Integration By Parts

Chapter 7

Class XII

Exercise 7.6

Q.No. 1. x.sinx

Solution:

From question it is clear that

I = ഽx.sinx.dx

Put First function = x
Second function = sinx
Now, applying integration by parts;
I = - x.cosx + ഽcosx.dx
or, - x.cosx + sinx + C Ans.

Q.No. 2. x.sin3x

Solution:

From question it clear that

I = ഽx.sin3x.dx
Put first function = x
Second function = sin3x
Now applying integration by parts
I = - x.cos3x/3 + ഽcos3x.dx/3
or, -x.cos3x/3 + sin3x/9 + C Ans.

Q.No. 3. x².ex

Solution:

From question it is clear that
 I = ഽx².ex.dx

Put first function = ex
Second function =  x²

Now applying integration by parts
I =  x².ex - ഽ2x.ex.dx
Again applying integration by parts
or,x².ex - 2x.ex +  ഽ2ex.dx
or,x².ex - 2x.ex + 2ex + C
or, ex(x² - 2x + 2) + C Ans.

Q.No. 4. x.logx

Solution:

From question it is clear that
I = ഽx.logx.dx
Put first function = logx
Second function = x

Now, applying integration by parts

I = x².logx/2 - ഽx.dx/2
or, x².logx/2 - x²/4 + C Ans.

Q.No. 5. x.log2x

Solution:

From question it is clear that
I = ഽx.log2x.dx

Put first function = log2x
Second function = x

Now applying integration by parts

I = log2x.x²/2 - ഽx².2/4x.dx
or, log2x.x²/2 - x²/4 + C Ans.

Q.No. 6. x².logx

Solution:

From question it is clear that
I = ഽ x².logx.dx

Put first function = logx
Second function = x²

Now applying integration by parts

I = logx. x³/3 - ഽx²dx/3
or, logx.x³/3 - x³/9 + C Ans.

Q.No. 7. xsin⁻¹x

Solution:

From question it is clear that
I =  ഽxsin⁻¹x.dx

Put first function = sin⁻¹x
Second function = x

Now applying integration by parts

I = sin⁻¹x. x²/2 - ഽx².dx/2√1-x²

or,  sin⁻¹x. x²/2+ഽ(1-x²-1).dx/√1-x²
or,  sin⁻¹x. x²/2+ ഽ√1-x².dx/2 -  ഽdx/2√1-x²
or,  sin⁻¹x. x²/2+ √1-x²/4 + sin⁻¹x./4 - sin⁻¹x./2 + C
or, (2x² - 1)sin⁻¹x./4 + √1-x²/4 +C Ans.

Q.No. 8 x.tan⁻¹x

Solution:

From question it is clear that
I = ഽx.tan⁻¹x

Put first function = tan⁻¹x
Second function = x

Now applying integration by parts

I = x².tan⁻¹x/2 - ഽx².dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽ(1 + x² - 1).dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽdx/2 + ഽdx/2(1 + x²)
or, x².tan⁻¹x/2 - x/2 + .tan⁻¹x/2 + C Ans.

Rest questions will be solved latter.

Similar Triangles

Exercise 6.4
Class X
NCERT Textbook

Q.No. 1. Let △ ABC ∼△DEF and there areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution:
ar(ABC)/ar(DEF) = (BC/EF)²
Now, putting the value of areas and EF
64/121 = (BC/15.4)²
Therefore,
BC = 8x15.4/11
Hence,
BC = 11.2 cm Ans.

Q.No. 2. Diagonals of trapezium ABCD with AB॥DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Solution:

Consider in triangles AOB and COD.
<OAB = <OCD     (Alternate angles)
<OBA = <ODC      (Alternate angles)
<AOB = <COD      (Vertically opposite angles)
△AOB∼ △COD    (AAA Similarity)

Therefore,
ar(AOB)/ar(COD) = (AB/CD)²
Now, putting the value of AB.
ar(AOB)/ar(COD) = (2CD/CD)²
ar(AOB)/ar(COD) = 4/1
Required ratio = 4:1 Ans.

