Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 10th Lesson Molecular Basis of Inheritance

Very Short Answer Questions

Question 1.
Distinguish between Heterochromatin and Euchromatin. Which of the two is transcriptionally active?
Answer:
The chromatin that is more densely packed and stains dark is called heterochromatin. The chromatin that is loosely packed and stain light is called Euchromatin. Euchromatin is transcriptionally active chromatin.

Question 2.
Who proved that DNA is Genetic Material? What is the organism they worked on?
Answer:
Alfred Hershey and Martha chase (1952). They worked with viruses that infect Bacteria, bacteriophages.

Question 3.
What is the function of DNA polymerase?
Answer:
DNA polymerise is a highly efficient enzyme which catalyse polymerisation of a large number of nucleotides in a very short time. It also catalyse the reaction with a high degree of accuracy.

Question 4.
What are the components of a nucleotide?
Answer:
A nucleotide has three components – a nitrogenous base, a pentose sugar and a phosphate group’.

Question 5.
Given below is the sequence of coding strand of DNA in a transcription unit.
5’AATGCAGCTATTAGG – 3 . Write the sequence of
a) Its complementary strand
b) The mRNA
Answer:
a) 5’TTACGTCGATAATCC-3′
b) 5’ AAUGCAGCUAUUAGG – 3’

Question 6.
Name any three viruses which have RNA as the Genetic Material.
Answer:
Tobacco mosaic virus, QB bacteriophage, HIV.

Question 7.
What are the components of a transcription unit?
Answer:
a) A promotor b) The structural Gene c) A terminator.

Question 8.
What is the difference between exons and Introns?
Answer:

Question 9.
What is meant by capping and tailing?
Answer:
Adding of an unusual nucleotide (methyl guanosine triphosphate) to the 5′ -end of heterogenous nuclear RNA [hnRNA) is called capping.

Adding of adenylate residues to the 3’ – end in a template independent manner is called tailing.

Question 10.
What is meant by point mutation? Give an example.
Answer:
Change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

Question 11.
What is meant by charging of tRNA?
Answer:
Activation of aminoacids in the presence of ATP and linked to their cognate tRNA is known as charging of tRNA or amino acylation of tRNA.

Question 12.
What is the function of the codon AUG?
Answer:
It has dual functions. It codes for methionine and also act as the initiator codon.

Question 13.
Define stop codon. Write the codons.
Answer:
Codons which do not code for any aminoacids are called stop codons. They are UAA, UAG, UGA.

Question 14.
What is the difference between the template strand and a coding strand in a DNA molecule?
Answer:
The two strands have opposite polarity and the DNA-dependant RNA polymerise also catalyses the polymerisation in only one direction that is, 5′ → 3′, the strand that has the polarity 3′ → 5′ acts as a template and is also called template strand. The other strand which has polarity 5′ → 3′ and does not code for anything is called coding strand.

Question 15.
Write any two differences between DNA and RNA.
Answer:

Question 16.
In a typical DNA molecule the proportion of thymine is 30% of the N bases. Find out the percentages of other N bases.
Answer:
Adenine = 30%
Guanine = 20%
Cytosine = 20%

Question 17.
The proportion of nucleotides in a given nucleic acid are Adenine 18%, Guanine 30%, Cytosine 42% and uracil 10%. Name the nucleic acid and mention the number of strands in it.
Answer:
RNA. Only one strand is present.

Short Answer Questions

Question 1.
Define transformation in Griffith’s Experiment. Discuss how it helps in the identifi¬cation of DNA as genetic material.
Answer:
Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with ‘s’ shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia.

He injected heat killed ‘s’ strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed ‘s’ strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

Question 2.
Discuss the significance of heavy isotope of Nitrogen in Meselson and Stahl’s experiment.
Answer:
Matthew Meselson and Franklin Stahl, grow E.Coli in a medium containing ¹⁵NH₄Cl and observed 15_N was incorporated in newly synthesized DNA. This heavy DNA molecule could be distinguished from themormal DNA by centrifugation in a cesium chloride density gradient.

They transformed the cells into a medium with ¹⁵NH₄Cl (normal) and took samples at various time intervals and extracted the DNA that remained as double stranded helices. The various samples were separated independently on cesium chloride (cscl) gradients to measure the densities of DNA. Thus the DNA that was extracted from the culture, one generation after the transfer from ¹⁵N to ¹⁴N medium had a hybrid density. DNA extracted from the culture after another generation (40 mts) was composed of equal amounts of this hybrid DNA and of ‘Light’ DNA.

Question 3.
A single base mutation in a gene may not always result in loss or gain of function. Do you think the statement is correct? Defined your answer.
Answer:
A single base mutation in a gene may result in loss or gain of a gene and so a function.
E.g.: A change of single base pair in the gene for betaglobin chain that results in the change of aminoacid residue glutamate to valine. It results in a diseased condition called sickle cell anaemia.

E.g.: 2) Consider a statement that is made like a genetic code is – RAM HAS RED CAP.
If we remove a letter ‘S’ in HAS, it will be RAM HAR EDC AP.

Question 4.
How many types of RNA polymerases exist in cells ? Write their names and functions.
Answer:
Three RNA polymerases exist in Cells. They are :

1. RNA polymerase I – It transcribes r RNAs (28S, 18S, and 5.8S)
2. RNA polymerase II – It transcribes the precursor of RNA, the heterogeneous nuclear RNA (hn RNA)
3. RNA polymerase III – It is responsible for transcription of tRNA 5 Sr RNA and Sn RNAs (small nuclear RNAs)

Question 5.
What are the contributions of George Gamow, H.G. Khorana, Marshall Nirenberg in deciphering the genetic code?
Answer:
George Ganow, suggested that, in order to code for all the 20 Amino’ acids, the code should be made up of three nucleotides. This was a very bold proposition, because a permutation and combination of 4³ would generate 64 codons, generating more codons than required.

H.G. Khorana developed chemical method in synthesising RNA molecules with defined combinations of bases.

Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. The enzyme polynucleotide phosphorylase was also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA)

Question 6.
On the diagram of the secondary structure of tRNA shown below label the location
of the following parts.
Answer:

a) Anticodon
b) Acceptor stem
c) Anticodon stem
d) 5′ end
e) 3′ end

Question 7.
Draw the schematic / diagrammatic presentation of the lac operon.
Answer:

Question 8.
What are the differences between DNA and RNA.
Answer:

Question 9.
Write the important features of Genetic code?
Answer:

1. The codon is triplet. 61 codons code for aminoacids and 3 codons donot code for any aminoacids called stop codons.
2. One codon codes for only one aminoacid, hence it is unambiguous and specific.
3. Some aminoacids are coded by more than one codon, hence the code is degenerated.
4. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal: For Ex: UUU code for phenylalanine (Phe) in Bacteria or Humans.
6. AUG has dual functions. It codes for methionine and also acts as initiator codon.

Question 10.
Write briefly on nucleosomes.
Answer:
Nucleosome is a bead like structure of chromosomes. It consists of eight histone molecules and a DNA segment of about 150 base pairs. Each Nucleosome is separated from one another by a linker DNA sequence of about 50 base pairs. Nucleosome helps to fold DNA into a compact form in the interphase nucleus. Otherwise the length of a chromosome, when linear is many orders of magnitude greater than the diameter of the nucleus.

Intext Questions

Question 1.
Group the following as nitrogenous bases and nucleosides : Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases : Adenine, Thymine, Uracil, Cytosine,
Nucleosides : Cytidine, Guanosine.

Question 2.
If a double stranded DNA has 20% of cytosine, calculate the percent of adenine in the DNA.
Answer:
Cytosine = 20%
Guamine = 20%
Adenine = 30%
Thymine = 30%

Question 3.
If the sequence of one strand of DNA is written as follows : Write down the sequence of complementary strand in 3′ → 5′ direction.
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3‘
Answer:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows.
5 – ATGCATGCATGCATGCATGCATGCATGC – 3 write down the sequence of m RNA
Answer:
3′ AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 5

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi conservative mode of DNA replication? Explain.
Answer:
The two strands would separate and acts as a template for the synthesis of new complementary strands. After completion of replication, each DNA would have one parental and one newly synthesised strand.

Question 6.
Depending upon the chemical nature of the template [DNA or RNA] and the nature of nucleic acids synthesized from it [DNA or RNA], List the types of nucleic acid polymerases.
Answer:
DNA polymerases
RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and chase purified biochemicals (proteins, DNA, RNA etc) from the heat killed ‘S’ cells. They discovered the DNA alone from S bacteria caused R bacteria to become transformed. They also discovered thal protein digesting enzymes and RNA digesting enzymes did not affect transformation. So the transforming principle was not a protein or RNA. Digestron with DNAse did inhibit transformation suggesting that the DNA caused the transformation.

Question 8.
Differentiate between the followings :
a) mRNA and tRNA
b) Template strand and Coding strand
Answer:

a)

b)

Question 9.
List two essential roles of ribosomes during translation.
Answer:

1. Ribosome acts as the site where protein synthesis takes place from individual aminoacids.
2. Ribosome acts a catalyst for forming peptide bond.
   E.g. : 23 S r-RNA in bacteria acts as ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added. Which induced the Lac operon. Then, why does Lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes namely, an operator gene, a promoter gene and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose. In Lac operon lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promotor region.

Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose or metabolise it into glucose and galactose. After sometime, when the level of the inducer decreases, it causes the synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence the transcription is stopped.

Question 11.
Explain (in one or two lines) the functions of the followings :
a) Promoter b) tRNA c) Exons.
Answer:
a) Promoter:
It is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase. :

b) tRNA :
It is a small RNA that reads the genetic code present on mRNA it carries specific aminoacids to mRNA on ribosome during translation of proteins.

c) Exons:
Exons are coding sequences of DNA in Eukaryotes that transcribe for proteins.

Question 12.
Briefly describe the following :
a) Transcription b) Translation.
Answer:
a) Transcription:
It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promotor region of the template DNA and terminates at the terminator region The segment of DNA between these two regions is known as transciption unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of nucleotides, and certain cofactors such as Mg²⁺

During transcription, three events occur. They are : a)Initiations b)Elongation, c) Termination. The DNA dependant RNA polymerase and certain initiation factors bind at the double stranded DNA at the promotor region of the template strand and initiate the process of transcription, RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex in two separate strands.

Then one of the strands, called sense strand acts as template for mRNA synthesis. The epzyme RNA polymerase utilises nucleoside triphosphates as raw material and polymerizes them to form m RNA according to the complimentary bases present on the template DNA. This process of opening of helix and elongation of polynucleotide chain continuous until the enzyme reaches terminator region. As RNA polymerase reaches the terminator region, the newly synthesized mRNA transcripted alpng with enzyme is released. Another factor called terminator factor (p) required for the termination of the transcription.

b) Translation :
It is the process of polymerizing amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain. This process invoivs 3 steps, a) Initiation b) Elongation c) Termination. During the initiation, tRNA gets charged when the aminoacid binds by using ATR The start codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in.a large subunit for the attachment of subsequent aminoacid.

The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon dowonstream along with mRNA so as to leave the space for binding of another charged tRNA, The aminoacid brought by tRNA get linked with the previous aminoacid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (UAA, UAG and UGA), the process of translation gets terminated. The polypeptide chain is released and ribosomes get detached from mRNA.

Question 13.
How the polymerization of nucleotides can be prevented in a DNA molecule?
Answer:
Due to Lack of Helicase enzyme, unwounding does not occurs. The single stranded Binding proteins cover the DNA strands preventing them from annealing into a double strand.

Question 14.
In an experiment, DNA is treated with a compound which trends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base pairs increases. From 0.34 nm to 0.44 nm calculate the length of DNA double helix (which has 2 × 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
2 × 10⁹ × 0.44 × 10^-9 bp.

Question 15.
Recall the experiments done by Frederick Griffith. Where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of pneumococcus have transformed the R – strain into virulent strain? Explain?
Answer:
RNA is more labile and prone to degradation (owing to the presence of 2’OH group in its ribose). Hence heat killed ‘s’ strain may not have retained its ability to transform the ‘R’ strain into virulent form if RNA was its genetic material.

Question 16.
You are repeating the Hershey – Chase experiment and are provided with two Isotopes : ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:
Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive Isotope while ¹⁵N is not radioactive but is the heavier Isotope of Nitrogen). Even if ¹⁵N was radioactive, them its presence would have been detected both inside the cell as well as in the supernatant because ¹⁵N would also get incorporated in amino group of aminoacids in proteins. Hence the use of ¹⁵N would not give any conclusive results.

