Explore topic-wise InterviewSolutions in Current Affairs.

Learning Based refers to the AI modelling where the machine learns by itself. 

Correct option: (c) Testing

In Metabolism cells are able to break down and use substances from food as fuel. 

Correct option: a) Computerized Technology

1. Be polite 

2. Be patient 

3. Be technically confident to solve problem. 

4. Ask the customer for elaborate explanation of problem 

5. Keep yourself updated about up gradation of any device

आरक्षण

1. भूमिका:

आरक्षण (Reservation) का अर्थ है सुरक्षित करना । हर स्थान पर अपनी जगह सुरक्षित करने या रखने की इच्छा प्रत्येक व्यक्ति को होती है, चाहे वह रेल के डिब्बे में यात्रा करने के लिए हो या किसी अस्पताल में अपनी चिकित्सा कराने के लिए विधानसभा या लोकसभा का चुनाव लड़ने की बात हो तो या किसी सरकारी विभाग में नौकरी पाने की । लोग अनेक तरह की दलीलें (Arguments) देकर अपनी जगह सुनिश्चित करवाने और सामान्य प्रतियोगिता (Common Competition) से अलग रहना चाहते हैं ।

2. आरक्षण के आधार:

आरक्षण पाने का कोई आधार होना आवश्यक है । रेल, बस आदि में सफर करना हो, तो आरक्षण के लिए किराये म्र छ द्व8 के अलावा कुछ अतिरिक्त राशि (Extra Amount) देना आवश्यक होता है, अन्यथा (Otherwise) आपको भीड़ में या पंक्ति (Queue) में रहना पड़ेगा ।

गारंटी नहीं कि आप यात्रा कर पायेंगे अथवा नहीं । यदि कहीं अस्पताल में आपको चिकित्सा करवानी हो या किसी डॉक्टर से इलाज कराना हो, तो अतिरिक्त राशि दीजिए नहीं तो आपको अच्छी सुविधा या अच्छी चिकित्सा नहीं मिल सकती ।

यदि आप चुनाव में जीतना चाहते हैं तो किसी बड़ी जनसंख्या (Population) वाली जाति या धर्म का होना जरूरी है अथवा किसी आरक्षित जाति का होना जरूरी है । किसी सरकारी विभाग में नौकरी कम अंकों पर (Lower Marks) मिल जाए, इसके लिए भी किसी खास जाति का होना आवश्यक होता है ।

इस प्रकार जाति, धर्म धन आदि आज हमारे समाज में सुविधाओं (Previleges) को सबसे पहले और सबसे अधिक पाने का आधार (Base) हैं । यहीं नहीं कुछ खास पदों (Posts) पर होना भी आरक्षण पाने के लिए जरूरी समझा गया है ।

3. लाभ-हानि:

आरक्षण वास्तव में समाज के उन्हीं लोगों के लिए हितकर (Beneficial) हो सकता है जो अपंग (Phisycally disabled) हैं किंतु शिक्षा और गुण (Quality) होते हुए भी अन्य लोगों से जीवन में पीछे रह जाते हैं । उन गरीब लोगों के लिए भी आरक्षण आवश्यक है जो गुणी होते हुए भी गरीबी में जीवन बिता रहे हैं ।

केवल जाति, धर्म और धन के आधार पर आरक्षण से गुणी व्यक्तियों को पीछे धकेल (Push) कर हम देश को नुकसान ही पहुँचा रहे हैं । यह देश जाति-धर्म, धनी-गरीब आदि आधारों पर और अधिक विभाजित (Divided) होता जा रहा है ।

4. उपसंहार:

आज यदि हम देश को उन्नति (Progress) की ओर ले जाना चाहते हैं और देश की एकता बनाये रखना चाहते हैं, तो जरूरी है कि आरक्षणों को हटाकर हम सबको एक समान रूप से शिक्षा दें और अपनी उन्नति का अवसर (Opportunity) पाने का मौका दें ।

सामाजिक सुरक्षा

सामाजिक सुरक्षा का अर्थ 

अन्तर्राष्ट्रीय श्रम संगठन के अनुसार ‘‘वह सुरक्षा जो समाज, उचित संगठनों के माध्यम से अपने सदस्यों के साथ घटित होने वाली कुछ घटनाओं और जोखिमों से बचाव के लिए प्रस्तुत करता है, ‘सामाजिक सुरक्षा’ है। ये जोखिमें बीमारी, मातृत्व, आरोग्यता, वृद्धावस्था तथा मृत्यु है। इन संदिग्धताओं की यह विशेषता होती है कि व्यक्ति को अपना तथा अपने परिवार का भरण-पोषण करने के लिए नियोक्ताओं द्वारा सुरक्षा प्रदान की जाये।’’

सर विलियम बैवरिज के अनुसार, ‘‘सामाजिक सुरक्षा योजना एक सामाजिक बीमा योजना है जो व्यक्ति को संकट के समय अथवा उस समय, जब उसकी कमार्इ कम हो जाय, तथा जन्म, मृत्यु या विवाह में होने वाले अतिरिक्त व्यय की पूर्ति के लिए लाभान्वित करती है।’’