Q.No.3. In Fig. 6.46 ABC and DBC are two triangles on the same base BC. If AD intersect BC at O, show that
ar(ABC)/ar(DBC) = AO/DO

Solution:

Draw perpendicular AP and DP on BC.
Consider in triangles APO and DQO.
<APO = <DQO (Each 90⁰)
<AOP = <DOQ  (Vertically opposite angles)

△APO ∼ △DQO
Therefore,
ar(APO)/ar(DQO) = (AP/DQ)² = (AO/DO)² = (PO/QO)² -------------- (i)

Again,
ar(ABC) = BCxAP/2
ar(DBC) = BCxDQ/2
ar(ABC)/ar(DBC) = AP/DQ

Hence from (i),
ar(ABC) = ar(DBC) = AO/DO Proved.

Q.No. 4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Consider about two triangles ABC and PQR.
Therefore,
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ------------------ (i)

Again,
ar(ABC) = ar(PQR)       (Given)        ----------------------  (ii)

Now, from (i) and (ii).

AB = PQ
BC = QR
AC = PR

Hence,
△ABC≅△PQR Proved.

Q.No. 5. D, E and F respectively the mid points of the sides AB, BC and CA of triangle ABC. Find the ratio of the areas of △ DEF and △ABC.

Solution:

D and F are the mid points of AB and CA.
Therefore,
AD/DB = AF/FC
So,
DF ॥ BC
and
DF = BC/2 ------------- (i)

Similarly,
DE = AC/2 ------------- (ii)
and
EF = AB/2 ----------- -- (iii)

Now, from (i), (ii) and (iii).
ADEF, BEFD and CFDE are parallelograms.

Now, consider in triangles ABC and DEF.
<A = <E ---------------- (ADEF is a parallelogram)
<B = <F ---------------- (BEFD is a parallelogram)
<C = <D ----------------- ( CFDE is parallelogram)

Hence,
△ABC ∼ △EFD
ar(ABC)/ar(DEF) = (AB/EF)²
Now, putting the value of EF.
ar(ABC)/ar(DEF) = 4:1

Therefore,
ar(DEF)/ar(ABC) = 1:4 Ans.


Q.No. 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Consider about two similar triangles ABC and PQR.
AX, BY and CZ are the medians of ABC. PK, QL and RM are medians of PQR.

Since,
ABC and PQR are similar triangles.
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ---------------------- (i)

Since,
AX and PK are medians of ABC and PQR.
Therefore,
Triangles ABX and PQK are similar.
ar(ABX)/ar(PQK) = (AB/PQ)² = (AX/PK)² ---------------------- (ii)

Now, from (i) and (ii).

The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians. Proved.

7. Prove that area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let the side of the square be X.
Diagonals of the square = ✔️2 X

Area of equilateral triangle on diagonal = 2✔️3X²/4 ------- (i)
Area of equilateral triangles on side = ✔️3X²/4 ------ (ii)

Now, from (i) and (ii)

Area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Proved

Areas of Similar Triangles

Class X

Theorem

"The ratios of areas of two triangles is equal to the square of the ratio of their corresponding sides".

Solution:-

Given

△ABC ~ △PQR

Therefore,

AB/PQ = BC/QR = CA/RP

To prove that

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²

Construction

Draw AM ⫡ BC and PN ⫡ QR

Proof

ar(ABC) = BCxAM/2 ---------------- (i)

ar(PQR) = QRxPN/2 ----------------- (ii)

Now, from (i) and (ii)

ar(ABC)/ar(PQR) = BCxAM/QRxPN ------------ (iii)

Now, consider in △ABM and △PQN

<B = <Q     (△ABC ~ △PQR)

<M = <N    ( Each 90⁰)

Hence,

△ABM ~ △PQN  (AA similarity)

Therefore,

AM/PN = AB/PQ

Hence,

AM/PN = AB/PQ = BC/QR = CA/RP

Now, from (iii)

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)² Proved.

