sin θ = 0 cos θ = 0 
θ = nπ, n ϵ lθ = (2n + 1) π/2, n ϵ l 
tan θ = 0 sin θ = sinα 
θ = nπ, n ϵ l θ = nπ + (1)ⁿ, α ϵ [π/2, π/2], n ϵ l 
cos θ = cosα tan θ = tan α 
θ = 2nπ ± α, α ϵ [0, π], n ϵ l θ = nπ + α, α ϵ (π/2, π/2), n ϵ l 
sin² θ = sin²α cos² θ = cos²α tan² θ = tan²α 
θ = nπ ± α, n ϵ l 
sin θ = 1 
θ = (4n + 1) π/2, n ϵ l 
cos θ = 1 
θ = 2nπ, n ϵ l 
cos θ = 1 
θ = (2n + 1) π, n ϵ l 
sinx = sinα and cosx = cosα 
x = 2nπ + α 
1) If θ is an odd multiple of π/2 i.e, when θ = (2n + 1) π/2, n ϵ l then secθ and tanθ are not defined
2) If θ is an even multiple of π/2 i.e, when θ = nπ, n ϵ l then cosec θ and cot θ are not defined
Example: Solve the equation tan θ + sec θ = √3
Solution: tan θ + sec θ = √3 … (1)
Then (sin θ)/ (cos θ) +1/ (cos θ) = √3 … (2) (∵ sin θ/cos θ = tan θ and 1/cos θ = secθ)
\(\frac{\sin \theta +1}{\cos \theta }=\sqrt{3}\)
Sinθ + 1 = √3 cosθ
½ Sinθ – √3/2 cosθ = 1/2
Sin(π/6) sinθ – cos(π/6) cosθ = cos(π/3)
We can use the cos (A + B) formula
cos (θ + π/6) = cos π/3
General solution θ + π/6 = 2nπ ± π/3, n ∈ I
Taking positive sign, θ + π/6 = 2nπ + π/3
⇒ θ = 2nπ + π/6
Taking negative sign, θ + π/6 = 2nπ – π/3
⇒ θ = 2nπ – π/2
i.e. θ = (4n – 1) π/2.
But the solution obtained is correct only if, cos θ ≠ 0 otherwise (2) is not defined.
i.e. θ ≠ odd multiple of π/2
⇒ θ ≠ (4n – 1) π/2.
Hence the general solution will be θ = 2nπ + π/6 only where n = 0, ±1, ±2, …
]]>A rectifier is a circuit which converts the Alternating Current (AC) input power into a Direct Current (DC) output power. The input power supply may be either a singlephase or a multiphase supply with the simplest of all the rectifier circuits being that of the Half Wave Rectifier.
A halfwave rectifier is a circuit that allows only one halfcycle of the AC voltage waveform to be applied to the load, resulting in one nonalternating polarity across it.
Half Wave Rectification: The power diode in a half wave rectifier circuit passes just one half of each complete sine wave of the AC supply in order to convert it into a DC supply. Then this type of circuit is called a “halfwave” rectifier because it passes only half of the incoming AC power supply.
Half – Wave Rectifier Circuit: During each “positive” half cycle of the AC sine wave, the diode is forward biased as the anode is positive with respect to the cathode resulting in current flowing through the diode.
Since the DC load is resistive (resistor, R), the current flowing in the load resistor is therefore proportional to the voltage (Ohm´s Law), and the voltage across the load resistor will therefore be the same as the supply voltage, V_{s} (minus V_{f}), that is the “DC” voltage across the load is sinusoidal for the first half cycle only so V_{out} = V_{s}.
During each “negative” half cycle of the AC sinusoidal input waveform, the diode is reverse biased as the anode is negative with respect to the cathode. Therefore, NO current flows through the diode or circuit. Then in the negative half cycle of the supply, no current flows in the load resistor as no voltage appears across it so therefore, V_{out} = 0.
]]>What is TNAU?
Tamil Nadu Agricultural University(TNAU) is an agricultural university located in Coimbatore, Tamil Nadu, India. TNAU released counselling notification2018 for admission into B.Sc/B.Tech Agriculture courses in Government & Private Colleges in Tamilnadu State.