समाजिक सुरक्षा के आवश्यक तत्व 

सामाजिक सुरक्षा योजना के लिए निम्नलिखित तत्व आवश्यक है :

1. योजना का उद्देश्य बीमारी की रोकथाम या इलाज करना होना चाहिए अथवा अनिच्छापूर्वक घटित हानि से सुरक्षा के लिए आय की गारण्टी देना जिससे श्रमिक पर निर्भर व्यक्ति लाभान्वित हो सके। 
2. यह प्रणाली एक निश्चित अधिनियम के अन्तर्गत लागू की जानी चाहिए जो व्यक्तिगत अधिकारों तथा उत्तरदायित्व के प्रति सरकार, अर्द्ध-सरकारी संस्थाओं, गैर-सरकारी संस्थाओं को सामाजिक सुरक्षा सुविधाएं प्रदान करने के लिए बाध्य करे।
3. यह प्रणाली सरकारी, अर्द्ध-सरकारी तथा गैर-सरकारी संस्थाओं द्वारा प्रशासित की जानी चाहिए। 
4. सुरक्षा को भली-भांति नियमित करने की दृष्टि से उपलब्ध सुविधाओं के प्रति कर्मचारियों का विश्वास होना आवश्यक है कि आवश्यकतानुसार उन्हें सामाजिक सुरक्षा सेवाओं के अन्तर्गत किये गये प्रावधान उपलब्ध होंगे तथा उनकी किस्म और मात्रा पर्याप्त होगी। 

सामाजिक सुरक्षा का महत्व 

विकसित देशों में सामाजिक सुरक्षा कार्यक्रमों को देश की गरीबी, बेरोजगारी तथा बीमारी का उन्मूलन करने की दृष्टि से राष्ट्रीय योजना का अभिन्न तथा महत्वपूर्ण अंग माना गया है। निम्नलिखित विचारों से सुरक्षा का महत्व अधिक स्पष्ट हो जाता है :

अ) ‘‘सामाजिक सुरक्षा का दर्शन तथा मूल विचार सामुदायिक आयोजन, सामुदायिक उत्तरदायित्व तथा नागरिकों के कर्त्तव्यों और अधिकारों का सामुदायिक स्तर पर विचार करना है। गरीबी हटाना, अभाव पर विजय तथा व्यक्तियों के रहन-सहन का वांछित स्तर उपलब्ध करना इसके उद्देश्य है। इसका मूल उद्देश्य अधिकांश व्यक्तियों के लिए, यथा सम्भव सभी के लिए तथा प्रत्येक की प्रसन्नता के लिए ऐसा प्रबन्ध करना है, जिससे व्यक्तित्व का विकास हो।’’ -जे. एस. क्लार्क

 ब) ‘‘वर्तमान रोजगार, रोजगार तथा कार्य की उचित दशाएं प्राप्त करने और सेवानिवृत्ति के लिए सुरक्षा, आत्म-विकास, चिकित्सा एवं स्वास्थ्य सहायता, बेरोजगारी तथा अयोग्यता की स्थिति में आय की निरन्तरता, दुर्घटना के समय व्यक्ति के परिवार की सुरक्षा, अयोग्यता, बीमारी या मृत्यु के समय परिवार की सुरक्षा’’ आदि सामाजिक सुरक्षा कार्यक्रम के अन्तर्गत सम्मिलित किये जाते हैं। -सामाजिक सुरक्षा समिति, सं. रा. अमरीका

सामाजिक सुरक्षा के उद्देश्य 

मनुष्य एक सामाजिक प्राणी है। सामाजिक प्राणी होने के कारण उसको अनेक आवश्यकताओं का सामना करना पड़ता है। वह कभी दूसरों को आश्रय प्रदान करता है तो कभी स्वयं ही उसे दूसरों पर आश्रित रहना पड़ता है। आधुनिक यांत्रिक युग में वह अनेक प्रकार की दुर्घटनाओं का शिकार हो सकता है। इन दुर्घटनाओं से मुक्ति दिलाने के लिए यह आवश्यक है कि व्यक्ति को सामाजिक सुरक्षा प्रदान की जाय। संक्षेप में, सामाजिक सुरक्षा के उद्देश्य के अन्तर्गत निम्न तीन तत्वों को सम्मिलित किया जाता है-

1. क्षतिग्रस्त व्यक्ति को क्षतिपूर्ति करना, 
2. क्षतिग्रस्त व्यक्ति के पुनरुत्थान का प्रयास करना, और 
3. खतरों की रोकथाम के लिए आवश्यक व्यवस्था, करना आदि।

आरक्षण- स्वतंत्रता प्राप्ति के पश्चात भारत में दलितों एवं आदिवासियों की दशा अति दयनीय थी| इसलिए हमारे संविधान निर्माताओं ने काफी सोच समझकर इनके लिए संविधान में आरक्षण की व्यवस्था की और वर्ष 1950 में संविधान के लागू होने के साथ ही सुविधाओं से वंचित वर्गों को आरक्षण की सुविधा मिलने लगी, ताकि देश के संसाधनों, अवसरों एवं शासन प्रणाली में समाज के प्रत्येक समूह की उपस्थिति सुनिश्चित हो सके उस समय हमारा समाज उच्च-नीच, जाति-पाति, छुआछूत जैसी कुरीतियों से ग्रसित था|