Relations and Functions

Class XII

Relations and Functions

All functions are relations but all relations are not functions.

Relations

A relation R in a set A is a subset of AxA.

Empty relation φ and  universal relation AxA are extreme relations.

Empty Relation

A relation R in a set A is said to be empty, if no element of A is related to any element of A.

R = φ ⊂AxA

Universal Relation

A relation R in a set A is said to be universal relation, if each element of A is related to  every element of A.

R = AxA

Method to Represent a Relation

There are two methods to represent a Relation.

1. Roster Method 2. Set Builder Method.

For Example

Relation R in the set

R = {1,2,3,4}

is a relation represent roster method.

R = {(a,b):b = a +1}

is a relation represent set builder method and read as R is a relation as a related to b or aRb if and only if b = a+1.

Reflexive, Symmetric and Transitive and Equivalence are the types of Relation.

A relation R in set A is said to be reflexive, Symmetric and Transitive if and only if 

1. Reflexive, if (a,a) ε R for every aεA.
2. Symmetric, if  (a,b)εR implies that (b,a)εR for all elements of A.
3. Transitive, if (a,b)εR and (b,c)εR implies (a,c)εR for all elements of A.
4. Any relations which satisfy all of these conditions are said to be Equivalence Relations or other words the relation which is reflexive, symmetric and transitive is said to be Equivalence Relation.

Exercise 1.1

NCERT Text Books

Question No. 1 Determine whether each of the following relations are reflexive, symmetric and transitive.

(i) Relation R in the set A = {1,2,3, ................. 13,14}

defined as 

R = {(x,y):3x - y = 0}

Solution:

Roster form 

R = { (1,3), (2,6),(3,9), (4,12)}
(a,a) is not the element of  R for every elements of  A.
Hence, it is not reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Therefore, this is none any one of these three.

(ii) Relation R in the set N of natural numbers defined as 
R = {(x,y):y = x + 5 and x<4}

Solution:
Roster form
R = { (1,6),(2,7), (3,8)}

(a,a) is not the element of  R for every elements of  A.
Hence, it is not reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Therefore, this is none any one of these three.

(iii) A = {1,2,3,4,5,6}
R = {(x,y):y is divisible by x}

Solution:
Roster form
R = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,2), (1,3), (1,4), (1,6), (2,4), (2,6), (3,6)}

From roster form it is clear that;
(a,a) is the element of R for every element of A.
Hence, it is reflexive.

(a,b)εR does not imply that (b,a)εR for all elements of A.
Hence, it is not symmetric.

 (a,b)εR and (b,c)εR does not imply (a,c)εR for all elements of A.
Hence, it is not transitive.

Similarly we can do remain problems of this section.

