## What is Quantum Computing ?

### Quantum Computing :

Quantum computing is a type of computing that utilizes the principles of quantum mechanics to perform mathematical calculations . Unlike classical computers that uses bits i.e binary digits to represent 0 or 1 , the quantum computers uses quantum bits , that can exist simultaneously in multiple states due to superposition property .
Quantum computers can solve the certain problems much faster than classical computers . This includes factoring large numbers and searching large databases .

Quantum computing techniques can enhance the machine learning algorithms to provide improvement in pattern recognition . Quantum computing have very much excellency in complex optimization problems , having applications in logistics , finance and various industries where finding best solution among various possibilities are important .

#### Superposition :

Quantum bits can exist in a combination of 0 and 1 states simultaneously . Due to this property quantum computers can process huge number of possibilities at once .

#### Quantum Gates :

Quantum Gates is analogous to classical logic gates but it takes the advantage of quantum properties . Quantum computers uses quantum gates to perform operations on quantum bits i.e qubits .

#### Quantum Parallelism :

Quantum computers can process many solutions to a problem simultaneously .
Therefore Quantum computers can solve complex problems more efficiently than classical computers , especially in those areas such as cryptography , optimization and simulation of quantum systems . However , quantum computers are still in the early stage of development and facing some challenges for implementing error correction and maintaining quantum bits .

## A particle moving on straight line whose velocity-time graph is shown in the figure. The average speed from t = 0 to t = 6 s is …

Q: A particle moving on straight line whose velocity-time graph is shown in the figure. The average speed from t = 0 to t = 6 s is v = 15/n m/s. Find the value of n. Solution : Area of v-t graph from zero to 5 sec is

A1 = Area of trapezium

$\displaystyle A_1 = \frac{1}{2}(5+3) \times 10$

$\displaystyle A_1 = 40 m$

Area of v-t graph from 5 sec to 6 sec is

A2 = Area of triangle

$\displaystyle A_2 = \frac{1}{2} \times 1 \times 10$

A2 = 5 m

Total distance traveled = 40 + 5 = 45 m

Average Speed = Total Distance / Total time

$\displaystyle V = \frac{45}{6} = \frac{15}{2}$

On comparing we get ,

n = 2

## What is Artificial Intelligence ?

Artificial Intelligence refers to the development of computer Systems that can perform tasks which require human intelligence . The Artificial Intelligence technology aim is to give human like functions by enabling machines to perform complex tasks in easy manner and adapt new situations . Artificial Intelligence includes several techniques and approaches by creating machines which are capable of doing intelligence tasks .

### Key Features of Artificial Intelligence :

Artificial intelligence have several key features to perform task , some of them are :

### Machine Learning (ML) :

Machine Learning is a part of Artificial intelligence that enables the machine to learn from data . It involves training , instructions on datasets and allowing the systems to improve their performance over time .

### Neural Networks :

Being inspired by the human brain , neural networks are the fundamental component of many AI Systems . Neural networks with multiple layers utilizes deep learning for performing more complex tasks .

### Natural Language Processing :

This feature enables machines to understand , interpret and generate human language . It is crucial for applications like speech recognition , language translation etc .

### Computer vision :

This feature enables machine to interpret and make decisions which are based on visual data . It can be used in facial recognition , object detection and many other applications .

### Robotics :

On integrating Artificial Intelligence with Robot , its ability can be enhanced to perceive the environment and perform the assigned task . This includes both the robots either physical robots or software driven robots .

### Expert Systems :

These are the computer systems designed to match or surpass the decision making ability of a human expert . It can be used to solve specific problems on the basis of their knowledge bases and reasoning abilities .

Conclusion : The purpose of Artificial intelligence is to create the systems that can adapt , learn and perform task intelligently & efficiently to provide the solutions for variety of difficulties that need to be addressed across various industries .