A very small aperture terminal (VSAT) is a two-way satellite ground station with a dish antenna that is smaller than 3.8 meters. The majority of VSAT antennas range from 75 cm to 1.2 m. Data rates, in most cases, range from 4 kbit/s up to 16 Mbit/s. VSATs access satellites in geosynchronous orbit or geostationary orbit to relay data from small remote Earth stations (terminals) to other terminals (in mesh topology) or master Earth station "hubs" (in star topology).

VSATs are used to transmit narrowband data (e.g., point-of sale transactions using credit cards, polling or RFID data, or SCADA), or broadband data (for the provision of satellite Internet access to remote locations, VoIP or video). VSATs are also used for transportable, on-the move (utilizing phased array antennas) or mobile maritime communications.

VSAT (Very Small Aperture Terminal) is a satellite communications system that serves home and business users. A VSAT end user needs a box that interfaces between the user's computer and an outside antenna with a transceiver.

The transceiver receives or sends a signal to a satellite transponder in the sky. The satellite sends and receives signals from an earth station computer that acts as a hub for the system. Each end user is interconnected with the hub station via the satellite in a star topology. For one end user to communicate with another, each transmission has to first go to the hub station which retransmits it via the satellite to the other end user's VSAT. VSAT handles data, voice, and video signals.

VSAT is used both by home users who sign up with a large service and by private companies that operate or lease their own VSAT systems. VSAT offers a number of advantages over terrestrial alternatives. For private applications, companies can have total control of their own communication system without dependence on other companies. Business and home users also get higher speed reception than if using ordinary telephone service or ISDN.

Faishon affect the Hindu Festival in following ways:

• It creates new style of celebrating festival.
• Wearing faishonable clothes affect the attitude of an individual during festival.

Correct answer is c) White rust or white blister 

Correct option is D) bigoted

Correct option is C) issue

Correct option is B) extensive

Correct option is A) task

Correct option is B) secondary

Radar is an object detection system that uses radio waves to determine the range, altitude, direction, or speed of objects. It can be used to detect aircraft, ships, spacecraft, guided missiles, motor vehicles, weather formations, and terrain. 

In aviation, aircraft are equipped with radar devices that warn of obstacles in or approaching their path and give accurate altitude readings.

The following are the forces acting on an aircraft in flight: - 

(a) Lift is a positive force caused by the difference in air pressure under and above a wing. The higher air pressure beneath a wing creates lift and is affected by the shape of the wing. Changing a wing's angle of attack affects the speed of the air flowing over the wing and the amount of lift that the wing creates. 

(b) Thrust is the force that propels an object forward. An engine spinning a propeller or a jet engine expelling hot air out the tailpipe are examples of thrust. In bats, thrust is created by muscles making the wings flap. 

(c) Drag is the resistance of the air to anything moving through it. Different wing shapes greatly affect drag. Air divides smoothly around a wing's rounded leading edge, and flows neatly off its tapered trailing edge this is called streamlining. 

(d) Weight is the force that causes objects to fall downwards. In-flight, the force of the weight is countered by the forces of lift and thrust. 

OR 

a) Total reaction: It is one single force representing all the pressures (force per unit area) over the surface of the Aerofoil. It acts through the center of pressure which is situated on the chord line. 

b) Angle of Attack: It is the angle between the chord line and the relative airflow undisturbed by the presence of Aerofoil. 

c) Aerofoil: It is a body designed to produce more lift than drag. A typical Aerofoil section is cambered on a top surface and is more or less straight at the bottom

Correct option is C) maintain

a) Operation Safed Sagar was the codename assigned to the Indian Air Force's strike to support the Ground troops during Operation Vijay that was aimed to flush out Regular and Irregular troops of the Pakistani Army from vacated Indian Positions in the Kargil sector along the Line of Control. 

b) The East Pakistan Rifles and East Bengal Regiment became the Mukti Fauj.

c) The angle between the chord line and the longitudinal axis of the aircraft is called angle of Incidence

Correct answer is a) Aricaceae 

(1) The next day, the trekkers left very early after the breakfast and reached Kalah in the afternoon after almost 11 km’s trek.

(2) From the trek, they could see Ravi for some time as they gained height. The also discovered some villages on the slopes that they couldn’t see before.

(3) At around 1 pm, the clouds started gathering on the horizon and soon they covered the whole sky. It started drizzling.

(4) Bawaji explained the young trekkers a few. characteristics of typical Himalayan weather.

(5) The words are :
(i) gained
(ii) characteristics

(1) The trekkers decided to halt early near a stream because the weather was not favourable and one of the team members was tired and unable to walk.

(2) Near the stream, the trekkers cleaned and leveled the ground and erected their tents.

