This is a 3 hour sample paper for cbse class 12 board exam. This covers all chapters of ncert 12th math book. For more such papers visit clay6.com/papers/.
This is a 3 hour sample paper for cbse class 12 board exam. This covers all chapters of ncert 12th math book. For more such papers visit clay6.com/papers/.
This learner's module discusses or talks about the topic of Quadratic Functions. It also discusses what is Quadratic Functions. It also shows how to transform or rewrite the equation f(x)=ax2 + bx + c to f(x)= a(x-h)2 + k. It will also show the different characteristics of Quadratic Functions.
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
UiPath Test Automation using UiPath Test Suite series, part 3DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 3. In this session, we will cover desktop automation along with UI automation.
Topics covered:
UI automation Introduction,
UI automation Sample
Desktop automation flow
Pradeep Chinnala, Senior Consultant Automation Developer @WonderBotz and UiPath MVP
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
Here is something new! In our next Connector Corner webinar, we will demonstrate how you can use a single workflow to:
Create a campaign using Mailchimp with merge tags/fields
Send an interactive Slack channel message (using buttons)
Have the message received by managers and peers along with a test email for review
But there’s more:
In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
But—if the “Reject” button is pushed, colleagues will be alerted via Slack message
Join us to learn more about this new, human-in-the-loop capability, brought to you by Integration Service connectors.
And...
Speakers:
Akshay Agnihotri, Product Manager
Charlie Greenberg, Host
4. Integration
Outcome 2
Higher
www.mathsrevision.com
Integrating is the opposite of differentiating, so:
f ( x)
differentiate
integrate
But:
f ( x) = 3x + 1
f ( x) = 3 x 2 + 4
f ( x) = 3 x 2 − 10
2
differentiate
6x
integrate
Integrating
f ′( x)
= f ′( x)
= g ′( x)
= h′( x)
6x….......which function do we get back to?
8. Just
Integration like we
differentiation,
www.mathsrevision.com
Higher
Outcome 2must arrange the
function as a series of
powers of x before
1 2 1
3x − dx
2
∫ 4 + x
we integrate.
3
x
⇒ ∫ (12 x − 4 x
2
−
2
3
3
2
−2
+ 3 x − x )dx
1
3
5
2
x3
x
x
x −1
⇒ 12 − 4 1 + 3 5 −
3
−1
3
2
1
3
5
2
−1
⇒ 4 x − 12 x + x + x + C
3
6
5
9. Integration
Outcome 2
www.mathsrevision.com
Higher
If F '( x) = 3 x 2 + 4 x − 1 and F (2) = 11, find F ( x).
To get the function F(x) from the derivative F’(x)
we do the opposite, i.e. we integrate.
F ( x) = ∫ (3x 2 + 4 x − 1)dx
Hence
:
F (2) = 11
⇒ 23 + 2.22 − 2 + C = 11
⇒ 8 + 8 − 2 + C = 11
x
x
=3 +4 − x+C
3
2
⇒ C = −3
3
2
= x + 2x − x + C
3
2
F ( x) = x + 2 x − x − 3
3
2
11. Area under a Curve
Outcome 2
www.mathsrevision.com
Higher
The integral of a function can be used to determine the
area between the x-axis and the graph of the function.
y
y = f ( x)
b
Area = ∫ f ( x) dx
a
a
If
b
∫ f ( x) dx = F ( x)
x
then
b
∫a f ( x) dx = F (b) − F (a)
NB: this is a definite integral.
It has lower limit
a and an upper limit b.
12. Area under a Curve
Outcome 2
Higher
www.mathsrevision.com
Examples:
∫
5
1
(3 + x) dx
F ( x) = ∫ (3 + x) dx
x2
⇒ F ( x) = 3 x +
2
( +c)
25
1
30 + 25 − 6 − 1
(3 + x) dx = F (5) − F (1) = 15 + − 3 + =
= 48
∫1
2
2
2
5
∫
2
−2
( x − 2) dx
2
F ( x) = ∫ ( x − 2) dx
2
x3
⇒ F ( x) = − 2 x ( + c)
3
16
−8
8
−8
∫−2 ( x − 2) dx = F (2) − F (−2) = 3 − 4 − 3 + 4 = 3 − 8 = 3
2
2
13. Area under a Curve
Outcome 2
Higher
www.mathsrevision.com
b
From the definition of
∫ f ( x) dx it follows that:
a
b
∫ f ( x) dx
a
a
= − ∫ f ( x) dx
b
F (b) − F (a ) = − ( F (a ) − F (b) )
Conventionally, the lower limit of a definite integral
is always less then its upper limit.
