Basic Proportionality Theorem

Basic Proportionality Theorem:-

For online classes yo can go through mention URL

https://www.youtube.com/watch?v=BXrpbl4jYBc&t=139s

Basic Proportionality Theorem states that "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio".

In the following figure, segment $DE$ is parallel to the side $BC$ of $\Delta ABC$. Note how $DE$ divides $AB$ and $AC$ in the same ratio:

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Proof of Basic Proportionality Theorem:

Given: 

1. $\Delta ABC$
2. $DE\parallel BC$

To prove: $\frac{AD}{DB}=\frac{AE}{EC}$

Construction: 

1. Join $BE$ and $CD$
2. Draw $DP\perp AC$
3. Draw $EQ\perp AB$

Proof: Consider $\Delta AED$. If you have to calculate the area of this triangle, you can take $AD$ to be the base, and $EQ$ to be the altitude, so that:

$$ar\left(\Delta AED\right)=\frac{1}{2}\times AD\times EQ$$

Now, consider $\Delta DEB$. To calculate the area of this triangle, you can take $DB$ to be the base, and $EQ$ (again) to be the altitude (perpendicular from the opposite vertex $E$ ).

Thus,

$$ar\left(\Delta DEB\right)=\frac{1}{2}\times DB\times EQ$$

Next, consider the ratio of these two areas you have calculated:

$$\frac{ar\left(\Delta AED\right)}{ar\left(\Delta DEB\right)}=\frac{\frac{1}{2}\times AD\times EQ}{\frac{1}{2}\times DB\times EQ}=\frac{AD}{DB}$$

In an exactly analogous manner, you can evaluate the ratio of areas of $\Delta AED$ and $\Delta EDC$:

$$\frac{ar\left(\Delta AED\right)}{ar\left(\Delta EDC\right)}=\frac{\frac{1}{2}\times AE\times DP}{\frac{1}{2}\times EC\times DP}=\frac{AE}{EC}$$

Finally, We know that "Two triangles on the same base and between the same parallels are equal in area". Here, $\Delta DEB$ and $\Delta EDC$ are on the same base $DE$ and between the same parallels – $DE\parallel BC$ .

$$\Rightarrow ar\left(\Delta DEB\right)=ar\left(\Delta EDC\right)$$

Considering above results, we can note,

$$\frac{ar\left(\Delta AED\right)}{ar\left(\Delta DEB\right)}=\frac{ar\left(\Delta AED\right)}{ar\left(\Delta EDC\right)}$$

$$\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}$$

This completes our proof of the fact that $DE$ divides $AB$ and $AC$ in the same ratio.

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Odd function & Even Functions

Even Functions

for video classes.: - https://www.youtube.com/watch?v=O178h_HkQss&t=111s

A function is "even" when:

f(x) = f(−x) for all x

This is the curve f(x) = x²+1

They got called "even" functions because the functions x², x⁴, x⁶, x⁸, etc behave like that, but there are other functions that behave like that too, such as cos(x):

Cosine function: f(x) = cos(x)
It is an even function

But an even exponent does not always make an even function, for example (x+1)² is not an even function.

https://www.youtube.com/channel/UCnSzn6hUZk-M7juvSAK2Eqw/videos

Odd Functions

A function is "odd" when:

−f(x) = f(−x) for all x

Note the minus in front of f(x): −f(x).

And we get 

This is the curve f(x) = x³−x

They got called "odd" because the functions x, x³, x⁵, x⁷, etc behave like that, but there are other functions that behave like that, too, such as sin(x):

Sine function: f(x) = sin(x)
It is an odd function

But an odd exponent does not always make an odd function, for example x³+1 is not an odd function.

Neither Odd nor Even

Don't be misled by the names "odd" and "even" ... they are just names ... and a function does not have to be even or odd.

In fact most functions are neither odd nor even. For example, just adding 1 to the curve above gets this:

This is the curve f(x) = x³−x+1

It is not an odd function, and it is not an even function either.
It is neither odd nor even

Even or Odd?

Example: is f(x) = x/(x²−1) Even or Odd or neither?

Let's see what happens when we substitute −x:

f(−x) =(−x)/((−x)²−1)

=−x/(x²−1)

=−f(x)

So f(−x) = −f(x) , which makes it an Odd Function

Even and Odd

The only function that is even and odd is f(x) = 0

Special Properties

Adding:

• The sum of two even functions is even
• The sum of two odd functions is odd
• The sum of an even and odd function is neither even nor odd (unless one function is zero).

For more videos.

https://www.youtube.com/channel/UCnSzn6hUZk-M7juvSAK2Eqw/videos

Tata Steel Recruitment 2019 for Assistant Engineer Trainee | Last Date: 17 February 2019

Tata Steel Recruitment 2019 scheduled to hire Assistant Engineer Trainee (AET) for Fresher Engineering graduates. The detailed eligibility and application process are given below.

Tata Steel Recruitment 2019:

Job Role	Assistant Engineer Trainee
Qualification	B.E/B.Tech/B.Sc(Engg)
Total Vacancies	Not disclosed
Experience	Freshers
Salary	1st year: Rs.28,000/Month 2nd year: Rs.30,000/Month
Job Location	Across India
Last Date	17 February 2019

Detailed Eligibility:

• B.E. / B. Tech. / B.Sc. (Engg) in the relevant disciplines only (Candidates with 0 years of experience post completing their engineering are also eligible to apply)
• Must have attended Full time AICTE/UGC approved Campus Program.
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Age limit (As on 17th February 2019): Below 30 years. (Candidates born on or after 17th February 1989)

Selection Procedure:

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Only shortlisted candidates would be invited for selection process. All selected candidate will undergo Background Verification as per company’s standard, before their appointment.

How to apply for Tata Steel Recruitment 2019 for AET?

All interested and eligible candidates can apply for AET vacancies through online latest by 17 February 2019.

For More details: Click here

To Apply: Click here

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• Visit careers.tatasteelindia.com
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