Tamil Nadu Agricultural University (TNAU) rated as the best Agricultural University in India by Indian Council of Agricultural Research, New Delhi, is an institute of excellence for higher education in Agricultural and allied subjects.
Eligibility:
Nativity:
For Govt Colleges:
Except Industry Sponsorship quota and NRI Quota, remaining all seats will be filled with Tamilnadu Native candidates only.
For Private Colleges:
Sixty five percent (65%) seats in the affiliated colleges are filled by TNAU in General Counseling. Only Tamilnadu Nativity candidates are eligible to admission for these quota seats.
Thirty five percent (35%) seats in the affiliated colleges are filled by the respective Management Quota.
i.e, for 65% Quota, only Tamilnadu Nativity candidates are eligible. For 35% Management Quota, All Over India candidates are eligible to apply
For minority institutions, fifty per cent seats are filled in through TNAU counseling and fifty per cent by management
Education:
Candidates should have passed Qualifying Examination with 10+2 years of schooling under Board of Higher Secondary Education of Government of Tamil Nadu/ Central Board of Secondary Education/ Council for the Indian School Certificate Examinations/ other State Government Boards
Qualifying marks for applying TNAU Counselling: For OC55%, OBC50%, MBC45% and for SC/STA pass in qualifying examination
Age:
Candidates should not have completed the age of 21 years on the first day of July of the admission year (i.e. as on 01.07.2018). However, for Scheduled Castes (SC), Scheduled CasteArunthathiars (SCA)and Scheduled Tribes (ST) there is no age limit.
Eligible Subjects of Study in the Qualifying Examination:
Sl. No  Degree  [+2] Subject requirements  
CATEGORY Ia  
1  B.Sc. (Agriculture) 
Group I: Mathematics, Physics, Chemistry and Biology OR Group II: Physics, Chemistry, Biology with a fourth (elective) subject viz., Biotechnology, Microbiology, Biochemistry and Home Science. OR Group II (A): Physics, Chemistry, Botany and Zoology 

2  B.Sc. (Horticulture)  
3  B.Sc. (Forestry)  
4  B.Sc. (Food, Nutrition and Dietetics)  
5  B.Sc. (Sericulture)  
6  B.Tech. (Biotechnology)  
7  B.Tech. (Bioinformatics)  
8  B.S. (Agribusiness Management)  
CATEGORY Ib  
1  B.Sc. (Food, Nutrition and Dietetics) 
Group I: Mathematics, Physics, Chemistry and Biology OR Group II: Physics, Chemistry, Biology with a fourth (elective) subject viz., Biotechnology, Microbiology, Biochemistry and Home Science / Nutrition and Dietetics OR Group II (A): Physics, Chemistry, Botany and Zoology 

CATEGORY Ic  
1  B.Tech. (Agricultural Information Technology) 
Group I: Mathematics, Physics, Chemistry and Biology OR Group II: Physics, Chemistry, Biology and Computer Science 

CATEGORY II  
1  B.Tech. (Agricultural Engineering) 
Group I: Mathematics, Physics, Chemistry and Biology OR Mathematics, Physics, Chemistry and Computer Science. 

2  B.Tech. (Food Process Engineering)  
3  B.Tech. (Energy and Environmental Engineering) 
Industry Sponsorship and NRI Quota: Candidate satisfying minimum mark requirements and maximum age limit can apply for NRI or Industry Sponsorship quota. Ten seats each for the Industry sponsorship and NonResident Indian quota in all the undergraduate programs of the constituent colleges, TNAU are allotted on merit basis. There will be a separate counseling for this quota.