## A particle is moving on a straight line whose velocity as function of time is (t-2) m/s . Find the distance traveled …

Q: A particle is moving on a straight line whose velocity as function of time is (t-2) ms-1. Find distance travelled by particle (in m) in 4 s.

Solution : Here , v= 0 at t = 2 sec

Distance travelled by particle in 4 sec is

$\displaystyle S = |\int_{0}^{2} v dt | + |\int_{2}^{4} v dt |$

$\displaystyle S = |\int_{0}^{2} (t-2) dt | + |\int_{2}^{4} (t-2) dt |$

$\displaystyle S = |[\frac{t^2}{2} – 2 t]_{0}^{2} | + |[\frac{t^2}{2} – 2 t]_{2}^{4} |$

$\displaystyle S = |[( \frac{2^2}{2} – 2 \times 2 ) – 0] | + |[( \frac{4^2}{2} – 2 \times 4 ) – (\frac{2^2}{2} – 2 \times 2)] |$

$\displaystyle S = |-2| + |2|$

S = 4 m

## Normal to x^2/3 + y^2/3 = a^2/3 in the form y cosθ – x sinθ = a cos2θ ….

Q: Normal to x2/3 + y2/3 = a2/3 in the form y cosθ – x sinθ = a cos2θ ; Where θ is the angle which the normal makes with the x – axis .

Solution : x2/3 + y2/3 = a2/3

Differentiating with respect to x

$\displaystyle \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = – \frac{x^{-1/3}}{y^{-1/3}} = – \frac{y^{1/3}}{x^{1/3}}$

Hence Slope of normal is

$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = tan\theta$ ; Since it is given that normal makes an angle θ with x -axis .

$\displaystyle = \frac{x^{1/3}}{y^{1/3}} = \frac{sin\theta}{cos\theta}$

$\displaystyle = \frac{x^{1/3}}{sin\theta} = \frac{y^{1/3}}{cos\theta} = [\frac{x^{2/3} + y^{2/3}}{sin^2 \theta + cos^2 \theta}]^{1/2} = a^{1/3}$

Hence , x = a sin3θ , y = a cos3θ

Therefore Equation of normal whose slope is tanθ will be

$\displaystyle y – a cos^3 \theta = \frac{sin\theta}{cos\theta} (x – a sin^3 \theta)$

$\displaystyle y cos\theta – a cos^4 \theta = x sin\theta – a sin^4 \theta)$

$\displaystyle y cos\theta – x sin\theta = a ( cos^4 \theta = – sin^4 \theta )$

$\displaystyle y cos\theta – x sin\theta = a cos 2 \theta$

## Find the equation of tangent and Normal to the curve at any point (x,y) x = a sin^3θ , y = a cos^3θ

Q: Find the equation of tangent and Normal to the curve at any point (x,y) x = a sin3θ , y = a cos3θ

Solution : x = a sin3θ , y = a cos3θ

$\displaystyle \frac{dx}{d\theta} = 3 a sin^2 \theta \; cos\theta$

$\displaystyle \frac{dy}{d\theta} = – 3 a cos^2 \theta \; sin\theta$

$\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta}/\frac{dx}{d\theta}$

$\displaystyle \frac{dy}{dx} = – \frac{- 3 a cos^2 \theta \; sin\theta}{3 a sin^2 \theta \; cos\theta}$

$\displaystyle \frac{dy}{dx} = – \frac{cos\theta}{sin\theta}$

= – cotθ

Equation of tangent is

$\displaystyle y – a cos^3 \theta = – \frac{cos\theta}{sin\theta} (x – a sin^3 \theta )$

$\displaystyle x cos\theta + y sin\theta = (a/2) sin2\theta$

Equation of Normal :

As Slope of tangent is – cotθ ; therefore Slope of normal is + tanθ .