14. Area under a Curve
Outcome 2
www.mathsrevision.com
Higher
y=f(x)
a
b
a
d
When calculating integrals:
areas above the x-axis are positive
d
b
∫
c
Very Important Note:
f ( x) dx > 0
∫
f ( x) dx < 0
areas below the x-axis are negative
c
When calculating the area between a curve and the x-axis:
•
make a sketch
•
calculate areas above and below the x-axis separately
•
ignore the negative signs and add
15. Area under a Curve
Outcome 2
Higher
www.mathsrevision.com
The Area Between Two Curves
To find the area between two
curves we evaluate:
Area = ∫ (top curve − bottom curve)
y = f ( x)
a
y = g ( x)
b
b
Area = ∫ ( f ( x) − g ( x) dx
a
16. Area under a Curve
Higher
Example:
Outcome 2
www.mathsrevision.com
Calculate the area enclosed by the lines x = 2, x = 4
and the curves y = x 2 and y = 4 − x 2
y
4
y = x2
4
Area = ∫ [ x − (4 − x )] dx = ∫ (2 x 2 − 4) dx
2
2
2
2 x3
∫ (2 x − 4) dx = F ( x) = 3 − 4 x
x=4
2
x
x=2
Area = F (4) − F (2)
128
16
− 16) − ( − 8)
3
3
112
=
−8
3
88
=
3
=(
y = 4 − x2
2
17. Area under a Curve
www.mathsrevision.com
Higher
Complicated Example:
Outcome 2
The cargo space of a small bulk carrier is 60m long. The shaded
part of the diagram represents the uniform cross-section of this
space.
x2
It is shaped like a parabola with y = , − 6 ≤ x ≤ 6,
4
between lines y = 1 and y = 9.
Find the area of this crosssection and hence find the
volume of cargo that this ship
can carry.
9
1
18. Area under a Curve
y
www.mathsrevision.com
y=9
Higher
y=
y = 1 −t
The shape is symmetrical about the
y-axis. So we calculate the area of
one of the light shaded rectangles
and one of the dark shaded wings.
The area is then double their sum.
2
x
4
−s
s
t
x
The rectangle:
let its width be s
The wing: extends from x = s to x = t
s2
y =1=
4
⇒ s=2
t2
y=9=
4
⇒ t =6
t
The area of a wing (W ) is given by:
x2
W = ∫ (9 − ) dx
4
s
19. Area under a Curve
Outcome 2
Higher
www.mathsrevision.com
6
x2
W = ∫ (9 − ) dx
4
2
F (6) = (9.6 −
x2
x3
∫ (9 − 4 ) dx = F ( x) = (9 x − 12 )
6.6.6
) = 54 − 18 = 36
12
W = F (6) − F (2) = 36 − 18 +
2
=
3
F (2) = (9.2 −
2.2.2
2
) = 18 −
12
3
56
3
The area of a rectangle is given by:
R = (9 − 1) s = 16
The area of the complete shaded area is given by:
A = 2(16 +
56
48 + 56
)=2
=
3
3
The cargo volume is:
A = 2( R + W )
208
3
60 A = 60.
208
= 20.208 = 4160m3
3
24. Calculus
Revision
Integrate
∫ (3x − 1)( x + 5) dx
Multiply out brackets
Integrate term by term
simplify
Back
∫ 3x
2
+ 14 x − 5 dx
3x 3 14 x 2
+
− 5x + c
3
2
x + 7 x − 5x + c
3
Quit
2
Next
39. Calculus
Given the acceleration a is:
1
Revision
a = 2(4 − t ) 2 , 0 ≤ t ≤ 4
a=
dv
dt
If it starts at rest, find an expression for the velocity v
3
where
3
1
2(4 − t ) 2
4
dv
→ v= 3
+c
→ v = − (4 − t ) 2 + c
= 2(4 − t ) 2
3
× ( −1)
dt
2
4
→ v=−
3
(
4
→ 0=−
3
( 4)
Back
4−t
3
)
3
+c
+c
Quit
Starts at rest, so
4
→ 0=−
3
( 4)
3
+c
v = 0, when t = 0
32
→ 0= − +c
3
3
4
32
→ v = − (4 − t ) 2 +
3
3
32
→ c=
3
Next
40. Calculus
(
Revision
dy
= 3sin(2 x) passes through the point
A curve for
dx
which
3
y = − cos(2 x) + c
Find y in terms of x.