Communication Systems

Class XII
Chapter 15

Pointwise Details of This Chapter

Communication Systems

1. J.C. Bose, F.B. Morse, G. Marconi and Alexander Graham Bell have outstanding contribution in the field of communication system.
2. Every communication system has three essential elements 1. Transmitter, 2. Medium/Channel and 3. Receiver.
3. Receiver and Transmitter are connected by Medium or Channel.
4. Medium or Channel is in the form of wires or cables or it may be wireless.
5. Point to Point and Broadcast are two basic modes of transmission.
6. In point to point communication, communication takes place over a link between a single transmitter and a receiver.
7. Telephony is an example of point to point communication.
8. In Broadcast communication, communication takes place between many receiver and a single transmitter.
9. Radio and Television are the examples of Broadcast communication.
10. Transducer is a device which converts one form of energy into another.
11. Electrical transducer is such device which converts pressure, displacement, force, temperature, etcetra into corresponding signals at its output.
12. Information converted in electrical form and suitable for transmission is called signal.
13. Analog and Digital are two types of signals.
14. Analog signals are continuous variation of voltage and current.
15. Sound and Picture signals in television are analog in nature.
16. Signals which can take only discrete stepwise values. Binary system is used in digital electronics employs just two levels of a signal. 0 corresponding to low level and 1 corresponding to high level of voltage and current.
17. There are several coding scheme in digital communication such as Binary Coded Decimal (BCD), American Standard Code for Information and Interchange (ASCII)
18. Nowadays one more signal that is Optical Signal.
19. Unwanted signals that tend to disturb the transmission and processing of message signals in a communication system is said to be Noise.
20. One which process the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception is said to be transmitter.
21. One which extracts the desired message signals from the received signals at the channel output.
22. The loss of strength of a signal while propagating through a medium is said to Attenuation.
23. The process of increasing the amplitude of a signal using an electronic circuit is said to be Amplifier.
24. Range is the largest distance between a source and a destination up to which the signal is received with sufficient strength.
25. Band width refers to the frequency range over which an equipment operates or the portion of the spectrum occupied by the signal.
26. Modulation is process in which original low frequency message or information signal cannot be transmitted to long distances because of reasons the transmitter, information contained in the low frequency message signal is superimposed on a high frequency wave, which acts as a carrier of the information.
27. Demodulation is the reverse process of modulation in which retrieval of information from the carrier wave at the receiver.
28. Repeater is the combination of a receiver and a transmitter which picks up the signal from the transmitter, amplifies and retransmits it to the receiver. Repeater are used to extend the range of a communication system.
29. Frequency range 300 Hz to 3100 Hz is adequate for Speech signal.
30. A bandwidth of 20 kHz is required for musical transmission.
31. A bandwidth of 4.2 MHz is required for video transmission.
32. A bandwidth of 6 MHz is required for TV signal because it transmits both voice and video.
33. Digital signals are in the form of rectangular waves.
34. Rectangular waves can be decomposed into a superposition of sinusodial waves of frequencies.

हमारे आस पास के पदार्थ

कक्षा नवम

NCERT 

पाठ्य पुस्तक के प्रश्नोत्तर

पदार्थ के भौतिक स्वरूप

1. निम्नलिखित में से कौन से पदार्थ है -

उत्तर-

कुर्सी, वायु, बादाम एवं नीबूं का पानी पदार्थ है क्योंकि यह स्थान घेरता है, द्रव्यमान एवं आयतन दोनों ही होता है।

2. गर्मा गरम खाने की गंध कई मीटर से ही आपके पास पहूँच जाती है लेकिन ठंढे खाने की महक लेने के लिए आपको उसके पास जाना पडता है।

उत्तर-

पदार्थ के कणों की गति तापमान के अनुक्रमाणुपाति होता है। पदार्थ का तापमान बढने से उसके कणों की गति बढती है जबकि घटने से घटती है। अतः गर्मा गरम खाने की गंध कई मीटर से ही हमारे पास पहूँच जाती है जबकी ठंढे खाने की महक लेने के लिए हमे उसके पास जाना पडता है।

3. स्वीमिंग पूल में गोताखोर पानी काट पाता है। इससे पदार्थ का कौन सा गुण प्रेक्षित होता है ?

उत्तर-

पदार्थ के कणों के बीच रिक्त स्थान होता है। पदार्थ का यही गुण प्रेक्षित होता है।

4. पदार्थ के कणों का क्या विशेषताएँ होती है ?

उत्तर-

पदार्थ के कणों की निम्नलिखित विशेषताएँ होती है जो नीचे लिखा गया है।

1. पदार्थ के कणों के बीच रिक्त स्थान होता है।
2. पदार्थ के कण निरंतर गतिशील होते है।
3. पदार्थ के कण एक दुसरे को आकर्षित करते है।

पदार्थ की अवस्थाएँ

1. किसी तत्व के द्रव्यमान प्रति इकाई आयतन को घनत्व कहते है।

घनत्व = द्रव्यमान / आयतन

बढते हुए घनत्व के क्रम में निम्नलिखित को व्यवस्थित करें

उत्तर-

वायु, चिमनी का धुआं, जल, शहद, रुई, चाॅक एवं लोहा।

2. (a) पदार्थ के विभिन्न अवस्थाओॆ के गुणों में  होने वाले अंतर को सारणीबद्ध कीजिए।