Equation of Normal is

$\displaystyle y – a cos^3 \theta = \frac{sin\theta}{cos\theta} (x – a sin^3 \theta )$

$\displaystyle y cos\theta – x sin\theta = a (cos^4 \theta – sin^4 \theta )$

$\displaystyle y cos\theta – x sin\theta = a cos2 \theta$

## Find the equation of the tangent to the curve at any point (x,y) $\frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Q: Find the equation of the tangent to the curve at any point (x,y) $\displaystyle \frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Solution : $\displaystyle \frac{x^m}{a^m} + \frac{y^m}{b^m} = 1$

Differentiating with respect to x we get ,

$\displaystyle m . \frac{x^{m-1}}{a^m} + m . \frac{y^{m-1}}{b^m} . \frac{dy}{dx} = 0$

$\displaystyle m . \frac{x^{m-1}}{a^m} = – m . \frac{y^{m-1}}{b^m} . \frac{dy}{dx}$

$\displaystyle \frac{x^{m-1}}{a^m} = – \frac{y^{m-1}}{b^m} . \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = -\frac{1}{a} (\frac{x}{a})^{m-1}.b (\frac{b}{y})^{m-1}$

Equation of tangent is

$\displaystyle Y – y = \frac{dy}{dx} (X-x)$

$\displaystyle Y- y = -\frac{1}{a} (\frac{x}{a})^{m-1}.b (\frac{b}{y})^{m-1} (X-x)$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} – (\frac{x}{a})^m = -\frac{Y}{b}(\frac{y}{b})^{m-1} + (\frac{y}{b})^m$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} + \frac{Y}{b}(\frac{y}{b})^{m-1} = (\frac{x}{a})^m + (\frac{y}{b})^m$

$\displaystyle \frac{X}{a} (\frac{x}{a})^{m-1} + \frac{Y}{b}(\frac{y}{b})^{m-1} = 1$

## Find the domain and range of the function $y = \frac{1}{\sqrt{x-[x]}}$

Q: Find the domain and range of the function $\displaystyle y = \frac{1}{\sqrt{x-[x]}}$

Solution : The function is defined for all x ∈ R except when x – [x] = 0

⇒ x is an integer i.e. x ∈ I

Domain = R – I , where I is set of integers .

Also x – [x] > 0

Again $\displaystyle x – [x] = \frac{1}{y^2}$ from given relation

By the definition of x-[x] we know that

0 < x-[x] < 1

$\displaystyle 0 < \frac{1}{y^2} < 1$

y2 > 1

⇒ y2 – 1 > 0

(y+1)(y-1) > 0

y < -1 , y > 1 , but y is +ve

Hence y > 1

Hence Range = (1 , ∞)

## Find the domain and Range of $y = \frac{1}{\sqrt{4 + 3 cosx}}$

Q:  Find the domain and Range of $\displaystyle y = \frac{1}{\sqrt{4 + 3 cosx}}$

Solution: Domain = R , Since  – 1 ≤ cosx ≤ 1

$\displaystyle y \in [\frac{1}{\sqrt{7}} , 1 ]$

The function is many-one but not one-one as for so many values of x we will have same image as

cos(2nπ+x) = cos x

## Find the domain and Range of the function y = (x+2)/(x^2 – 8x -4)

Q: Find the domain and Range of the function $\displaystyle y = \frac{x+2}{x^2 – 8x -4}$

Solution : As we know that denominator should not be zero .

x2 – 8x -4 = 0

$x = 4 \pm 2 \sqrt{5}$

Domain $R – ( 4 \pm 2 \sqrt{5})$

Let $\displaystyle y = \frac{x+2}{x^2 – 8x -4}$

$\displaystyle x^2 y – (8y+1 )x – (4y+2) = 0$

Since x is real Δ = B2 – 4 A C ≥ 0

80 y2 + 24 y + 1 ≥ 0

(20 y + 1)(4y+1) = +ve

Hence , y ≤-1/4 or ≥ -1/20

Also when y = 0 , x = -2 ∈ R

Y = 0 also belongs to Range

Range = (-∞ , -1/4) U (-1/20 , ∞)