Question 17.
Do you think that the alternate splicing of exons may enable a structural gene to code for several Isoproteins from one and the same gene? If yes, how? If not, why so?
Answer:
Functional m RNA of structural genes need not always include all of its exons. This alternate splicing of exons in sex-specific, tissue – specific, and even developmental stage specific. By such alternate splicing of exons a single gene may encode for several Isoproteins and/or proteins of similar class’. In absence of such a kind of splicing, there should have been new genes for every protein/Isoprotein. Such an extravagency has been avoided in natural phenomena by way of alternate splicing.

Question 18.
Can you recall what centrifugal force is, and think why a molecule with higher mass/ Density would sediment faster?
Answer:
Proteins have lower density when compared to others. So a molecule with higher mass would sediment faster than molecules with light weight (density).

Question 19.
Do Retroviruses follow central Dogma? Give one example.
Answer:
Genetic material of Retroviruses in RNA. At the time of synthesis of protein RNA is reverse transcribed to its complementary DNA first/which is opposite to the central Dogma. Hence Retroviruses are not known to follow central Dogma.

Question 20.
If there are 2.9 × 10⁹ complete turns in a DNA molecule. Estimate the length of the molecule (1 angstrom = 10^-8 cm).
Answer:
1 turn of DNA = 3.4 nm.
Number of turns are = 2.9 × 10⁹
The length of the DNA molecule = 2.9 × 10⁹ × 3.4 nm.

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d)

I. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=-\frac{(12x+5y-9)}{5x+2y-4}\)
Solution:
A non-homogenous equation
\(\frac{dy}{dx}=-\frac{(ax+by-9)}{a’x+b’y-c’}\) where b = -a’
b = -5, a = 5 ⇒ b = -a
(5x + 2y-4)dy = -(12x + 5y-9) dx
(5x + 2y – 4)dy + (12x + 5y – 9) dx = 0
5 (x dy + y dx) + 2y dy – 4 dy + 12x dx – 9 dx = 0
integrating 5xy + y² – 4y + 6x² – 9x = c

Question 2.
\(\frac{dy}{dx}=-\frac{-3x-2y+5}{2x+3y+5}\)
Solution:
b = – 2, a = 2 ⇒ b = -a
(2x + 3y + 5) dy = (- 3x – 2y + 5) dx
2x dy + 3y dy + 5 dy = -3x dx- 2y dx + 5 dx
2(x.dy + y dx) + By dy + 3x dx + 5 dy – 5 dx = 0
Integrating
2xy + \(\frac{3}{2}\)y² + \(\frac{3}{2}\)x² + 5y – 5x = c
4xy + 3y² + 3x² – 10x + 10y = 2c = c’
Solution is
4xy + 3(x² + y²)- 10(x – y) = c

Question 3.
\(\frac{dy}{dx}=\frac{-3x-2y+5}{2x+3y-5}\)
Solution:
\(\frac{dy}{dx}=\frac{-(3x-2y+5)}{2x+3y-5}\)
Here b = – 2, a¹ = 2
∵ b = -a¹
(2x + 3y – 5) dy = (-3x – 2y + 5) dx „
⇒ 2(x dy + y dx) + (3y – 5) dy + (3x – 5) dx – 0
⇒ 2d (xy) + (3y- 5) dy + (3x- 5) dx = 0
Now integrating term by term, we get
⇒ 2 ∫d (xy) + ∫(3y – 5)dy + ∫(3x – 5)dx = 0
⇒ 2xy + 3.\(\frac{y^2}{2}\) – 5y + 3\(\frac{x^2}{2}\) – 5x = \(\frac{c}{2}\)
or) 3x² + 4xy + 3y² – 10x – 10y = c
Which is the required solution.

Question 4.
2(x – 3y + 1) \(\frac{dy}{dx}\) = 4x – 2y + 1
Solution:
(2x – 6y + 2) dy = (4x – 2y + 1) dx
(2x – 6y + 2) dy – (4x – 2y + 1) dx = 0
2 (x dy + y dx) – 6y dy + 2 dy – 4x dx – dx = 0
Integrating
2xy – 3y² – 2x² + 2y – x = c

Question 5.
\(\frac{dy}{dx}=\frac{x-y+2}{x+y-1}\)
Solution:
b = -1, a’ = 1 ⇒ b = -a’
(x + y – 1) dy = (x – y + 2) dx
(x + y – 1) dy = (x – y + 2) dx = 0
(x dy + y dx) + y dy – dy – x dx – 2 dx = 0
integrating
xy + \(\frac{y^2}{2}\) – \(\frac{x^2}{2}\) – y – 2x = c
2xy + y² – x² – 2y – 4x = 2c = c’

Question 6.
\(\frac{dy}{dx}=\frac{2x-y+1}{x+2y-3}\)
Solution:
b = -1, a = 1 ⇒ b = -a’
(x + 2y – 3) dy = (2x – y + 1) dx
(x + 2y – 3) dy – (2x – y + 1) dx = 0
(x dy + y dx) 4- 2y dy – 3 dy – 2x dx – dx = 0
Integrating
xy + y² – x² – 3y – x = c

II. Solve the following differential equations.

Question 1.
(2x + 2y + 3) \(\frac{dy}{dx}\) = x + y + 1
Solution:
\(\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}\)

Multiplying with 9
6v + log (3v + 4) = 9x + 9c
6(x + y) + log [3(x + y) + 4] = 9x + c
i.e., log (3x + 3y + 4) = 3x – 6y + c

Question 2.
\(\frac{dy}{dx}=\frac{4x+6y+5}{3y+2x+4}\)
Solution:

Multiplying with 64
8v + 9log (8v + 23) = 64x + 64c
8 (2x + 3y) – 64x + 9 log (16x + 24y + 23) = c’
Dividing with 8
2x + 3y – 8x + \(\frac{9}{8}\) log (16x + 24y + 23) = c”
3y – 6x + \(\frac{9}{8}\) log (16x + 24y + 23) = c”
Dividing with 3, solution is 3
y – 2x + \(\frac{3}{8}\) log (16x + 24y + 23) = k

Question 3.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:

∫(2 + \(\frac{1}{v-1}\))dv = 3∫dx
2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c

Question 4.
\(\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+3}\)
Solution:

Multiplying with 8
4v + log (4v + 5) = 8x + 8c
4(x + 2y) – 8x + log [4(x + 2y) + 5] = c’
Solution is
4x + 8y – 8x + log (4x + 8y + 5) = c’
8y – 4x + log (4x + 8y + 5) = c’

Question 5.
(x + y – 1) dy = (x + y + 1)dx
Solution:

v – log v = 2x + c
x + y – log (x + y) = 2x – c
(x – y) + log (x + y) = c is the required
solution.

III. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{3y-7x+7}{3x-7y-3}\)
Solution:
Let x = x + h, y = y + k so that \(\frac{dy}{dx}=\frac{dy}{dx}\)


3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10ln (v + 1) – 4 ln (v – 1)
ln (v + 1)⁵ + ln (v – 1)² + ln x⁷ = ln c
(v +1)⁵. (v – 1)². x⁷ = c
(\(\frac{y}{x}\) + 1)⁵ (\(\frac{y}{x}\) – 1)².x⁷ = c
(y – x)² (y + x)⁵ = c
[y – (x – 1 )]² (y + x – 1 )⁵ = c
Solution is [y-x + 1 ]² (y + x – 1)⁵ = c.

Question 2.
\(\frac{dy}{dx}=\frac{6x+5y-7}{2x+18y-14}\)
Solution:



Multiplying with (3V – 2)(2V + 1)
2 + 18V = A(2V + 1) + B(3V – 2)

2 log (3V- 2)+ log (2V+ 1) = – 3 log X + log c
log (3V – 2)².(2V + 1) + log X³ = log c
log X³(3V – 2)² (2V + 1) = log c
x³(3V – 2)² (2V + 1) = c

Solution is (3y – 2x – 1)² (x + 2y – 2) = 343c = c”.

Question 3.
\(\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}\) = 0
Solution:
\(\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}\) = 0
x = X + h, y = Y + k


5V + 7 = A(V + 2) + B (V + 1)
V = -1 ⇒ 2 = A(-1 + 2) = A ⇒ A = 2
V = -2 ⇒ -3 = B(-2 + 1) = -B, B = 3
∫(\(\frac{2}{(V+1)}+\frac{3}{(V+2)}\))dv = ∫\(\frac{dx}{X}\)
2 log (V + 1) + 3 log (V + 2) = – 5 log X + c
c = 2 log (V + 1) + 3 log (V + 2) + 5 log X
= log (V + 1)². (V + 2)³. X⁵
= log(\(\frac{2}{(V+1)})\))².(\(\frac{3}{(V+2)}\))³. X⁵
= log\(\frac{(Y+X)^2}{X^2}\) \(\frac{(Y+2X)^3}{X^3}\) . X⁵
⇒ (Y + X)² . (Y + 2X)³ = e^c = c’
(Y + 1 – X – 2)² (Y + 1 – 2x – 4)³ = c
Solution is (x + y – 1)² (2x + y – 3)³ = c.

Question 4.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation is \(\frac{dy}{dx}=\frac{-x+y+2}{x-2y-3}\)
Let x = X + h, y = Y + k




is the required solution.

Question 5.
(x – y) dy = (x + y + 1) dx
Solution:

Question 6.
(2x + 3y – 8) dx = (x + y – 3) dy
Solution:

Question 7.
\(\frac{dy}{dx}=\frac{x+2y+3}{2x+3y+4}\)
Solution:
Let x = X + h, y = Y + k so that \(\frac{dY}{dX}=\frac{dy}{dx}\)

Choose h and k so that
h + 2k + 3 = 0
2h + 3k + 4 = 0

This is a homogeneous equation

Question 8.
\(\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}\)
Solution:
Given equation is \(\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}\)
Let x = X + h, y = Y + k so that \(\frac{dY}{dX}=\frac{dy}{dx}\)

∴ \(\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}\)
This is a homogeneous equation

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(b)

I.

Question 1.
Find the general solution of \(\sqrt{1-x^2}\) dy + \(\sqrt{1-y^2}\) dx = 0.
Solution:
Given differential equation is

sin^-1 y = – sin^-1 x + c
Solution is sin^-1 x + sin^-1 y = c, where c is a constant.

Question 2.
Find the general solution of \(\frac{dy}{dx}=\frac{2y}{x}\).
Solution:
\(\frac{dy}{dx}=\frac{2y}{x}\)
∫\(\frac{dy}{dx}\) = 2∫\(\frac{2y}{x}\)
log c + log y = 2 log x
log cy = log x²
Solution is cy = x², where c. is a constant.

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
Solution:
\(\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\)
∫\(\frac{dy}{1+y^2}\) = ∫\(\frac{dx}{1+x^2}\)
tan^-1 y = tan^-1 x + tan^-1c where c is a constant.

Question 2.
\(\frac{dy}{dx}\) = e^y-k
Solution:
\(\frac{dy}{dx}=\frac{e^y}{e^x}\)
\(\frac{dy}{e^y}=\frac{dx}{e^x}\)
∫e^-xdx = ∫e^-ydy
-e^-x = -e^-y + C
e^-y = e^-x + c where c is a constant.