उत्तर-

पदार्थ के गुणों में होने वाले अंतर को नीचे सारणीबद्ध किया गया है।

ठोस के आकार एवं आयतन निश्चित होता है।

द्रव्य के आकार अनिश्चित एवं आयतन निश्चित होता है।

गैस के आकार एवं आयतन दोनों ही अनिश्चित होता है।

ठोस अवस्था में पदार्थ के कण एक दुसरें के काफी निकट होते हैं।

द्रव्य अवस्था में पदार्थ के कण ठोस के अपेक्षा दूर लेकिन गैस के अपेक्षा निकट होते है।

गैसीय अवस्था में पदार्थ के कण ठोस एवं द्रव्य के अपेक्षा काफी दूरी पर होते है।

ठोस में कणों की संपीड्यता सबसे  कम होता है अतः ये दृढ होते है।

द्रव्य में कणों की संपीड्यता ठोस के अपेक्षा अधिक एवं गैस के अपेक्षा कम होता है अतः ये तरल होते है।

गैस में कणों की संपीड्यता सबसे अधिक होता है।

ठोस के कणों की गतिज ऊर्जा सबसे कम होता है।
द्रव्य के कणों की गतिज ऊर्जा ठोस के अपेक्षा अधिक लेकिन गैस के अपेक्षा कम होता है।
गैस के कणों की गतिज ऊर्जा सबसे अधिक होता है।

(b) निम्नलिखित पर टिप्पणी कीजिए

उत्तर-
दृढता
वैसे पदार्थ जिसका आकार एवं आयतन निश्चित होता है, जिसमें कण एक दुसरे बहुत नजदीक होते है। इसमें वाह्य बल लगाने पर टूट सकते है लेकिन अपना आकार नही बदलता हो दृढ होते है। पदार्थ के इस गुण को दृढता कहा जाता है। ठोस में ऐसा गुण पाया जाता है।

संपीड्यता

पदार्थ का वह गुण जिसके कारण पदार्थ के अत्यधिक आयतन को एक कम आयतन वाले वर्तन में संपीडित किया जा सकता है, पदार्थ के इस गुण को ही संपीड्यता कहा जाता है।

तरलता

पदार्थ का वह गुण जिसके कारण वह अपना आकार असानी से बदलता है तरलता कहलाता है। द्रव्य में तरलता का गुण पाया जाता है।

बर्तन में गैस का भरना
गैस के कणों की गति अनियमित एवं अत्यधिक तिव्र होती है। इस अनियमित गति के कारण इसके कण बर्तन के दीवारों से से टकराते है जिसके कारण गैस का दबाव बनता है।

गतिज ऊर्जा
ठोस में कणों की गतिज ऊर्जा सबसे कम, द्रव्य में ठोस से ज्यदा एवं गैस से कम जबकि गैस में सबसे अधिक होता है।

Matter in Our Surroundings

Class IX
Chemistry
Questions Answers

Any thing which occupy space or having both mass and volume is said to be matter.

Q.No. 1 Which of the following are matter

Answer

Chair, Air, Almond cold drink and smell of perfume are matter because they occupy space and having both mass and volume.

Q.No. 2 Give reasons for the following observation:

We know that particles of matter are continuously moving. As temperature rises, particles move faster. Therefore, the smell of hot sizzling food reaches us several metres away.

But to get the the smell from cold food we have to go close because the temperature of cold food is less.

Q.No. 3 Answer

A diver is able to cut through water in swimming pool. This observation show the property of matter "Particles of matter have space between them".

Q.No. 4

There are following characteristics of particles of matter

1. Particles of matter have space between them.

2. Particles of matter are continuously moving.

3. Particles of matter attract each other.

Q.No. 5. The mass per unit volume of a substance is called density. density = mass /volume. Arrange the following in order of increasing density

Answer

Iron, Chalk, cotton, honey, water exhaust from chimney and air are written in decreasing order of density.