(3) The trekkers were surprised to know from Gaddi that they live in such cold and hard Himalayan weather without a tent or even a sleeping bag for months together.

(4) The dog helped the Gaddi in protecting and managing the sheep.

(5) Generally, the snow melts by late May but due to late winter heavy snowfall there was still heavy snow in Sukhdali.

Concept:

We have the formula of arc length in parametric form as,

\(\hbox{Arc length }=\int_a^b\; \sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt\)

Calculation:

Given:

\(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\)

\(\frac{dx}{dt}\) = 3t2 and \(\frac{dy}{dt}\) = 3t

Placing, these values in the above equation, we get

\({ arc\;length }=\int_0^1\; \sqrt{\left({3t^2}\right)^2+\left({3t}\right)^2}\;dt\)

\({ arc\;length }=\mathop \smallint \limits_0^1 3t\sqrt {{t^2} + 1} dt\)

taking u = t2 + 1, 

\(\frac{du}{dt}=2t\) and \(dt=\frac{1}{2t}du\)

\(arc\;length = \frac{3}{2}\mathop \smallint \limits_1^2 \sqrt u du\)

\(arc\;length = |{u^{{3}/{2}}}|_1^2\)

\(arc\;length = {2^{{3}/{2}} }- 1\)

Hence the required arc length will be (23/2 - 1) unit.

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

I = \(\rm \int (x-2)(x-3)dx\)

= \(\rm \int (x^2-2x - 3x + 6)dx\)

= \(\rm \int (x^2-5x + 6)dx\)

= \(\rm \frac{x^3}{3} - \frac{5x^2}{2}+6x +c\)

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

Let I = \(\rm \int x^2\sqrt x dx\)

= \(\rm \int x^2 x^{\frac{1}{2}} dx \)

= \(\rm \int x^{2+\frac{1}{2}} dx \)          (∵ x^ax^b = x^a+b)

= \(\rm \int x^{\frac{5}{2}} dx \)

= \(\rm \frac{ x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} +c\)

= \(\rm \frac{2}{7} x^{\frac{7}{2}} +c\)

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm\frac{d\sin x}{dx} = \cos x\)

\(\rm\frac{d\cos x}{dx} =- \sin x\)

Calculation:

We have to find the value of \(\rm \frac{d^2y}{dx^2}\)

Given: y = cos x

\(\rm \frac{d^2y}{dx^2} =\rm \frac{d^2\cos x}{dx^2} = \frac{d}{dx} \left(\frac{d\cos x}{dx} \right )\)

\(= \rm\frac{d(-\sin x)}{dx} =- \cos x\)

\(\rm \frac{d^2y}{dx^2}\) = -y

∴ \(\rm \frac{d^2y}{dx^2} + y = 0\)

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm\frac{d\sin x}{dx} = \cos x\)

\(\rm\frac{d\cos x}{dx} =- \sin x\)

Calculation:

We have to find the value of \(\rm \frac{d^2y}{dx^2}\)

Given: y = sin x

\(\rm \frac{d^2y}{dx^2} =\rm \frac{d^2\sin x}{dx^2} = \frac{d}{dx} \left(\frac{d\sin x}{dx} \right )\)

\(= \rm\frac{d(\cos x)}{dx} =- \sin x\)

∴ \(\rm \frac{d^2y}{dx^2} = -y\)

Concept:

\(\rm \frac{dx^{n}}{dx}= nx^{n-1}\)

Calculation:

To Find: \(\rm \frac{d^2 (x^{10})}{dx^2}\)

\(\rm \frac{d^2 (x^{10})}{dx^2} = \frac{d}{dx} \left(\frac{dx^{10}}{dx} \right )\)

\(\rm = \frac{d}{dx} (10x^{9})= 10 \frac{dx^{9}}{dx}\)

= 10 × 9 × x8

= 90(x8)

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formulas:

\(\rm\frac{d\tan x}{dx} = \sec^2 x\)

\(\rm\frac{d\sec x}{dx} =\sec x \tan x\)

\(\rm \frac{dx^{n}}{dx}= nx^{n-1}\)

Calculation:

We have to find the value of \(\rm \frac{d^2\tan x}{dx^2}\)

\(\rm \frac{d^2\tan x}{dx^2} = \frac{d}{dx} \left(\frac{d\tan x}{dx} \right )\)

\(\rm = \frac{d}{dx} \left(sec^2 x \right )\)

Apply chain rule, we get

\(\rm = \frac{d\sec^2 x}{d\sec x} × \frac{d\sec x}{dx}\)

= 2sec x . sec x tan x

= 2sec² x tan x

Concept:

Second-order Derivative.