Question 3.
(e^x + 1) y dy + (y + 1) dx = 0
Solution:
(e^x + 1 )y. dy = – (y + 1) dx

y – log (y + 1) = log (e^-x + 1) + log c
⇒ y – log (y + 1) = log c (e^-x + 1)
⇒ y = log (y + 1) + log c (e^-x + 1)
y = log c (y + 1) (e^-x +1)
Solution is
e^y = c(y + 1) (e^-x +1)

Question 4.
\(\frac{dy}{dx}\) = e^x-y + x²e^-y
Solution:
\(\frac{dy}{dx}\) = e^x-y + x² . e^-y
= \(\frac{e^x}{e^y}=\frac{x^2}{e^y}\)
∫e^y . dy = ∫(e^x + x²) dx
Solution is
e^y = e^x + \(\frac{x^3}{3}\) + c

Question 5.
tan y dx + tan x dy = 0
Solution:
tan y dx = – tan x dy

log sin x = – log sin y + log c
log sin x + log sin y = log c
log (sin x . sin y) = log c
⇒ sin x . sin y = c is the solution

Question 6.
\(\sqrt{1+x^2}\)dx + \(\sqrt{1+y^2}\)dy = 0
Solution:
\(\sqrt{1+x^2}\)dx = –\(\sqrt{1+y^2}\)dy
Integrating both sides we get
∫\(\sqrt{1+x^2}\)dx = -∫\(\sqrt{1+y^2}\)dy
Integrating both sides we get

Question 7.
y – x\(\frac{dy}{dx}\) = 5(y² + \(\frac{dy}{dx}\))
Solution:
y – 5y² = (x + 5)\(\frac{dy}{dx}\)
\(\frac{dx}{x+5}=\frac{dy}{y(1-5y}\)
Integrating both sides

Question 8.
\(\frac{dy}{dx}=\frac{xy+y}{xy+x}\)
Solution:

III. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{1+y^2}{(1+x^2)xy}\)
Solution:

log (1 + y²) = log x² – log (1 + x²) + log c
log (1 + x²) + log (1 + y²) = log x² + log c
Solution is (1 + x²) (1 + y²) = cx²

Question 2.
\(\frac{dy}{dx}\) + x² = x² e^3y
Solution:

log(1 – e^-3y) = x³ + c'(c’ = 3c)
Solution is
1 – e^-3y = e^x³ . k(k = e^c’)

Question 3.
(xy² + x)dx+(yx²+y)dy = 0.
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
\(\frac{x dx}{1+x^2}+\frac{y dy}{1+y^2}\) = 0
Integrating
∫\(\frac{x dx}{1+x^2}\) + ∫\(\frac{y dy}{1+y^2}\) = 0
\(\frac{1}{2}\) [(log (1 + x²) + log (1 + y²)] = log c
log (1 + x²) (1 + y²) = 2 log c = log c²
Solution is (1 + x²) (1 + y²) = k when k = c².

Question 4.
\(\frac{dy}{dx}\) = 2y tanh x
Solution:
\(\frac{dy}{dx}\) = 2y tanh x
\(\frac{dy}{y}\) = 2 tanh x dx
Integrating both sides we get
∫\(\frac{dy}{y}\) = 2 ∫ tanh x dx
log y = 2 log |cosh x| + log c
lny = 2ln cosh x + In c
y = c cos²h x

Question 5.
sin^-1 \(\frac{dy}{dx}\) = x + y
Solution:
\(\frac{dy}{dx}\) = sin(x + y)
x + y = t
1 + \(\frac{dy}{dx}=\frac{dt}{dx}\)
\(\frac{dt}{dx}\) – 1 = sin t
\(\frac{dt}{dx}\) = 1 + sin t
Integrating both sides we get
∫\(\frac{dt}{1+\sin t}\) = ∫dx
∫\(\frac{1-\sin t}{\cos^2 t}\) dt = x + c
∫sec² t dt – ∫tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 6.
\(\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}\) = 0
Solution:
\(\frac{-dy}{y^2+y+1}=\frac{dx}{x^2+x+1}\)
Integrating both sides dy

Question 7.
\(\frac{dy}{dx}\) = tan² (x + y)
Solution:
\(\frac{dy}{dx}\) = tan² (x + y)
put v = x + y

2v + sin 2v = 4x + c’
2(x + y) + sin 2(x + y) = 4x + c’
x – y – \(\frac{1}{2}\)sin [2(x + y)] = c

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a)

I.

Question 1.
Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = ce^x – be^-x + x².
Solution:
Given equation is xy = ce^x – be^-x + x²
Differentiating w.r.to x, we get
xy₁ + y = ce^x – be^-x + 2x.
Again differentiating w.r.to x, we get
xy₂ + y₁ + y₁ = ce^x – be^-x + 2
xy₂ + 2y₂ = xy – x² + 2
Arbitary constants a and b are eliminated.
∴ The order is 2.

Question 2.
Find the order of the differential equation of the family of all circles with their centres at the origin.
Solution:
Equation of the circle with centre at origin is x² + y² = r²
Order = no .of arbitrary constants = 1

II.

Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets.
i) y = c(x – c)² ; (c)
Solution:
y = c(x – c)² ………….. (1)
Differentiating w.r. to x
y₁ = c. 2(x – c) ………….. (1)
Dividing (2) by (1)
\(\frac{y_2}{y}=\frac{2c(x-c)}{c(x-c)^2}\)
x – c = \(\frac{2y}{y_1}\)
c = x – \(\frac{2y}{y_1}\)
Substituting in (1)
y = x – \(\frac{2y}{y_1}\)(x – \(\frac{2y}{y_1}\))²
= \(\frac{xy_1-2y}{y_1}.\frac{4y^2}{y_1^2}\)
y.y³₁ = 4y²(xy₁ – 2y)
i.e., y³₁ = 4y (xy₁ – 2y)
= 4xyy₁ – 8y²
(\(\frac{dy}{dx}\))³ – 4xy\(\frac{dy}{dx}\) + 8y² = 0

ii) xy = ae^x + be^-x; (a, b)
Solution:
xy = ae^x + b.e^-x
Differentiating w.r.t. x
x . y₁ + y = ae^x – b . e^-x
Differentiating again w.r.t. x
xy₂ + y₁ + y₁ = ae^x + be^-x = xy
\(\frac{d^2y}{dx^2}\) + 2\(\frac{dy}{dx}\) – xy = 0

iii) y = (a + bx)e^kx ; (a, b)
Solution:
y = (a + bx)e^kx
Differentiating w.r.t. x
y₁ = (a + bx) e^kx. k + e^kx . b
= k . y + b.e^kx
y₁ – ky = b.e^kx …………. (1)
Differentiating again w.r.t. x
y₂ – ky₁ = kb e^kx
= k(y₁ – ky) ………… (2)
= ky₁ – k²y
\(\frac{d^2y}{dx^2}\) – 2k\(\frac{dy}{dx}\) + k²y = 0

iv) y = a cos (nx + b); (a, b)
Solution:
y = a cos (nx + b)
y₁ = – a sin (nx + b) n
y₂ = – an. cos (nx + b) n
= – n² . y
\(\frac{d^2y}{dx^2}\)+n².y = 0

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
i) The rectangular hyperbolas which have the co-ordinate axes as asymptotes.
Solution:
Equation of the rectangular hyperbolas is xy = c² where c is arbitrary constant
Differentiating w.r.t. x
x\(\frac{dy}{dx}\) + y = 0

ii) The ellipses with centres at the origin and having co-ordinate axes as axes.
Solution:
Equation of ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Differentiating w.r.to ‘x’ we get

Multiply (ii) by x and subtract from (i)

III.

Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets :
i) y = ae^3x + be^3x; (a, b)
Solution:
Differentiating w.r. to x
y₁ = 3ae^3x + 4be^4x
y₁ – 3a. e^3x = 4b.e^4x
= 4(y – a. e^3x)
= 4y – 4a. e^3x
y₁ – 4y = – a.e^3x ………… (1)
Differentiating again w.r.t. x
y₂ – 4y₁ = – 3a. e^3x
= 3 (y₁ – 4y) by (1)
= 3y₁ – 12y
\(\frac{d^2y}{dx^2}\) – 7\(\frac{dy}{dx}\) + 12y = 0

ii) y = ax² + bx; (a, b)
Solution:

Adding all three equations we get
x²\(\frac{d^2y}{dx^2}\) – 2x\(\frac{dy}{dx}\) + 2y = 0

iii) ax² + by² = 1; (a, b)
Solution:
ax² + by² = 1
by² = 1 – ax² ………….. (1)
Differentiating w.r.t. x
2by. y₁ = – 2ax ………….. (2)
Dividing (2) by (1)
\(\frac{by.y_1}{by^2}=\frac{-ax}{1-ax^2}\)

Differentiating w.r.t. x

iv) xy = ax² + \(\frac{b}{x}\); (a, b)
Solution:
xy = ax² + \(\frac{b}{x}\)
x²y = ax³ + b
Differentiating w.r.t. x
x²y₁ + 2xy = 3ax²
Dividing with x
xy₁ + 2y = 3ax ………… (1)
Differentiating w.r.t. x
xy₂ + y₁ + 2y₁ = 3a
xy₂ + 3y₁ = 3a ………… (2)
Dividing (1) by (2)
\(\frac{xy_1+2y}{xy_2+3y_1}=\frac{3ax}{3a}=x\)
Cross multiplying
xy₁ + 2y = x²y₂ + 3xy
x²y₂ + 2xy₁ – 2y = 0
x²\(\frac{d^2y}{dx^2}\) + 2x\(\frac{dy}{dx}\) – 2y = 0

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
i) The circles which touch the Y – axis at the origin.
Solution:
Equation of the required circle is
x² + y² + 2gx = 0
x² + y² = – 2gx …………. (1)
Differentiating w.r. t x
2x + 2yy₁ = – 2g ……….. (2)
Substituting in (1)
x² + y² = x(2x + 2yy₁) by (2)
= 2x² + 2xyy₁
yy² – 2xyy₁ – 2x² = 0
y² – x² = 2xy\(\frac{dy}{dx}\)

ii) The parabolas each of which has a latus rectum 4a and whose axes are parallel to X – axis.
Solution:

Equation of the required parabola is
(y – k)² = 4a (x – h) …………. (1)
Differentiating w.r.t. x
2(y – k)y₁ = 4a …………. (2)
Differentiating w.r.t. x
(y – k) y₂ + y²₁ = 0 …………. (3)
From (2), y – k = \(\frac{2a}{y_1}\)
Substituting in (3)
\(\frac{2a}{y_1}\).y₂ = y²₁ = 0
2ay₂ + y³₁ = 0

iii) The parabolas have their foci at the origin and axis along the X – axis.
Solution:
Equation of parabola be y² = 4a(x + a)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(c)

I. Evaluate the following definite integrals.

Question 1.
\(\int_{\pi/2}^{\pi/2}\)sin¹⁰ x dx
Solution:

Question 2.
\(\int_0^{\pi/2}\)cos¹¹ x dx
Solution:

Question 3.
\(\int_0^{\pi/2}\)cos⁷ x . sin²x dx.
Solution:

Question 4.
\(\int_0^{\pi/2}\)sin⁴ x . cos⁴ x dx.
Solution:

Question 5.
\(\int_0^{\pi/2}\)sin³ x cos⁶ x dx.
Solution:
\(\int_0^{\pi/2}\)sin³ x cos⁶ x dx.
\(\int_0^{\pi/2}\)(1 – cos² x) cos⁶ x.sin x dx

Question 6.
\(\int_0^{2\pi}\)sin² x cos⁴ x dx.
Solution:

Question 7.
\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos⁷ θ dθ.
Solution:
sin² θ cos⁷ θ is even function
f(θ) = sin² θ . cos⁷ θ dθ
f(-θ) = sin² (-θ) . cos⁷ (-θ)
= f(θ)
= 2\(\int_{-\pi/2}^{\pi/2}\)sin² θ cos⁷ θ dθ

\(\int_0^{\pi/2}\)sin^m x cosⁿx dx
n is odd n = 7

Question 8.
\(\int_{-\pi/2}^{\pi/2}\)sin³ θ .cos³ θ dθ.
Solution:
f(θ) = sin³ θ . cos³ θ dθ
f(-θ) = sin³ (-θ) . cos³ (-θ)
= -sin³ θ cos³ θ = -f(θ)
f(θ) is odd
∴ \(\int_{-\pi/2}^{\pi/2}\)sin³ θ.cos³ θ dθ = 0

Question 9.
\(\int_0^a\)x(a² – x²)^7/2 dx
Solution:
x = a sin θ, a = a sin θ
dx = a cos θ dθ, θ = π/2
= \(\int_0^{\pi/2}\)a sin θ(a² – a²sin²θ)^7/2 a cos θ dθ
= \(\int_0^{\pi/2}\)a⁹ cos⁸ θ sin θ dθ
= a⁹\(\int_0^{\pi/2}\)cos⁸ . θ sin θ dθ

Question 10.
\(\int_0^2\)x^3/2.\(\sqrt{2-x}\)dx
Solution:
x = 2 cos² θ
dx = 4 cos θ sin θ dθ

II. Evaluate the following integrals.

Question 1.
\(\int_0^1\)x⁵(1 – x)^3/2 dx
Solution:
x = sin² θ
dx = 2 sin θ . cos θ . dθ

Question 2.
\(\int_0^4\)(16 – x²)^5/2 dx
Solution:

Question 3.
\(\int_{-3}^3\)(9 – x²)^3/2 x dx
Solution:
Let f(x) = (9 – x²)^3/2_x
f(x) = (9 – (-x²))^3/2(-x)
= (9 – x²)^3/2 . x
= -f(x)
∴ f is odd function
∴ \(\int_{-3}^3\)(9 – x²)^3/2 x dx = 0