Mode of Application: Online Only
Application Fee: 600/
Important Dates:
Opening of online application  18.05.2018 
Last date for online application  17.06.2018 
Document Verification for Special Reservations  18.06.2018 to 20.06.2018 
Publication of Rank List  22.06.2018 
Counseling for Sports, Exservicemen and Decedents of freedom fighter and Differently abled quota  07.07.2018 
I Phase Counseling (Online)  09.07.2018 to 13.07.2018 
Counseling for vocational stream  16.07.2018 
Counseling for NRI and Industry sponsorship quota  17.07.2018 & 18.07.2018 
II Phase Counseling (Online)  23.07.2018 to 27.07.2018 
Registration for I semester  01.08.2018 
Closure of Admission  31.08.2018 
For further details: http://myrank.co.in/notifications/notification.php?notice_id=213
]]>The process by which a p – n junction diode allows the electric current in the presence of applied voltage is called forward biased p – n junction diode. In forward biased p – n junction diode, the positive terminal of the battery is connected to the semiconductor material and the negative terminal of the battery is connected to the n – type semiconductor material.
Unbiased Diode and Forward Biased Diode: Under no voltage or unbiased condition, the p – n junction diode does not allow the electric current. If the external forward voltage applied on the p – n junction diode is increased from zero to 0.1 volts, the depletion region slightly decreases. Hence, very small electric current flows in the p – n junction diode. However, this small electric current ion the p – n junction diode is considered as negligible. Hence, they do not used for any practical applications.
If the voltage applied on the p – n junction diode is further increased, then even more number of free electron and holes are generated in the p – n junction diode. This large number of free electron and holes further reduces the depletion region. Hence, the electric current in the p – n junction diode increase. Thus, the depletion region of a p – n junction diode decreased with increase in voltage. In other words, the electric current in the p – n junction diode increases with the increase in voltage.
]]>i. Area of triangle: If (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the verticals of a triangle then Area of the triangle =\(\left \frac{1}{2}\left \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right \right\) (or) = ½ x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂).
Example: The area of the triangle with vertices (2, 3) (3, 2) and (1, 8).
Solution: Given that (2, 3) (3, 2) and (1, 8)
We can use the formula
Area of the triangle =\(\left \frac{1}{2}\left \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right \right\),
Area of the triangle = \(\left \frac{1}{2}\left \begin{matrix} 2 & 3 & 1 \\ 3 & 2 & 1 \\ 1 & 8 & 1 \\\end{matrix} \right \right\),
= ½ 2 (2 + 8) + 3 (3 + 1) + 1 ( 24 + 2) 
= ½20 + 12 22
= ½30
= 15 sq. Unit
ii. Condition of Collinearity of Three Points: Let tree points are (x₁, y₁) (x₂, y₂) and (x₃, y₃) then these points will be collinear, if Area of triangle = \(\left \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right\) = 0.
Example: The points (a, b + c), (b, c + a) and (c, a + b).
Solution: Area of triangle = \(\left \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right\) = 0,
Area of triangle = \(\left \begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \\\end{matrix} \right=0\),
We can apply the column transformations: C₁ → C₁ + C₂.
\(\left \begin{matrix} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \\\end{matrix} \right=0\),
\((a+b+c)\left \begin{matrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \\\end{matrix} \right=0\) (∵ Two columns are similar that matrix det is zero)
]]>What is ICAR?
23rd All India Entrance Examination for Admission (AIEEAUG2018) to 15% seats (100% seats at NDRI Karnal, RLB CAU Jhansi and Dr. RP CAU PUSA) in bachelor degree programmes and award of National Talent Scholarships in Agriculture & Allied Science subjects (Other than Veterinary science) in accredited Agricultural Universities for academic session 201819.
Eligibility:
Candidate must have passed 10+2 Senior Secondary Examination of the Central Board of Secondary Education or any other examination within scope and standard found to be equivalent to the Senior Secondary Examination of a recognised Indian Board/university after a period of 12 years of study.
Candidate must have passed any one of the qualifying examinations enumerated above securing not less than 50% marks in aggregate for General, OBC(NCL) and UPS categories and 40% marks in aggregate for SC, ST and PC categories.
There will be no roundingoff of the OGPA/percentage of marks of qualifying examination while deciding the basic eligibility of any candidate for admission e.g. if a candidate obtained 49.99% marks in his/her qualifying examination, then it will not be roundedoff to 50%.