\(\rm \frac {d^2 f(x)}{dx^2} = \) \(\rm \frac {d}{dx} [\frac {d}{dx} f(x)]\)

\(\rm \frac {d}{dx}e^{f(x)} = e^{f(x)} f'(x)\)

Calculations:

Let y = ekx 

Differentiating w.r. to x on both side, we get

⇒\(\rm \dfrac {dy}{dx} = ke^{kx}\)

Again Differentiating w.r. to x on both sides, we get

⇒\(\rm \dfrac {d^2y}{{dx}^2} = k^2e^{kx}\)

⇒\(\rm \dfrac {d^2y}{{dx}^2} = k^2y\)

Hence, If y = ekx then \(\rm \dfrac {d^2y}{{dx}^2} = k^2y\) 

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a. Biogas is a cheap, safe and renewable source of energy. 

b. It burns with a blue flame and without smoke. 

c. It can be used for cooking, domestic lighting, street lighting and small scale industries. 

d. It is eco-friendly and does not cause pollution and imbalance of the environment. 

e. It helps to improve sanitation of the surroundings. 

f. It can be easily generated, stored and transported.

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x^2+1)}dx=tan^{-1}x+C\)

Calculation:​

To solve \(\rm \int \frac{2x^3}{(x^2+2)(x^4+1)}dx\) 

Let x² = t ⇒ 2x dx = dt

Let us put this in given integration, then our integration becomes

\(⇒ \rm \int \frac{2x^3}{(x^2+2)(x^4+1)}dx= \rm \int \frac{t}{(t+2)(t^2+1)}dt\)

This integrand is a proper rational fraction. So, by using the form of partial fraction, we write

⇒\(\rm \frac{t}{(t+2)(t^2+1)}=\frac{A}{t+2}+\frac{Bt+C}{t^2+1}\)

⇒ t = A(t² + 2) + (Bt + C)(t + 2)

 ⇒ t = (A + B)t2 + (2B + C)t + 2(A + C)

Comparing coefficient of t2, t and constant terms on both sides, we get A + B = 0, 2B + C = 1 and A + 2C = 0  

By solving these equation, we get  A = -2/5,  B = 2/5 and C = 1/5

Thus  \(\rm \frac{t}{(t+2)(t^2+1)}=\frac{-2/5}{t+2}+\frac{1}{5}\frac{2t+1}{(t^2+1)}\) we can re-write it as 

⇒ \(\rm \frac{t}{(t+2)(t^2+1)}=-\frac{2}{5}\frac{1}{t+2}+\frac{1}{5}\frac{2t}{(t^2+1)}+\frac{1}{5}\frac{1}{(t^2+1)}\)

Therefore, \(\rm \int\frac{t}{(t+2)(t^2+1)}dt=-\frac{2}{5}\int\frac{1}{t+2}dt+\frac{1}{5}\int\frac{2t}{(t^2+1)}dt+\frac{1}{5}\int\frac{1}{(t^2+1)}dt\)

⇒ \(\rm \int\frac{t}{(t+2)(t^2+1)}dt=\frac{-2}{5}log\left | t+2 \right |+\frac{1}{5}log\left | t^2+1 \right |+\frac{1}{5}tan^{-1}t+C\) 

Now put t = x² in the above equation we get

⇒ \(\rm \int\frac{2x^3}{(x^2+2)(x^4+1)}dx=\frac{-2}{5}log\left | x^2+2 \right |+\frac{1}{5}log\left | x^4+1 \right |+\frac{1}{5}tan^{-1}x^2+C\)   

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b. Mutational breeding helps in producing disease resistant varieties.

c. Plants are produced by inducing mutations. Chemicals or physical mutagens are used for bringing about mutation.

d. Varieties of moong beans resistant to yellow mosaic virus and powdery mildew have been developed by the technique of mutational breeding.

Concept:

• \({\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
• \(\sin 2{\rm{x}} = 2\sin {\rm{x}}\cos {\rm{x}}\)

Calculation:

Let, \({\rm{I}} = \mathop \smallint \limits_0^{1} \sqrt {1 + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + \sin \frac{{\rm{x}}}{2}} {\rm{dx}}\)                  ( \({\sin ^2}\left( {\frac{{\rm{x}}}{4}} \right) + {\cos ^2}\left( {\frac{{\rm{x}}}{4}} \right)\) = 1)

\(= \mathop \smallint \limits_0^1 \sqrt {{{\sin }^2}\left( {\frac{{\rm{x}}}{4}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{4}} \right) + 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} {\rm{dx}}\)             \(\left( {\sin \frac{{\rm{x}}}{2} = 2\sin \left( {\frac{{\rm{x}}}{4}} \right)\cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1 \sqrt {{{\left( {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right)}^2}} {\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^1\left| {\sin \left( {\frac{{\rm{x}}}{4}} \right) + \cos \left( {\frac{{\rm{x}}}{4}} \right)} \right|{\rm{dx}}\)

Let \(\frac{{\rm{x}}}{4} = {\rm{t}}\)

Differentiating with respect to x, we get

\(\Rightarrow \frac{{{\rm{dx}}}}{4} = {\rm{dt}}\)                      

\(\Rightarrow {\rm{dx}} = 4{\rm{dt}}\)

\(\therefore {\rm{I}} = 4\mathop \smallint \limits_0^1 \left| {\sin \left( {\rm{t}} \right) + \cos \left( {\rm{t}} \right)} \right|{\rm{dt}}\)

\(\Rightarrow {\rm{I}} = 4\left[ { - \cos {\rm{t}} + \sin {\rm{t}}} \right]_0^1\)

\(\rm = 4\left[ { - \cos {\rm{\frac x 4}} + \sin {\rm{\frac x 4}}} \right]_0^1\)

\(= 4\left[ {\left( { - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right) - \left( { - \cos 0 + \sin 0} \right)} \right]\)

\(= 4\left[ { 1 - \cos \left( {\frac{{\rm{1 }}}{4}} \right) + \sin \left( {\frac{{\rm{1 }}}{4}} \right)} \right]\)

Hence, option (4) is correct.