Question 4.
\(\int_0^5\)x³(25 + x²)^7/2 dx
Solution:
Let I = \(\int_0^5\)x³(25 + x²)^7/2 dx
Put x = 5 sin θ
dx = 5 cosθ dθ

Question 5.
\(\int_{-\pi}^{\pi}\)sin⁸ x cos⁷ x dx
Solution:
Let f(x) = sin⁸ x. cos⁷ x
f(-x) = sin⁸ (-x) . cos⁷ (-x)
= sin⁸ x. cos⁷ x
∴ f is even function.
∴ \(\int_{-\pi}^{\pi}\)sin⁸ x cos⁷ x dx = 2\(\int_0^{\pi}\)sin⁸ x cos⁷ x = 0

Question 6.
\(\int_3^7 \sqrt{\frac{7-x}{x-3}}\)dx
Solution:
Put x = 3 cos²θ + 7 sin²θ
dx = (7 – 3)sin2θ dθ
dx = 4 sin 2θ dθ
U.L.
x = 3 cos²θ + 7 sin²θ
7 = 3 cos²θ + 7 sin²θ
4 cos²θ = 0
θ = \(\frac{\pi}{2}\)
L.L
x = 3 cos²θ + 7 sin²θ
3 = 3 sin²θ + 7 sin²θ
4 sin²θ = 0
sinθ = 0
θ = 0
7 – x = 7 – (3 cos²θ + 7 sin²θ)
= (7 – 3)cos²θ
= 4 cos²θ
x – 3 = 3 cos²θ + 7 sin²θ – 3
= (7 – 3)sin²θ
= 4 sin²θ

Question 7.
\(\int_2^6\sqrt{(6-x)(x-2)}\)dx
Solution:
Put x = 2 cos²θ + 6 sin²θ
dx = (6 – 2) sin2θ dθ
dx = 4 sin2θ dθ
U.L
x = 2 cos²θ + 6 sin²θ
6 = 2 cos²θ + 6 sin²θ
4 cos²θ = 0
cos θ = 0
θ = \(\frac{\pi}{2}\)

L.L
x = 2 cos²θ + 6 sin²θ
2 = 2 cos²θ + 6 sin²θ
4 sin²θ = 0
θ = 0
6 – x = 6 – (2 cos²θ + 6 sin²θ)
= (6 – 2) cos²θ
= 4 cos²θ
x – 2 = 2 cos²θ + 6 sin²θ – 2
= (6 – 2)sin²θ
= 4 sin²θ

Question 8.
\(\int_0^{\pi}\)tan⁵x cos⁸x dx
Solution:

III. Evaluate the following integrals.

Question 1.
\(\int_0^1\)x^7/2 (1 – x)^5/2 dx
Solution:
Put x = sin²θ
dx = 2 sin θ cos θ dθ
U.L
x = sin²θ
1 = sin²θ
θ = \(\frac{\pi}{2}\)

L.L
x = sin²θ
0 = sin²θ
θ = 0

Question 2.
\(\int_0^{\pi}\)(1 + cos x)³ dx
Solution:

Question 3.
\(\int_4^9\frac{dx}{\sqrt{(9 – x)(x – 4)}}\)
Solution:
Put x = 4 cos²θ + 9 sin²θ
dx = (9 – 4)sin2θ dθ
dx = 5 sin2θ dθ

U.L
x = 4 cos²θ + 9 sin²θ
9 = 4 cos²θ + 9 sin²θ
5 cos²θ = 0
θ = \(\frac{\pi}{2}\)

L.L
x = 4 cos²θ + 9 sin²θ
4 = 4 cos²θ + 9 sin²θ
5 sin²θ = 0
θ = 0

9 – x = 9 – (4 cos²θ + 9 sin²θ)
= (9 – 4) cos²θ
= 5 cos²θ

x – 4 = 4 cos²θ + 9 sin²θ – 4
= (9 – 4) sin²θ
= 5 sin²θ

Question 4.
\(\int_0^5\)x²(\(\sqrt{5-x}^7\) dx
Solution:
Put x = 5 sin²θ
dx = 10 sinθ cosθ dθ

U.L
x = 5 sin²θ
5 = 5 sin²θ
sin²θ = 1
θ = \(\frac{\pi}{2}\)

L.L
x = 5 sin²θ
0 = 5sin²θ
sin²θ = 0
θ = 0

Question 5.
\(\int_0^{2\pi}\)(1 + cos x)⁵(1 – cos x)³ dx.
Solution:
\(\int_0^{2\pi}\)(1 + cos x)⁵(1 – cos x)³ dx . (1 + cos x)³(1 + cos x)²(1 – cos x)³

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(b)

I. Evaluate the following definite integrals.

Question 1.
\(\int_0^a(a^2x-x^3) d x\)
Solution:

Question 2.
\(\int_2^3 \frac{2 x}{1+x^2} d x\)
Solution:
I = [ln|1 + x²|]³₂
= ln 10 – ln 5
= ln(10/5)
= ln 2

Question 3.
\(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\)
Solution:

Question 4.
\(\int_0^\pi\sin^3x\cos^3xd x\)
Solution:

Question 5.
\(\int_0^2|1-x|d x\)
Solution:

Question 6.
\(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
Solution:
\(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x dx}{1+e^x}\) ………….. (i)
cos x is even function
e^x is neither even nor odd.

Adding (i) and (ii) we get

Question 7.
\(\int_0^1\frac{dx}{\sqrt{3-2x}}\)
Solution:
3 – 2x = t²
-2dx = 2t dt
dx = -t dt
3 – (2.1) = t²
1 = t²
3 – 2.0 = t²

Question 8.
\(\int_0^a(\sqrt{a}-\sqrt{x})^2\)
Solution:

Question 9.
\(\int_0^{\pi / 4} \sec^4\theta d\theta\)
Solution:
Let ∫sec⁴ θ dθ = \(\int_0^{\pi / 4} \sec^2\theta(1+\tan^2\theta)d\theta\)
Put tan θ = y
sec² θ dθ = dy
θ = \(\frac{\pi}{4}\) ⇒ y= 1
θ = 0 ⇒ y = 0
I = \(\int_0^a\)(1 + y²)dy = [y + \(\frac{y^3}{3}\)]¹₀
= 1 + \(\frac{1}{3}=\frac{4}{3}\)

Question 10.
\(\int_0^3\frac{x}{\sqrt{x^2+16}}\)
Solution:

Question 11.
I = \(\int_0^1\)x.e^-x²dx
Solution:
I = \(\int_0^a\)x.e^-x²dx
⇒ -x² = t
⇒ -2x dx = dt
2x dx = -dt
x = 1 ⇒ t = 0
x = 0 ⇒ t = 1
I = \(\frac{1}{2}\)\(\int_0^a\)-e^t dt
= \(\frac{1}{2}\)[-e^t]^-1₀
= \(\frac{1}{2}\)[e⁰ – e^-1]
= \(\frac{1}{2}\)(1 – \(\frac{1}{e}\))

Question 12.
I = \(\int_1^5\frac{dx}{\sqrt{2x-1}}\)
Solution:

II. Evaluate the following integrals:

Question 1.
I = \(\int_0^4\frac{x^2}{1+x}\)
Solution:

Question 2.
\(\int_{-1}^2 \frac{x^2}{x^2+2}\)
Solution:

Question 3.
I = \(\int_0^1 \frac{x^2}{x^2+2}\)
Solution:

Question 4.
\(\int_0^{\pi / 2}\)x² sin x dx
Solution:

Question 5.
\(\int_0^4\)|2-x|dx
Solution:

Question 6.
\(\int_0^{\pi / 2}\frac{\sin^5 x}{\sin^5 x+\cos^5 x}\)
Solution:

Question 7.
\(\int_0^{\pi / 2}\frac{\sin^2 x-\cos^2 x}{\sin^3 x+\cos^3 x}\)dx
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

III. Evaluate the following integrals :

Question 1.
\(\int_0^{\pi / 2}\frac{dx}{4+5\cos x}\)
Solution:

Question 2.
\(\int_a^b\sqrt{(x-a)(b-x)}\)dx
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.
\(\int_0^a\)x(a – x)ⁿdx
Solution:
I = \(\int_0^a\)x(a – x)ⁿdx ………… (i)
\(\int_0^a\)f(x) dx = \(\int_0^a\)f(a – x)dx
I = \(\int_0^a\)(a – x).(x)ⁿdx ………… (ii)
I = \(\int_0^a\)axⁿ dx – xⁿ⁺¹ dx

Question 7.
\(\int_0^2x\sqrt{2-x}\)dx
Solution:
I = \(\int_0^2x.\sqrt{2-x}\)dx
\(\int_0^2\)f(x)dx = \(\int_0^2\)f(a – x)dx
= \(\int_0^2\)(2 – x).√x dx
= \(\int_0^2\)((2√x – x√x)) dx

Question 8.
\(\int_0^{\pi}\)x sin³ x dx
Solution:

Question 9.
\(\int_0^{\pi}\frac{x}{1+\sin x}\)dx
Solution:
\(\int_0^{\pi}\frac{x}{1+\sin x}\)dx …………… (i)

Question 10.

Solution:

Question 11.
\(\int_0^1\frac{log(1+x)}{1+x^2}\)dx
Solution:
Put x = tan θ
dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = x ⇒ θ = \(\frac{\pi}{4}\)

Question 12.
\(\int_0^{\pi}\frac{x \sin x}{1+\cos^2 x}\)dx
Solution:

Question 13.
\(\int_0^{\pi/2}\frac{\sin^2 x}{\cos x+\sin x}\)dx
Solution:

Question 14.
\(\int_0^{\pi}\frac{1}{3+2\cos x}\)dx
Solution:

Question 15.
\(\int_0^{\pi/4}\)log(1 + tan x)dx
Solution:

Question 16.
\(\int_{-1}^{3/2}\)|x sin πx| dx
Solution:
We know that |x. sin πx| = x . sin πx
where -1 ≤ x ≤ 1
and |x . sin πx| = – x.sin πx where 1 < x ≤ 3/2

Question 17.
\(\int_0^1sin^{-1}\frac{2x}{1+x^2}\)dx
Solution:
\(\int_0^1sin^{-1}\frac{2x}{1+x^2}\)dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0
x = 1 ⇒ θ π/4

Question 18.
\(\int_0^1\)x tan^-1x dx
Solution:
\(\int_0^1\)x. tan^-1x dx
Put x = tan θ ⇒ dx = sec² θ dθ
x = 0 ⇒ θ = 0;

Question 19.
\(\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}\)dx
Solution:

Question 20.
Suppose that f : R → R is continuous periodic function and T is the period of it. Let a ∈ R. Then prove that for any positive integer n,

Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(a)

I. Evaluate the following integrals as the limit of a sum.

Question 1.
\(\int_0^5(x+1) d x\)
Solution:

Question 2.
\(\int_0^4(x^2+1) d x\)
Solution:

II. Evaluate the following integrals as limit of a sum.

Question 1.
\(\int_0^4(x+e^{2x}) d x\)
Solution:

Question 2.
\(\int_0^1(x-x^2) d x\)
Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(f) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(f)

I. Evaluate the following integrals.

Question 1.
∫e^x(1 + x²) dx
Solution:
∫e^x(1 + x²) dx = ∫e^x + ∫x² . e^x dx
= e^x + (x².e^x – 2 ∫x.e^xdx)
= e^x + x².e^x – 2(x.e^x – ∫e^x dx
= e^x + x².e^x – 2x. e^x + 2e^x + C
= e^x(x² – 2x + 3) + C

Question 2.
∫x² e^-3x dx
Solution:

Question 3.
∫x³ e^ax dx
Solution:

II.

Question 1.
Show that
∫xⁿ.e^-x dx = -xⁿ e^-x + n∫x^n-1.e^-x dx
Solution:
∫xⁿ.e^-x dx = \(\frac{x^ne^{-x}}{(-1)}\) + ∫e^-x. nx^n-1 dx
= -xⁿ . e^-x + n∫x^n-1e^-x dx

Question 2.
If I_n = ∫cosⁿ x dx, than show that
I_n = \(\frac{1}{n}\) cos^n-1 x sin x + \(\frac{n-1}{n}\) I_n-2.
Solution:
I_n = ∫cosⁿ x dx = ∫cos^n-1 x. cos x dx
= cos^n-1x . sin x – ∫sin x.(n – 1)cos^n-2x (-sin x)dx
= cos^n-1x . sin x + (n – 1) ∫cos^n-2x(1 – cos² x)dx
= cos^n-1x . sin x + (n – 1)I_n-2 – (n – 1) I_n
∴ I_n(1 + n – 1) = cos^n-1x. sin x + (n – 1) I_n-2
I_n = \(\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n}\)I_n-2

III.