Mode of Application: Online Mode only
Application Fee:
For General, OBC (NCL) & UPS Categories: Rs.700/
For SC, ST and PC Categories: Rs.350/
With additional Payment Gateway charges plus GST, as applicable
Mode of Exam: Online Mode
Exam Pattern:
Subject  No. of Questions  Marks  
Physics  60  240  
Chemistry  60  240  
Biology/ Maths/ Agriculture/ Forestry  60  240  
Total  180  720  
Negative Marking: 1 mark will be deducted for each wrong answer 
Important Dates:
Event  Date 
Date and Time of Online Examination (CBT) 
23.06.2018 (Saturday) 10:00 A.M to 12:30 P.M. (2 ½ hrs) 
Commencement of Online Application submission  18.05.2018 (Friday) 
Last date of online application  31.05.2018 (Thursday up to 11.59 P.M.) 
eAdmit card information (on ICAR website)  14.06.2018 (Thursday) 
Online Mock Test  14.06.2018 (will be available on www.aieea.net) 
Declaration of result  Last week of June, 2018 
Schedule for online Counseling 1. Choice Filling 2. 1^{st} Counseling Result 3. 1^{st} Round Reporting 4. 2^{nd} Counseling Result 5. 2^{nd} Round Reporting 6. 3^{rd} Counseling Result 7. 3^{rd} Round Reporting 8. 4^{th} Round (If need be) 
01^{st} July 2018 to 03^{rd} July 2018 06^{th} July 2018 06^{th} July 2018 to 10^{th} July 2018 12^{th} July 12^{th} July 2018 to 16^{th} July 2018 18^{th} July 18^{th} July 2018 to 22st July 2018 
For further details: http://myrank.co.in/notifications/notification.php?notice_id=214
]]>A diode is a fundamental electronic component that allows the flow of electricity in one direction. A diode has low resistance to the flow of current in one direction and a very high resistance in the other. The diode most commonly used is the semiconductor diodes.
What is a Zener Diode?
Zener diode is a unique type of diode which allows the current to flow in one direction like a regular diode but it also permits it to flow in the opposite direction. It allows the flow in the opposite direction when the voltage is above a certain value known as Zener Voltage or Avalanche Point or Breakdown Voltage. A Zener diode consists of a highly doped, reverse biased (refer to an article on pn junction diode) pn junction diode while operating in the breakdown region.
Symbol of Zener Diode: The symbol of Zener diode is shown in below figure. Zener diode consists of two terminals: cathode and anode.
In Zener diode, electric current flows from both anode to cathode and cathode to anode. The symbol of Zener diode is similar to the normal pn junction diode, but with bend edges on the vertical bar.
Avalanche Breakdown:
Advantages of Zener Diode:
Applications of Zener Diode:
If f(x) and g(x) are two polynomials, then \(\frac{f\left( x \right)}{g\left( x \right)}\) defines a rational algebraic function or a rational function of x.
Case I: when denominator is expressed is expressible as the product of nonrepeating linear factors.
Let g(x) = (x – a₁) (x – a₂) … (x – a_{n}) then we assume that \(\frac{f\left( x \right)}{g\left( x \right)}=\frac{{{A}_{1}}}{x{{a}_{1}}}+\frac{{{A}_{2}}}{x{{a}_{2}}}+…+\frac{{{A}_{n}}}{x{{a}_{n}}}\).
1. Resolve \(\frac{5x+1}{\left( x+2 \right)\left( x1 \right)}\) into Partial fractions.
Solution: Given that \(\frac{5x+1}{\left( x+2 \right)\left( x1 \right)}\).
Let \(\frac{5x+1}{\left( x+2 \right)\left( x1 \right)}=\frac{A}{x+2}+\frac{B}{x1}\).
Then \(\frac{5x+1}{\left( x+2 \right)\left( x1 \right)}=\frac{A\left( x1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x1 \right)}\).
5x + 1 = A (x – 1) + B (x + 2) … (1)
Put x = 1 in (1)
5 (1) + 1 = A (0) + B (1 + 2)
3B = 6 ⇒ B = 2
Put x = – 2 in (1)
5 (2) + 1 = A ( 2 – 1) + B (0)
– 9 = – 3A ⇒ A = 3
∴ \(\frac{5x+1}{\left( x+2 \right)\left( x1 \right)}=\frac{3}{x+2}+\frac{2}{x1}\).