5
π,
12
3
2
Use the point
→
3=−
3
2
(
5
π,
12
5
cos( π ) + c
6
4 3 3 3
→
−
=c
4
4
Back
3
)
3
2
3 = − cos(2 ×
→
3=−
3
→ c=
4
Quit
5
π)+c
12
3
3
−
÷+ c
2 2
3
2
→
3=
y = − cos(2 x) +
3 3
+c
4
3
4
Next
)
41. Calculus
Integrate
∫
(x
2
Revision
− 2) ( x + 2)
2
x
2
dx, x ≠ 0
→
Multiply out brackets
Split into
→
separate fractions
∫
x4 4
− 2 dx
2
x
x
x 3 4 x −1
→
−
+c
3
−1
Back
∫
Quit
x4 − 4
dx
2
x
→
→
∫
1
3
x 2 − 4 x −2 dx
4
x + +c
x
3
Next
42. Calculus
Revision
passes through the point
f ′( x) = sin(3 x)
y = f ( x)
If
f ( x) =
1=
→
7
6
=c
Back
)
, 1
express y in terms of x.
1
− cos(3 x) + c
3
π
1
− cos 3 ×
3
9
(
π
9
÷+ c
Use the point
→ 1=
π
1
− cos
3
3
1
3
y = − cos(3x ) +
Quit
(
÷+ c
π
9
)
, 1
→ 1=
1 1
− ×
3 2
7
6
Next
+c
44. Calculus
Revision
The graph of y = g ( x ) passes through the point (1, 2).
If
dy
1 1
3
=x + 2−
dx
x 4
dy
1
3
−2
=x +x −
dx
4
express y in terms of x.
x 4 x −1 1
→ y= +
− x+c
4 −1 4
4
x 1 1
y = − − x+c
4 x 4
Evaluate c
Back
Use the point
c=3
Quit
( 1)
2=
simplify
4
1 1
−
− ( 1) + c
4
( 1) 4
1 1
→ y = x − − x+3
x 4
1
4
4
Next
46. Calculus
Revision
dy
= 6 x 2 − 2 x passes through the point (–1, 2).
A curve for which
dx
Express y in terms of x.
6 x3 2 x 2
y=
−
+c
3
2
Use the point
→ y = 2 x3 − x 2 + c
→ 2 = 2(−1)3 − (−1) 2 + c
→ c=5
→ y = 2 x3 − x 2 + 5
Back
Quit
Next
47. Calculus
Evaluate
∫
2
1
2
2 1
x + dx
x
Revision
→
→
→
2
1
→ x5 + x 2 − x −1
5
1
(
+ 4−
Back
1
2
) ()
−
1
5
→
64
10
∫
x 2 + x −1
)
2
dx
2
x 4 + 2 x + x −2 dx
1
So multiply out
32
5
∫(
1
Cannot use standard integral
→
2
+
(
40
10
−
1
5
25 + 2 2 −
20
10
Quit
−
2
10
1
2
) (
−
1 5
1
5
→
82
10
+ 12 −
1
1
)
→ 81
5
Next
50. Calculus
Revision
π
,1
The curve y = f ( x) passes through the point 12 ÷
f ′( x) = cos 2 x
→ f ( x) =
→ 1=
→ 1=
Find
1
sin 2 x
2
1
π
sin 2 ×
2
12
1 1
×
2 2
Back
+c
f(x)
+c
use the given point
→ 1=
+c
→ c=
3
4
Quit
1
π
sin
2
6
→ f ( x) =
π
,1 ÷
12
+ c sin π = 1
6
2
1
sin 2 x
2
Next
+
3
4