Question 1.
Obtain reduction formula for I_n = ∫cotⁿ x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cot⁴ x dx.
Solution:
I_n = ∫cotⁿ dx = ∫cot^n-2 x . cot² x dx
= ∫cot^n-2 x . (cosec² x – 1) dx
= ∫cot^n-2 x . cosec² x dx – I_n-2
= – \(\frac{cot^{n-1}x}{n-1}\) – I_n-2
n = 4 ⇒ I₄ = –\(\frac{cot^{3}x}{3}\) – I₂
n = 2 ⇒ I₂ = – cot x – I₀ where I₀ = ∫dx = x
I₂ = – cot x – x
I₄ = –\(\frac{cot^{3}x}{3}\) – (-cot x – x) + C
= –\(\frac{cot^{3}x}{3}\) + cot x + x + C

Question 2.
Obtain the reduction formula for I_n = ∫cosecⁿ x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec⁵ x dx.
Solution:
I_n = ∫cosecⁿ x dx
= cosec^n-2 x(-cot x) + ∫cot x. (n – 2)cosec^n-3 x. (cot x)dx
= -cosec^n-2x. cot x + (n – 2)∫cosec^n-2x. (cosec² x – 1)dx

Question 3.
If I_{m, n} = ∫sin^m xcosⁿ xdx, then show that
for a positive integer m ≥ 2.
Solution:
I_{m, n} = ∫(sin^m x) (cosⁿ x) dx
= ∫sin^m-1 (x).(cos x)ⁿ. sin x dx
= ∫sin^m-1 (x)(cos x)ⁿ(-sin x) dx

Question 4.
Evaluate ∫sin⁵ x cos⁴ x dx
Solution:
Reduction formula
I_{m, n} = \(\frac{-\sin ^{m-1} x \cdot \cos ^{n+1} x}{m+n}+\frac{m-1}{m+n}\).I_{m-2, n}

Question 5.
If I_n = ∫(log x)ⁿ dx, then show that I_n = x(log x)ⁿ – nI_n-1, and hence find
∫(log x)⁴ dx.
Solution:
I_n = ∫(log x)ⁿ dx
= (log x)ⁿ. x – ∫x . n . (log x)^n-1 . \(\frac{1}{x}\)dx
= x.(log x)ⁿ – n∫(log x)^n-1 dx
= x(log x)ⁿ – n . I_n-1
I₄ = x(log x)⁴ – 4 . I₃
I₃ = x(log x)³ – 3 . I₂
I₂ = x(log x)² – 2 . I₁
I₁ = x log x I₀ where I₀ = ∫dx = x
I₁ = x log x – x
I₂ = (x (log x)² – 2x log x = 2x
I₃ = x(log x)³ – 3(x (log x)² – 2x log x + 2x)
= x . (log x)³ – 3x (log x)² + 6x(log x) – 6x
I₄ = x(log x)⁴ – 4[x . (log x)³ – 3x(log x)² + 6x(log x) -6x] + C
= x[(log x)⁴ – 4(log x)³ + 12(log x)² – 24(log x) + 24] + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e^2x dx, x ∈ R.
Solution:
∫e^2x dx = \(\frac{e^{2x}}{2}\) + C

Question 2.
∫sin 7x dx, x ∈ R.
Solution:
∫sin 7x dx = \(\frac{\cos 7x}{7}\) + C

Question 3.
∫\(\frac{x}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{xdx}{1+x^2}\) dx = \(\frac{1}{2}\)∫\(\frac{2xdx}{1+x^2}\) = \(\frac{1}{2}\) log(1+x²) + C

Question 4.
∫2xsin(x²+1) dx, x ∈ R.
Solution:
∫2x.sin(x²+1) dx, x ∈ R.
t = x² + 1 ⇒ dt = 2x dx
∫2x. sin(x²+1) dx = ∫ sin t dt = -cos t + C
= -cos (x²+1) + C

Question 5.
∫\(\frac{(logx)^2}{x}\) dx on I ⊂ (0, ∞).
Solution:
∫\(\frac{(logx)^2}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{(logx)^2}{x}\) dx = ∫t² dt
= \(\frac{(t^3}{3}\) + C = \(\frac{(logx)^3}{3}\) + C

Question 6.
dx on I ⊂ (1, ∞).
Solution:

Question 7.
∫\(\frac{\sin(Tan^{-1}x)}{1+x^2}\) dx, x ∈ R.
Solution:
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\) dx
t = tan^-1 x ⇒ dt = \(\frac{dx}{1+x^2}\)
∫\(\frac{\sin(tan^{-1}x)}{1+x^2}\)dx = ∫ sin t
= -cos t + t
= -cos (tan^-1 x) + C

Question 8.
∫\(\frac{1}{8+2x^2}\) dx on R.
Solution:

Question 9.
∫\(\frac{3x^2}{1+x^6}\)dx, on R.
Solution:

Question 10.
∫\(\frac{2}{\sqrt{25+9x^2}}\)dx on R.
Solution:

Question 11.
∫\(\frac{3}{\sqrt{9x^2-1}}\)dx on (\(\frac{1}{3}\), ∞).
Solution:

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx cos nx = \(\frac{1}{2}\)∫ 2 sin mx. cos nx dx
= \(\frac{1}{2}\)∫sin(m+n)x + sin(m-n)x) dx
= –\(\frac{1}{2}\)(\(\frac{\cos (m+n) x}{m+n}+ cos\frac{(m-n)x}{m-n}\)) + C

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx. sin nx dx = \(\frac{1}{2}\) ∫2 sin mx. sin nx dx
= \(\frac{1}{2}\)∫cos (m – n)x – cos (m + n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m-n) x}{m-n}+\frac{\sin (m+n)x}{m+n}\)) + C

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫cos mx. cos nx dx = \(\frac{1}{2}\) ∫2 cos mx.cos nx dx
= \(\frac{1}{2}\)∫cos (m + n)x – cos (m – n)x dx
= \(\frac{1}{2}\)(\(\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n)x}{m-n}\)) + C

Question 15.
∫sin x sin 2x. sin 3x dx on R.
Solution:
sin 2x . sin 3x = \(\frac{1}{2}\)(2 sin 3x. sin 2x)
= \(\frac{1}{2}\)(cos x – cos 5x)
sin x sin 2x sin 3x
= \(\frac{1}{2}\)(sin x . cos x – cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1.2}{2}\)cos 5x . sin x)
= \(\frac{1}{2}\)(\(\frac{1}{2}\) sin 2x – \(\frac{1}{2}\) – (sin 6x – sin 4x)

Question 16.
∫\(\frac{\sin x}{\sin(a+x)}\) dx on I ⊂ R\{nπ – a : n ∈ Z}.
Solution:
sin x = sin (a + x – a)
= sin (a + x) . cos a – cos (a + x) sin a
∫\(\frac{\sin x}{\sin(a+x)}\)dx
= cos a ∫ dx – sin a ∫\(\frac{\cos (a+x)}{\sin(a+x)}\) dx
= x . cos a – sin a . log |sin (a + x)| + c

II. Evaluate the following integrals.

Question 1.
∫(3x – 2)^½ dx on (\(\frac{2}{3}\), ∞)
Solution:
t = 3x – 2 ⇒ dt = 3 dx
∫(3x – 2)^½ dx = \(\frac{1}{3}\) ∫t^½ dt = \(\frac{1}{3}\) \(\frac{t^{3/2}}{3/2}\) + C
= \(\frac{2}{9}\)(3x – 2)^3/2 + C

Question 2.
∫\(\frac{1}{7x+3}\) dx on I ⊂ R\{-\(\frac{3}{7}\)}.
Solution:
∫\(\frac{1}{7x+3}\) dx
t = 7x + 3
⇒ dt = 7 dx
= ∫\(\frac{1}{7x+3}\) dx = \(\frac{1}{7}\) ∫\(\frac{dt}{t}\)
= \(\frac{1}{7}\) log |t| + C = \(\frac{1}{7}\) log|7x + 3| + C

Question 3.
∫\(\frac{log(1+x)}{1+x}\) dx on (-1, ∞).
Solution:
∫\(\frac{log(1+x)}{1+x}\) dx
t = 1 + x ⇒ dt = dx

Question 4.
∫(3x² – 4)x dx on R.
Solution:
∫(3x² – 4)x dx
t = 3x² – 4 ⇒ dt = 6x dx
∫(3x² – 4)x dx = \(\frac{1}{6}\)∫t dt = \(\frac{1}{6}\).\(\frac{t^2}{2}\) + C
= \(\frac{(3x^2-4)^2}{12}\) + C

Question 5.
∫\(\frac{dx}{\sqrt{1+5x}}\) dx on (-\(\frac{1}{5}\), ∞).
Solution:

Question 6.
∫(1 – 2x³)x² dx on R.
Solution:
∫(1 – 2x³)x² dx
t = 1 – 2x³ ⇒ dt = -6x² dx
∫(1 – 2x³)x² dx = –\(\frac{1}{6}\)∫t dt
= –\(\frac{1}{6}\) . \(\frac{t^2}{2}\) + C
= \(\frac{-(1-2x^3)^2}{12}\) + C

Question 7.
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx on I ⊂ R \ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫\(\frac{\sec^2 x}{(1+tan x)^3}\) dx
t = 1 + tan x ⇒ dt = sec² x dx

Question 8.
∫x³ sinx⁴ dx on R.
Solution:
∫x³ . sinx⁴ dx
t = x⁴ ⇒ dt = 4x³ dx
∫x³ . sinx⁴ dx = \(\frac{1}{4}\)∫sin t. dt
= –\(\frac{1}{4}\)cos t + C
= –\(\frac{1}{4}\). cos x⁴ + C

Question 9.
∫\(\frac{\cos x}{(1+sin x)^2}\) dx on I ⊂ R\{2nπ + \(\frac{3 \pi}{2}\) : n ∈ Z}.
Solution:

Question 10.
∫\(\sqrt[3]{sin x}\) cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:

Question 11.
∫ 2x e^x² dx on R.
Solution:
∫ 2x. e^x² dx
t = x² ⇒ dt = 2x dx
∫ 2xe^x² dt = ∫ e^t dt = e^t + C
= e^x² + C

Question 12.
∫\(\frac{e^{log x}}{x}\) dx on (0, ∞).
Solution:
∫\(\frac{e^{log x}}{x}\) dx
t = log x ⇒ dt = \(\frac{1}{x}\).dx
∫\(\frac{e^{log x}}{x}\) dx = ∫e^t. dt
= e^t + C
= e^{log x} + C
= x + C

Question 13.
∫\(\frac{x^2}{\sqrt{1-x^6}}\) dx on I = (-1, 1).
Solution:

Question 14.
∫\(\frac{2x^3}{1+x^8}\) dx on R.
Solution:
t = x⁴ ⇒ dt = 4x³ dx
∫\(\frac{2x^3}{1+x^8}\) = \(\frac{1}{2}\)∫\(\frac{dt}{1+t^2}\)
= \(\frac{1}{2}\) tan^-1 t + C
= \(\frac{1}{2}\) tan^-1(x⁴) + C

Question 15.
∫\(\frac{x^8}{1+x^18}\) dx on R.
Solution:
t = x⁹ ⇒ dt = 9x⁸ dx

Question 16.
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx on I ⊂ R/{x ∈ R : cos (xe^x) = 0}.
Solution:
t = x . e^x
dt = (x . e^x + e^x)dx = e^x(1 + x)dx
∫\(\frac{e^x(1+x)}{\cos^2(xe^x)}\) dx = ∫\(\frac{dt}{\cos^2 t}\) dx
= ∫sin² t dt
= tan t + C
= tan (x. e^x) = C

Question 17.
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx on I ⊂ R\ {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
Let t = a + b cot x
dt = -b cosec² x dx
∫\(\frac{cosec^2 x}{(a+b cot x)^5}\) dx = –\(\frac{1}{b}\) ∫\(\frac{bt}{t^5}\)
= –\(\frac{1}{b}\) ∫t^-5 dt
= –\(\frac{1}{b}\) \(\frac{t^{-4}}{-4}\) + C
= \(\frac{1}{4b t^{4}}\) + C
= \(\frac{1}{4b(a+b cotx)^{4}}\) + C

