resource Locator

(a) Hyper Text Transfer Protocol

(a) It is system in the internet that helps computers to convert web addresses to IP addresses and vice versa

(b) Domain Name Server

(c) Host names

(d) google.co.in

Answer is a and c

Answer is a and c

Answer is c and d

Answer is b and d

Answer is (d) one

Terminal is using to find IP address using host name.

(a) networking

(d) more than one

(b) hosting a website

Diaspora is a medium for open-source communication system.

Answer is (c) Email

दिया है- R(x) = 3x² + 36x + 5 

सीमान्त । 

सीमान्त आय = d/dxR(x) = d/dx(3x² + 36x + 5) = 6x + 36 = 6(x + 6)

अब, x = 15, सीमान्त आय = 6 × 21 = Rs 126 

अत: विकल्प (d) सत्य है।

Correct answer is C.

Let f(x) = ax³ + bx² + cx + d --------- (i)

f(0) = d

f(2) = a(2)³ + b(2)² + c(2) + d

= 8a + 4b + 2c + d

= 2(4a + 2b + c) + d

\(\because\) 4a + 2b + c = 0 {Given}

= 2 (0) + d

= 0 + d

= d

f is continuous in closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

As per Rolle’s Theorem,

f’(α) = 0 for 0 < α < 2

f’(x) = 3ax² + 2bx + c

f’(α) = 3aα² + 2b(α) + c

3aα² + 2b(α) + c = 0

Hence equation (i) has at least one root in the interval (0, 2).

Thus, f’(x) must have one root in the interval (0, 2).

Given, as f(x) = x³ – 3x

On differentiating with respect to x then we get,

f’(x) = 3x² – 3

f‘(x) =0

= 3x² = 3 ⇒ x = ±1

Now, again differentiate f’(x) = 3x² – 3

f’’(x)= 6x

f’’(1)= 6 > 0

f’’ (– 1)= – 6 > 0

On second derivative test, x = 1 is a point of local minima and local minimum value of g at

x = 1 is f (1) = 1³ – 3 = 1 – 3 = – 2

However, x = – 1 is a point of local maxima and local maxima value of g at

x = – 1 is f (– 1) = (– 1)³ – 3(– 1)

= – 1 + 3

= 2

Thus, the value of minima is – 2 and maxima is 2.

Given as f(x) = (x – 5)⁴

On differentiating with respect to x

f’(x) = 4(x – 5)³

For the local maxima and minima

f‘(x) = 0

= 4(x – 5)³ = 0

= x – 5 = 0

x = 5

f‘(x) the changes from negative to positive as passes through 5.

Therefore, x = 5 is the point of local minima

Hence, local minima value is f (5) = 0

Given as f(x) = |x + 2| ≥ 0 for x ∈ R

= f(x) ≥ 0 for all x ∈ R

Therefore, the minimum value of f(x) is 0, which attains at x =2

Thus, f(x) = |x + 2| does not have the maximum value.

f(x) = 4x² – 4x + 4 on R 

= 4x² – 4x + 1 + 3 

= (2x – 1)² + 3 

Since, 

(2x – 1)² ≥ 0 

= (2x – 1)² + 3 ≥ 3 

= f(x) ≥ f(\(\frac{1}{2}\))

Thus, the minimum value of f(x) is 3 at x = \(\frac{1}{2}\)

Since, 

f(x) can be made large. 

Therefore maximum value does not exist.

Given as f(x) = |sin 4x + 3| on R

As we know that – 1 ≤ sin 4x ≤ 1

= 2 ≤ sin 4x + 3 ≤ 4

= 2 ≤ |sin 4x + 3| ≤ 4

Thus, the maximum and minimum value of f are 4 and 2 respectively.

Given as f(x) = – (x – 1)² + 2

It can be observed that (x – 1)² ≥ 0 for every x ∈ R

So, f(x) = – (x – 1)² + 2 ≤ 2 for every x ∈ R

The maximum value of f is attained when (x – 1) = 0

(x – 1) = 0, x = 1

Here, Maximum value of f = f (1) = – (1 – 1)² + 2 = 2

Thus, function f does not have minimum value.

Given as f(x) = 4x² – 4x + 4 on R

= 4x² – 4x + 1 + 3

On grouping the above equation we get,

= (2x – 1)² + 3

Here, (2x – 1)² ≥ 0

= (2x – 1)² + 3 ≥ 3

= f(x) ≥ f (1/2)

Hence, the minimum value of f(x) is 3 at x = 1/2

Clearly, f(x) can be made large. So maximum value does not exist.