Concept:         

Second Derivative Test:

• Calculate f’(x)
• Solve f’(x) = 0 and find the roots. Suppose x = c is the root of f’(x) = 0.
• Calculate f’’(x) and put x = c to get the value of f’’(c).

• If f’’(c) < 0 then x = c is a point of maxima.
• If f’’(c) > 0 then x = c is a point of minima.
• If f’’(c) = 0 then we need to use the first derivative test.

Calculation:      

Given: \(f(x)={(x^2-x+1) \over(x^2+x+1) } \)

\(\frac{{df}}{{dx}} = \frac{{({x^2} + x + 1)(2x - 1) - ({x^2} - x + 1)(2x + 1)}}{{{{({x^2} + x + 1)}^2}}}\)

\(\frac{{df}}{{dx}} = \frac{{{x^2} - 1}}{{{{({x^2} + x + 1)}^2}}}\)

\(\frac{{df}}{{dx}} = 0 ⇒ {x^2} - 1 = 0 ⇒ x = \pm 1\)

\(⇒ \frac{{{d^2}f}}{{d{x^2}}} = \frac{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) - 2\left( {{x^2} - 1} \right)\left( {2x + 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^3}}}\)

⇒ f''(1) = 1/9 > 0

So, x = 1 is  a point of minima

⇒ f''(-1) = - 1/9 < 0

So, x = - 1 is a point of maxima

So minimum value of the given function is attained at x = 1

By substituting x = 1 in f(x) we get f(1) = 1/3.

Hence, option D is the correct answer.

Concept:

For a function y = f(x):

• Relative (local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x) < 0.
• At the points of relative (local) minima, f''(x) > 0.

Calculation:

y = (x + 5)(x + 7)

⇒ y = x2 + 5x + 7x + 35

⇒ y = x2 + 12x + 35      ----(i)

By differentiating equation (i),

dy/dx = 0

⇒ 2x + 12 = 0      ----(ii)

⇒ x = - 6

Now by double differentiating equation (ii),

d2y/dx2 = 2 > 0

that means, the minimum value of equation (i) is at x = - 6

Minimum value of equation (i)

⇒ y = 36 - 72 + 35

⇒ y = - 1

∴ The minimum value is - 1.

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Integration by substitution:

If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.

• ∫ e^x dx = e^x + C.

Calculation:

Let \(\rm I=\int \sqrt{x}e^{\sqrt{x}}\ dx\)

Substituting \(\rm \sqrt x=t\), we get:

\(\rm \frac{1}{2\sqrt x}dx=dt\)

⇒ dx = 2t dt

∴ \(\rm I=\int t\times e^t\times2t\ dt\)

Integrating by parts, taking t² as the first function and e^t as the second function, we get:

⇒ \(\rm I=2\left[t^2 \int e^t\ dt-\int \left(\frac{d}{dt}t^2\int e^t\ dt \right)\ dt \right]+C\)

⇒ \(\rm I=2t^2 e^t-4\int t e^t\ dt+C\)

Integrating \(\rm \int t e^t\ dt\) by parts, we get:

⇒ \(\rm I=2t^2 e^t-4\left[t \int e^t\ dt-\int \left(\frac{d}{dt}t\int e^t\ dt \right)\ dt \right]+C\)

⇒ \(\rm I=2t^2 e^t-4\left(t e^t-e^t\right)+C\)

⇒ \(\rm I=\left(2t^2-4t+4\right)e^t+C\)

Back substituting \(\rm \sqrt x=t\), we get:

⇒ \(\rm I=(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C\).

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x^2+1)}dx=tan^{-1}x+C\)

Calculation:​

\(\rm \int \frac{x^2-x+1}{(x+1)(x^2+1)}dx\)

This integrand is a proper rational fraction therefore, by using the form of partial fraction, we can write the integrand as

⇒\(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\)

⇒ x2 - x + 1 = A(x² + 1) + (Bx + C) × (x + 1)

⇒ x2 - x + 1 = (A + B)x2 + (B + C)x + (A + C)

Comparing coefficient of x2, x and constant terms on both sides, we get A + B = 1, B + C = -1 and A + C = 1  

By solving these equation, we get  A = 3/2,  B = -1/2 and C = -1/2

⇒ \(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{3/2}{x+1}-\frac{1}{2}\frac{x+1}{(x^2+1)}\)  We can re-write it as 

⇒   \(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{3/2}{x+1}-\frac{1}{4}\frac{2x}{(x^2+1)}-\frac{1}{2}\frac{1}{(x^2+1)}\)

⇒\(\rm ​​\int\frac{x^2-x+1}{(x+1)(x^2+2)}dx=\frac{3}{2}\int\frac{1}{x+1}dx-\frac{1}{4}\int\frac{2x}{(x^2+1)}dx-\frac{1}{2}\int\frac{1}{(x^2+1)}dx\)

⇒\(\rm \int\frac{x^3-x+1}{(x+1)(x^2+1)}dx=\frac{3}{2}log\left | x+1 \right |-\frac{1}{4}log\left | x+1 \right |-\frac{1}{2}tan^{-1}x+C\)   

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Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm\int_a^b f(x)\ dx= [ g(x)]_a^b\) = g(b) - g(a).