Question 18.
∫e^x sin e^x dx on R.
Solution:
t = e^x ⇒ dt = e^x dx
∫e^x . sin e^x dx = ∫ sin t dt
= -cos t + C
= -cos (e^x) + C

Question 19.
∫\(\frac{\sin(log x)}{x}\) dx on (-1, ∞).
Solution:
t = log x ⇒ dt = \(\frac{1}{x}\) dx
∫\(\frac{\sin(log x)}{x}\)dx = ∫sin t dt
= -cos t + C
= -cos (log x) + C

Question 20.
∫\(\frac{1}{x log x}\) dx on (-1, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) . dx
∫\(\frac{1}{x log x}\) dx = ∫\(\frac{1}{t}\)dt = log t + C = log(log x + C)

Question 21.
∫\(\frac{(1+log x)^n}{x}\) dx on (e^-1, ∞). n ≠ -1.
Solution:
t = 1 + log x
dt = \(\frac{1}{x}\) dx

Question 22.
∫\(\frac{\cos(log x)}{x}\) dx on (0, ∞).
Solution:
t = log x
dt = \(\frac{1}{x}\) dx
∫\(\frac{\cos(log x)dx}{x}\) = ∫cos t dt
= sin t + C
= sin (log x) + C

Question 23.
∫\(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx on (0, ∞).
Solution:

Question 24.
∫\(\frac{2x+1}{x^2+x+1}\) dx on R.
Solution:

Question 25.
∫\(\frac{ax^{n-1}}{bx^n+c}\) dx, where n ∈ N, a, b, c are real nembers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xⁿ ≠ –\(\frac{c}{b}\)}.
Solution:
t = bxⁿ + C
dt = nbx^n-1dx

Question 26.
∫\(\frac{1}{x log x[log(log x)]}\) dx on (1, ∞).
Solution:

Question 27.
∫cot hx dx on R.
Solution:
t = sinh x ⇒ dt = cosh x dx
∫cot hx dx = ∫\(\frac{dt}{t}\)
= log |t| +C
= log |log(log x)| + C

Question 28.
∫\(\frac{1}{\sqrt{1-4x^2}}\) dx on (-\(\frac{1}{2}\), \(\frac{1}{2}\)).
Solution:

Question 29.
∫\(\frac{dx}{\sqrt{25+x^2}}\) dx on R.
Solution:

Question 30.
∫\(\frac{1}{(x+3\sqrt{x+2}}\) dx on I ⊂ (-2, ∞).
Solution:
x + 2 = t²
dx = 2t dt

Question 31.
∫\(\frac{1}{1+\sin 2x}\) dx on I ⊂ R\{\(\frac{n \pi}{2}\) + (-1)ⁿ \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:

Question 32.
∫\(\frac{x^2+1}{x^4+1}\) dx on R.
Solution:
∫\(\frac{x^2+1}{x^4+1}\) dx
Dividing Nr and Dr by x²

Question 33.
∫\(\frac{dx}{\cos^2+\sin 2x}\) on I ⊂ R\({(2n+1)\(\frac{\pi}{2}\) : n ∈ Z}∪{2nπ + tan^-1 \(\frac{1}{2}\) : n ∈ Z})
Solution:
∫\(\frac{dx}{\cos^2+\sin 2x}\)
Dividing Nr and Dr by cos² x

Question 34.
∫\(\sqrt{1-sin 2x}\) dx on I ⊂ {2nπ – \(\frac{3 \pi}{4}\), 2nπ + \(\frac{\pi}{4}\)}, n ∈ Z.
Solution:

Question 35.
∫\(\sqrt{1+cos 2x}\) dx I ⊂ {2nπ – \(\frac{\pi}{2}\), 2nπ + \(\frac{\pi}{2}\)}, n ∈ Z.
Solution:
∫\(\sqrt{1+cos 2x}\) dx = ∫\(\sqrt{2cos^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

Question 36.
∫\(\frac{\cos x + \sin x}{\sqrt{1+\sin 2x}}\) dx on I ⊂ {2nπ – \(\frac{\pi}{4}\), 2nπ + \(\frac{3\pi}{4}\)}, n ∈ Z.
Solution:

Question 37.
∫\(\frac{\sin 2x}{(a+b\cos x)^2}\) dx on {R, if |a| > |b| I ⊂ {x ∈ R : a + b cos x ≠ x}, if |a| < |b|.
Solution:
Put a + b cos x = t ⇒ cos x = \(\frac{t-a}{b}\)
Then b(-sin x)dx = dt
⇒ sin dx = \(\frac{-1}{b}\) dt

Question 38.
∫\(\frac{\sec x}{(\sec x+\tan x)^2}\) dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
Put sec x + tan x = t
Then (sec x tan x + sec² x) dx = dt
sec x (sec x + tan x) dx = dt

Question 39.
∫\(\frac{dx}{a^2\sec^2 x+b^2\cos^2 x}\) on R, a ≠ 0, b ≠ 0.
Solution:

Question 40.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\({a + nπ : n ∈ Z} ∪ {b + nπ n ∈ Z}).
Solution:

Question 41.
∫\(\frac{dx}{\sin(x-a)\sin(x-b)}\) on I ⊂ R\ ({a + \(\frac{(2n+1)\pi}{2}\) : n ∈ Z} ∪ {b + \(\frac{(2n+1)}{2}\)π : n ∈ Z}).
Solution:

III. Evaluate the following integers.

Question 1.
∫\(\frac{\sin 2x}{a \cos^2 x + b \sin^2 x}\) dx on I ⊂ R\ {x ∈ R |a cos² x + b sin² x = 0}.
Solution:
t = a cos² x + b sin² x
dt = (a(2cos x)(-sin x) + b(2sin x cos x))dx

Question 2.
∫\(\frac{1-\tan x}{1+\tan x}\) dx for x ∈ I ⊂ R\ {nπ – \(\frac{\pi}{4}\) : n ∈ Z}.
Solution:

Question 3.
∫\(\frac{\cot(log x)}{x}\) dx, x ∈ I ⊂ (0, ∞)\{e^nπ : n ∈ Z}.
Solution:
t = log x ⇒ dt = \(\frac{dx}{x}\)
∫\(\frac{1-\tan x}{1+\tan x}\) dx = ∫cot t dt
= log(sin t) + C
= log |sin (log x)| + C

Question 4.
∫e^x . cot e^x dx, x ∈ I ⊂ R\ {log n π : n ∈ Z}.
Solution:
t = e^x ⇒ dt = e^x dx
∫e^x . cot e^x dx = ∫cot t dt
= log (sin t) + C
= log |sin(log x)| + C

Question 5.
∫sec (tan x) sec² x dx, on I ⊂ {x ∈ E : tan x ≠ \(\frac{(2k+1)\pi}{2}\) for any k ∈ Z}. where E = R / {\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫sec(tan x) sec² x dx = ∫sec t. dt
= log tan (\(\frac{\pi}{4}+\frac{t}{2}\) + C
= log (tan(\(\frac{\pi}{4}+\frac{\tan x}{2}\))) + C

Question 6.
∫\(\sqrt{sin x}\)cos dx on {2nπ, (2n + 1)π}, (n ∈ Z).
Solution:
t = sin x ⇒ dt = cos x dx
∫\(\sqrt{sin x}\)cos dx = ∫√t dt
= \(\frac{2}{3}\) = t^3/2 + C
= \(\frac{2}{3}\) = (sin x)^3/2 + C

Question 7.
∫tan⁴ x sec² x dx, x ∈ I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫tan⁴ x sec² x dx = ∫t⁴dt
= \(\frac{t^5}{5}\) + C = \(\frac{(\tan x)^2}{5}\) + C

Question 8.
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) dx, x ∈ T ⊂ \{-4, 1}
Solution:
t = x² + 3x – 4
dt = (2x + 3)dx
∫\(\frac{2x+3}{\sqrt{x^2+3x-4}}\) = ∫\(\frac{dt}{\sqrt{t}}\)
= 2√t + C
= \(\sqrt{x^2+3x-4}\) + C

Question 9.
∫cosec² x\(\sqrt{\cot x}\) dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = cot x ⇒ dt = -cosec² x dx
∫cosec² x\(\sqrt{\cot x}\) dx = -∫√t dt
= –\(\frac{2}{3}\) t√t + C
= –\(\frac{2}{3}\) cot(x)^3/2 + C

Question 10.
∫sec x log(sec x + tan x)dx on (0, \(\frac{\pi}{2}\)).
Solution:
t = log(sec x + tanx)
dt = \(\frac{(\sec x.\tan x+\sec^2 x)}{(\sec x +\tan x}\) = sec x dx
∫sec x .log(sec x + tan x)dx = ∫t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{(log(\sec x+\tan x)^2}{2}\) + C

Question 11.
∫sin³ x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin³ x
sin³ x = \(\frac{1}{4}\)(3 sin x – sin 3x)
∫sin³ x dx = \(\frac{3}{4}\)∫sin x – \(\frac{1}{4}\)∫sin 3x dx
= –\(\frac{3}{4}\) cos x + \(\frac{1}{12}\) cos 3x + C
= \(\frac{1}{12}\)(cos 3x – 9 cos x) + C

Question 12.
∫cos³ x dx on R.
Solution:
cos 3x = 4 cos³ x – 3 cos x
∫cos³ x dx = \(\frac{3}{4}\)∫cos x + \(\frac{1}{4}\)∫cos 3x dx
= –\(\frac{3}{4}\) sin x + \(\frac{1}{12}\) sin 3x + C
= \(\frac{1}{12}\)(9 sin x + 9 sin 3x) + C

Question 13.
∫cos x cos 2x dx on R.
Solution:
cos 2x cos x = \(\frac{1}{2}\)(2cos 2x cos x)
∫cos x cos 2x dx = \(\frac{1}{2}\)∫(cos 3x + cos x) dx
= \(\frac{1}{2}\)∫cos 3x + \(\frac{1}{2}\)∫cos x
= \(\frac{1}{2}\)(\(\frac{\sin 3x}{3}\) + sin x) + C
= \(\frac{\sin 3x+3\sin x}{6}\) + C

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos 3x cos x = \(\frac{1}{2}\)(2cos 3x . cos x)
= \(\frac{1}{2}\)(cos 4x + cos 2x)
∫cos x cos 3x dx = \(\frac{1}{2}\)∫cos 4x dx + \(\frac{1}{2}\)∫cos 2x dx
= \(\frac{1}{2}\)(\(\frac{\sin 4x}{4}+\frac{\sin 2x}{2}\) + C
= \(\frac{1}{8}\)(sin 4x + 2sin 2x) + C

Question 15.
∫cos⁴ x dx on R.
Solution:
cos⁴ x = (cos² x)² = (\(\frac{1+\cos 2x}{2}\))²
= \(\frac{1}{4}\) (1 + 2 cos 2x+ cos² 2x)
= \(\frac{1}{4}\) (1 + 2 cos 2x + = \(\frac{1+\cos 4x}{2}\))
= \(\frac{1}{8}\) (2 + 4 cos 2x + 1 + cos 4x)
= \(\frac{1}{8}\) (3 + 4 cos 2x + cos 4x)
= \(\frac{1}{8}\)(3∫dx + 4∫cos 2x dx + ∫cos 4x dx)
= \(\frac{1}{8}\)(3x + 4\(\frac{\sin 2x}{2}+\frac{\sin 4x}{4}\)) + C
= \(\frac{1}{32}\)(12x + 8 sin 2x + sin 4x) + C

Question 16.
∫x \(\sqrt{4x+3}\) dx on (-\(\frac{3}{4}\), ∞).
Solution:

Question 17.
∫\(\frac{dx}{\sqrt{a^2-(b+cx)^2}}\) on {x ∈ R : |b + cx| < a}. where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:

Question 18.
∫\(\frac{dx}{a^2+(b+cx)^2}\) on R, a, b, c are real numbers, c ≠ 0 and a > 0.
Solution:

Question 19.
∫\(\frac{dx}{1+e^x}\), x ∈ R
Solution:

Question 20.
∫\(\frac{x^2}{(a+bx)^x}\) dx, x ∈ I ⊂ R\{-\(\frac{a}{b}\)}, where a, b are real nembers, b ≠ 0.
Solution:
Put t = a + bx
dt = b dx ⇒ dx = \(\frac{1}{b}\). dt

Question 21.
∫\(\frac{x^2}{\sqrt{1-x}}\) dx, x ∈ (-∞, 1).
Solution:

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(a)