2. Resolve \(\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}\) into Partial fractions.
Solution: Given that \(\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}\).
Let \(\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}=\frac{A}{x+2}+\frac{B}{2x+1}\).
\(\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{A\left( 2x+1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( 2x+1 \right)}\).
2x + 3 = A (2x + 1) + B (x + 2) … (1)
Put x = 2 in (1)
2 (2) + 3 = A [2 (2) + 1] + B (0)
1 = A (3) ⇒ A = ⅓
Put x = – ½ in (1)
2 ( ½) + 3 = A ( 1 + 1) + B ( ½ + 2)
2 = B (3/2)
B = 4/3
∴ \(\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{1/3}{x+2}+\frac{4/3}{2x+1}\).
∴ \(\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{1}{3\left( x+2 \right)}+\frac{4}{3\left( 2x+1 \right)}\).
Hence \(\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}=\frac{1}{15\left( x+2 \right)}+\frac{4}{15\left( 2x+1 \right)}\).
3. Resolve \(\frac{{{x}^{2}}+5x+7}{{{\left( x3 \right)}^{3}}}\) into Partial fractions.
Solution: Given that \(\frac{{{x}^{2}}+5x+7}{{{\left( x3 \right)}^{3}}}\).
Let x – 3 = y ⇒ x = y + 3
\(\frac{{{x}^{2}}+5x+7}{{{\left( x3 \right)}^{3}}}=\frac{{{\left( y+3 \right)}^{2}}+5\left( y+3 \right)+7}{{{y}^{3}}}\).
\(=\frac{{{y}^{2}}+6y+9+5y+15+7}{{{y}^{3}}}\).
\(=\frac{{{y}^{2}}+11y+31}{{{y}^{3}}}=\frac{1}{y}+\frac{11}{{{y}^{2}}}+\frac{31}{{{y}^{3}}}\).
∴ \(\frac{{{x}^{2}}+5x+7}{{{\left( x3 \right)}^{3}}}=\frac{1}{x3}+\frac{11}{{{\left( x3 \right)}^{2}}}+\frac{31}{{{\left( x3 \right)}^{3}}}\).
]]>Functional Applications
(i) The number of all permutations (arrangements) of n different objects taken r at a time,
(ii) If the sets A has m elements and B has n elements, then
Example: Let A = {x₁, x₂, x₃, x₄, x₅}, B = {y₁, y₂, y₃, y₄, y₅}. Then, the number of oneone mappings, from A to B such that f (x_{i}) ≠ y_{i}, i = 1, 2, …, 5 is
Interpret: Here, we use the formula, the number of bijections from A to A such that f (x) ≠ x, ” x ϵ A is \(m!\left[ \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}…..+\frac{{{(1)}^{m}}}{m!} \right]\)
Here required number of ways \(=5!\left( \frac{1}{2!}\frac{1}{3!}+\frac{1}{4!}\frac{1}{5!} \right)\)
\(=120\left( \frac{1}{2}\frac{1}{6}+\frac{1}{24}\frac{1}{120} \right)\)
\(=120\left( \frac{6020+51}{120} \right)=44\).
Geometrical Applications:
Diode: A diode is a twoterminal electronic component that has a low resistance to the flow of current in one direction thus allowing the passage of current in one direction whereas a high resistance in the other, thus restricting the flow of current in that direction. Semiconductor diodes are two terminal devices that consist of a pn junction and metallic contacts at their two ends.
A pn junction is denoted by the symbol shown. Here, the direction of the arrow indicates the permissible direction of the current.
Semiconductor Diode: A semiconductor diode is basically a pn junction diode. It is a two terminal device which conducts current only in one direction.
The above figure represents the symbol for pn junction diode which symbolizes the direction of the current. By applying an external voltage V we can vary the potential barrier.
Diodes may be considered to be in two states when in a closed circuit:
Forward Biased: This simply means that the diode is arranged in the direction of the current. Hence the current flows from positive to negative part inside the diode.
Reversed Biased: This term denotes that the diode is arranged in a manner so that the conventional current, if it were possible, should move from negative part of diode to positive part. However, as it is not possible, practically no current flows in this situation.
]]>