• \(\rm \int_a^bf(x)\ dx=\int_a^cf(x)\ dx+\int_c^bf(x)\ dx\).
• \(\rm \sum n=\frac{n(n+1)}{2}\).

Calculation:

[x] is a multi-valued function in the interval 0-9, with:

[x] = 0, x ∈ (0, 1)

[x] = 1, x ∈ (1, 2)

[x] = 2, x ∈ (2, 3)

And so on.

∴ \(\rm \int_0^9 [x]\ dx\)

= \(\rm \int_0^1 0\ dx+\int_1^2 1\ dx\ +\ ...\ +\int_8^9 8\ dx\)

= \(0[x]_0^1+1[x]_1^2\ +\ ...\ +8[x]_8^9\)

= 0[1 - 0] + 1[2 - 1] + ... + 8[9 - 8]

= 0 + 1 + 2 + ... + 8

= \(\frac{8(8+1)}{2}\)

= 36.

Concept:

If the tangent of the curves touches orthogonally then the product of their slopes = -1.

Calculations:

Given curves are y - ax3 = 4 and x2 = y

slope of tangent of the curve y - ax3  = 4.is \(\rm m_1 = \dfrac {dy}{dx} = 3x^2a\)

⇒ slope of tangent of the curve y - ax3  = 4 at the point (- 1, 1) is \(\rm m_1 =\) 3a.

Now, slope of tangent of the curve x2 = y is \(\rm m_2 = \dfrac {dy}{dx} = 2x\)

⇒ slope of tangent of the curve x2 = y at the point (- 1, 1) is \(\rm m_2 = - 2\) 

Given, the curve y - ax3 = 4 and x2 = y, cut orthogonally at (-1, 1).

⇒ \(\rm m_1 .m_ 2 = -1\)

⇒ (3a)(-2) = - 1

⇒ a = \(\dfrac 16\)

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

To solve: \(\rm \int \frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta\)

let tan θ = x  ⇒ sec²θ dθ = dx

⇒ \(\rm \int \frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta = \rm \int \frac{1}{(x)(x-1)(x-2)}dx\)

This integrand is a proper rational fraction. So, by using the form of partial fraction, we can write it as

⇒ \(\rm \frac{1}{(x)(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\)

⇒ 1 = A(x - 1)(x - 2) + B(x)(x - 2) +C(x)(x - 1)

⇒ 1 = (A + B +C)x2 + (-3A - 2B - C)x + 2A

Comparing coefficient of x2, x and constant terms on both sides, we get A + B +C = 0, -3A - 2B - C = 0 and 2A = 1

By solving these equation, we get  A = 1/2,  B = -1 and C = 1/2

⇒ \(\rm \frac{1}{(x)(x-1)(x-2)}=\frac{1/2}{x}-\frac{1}{x-1}+\frac{1/2}{x-2}\)  

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}\int\frac{1}{x}dx-\int\frac{1}{x-1}dx+\frac{1}{2}\int\frac{1}{x-2}dx\)

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}log\left | x \right |-log\left | x-1 \right |+\frac{1}{2}log\left | x-2 \right |+C\)   

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=\frac{1}{2}log\left | x(x-2) \right |-log\left | x-1 \right |+C\)

⇒ \(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=log\sqrt{|x(x-2)|}-log\left | x-1 \right |+C\)

⇒\(\rm \int\frac{1}{(x)(x-1)(x-2)}dx=log|\frac{\sqrt{|x(x-2)|}}{ x-1}|+C\)

Now put x = tanθ in the above equation we get

⇒ \(\rm \int\frac{sec^2\theta}{(tan\theta)(tan\theta-1)(tan\theta-2)}d\theta=log|\frac{\sqrt{|tan\theta(tan\theta-2)|}}{ tan\theta-1}|+C\)

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

To solve: \(\rm \int \frac{2x^3}{(x^2+7)^2}dx\) 

Let us put x² = t ⇒2xdx = dt in \(\rm \int \frac{2x^3}{(x^2+7)^2}dx\)  

⇒ \(\rm \int \frac{2x^3}{(x^2+7)^2}dx = \rm \int \frac{t}{(t+7)^2}dt\)

This integrand is a proper rational fraction. therefore, by using the form of partial fraction, we write

⇒ \(\rm \frac{t}{(t+7)^2}=\frac{A}{t+7}+\frac{B}{(t+7)^2}\)