I. Evaluate the following integrals.

Question 1.
∫(x³ – 2x² + 3) dx on R.
Solution:
∫(x³ – 2x² + 3) dx = \(\frac{x^4}{4}-\frac{2}{3}\)x³ + 3x + c

Question 2.
∫2x√x dx on (0, ∞).
Solution:
∫2x√x dx = 2 ∫ x^3/2 dx = \(\frac{2x^{5/2}}{5/2}\)
= \(\frac{4}{5}\)x^5/2 + c

Question 3.
∫\(\sqrt[3]{2 x^2}\) dx’on (0, ∞).
Solution:
∫\(\sqrt[3]{2 x^2}\) dx = ∫ 2^1/3. x^2/3 dx
= 2^1/3. \(\frac{x^{5/3}}{5/3}\) + c
= \(\sqrt[3]{2}\).\(\frac{3}{5}\)x^5/3 + c

Question 4.
∫\(\frac{x^2+3x-1}{2x}\)dx, x ∈ I ⊂ R\{0}.
Solution:

Question 5.
∫\(\frac{1-\sqrt{x}}{x}\)dx on (0, ∞).
Solution:

Question 6.
∫(\(1+\frac{2}{x}-\frac{3}{x^2}\)) dx on I⊂R\{0}
Solution:

Question 7.
∫(x + \(\frac{4}{1+x^2}\))dx on R.
Solution:

Question 8.
∫(e^x \(\frac{1}{x}-\frac{2}{\sqrt{x^2+1}}\))dx on I⊂R\[-1, 1].
Solution:

Question 9.
∫(\(\frac{1}{1-x^2}+\frac{1}{1+x^2}\))dx on (-1, 1).
Solution:

Question 10.
∫(\(\frac{1}{1-x^2}+\frac{2}{1+x^2}\))dx on (-1, 1).
Solution:

Question 11.
∫e^{log(1+tan²x)} dx on I ⊂ R \{\(\frac{(2n+1)\pi}{2}\):n ∈ Z}
Solution:
∫e^{log(1+tan²x)} dx = ∫e^log(sec²x) dx
= ∫sec²x dx = tan x + c

Question 12.
∫\(\frac{\sin^{2}x}{1+\cos2x}\) dx on I ⊂ R \{(2n ± 1)π : n ∈ Z}
Solution:

II. Evaluate the following intergrals.

Question 1.
∫(1 – x²)³ dx on (-1, 1).
Solution:
∫(1 – x²)³ dx = ∫(1 – 3x² + 3x⁴ – x⁶)dx
= x – x³ + \(\frac{3}{5}\)x⁵ – \(\frac{x^7}{7}\) + c

Question 2.
∫(\(\frac{3}{\sqrt{x}}-\frac{2}{x}+\frac{1}{3x^2}\)) dx on (0, ∞).
Solution:

Question 3.
∫(\(\frac{\sqrt{x}+1}{x}\))² dx on (0, ∞).
Solution:

Question 4.
∫(\(\frac{(3x+1)^2}{2x}\)) dx, x ∈ I ⊂ R\ {0}.
Solution:

Question 5.
∫(\(\frac{2x-1}{3\sqrt{x}}\))² dx on (0, ∞).
Solution:

Question 6.
∫(\(\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x^2-1}}-\frac{3}{2x^2}\))² dx on (0, ∞).
Solution:

Question 7.
∫(sec² x – cos x + x²) dx, x ∈ I ⊂ R/{\(\frac{n \pi}{2}\) : n is an odd integer}.
Solution:
∫(sec² x – cos x + x²) dx
= ∫sec² x dx – ∫cos x + ∫x² dx
= tan x – sin x + \(\frac{x^3}{3}\) + C

Question 8.
∫(sec x tan x + \(\frac{3}{x}\) – 4) dx, x ∈ I ⊂ R\ ({\(\frac{n \pi}{2}\) : n is an odd integer} ∪ {0}).
Solution:
∫(sec x tan x + \(\frac{3}{x}\) – 4) dx
= sec x tan x dx + 3∫\(\frac{dx}{x}\) – 4 ∫dx
= sec x + 3 log |x| – 4x + c

Question 9.
∫(√x – \(\frac{2}{1-x^2}\)) dx on (0, 1).
Solution:

Question 10.
∫(x³ – cos x + \(\frac{4}{\sqrt{x^2+1}}\)) dx
Solution:
∫(x³ – cos x + \(\frac{4}{\sqrt{x^2+1}}\)) dx
= ∫ x³ dx – ∫cos x dx + 4 ∫\(\frac{dx}{\sqrt{x^2+1}}\)
= \(\frac{x^4}{4}\) – sin x + 4 sinh^-1 x + C

Question 11.
∫(cosh x + \(\frac{1}{\sqrt{x^2+1}}\))dx, x ∈ R.
Solution:
∫(cosh x + \(\frac{1}{\sqrt{x^2+1}}\))dx
= ∫cosh x dx + ∫\(\frac{dx}{\sqrt{x^2+1}}\)
= sinh x + sinh^-1 x + c

Question 12.
∫(sinh x + \(\frac{1}{(x^2-1)^{1/2}}\)) dx, x ∈ I ⊂ (-∞, -1) ∪ (1, ∞).
Solution:
∫(sinh x + \(\frac{1}{(x^2 – 1)^{1/2}}\)) dx
= ∫sinh x dx + ∫\(\frac{dx}{\sqrt{x^2-1}}\)
= cosh x + log(x + \(\sqrt{x^2-1}\)) + C

Question 13.
∫\(\frac{a^{x}-b^{x}}{a^{x}b^{x}}\) dx (a > 0, a ≠ 1 and b > 0, b ≠ 1) on R.
Solution:

Question 14.
∫sec² x cosec² x dx on I ⊂ R\ (nπ : n ∈ Z} ∪ { (2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}).
Solution:
∫sec² x cosec² x dx

= ∫\(\frac{1}{\cos^{2}x}\)dx + ∫\(\frac{1}{(\sin^{2}x}\)dx
= ∫sec² x dx + ∫cosec² x dx
= tan x – cot x + C

Question 15.
∫\(\frac{1+\cos^{2}x}{1+\cos2x}\) dx on I ⊂ R\{nπ :n ∈ Z}
Solution:

Question 16.
∫\(\sqrt{1-cos2x}\)dx on I ⊂ [2nπ, (2n + 1)π], n ∈ Z.
Solution:

Question 17.
∫\(\frac{1}{\cosh x+\sinh x}\) dx on R.
Solution:
∫\(\frac{1}{\cosh x+\sinh x}\) dx
= ∫\(\frac{\cosh x-\sinh x}{\cosh^{2}x-\sinh^{2}x}\) dx
= ∫(cosh x – sinh x) dx
= sinh x – cosh x + C

Question 18.
∫\(\frac{1}{1+\cos x}\) dx on I ⊂ R \{(2n + 1)π : n ∈ Z}.
Solution:

= ∫cosec² (x) dx – ∫cosec x cot x dx
= -cot x + cosec x + C

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Hyperbola Solutions Exercise 5(a)

I.

Question 1.
One focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is \(\frac{3}{2}\).
Solution:
S(1 – 3) is thefocus, equation of the directrix is y – 2 = 0

P(x₁, y₁) is any point on the hyperbola join SP and draw PM perpendicular to the directrix S.P = e. PM ⇒ SP² = e². PM²
(x₁ – 1)² + (y₁ + 3)² = \(\frac{9}{4}\)|\(\frac{(y_1-2}{\sqrt{1+0}}\)|²
x²₁ + 1 – 2x₁ + y²₁ + 9 + 6y₁ = \(\frac{9}{4}\) (y₁ – 2)²
4x²₁ + 4y²₁ – 8x₁ + 24y₁ + 40 = 9(y²₁ + 4 – 4y₁) = 9y²₁ – 36y₁ + 36
4x²₁ – 5y²₁ – 8x₁ + 60y₁ + 4 = 0
Focus of P(x², y²) is
4x² – 5y² – 8x + 60y + 4 = 0
This is the equation of the required hyperbola.

Question 2.
If the lines 3x – 4y = 12 and 3x + 4y = 12 meets on a hyperbola S = 0 then find the eccentricity of the hyperbola S = 0.
Solution:
The given lines as
3x – 4y = 12
3x + 4y = 12
The combined equation of the lines
(3x – 4y) (3x + 4y) = 144
9x² – 16y² = 144

Question 3.
Find the equations of the hyperbola whose foci are (±5, 0) the transverse axis is of length 8.
Solution:
Foci are S(±5, 0) ∴ ae = 5
Length of transverse axis = 2a = 8
a = 4
e = \(\frac{5}{4}\)
b² = a² (e² – 1) = 16(\(\frac{25}{16}\) – 1)
Equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1
9x² – 16y² = 144

Question 4.
Find the equation of the hyperbola, whose asymptotes are the straight lines (x + 2y + 3) = 0, (3x + 4y + 5) = 0 and which passes through the point (1, -1).
Solution:
Combined equation of the asymptotes as
(x + 2y + 3) (3x + 4y + 5) = 0
∴ Equation of the hyperbola can be taken as
(x + 2y + 3) (3x + 4y + 5) + k = 0
The hyperbola passes through P(1, -1)
(1 – 2 + 3) (3-4 + 5) + k = 0
8 + k = 0 ⇒ k = -8
Equation of the hyperbola is
(x + 2y + 3) (3x + 4y + 5) – 8 = 0
3x² + 6xy + 9x + 4xy + 8y² + 12y + 5x +10y + 15 – 8 = 0
3x² + 10xy + 8y² + 14x + 22y + 7 = 0

Question 5.
If 3x – 4y + k = 0 is a tangent to x² – 4y² = 5, find value of k.
Solution:
Equation of the hyperbola x² – 4y² = 5
\(\frac{x^2}{5}-\frac{y^2}{\left(\frac{5}{{4}}\right)}\) = 1
a² = 5, b² = \(\frac{5}{4}\)
Equation of the given line is 3x – 4y + k = 0
4y = 3x + k
y = \(\frac{3}{4}\)x + \(\frac{k}{4}\)
m = \(\frac{3}{4}\) c = \(\frac{k}{4}\)
Condition for tangency is c² = a²m² – b²
\(\frac{k^2}{16}\) = 5. \(\frac{9}{16}-\frac{5}{4}\)
k² = 45 – 20 = 25
k = ±5.

Question 6.
Find the product of lengths of the perpen-diculars from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1 to its asymptotes.
Solution:
Equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1
a² = 16, b² = 9
Product of the perpendiculars from any point as the hyperbola to its asymptotes

Question 7.
If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate-hyperbola.
Solution:
If e and er, are the eccentricity of a hyperbola and its conjugate hyperbola, then

Question 8.
Find the equation of the hyperbola whose asymptotes are 3x = ± 5y and the vertices are (± 5, 0).
Solution:
The equation of the asymptotes are
3x = ±5y
3x – 5y = 0, 3x + 5y = 0
Combined equation of the asymptotes is
(3x – 5y) (3x + 5y) = 0
9x² – 25y² = 0
Equation of the hyperbola is 9x² – 25y² = k
The hyperbola passes through the vertex (5, 0)
9(5)² – 0 = k ⇒ k = 225
Equation of the hyperbola is
9x² – 25y² = 225

Question 9.
Find the equation of the normal at θ = \(\frac{\pi}{3}\) to the hyperbola 3x² – 4y² = 12.
Solution:
Equation of the hyperbola is 3x² – 4y² = 12
\(\frac{x^2}{4}+\frac{y^2}{3}\) = 1
Equation of the normal is

Question 10.
If the angle between the asymptotes is 30° then find its eccentricity.
Solution:
Angle between the asymptotes = 2θ = 30°
θ = 15°

Eccentricity = e = √6 – √2

II.