Given as x = at², y = 2at, at t = 1

Differentiate with respect to t, to get slope of tangent 

dx/dt = 2at

dy/dt = 2a

Dividing dy/dt and dx/dt to obtain the slope of tangent 

dy/dx = 1/t

m(tangent) at t = 1 is 1

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at t = 1 is -1

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - 2a = 1(x - a)

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - 2a = -1(x - a)

finding the slope of the tangent by differentiating the curve

\(2y\frac{dy}{dx}=4a\)

\(\frac{dy}{dx}=\frac{2a}{y}\)

m(tangent) at (\(\frac{a}{m^2},\frac{2a}{m}\))

m(tangent) = m

normal is perpendicular to tangent so, m₁m₂ = – 1

\(m(normal)=-\frac{1}{m}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{2a}{m}=m(x-\frac{a}{m^2})\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-\frac{2a}{m}=-\frac{1}{m}(x-\frac{a}{m^2})\)

Given as y = x² at (0, 0)

Differentiate the given curve, to get the slope of the tangent

dy/dx = 2x

m (tangent) at (x = 0) = 0

Normal is perpendicular to tangent so, m₁m₂ = – 1

n(normal) at (x = 0) = 1/10

As we can see that the slope of normal is not defined

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y = 0

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 0

Given as x = a sec t, y = b tan t at t

Differentiate with respect to x, to get the slope of tangent

dx/dt = a sec t tan t

dy/dt = b sec² t

Dividing dy/dt and dx/dt to obtain the slope of tangent

dy/dx = b cosec t/a

m(tangent) at t = b cosec t/a

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at t = (-a/b)sin t

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - b tan t = (b cosec t/a)(x - a sec t)

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - b tan t = (-a sin t/b)(x - a sec t)

Given as y = 2x² – 3x – 1 at (1, – 2)

Differentiate the given curve, to get the slope of the tangent

dy/dx = 4x - 3

m(tangent) at (1, – 2) = 1

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at (1, – 2) = – 1

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y + 2 = 1(x – 1)

y = x – 3

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y + 2 = –1(x – 1)

y + x + 1 = 0

finding slope of the tangent by differentiating the curve

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

m(tangent) at (3,0) = 10

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (3, 0) = \(-\frac{1}{10}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y at x = 3

y = 3² + 4×3 + 1

y = 22

y – 22 = 10(x – 3)

y = 10x – 8

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-22=-\frac{1}{10}(x-3)\)

x + 10y = 223

Given as x = a(θ + sinθ), y = a(1 – cosθ) at θ

Differentiate with respect θ, to get the slope of the tangent

dx/dθ = a(1 + cosθ)

dy/dθ = a(sinθ)

Dividing dy/dθ and dx/dθ to obtain the slope of tangent 

dy/dx = sinθ/(1 + cosθ)

m(tangent) at θ is sinθ/(1 + cosθ)

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at θ is -sinθ/(1 + cosθ)

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - a(1 - cosθ) = (sinθ/(1 + cosθ))(x - a(θ + sinθ))

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - a(1 - cosθ) = ((1 + cosθ)/sinθ)(x - a(θ + sinθ)

Given as x² + 2y² - 4x - 6y + 8 = 0

Differentiate with respect to x, to get the slope of tangent

2x + (4y(dy/dx)) - 4 - (6dy/dx) = 0

dy/dx = (4 - 2x)/(4y - 6)

Finding the y co–ordinate by substitute x in the given curve

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m(tangent) at x = 2 is 0

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at x = 2 is 1/0, which is undefined

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 2

finding slope of the tangent by differentiating the curve

\(2x+4y\frac{dy}{dx}-4-6\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=\frac{4-2x}{4y-6}\)

Finding y co – ordinate by substituting x in the given curve

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m(tangent) at x = 2 is 0

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at x = 2 is \(\frac{1}{0}\), which is undefined

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 2

Correct Answer is \(\frac{5\pi}{6}\)

Let the principle value be given by x 

Now, let x = \(sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)\)

⇒ sec x = \(\frac{-2}{\sqrt{3}}\)

⇒ sec x= - sec( \(\frac{\pi}{6}\)) ( \(\because sec\left(\frac{\pi}{6}\right)=\left(\frac{2}{\sqrt{3}}\right)\)

⇒ sec x=sec( \(\pi-\frac{\pi}{6}\)) ( \(\because -sec(\theta)=sec(\pi-\theta\))) 

⇒ x =\(\frac{5\pi}{6}\)

Given as ay² = x³

Differentiate with respect to x, to get slope of tangent 2ay(dy/dx) = 3x² 

dy/dx = 3x²/2ay

m(tangent) at (am², am³) is 3m/2

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at (am², am³) is -2/3m

The equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - am³ = (-2/3m)(x - am²)