⇒ t = At + 7A + B 

By comparing coefficient of t and constant terms on both sides, we get A = 1 and 7A + B = 0

By solving these equation, we get  A = 1 and B = -7

\(⇒ \rm \frac{t}{(t+7)^2}=\frac{1}{t+7}-\frac{7}{(t+7)^2}\)

⇒ \(\rm \int\frac{t}{(t+7)^2}dt=\int\frac{1}{t+7}dt-\int\frac{7}{(t+7)^2}dt\)

⇒ \(\rm \int\frac{t}{(t+7)^2}dt=log\left | t+7 \right |+\frac{7}{(t+7)}+C\) 

Now put t = x² in the above equation we get

⇒ \(\rm \int\frac{2x^3}{(x^2+7)^2}dx=log\left | x^2+7 \right |+\frac{7}{(x^2+7)}+C\)

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

I = \(\rm \int_0^1 x^2(1+x^3)dx\)

Let 1 + x³ = t

Differentiating with respect to x, we get

⇒ (0 + 3x²)dx = dt

⇒ x² dx = \(\rm \frac {dt}{3}\)

Now,

I = \(\rm \frac{1}{3}\int_1^2 tdt\)

= \(\rm \frac{1}{3} \left[\frac{t^2}{2} \right ]_1^2\)

= \(\rm \frac{1}{6} [4-1] = \frac{3}{6} = \frac{1}{2}\)

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

\(\rm \int \frac{x+4}{(x+5)^2}dx\)

This integrand is a proper rational fraction. So, by using the form of a partial fraction, we can write it as:

⇒\(\rm \frac{x+4}{(x+5)^2}=\frac{A}{x+5}+\frac{B}{(x+5)^2}\)

This gives,  x + 4 = Ax + 5A + B 

By comparing the coefficient of x and constant terms on both sides, we get A = 1 and 5A + B = 4

By solving these equation, we get  A = 1 and B = -1

 \(⇒ \rm \frac{x+4}{(x+5)^2}=\frac{1}{x+5}-\frac{1}{(x+5)^2}\)  

⇒\(\rm \int \frac{x+4}{(x+5)^2}dx=\int\frac{1}{x+5}dx-\int\frac{1}{(x+5)^2}dx\)

⇒ \(\rm \int\frac{x+4}{(x+5)^2}dx=log\left | x+5 \right |+\frac{1}{(x+5)}+C\)   

Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x+a)^n}dx=\frac{1}{(1-n)(x+a)^{n+1}}+C\)

Calculation:​

To solve: \(\rm \int \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx\)

⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\)

⇒ x² - 3x + 2 = A(x + 2)(x + 3) + B(x + 1)(x + 3) +C(x + 1)(x + 2)

⇒ x2 - 3x + 2 = (A + B +C)x2 + (5A + 4B + 3C)x + (6A + 3B + 2C)

By comparing coefficient of x2, x and constant terms on both sides, we get A + B +C = 1, 5A + 4B + 3C = -3 and 6A + 3B + 2C = 2

By solving these equation, we get  A = 3,  B = -12 and C = 10

⇒\(\rm \frac{x^2-3x+2}{(x+1)(x+2)(x+3)}=\frac{3}{x+1}-\frac{12}{x+2}+\frac{10}{x+3}\)  

⇒\(\rm \int\frac{x^2-3x+2}{(x+1)(x+2)(x+3)}dx=\int\frac{3}{x+1}dx-\int\frac{12}{x+2}dx+\int\frac{10}{x+3}dx\)

⇒ \(\rm \int\frac{x^3-3x+2}{(x+1)(x+2)(x+3)}dx=3log\left | x+1 \right |-12log\left | x+2 \right |+10log\left | x+3 \right |+C\)   

Concept:

\(\rm \int \frac{1}{x} dx = \log x + c\)

Calculation:

I = \(\rm \int \frac{x}{3x^2+4} dx \)

Let 3x² + 4 = t

Differentiating with respect to x, we get

⇒ 6xdx = dt

⇒ xdx = \(\rm \frac {dt}{6}\)

Now,

I = \(\rm \frac{1}{6}\int \frac{1}{t} dt\)

= \(\rm \frac{1}{6}\log t + c\)

= \(\rm \frac{1}{6}\log (3x^2+4)+ c\)

1) Avoid delay or postponing any planned activity 

2) Organise your room and school desk 

3) Develop a ‘NO DISTURBANCE ZONE’, where you can sit and complete important tasks 

4) Use waiting time productively 

5) Prepare a ‘To-do’ list 

6) Prioritise 

7) Replace useless activities with productive activities

Attitude decides our focus. What we focus on becomes a habit, habits form our behaviour, behaviour becomes part of our psyche, our psyche is reflected in our activities and finally that becomes the core of personality. Thus, by changing our focus (i.e. attitude) we can change our personality.

Personality Development is a continuous and multi-faceted process which requires a set of skills that need to be learned and at times unlearned. In order to cope up with the challenges offered by external environment an individual strives to change and develop personality in meaningful ways throughout his/her lifespan.