Question 1.
Find the centre, foci, eccentricity equation of the directrices, length of the latus rectum of the following hyperbola.
i) 16y² – 9x² = 144
Solution:
Equation of the hyperbola is
16y² – 9x² = 144
\(\frac{y^2}{9}-\frac{x^2}{16}\) = 1
a² = 16, b² = 9
Centre C(0, 0)
b²e² = a² + b² = 16 + 9
= 25 ⇒ be = 5
Foci are 5(0, ±ae) = (0, ±5)
Eccentricity = \(\frac{be}{b}=\frac{5}{3}\)
Equation of the directrices are y = ±b/e
± 3. \(\frac{5}{3}\)
5y = ± 9
Length of the latus return = 2. \(\frac{a^2}{b}\)
= 2. \(\frac{16}{3}\) = \(\frac{32}{3}\)

ii) x² – 4y² = 4
Solution:
Equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{1}\) = 1
a² = 4, b² = 1
Centre is c (0,0)
a²e² = a² + b² = 4 + 1 = 5
ae = √5
Foci are (±ae, 0) = (± √5, 0)
Eccentricity = \(\frac{ae}{a}=\frac{\sqrt{5}}{2}\)
Equations of directrices are x = ±\(\frac{a}{e}\)
= ± 2. \(\frac{2}{\sqrt{5}}\)
⇒ √5 = ± 4
⇒ √5x ± 4 = 0
Length of the latus rectum = \(\frac{2b^2}{a}=\frac{2.1}{2}\) = 1

iii) 5x² – 4y² + 20x + 8y = 4
Solution:
5(x² + 4x + 4) -4 (y² – 2y + 1) = 4 + 20 – 4
5(x + 2)² – 4(y – 1)² = 20
\(\frac{(x+2)^2}{4}+\frac{(y-1)^2}{5}\) = 1
a² = 4, b² = 5 => a < b
Centre C(-2, +1)
a²e² = a² + b² = 4 + 5 = 9
ae = 3
Eccentricity = \(\frac{ae}{a}=\frac{3}{2}\)
Foci are (h ± ae, k) = (-2 ± 3, 1)
= (-5, 1) and (1, 1)
Equations of the directrices are x – h = ±\(\frac{a}{e}\)
x + 2 = ± 2. \(\frac{2}{3}\)
3x + 6 = ± 4
3x+ 10 = 0 or 3x + 2 = 0
Length of the latus rectum = \(\frac{2b^2}{a}=\frac{2.5}{2}\) = 5

iv) 9x² – 16y² + 72x – 32y – 16 = 0
Solution:
Equation of the hyperbola is
9x² – 16y² + 72x – 32y -16 = 0
⇒ 9(x² +8x) – 16(y² + 2y) = 16
⇒ 9(x² + 8x +16) – 16 (y² + 2y + 1)
= 16 + 144 – 16
⇒ 9(x + 4)² – 16(y + 1)² = 144
\(\frac{(x+4)^2}{16}+\frac{(y+1)^2}{9}\) = 1
Comparing with \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}\) = 1
a² = 16, b² = 9, h = -4, k = -1
Centre (h, k) = (-4, -1)

Foci = (h ± ae, k) = (- 4±4.\(\frac{5}{4}\), 1)
= (-4 ± 5, -1)
= (1, -1) and (-9, -1)
Equation of the directrices are
x + 4 = ± 4. \(\frac{4}{5}\)
= ± \(\frac{16}{5}\)
5x + 20 = ± 16
Equation of the directices are 5x + 4 = 0
and 5x + 36 = 0
Length of Latus rectum = \(2\frac{b^2}{a}=\frac{2.9}{4}=\frac{9}{2}\)

Question 2.
Find the equation to the hyperbola whose foci are (4, 2) and (8, 2) and eccentricity is 2.
Solution:
Foci are (4, 2) and (8, 2) Centre C is the mid point of the foci
∴ Centre is (\(\frac{4+8}{2}\), \(\frac{2+2}{2}\)) = (6, 2)
ae = 6 – 4 = 2
Given e = 2 ⇒ a = \(\frac{ae}{e}=\frac{2}{2}\) = 1
b² = a² (e² – 1) = 1(4 – 1) = 3
Equation of the hyperbola is

Multiplying with 3
3(x – 6)² – (y – 2)² = 3
⇒ 3(x² – 12x + 36) – (y² – 4y + 4) = 3
⇒ 3x² – 36x + 108 – y² + 4y – 4 -3 = 0
3x² – y² – 36x + 4y + 101 = 0

Question 3.
Find the equation of the hyperbola of given length of transverse axis 6 whose vertex bisects the distance between the centre and the focus.
Solution:
Given CA = AS

a = ae – a
2a = ae ⇒ e = 2
Length of transverse axis = 2a = 6 ⇒ a = 3
b² = a²(e² – 1) = 9(4 – 1) = 27
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
\(\frac{x^2}{9}-\frac{y^2}{27}\) = 1
3x² – y² = 27

Question 4.
Find the equations of the tangents to the hyperbola x² – 4y² = 4 Which are i) Parallel ii) Perpendicular to the line x + 2y = 0.
Solution:
Equation of the hyperbola is x² – 4y² = 4
\(\frac{x^2}{4}-\frac{y^2}{1}\) = 1
a² = 4, b² = 1

i) The+anget is parallel to x + 2y = 0
m = –\(\frac{1}{2}\)
c² = a²m² – b² = 4.\(\frac{1}{4}\) – 1 = 1 – 1 = 0
c = 0
Equation of the parallel tangent is
y = mx + c
= –\(\frac{1}{2}\)x
2y = -x
x + 2y = 0.

ii) The tangent is perpendicular to x +2y = 0
Slope of the tangent = m = \(\frac{-1}{\left(-\frac{1}{2}\right)}\) = 2
c² = a²m² – b² = 4. 4 – 1 = 16
c = ±√15
Equation of the perpendicular tangent is
y = 2x ± √15

Question 5.
Find the equations of tangents drawn to the hyperbola 2x² – 3y² = 6 through (-2, 1).
Solution:
Equation of the hyperbola is 2x² – 3y² = 6
\(\frac{x^2}{3}-\frac{y^2}{2}\) = 1
Suppose the slope of the tangent is m¹
The tangent passes through P (-2, 1)
Equation of the tangent is
y – 1 = m(x + 2) = mx + 2m
y = mx + (2m + 1) ………. (1)
Condition for tangency is c² = a²m² – b²
(2m + 1)² = 3m² – 2
4m² + 4m + 1 = 3m² – 2
m² + 4 m + 3 = 0
(m + 1) (m + 3) = 0
m = -1 or – 3

Case (i): m = -1
Substituting in (1) equation of the tangent is
y = -x – 1
x + y + 1 =0

Case (ii) : m = – 3
Equation of the tangent is
y = – 3x – 5
3x + y + 5 = 0

Question 6.
Prove that the product of the perpendicular distance from any point on a hyperbola to its asymptotes is constant.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Any point on the hyperbola is P(a sec θ, b tan θ)
Equations of the asymptotes are \(\frac{x}{a}\) = ± \(\frac{y}{b}\)
i.e., \(\frac{x}{a}-\frac{y}{b}\) = 0 and \(\frac{x}{a}+\frac{y}{b}\) = 0
PM = Perpendicular distance from P on

PN = Perpendicular distance from P on

III.

Question 1.
Tangents to the hyperbola make angles \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 make angles θ₁, θ₂ with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x² – a²) when tan θ₁ + tan θ₂ = k.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Equation of the tangent to the hyperbola can be taken as
y = mx ± \(\sqrt{a^2m^2-b^2}\)
Suppose p(x₁, y₁) is the point of intersection of the tangents
y₁ = mx₁ ± \(\sqrt{a^2m^2-b^2}\)
y₁ – mx₁ = ±\(\sqrt{a^2m^2-b^2}\)
Squaring both sides
(y₁ – mx₁)² = a²m² – b²
y²₁ + m²x²₁ – 2mx₁y₁ – a²m² + b² = 0
m² (x²₁ – a²) – 2mx₁y₁ + (y²₁ + b²) = 0
This is a quadratic equation in m giving the values for m say m₁, m₂

i.e.,k = \(\frac{2x_1y_1}{x^2_1-a^2}\)
or 2x₁y₁ = k(x²₁ – a²₁)
p(x₁, y₁) lies on the curve 2xy = k(x² – a²)

Question 2.
Show that the locus of feet of the perpen-diculars drawn from foci to any tangent of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 is the auxiliary circle of the hyperbola.
Solution:
Equation of the hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Equation of the tangent to the hyperbola is

y = mx ± \(\sqrt{a^2m^2-b^2}\)
y – mx = ± \(\sqrt{a^2m^2-b^2}\) ………… (1)
Equation to the perpendicular from foci (±ae, 0) on this tangent is 1
y = –\(\frac{1}{m}\)(x ± ae)
⇒ my = – (x ± ae)
x + my = ±ae …………. (2)
Squaring and adding (1) and (2)
(y – mx)² + (x + my)² = a²m² – b² + a²e²
⇒ y² + m²x² – 2mxy + x² + m²y² + 2mxy
= a²m² – a²(e² – 1) + a²e² .
⇒ (x² + y²) (1 + m²) = a²m² – a²e² + a² + a²e²
= a²(1 + m²)
or x² + y² = a² which is the auxiliary circle.
Locus of the feet of the perpendiculars drawn from the foci to any tangent to the hyperbola lies on the auxiliary circle.

Question 3.
Show that the equation \(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1 represents.
i) An ellipse if ‘c’ is a real constant less than 5.
ii) A hyperbola if ‘c’ is any real constant between 5 and 9.
iii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (±2, 0) independent of the value of ‘c’.
i) An ellipse if ‘c’ is a real constant less than 5.
Solution:
Given equation is
\(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1
This equation represents an ellipse if 9 – c >0, 5 – c > 0
∴ c < 9, c < 5
⇒ c < 5

ii) A hyperbola if ‘c’ is any real constant between 5 and 9.
Solution:
Given equation is \(\frac{x^2}{9-c}-\frac{y^2}{5-c}\) = 1
This equation represents a hyperbola if
9 – c > 0 and 5 – c < 0
9 > c and 5 < c
ic. 5 < c < 9

iii) Show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (±2, 0), independent of the value ‘c’.
Solution:
In both cases
Case (i): a² = 9 – c, b² = 5 – c
a² – b² = (9 – c) – (5 – c)
= 9 – c – 5 + c = 4
a²e² = 4 ⇒ ae = 2
Foci are (±ae, 0) (±2, 0)

Case (ii) : a² = 9 – c, b² = c – 5
a² + b² = 9 – c + c – 5 = 4
a²e² = 4 ⇒ ae = 2
Foci are (± ae, 0) = (± 2, 0)

Question 4.
Show that the angle between two asymptotes of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 is 2 Tan^-1(\(\frac{b}{a}\)) Or 2 Sec^-1(e).
Solution:
Equations of the asymptotes are \(\frac{x}{a}-\frac{y}{b}\) = 0
and \(\frac{x}{a}+\frac{y}{b}\) = 0

If 2 θ is the angle between the asymptotes then tan θ = \(\frac{b}{a}\) = slope of the asymptotes of θ = tan^-1\(\frac{b}{a}\)
The angle between the asymptotes

Angle between the asymptotes
= 2tan^-1\(\frac{b}{a}\) or 2 Sec^-1(e)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(b)

I.

Question 1.
Find the equation of tangent and normal to the ellipse x + 8y = 33 at (-1, 2).
Solution:
Equation of the tangent is
\(\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\)
x(-1) + 8y(2) = 33
⇒ -x +16y = 33
⇒ x – 16y + 33 = 0
Equation of the normal is
16x + y + k = 0
It passes through P(-1, 2)
-16 + 2 + k = 0 ⇒ k =14
Equation of the normal is 16x + y + 14 = 0.

Question 2.
Find the equation of tangent and normal to the ellipse
x² + 2y² – 4x + 12y + 14 = 0 at (2, – 1).
Solution:
xx₁ + 2yy₁ – 2(x + x₁) + 6(y + y₁) + 14 = 0
⇒ 2x – 2y – 2(x + 2) + 6(y – 1) + 14 = 0
⇒ 4y + 4 = 0
y = – 1 required equation of tangent.
Slope of tangent is ‘0’
Equation of normal be
y + 1 = \(\frac{-1}{0}\) (x – 2)
x = 2 equation of normal.

Question 3.
Find the equation of the tangents to 9x² + 16y² = 144, which makes equal intercepts on the co-ordianate axis.
Solution:
Equation of the ellipse is 9x² + 16y² = 144
⇒ \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
Equation of the tangent is
\(\frac{x}{a}\). cos θ + \(\frac{y}{b}\) sin θ = 1
Slope of the tangent = –\(\frac{b \cos \theta}{a \sin \theta}\) = -1
cot θ = \(\frac{a}{b}=\frac{4}{3}\)
cos θ = ± \(\frac{4}{5}\), sin 0 = ± \(\frac{3}{5}\)
Equation of the tangent is
\(\frac{x}{4}\)(±\(\frac{4}{5}\)) + \(\frac{y}{3}\)(±\(\frac{3}{5}\)) = 1
x ± y ± 5 = 0.

Question 4.
Find the co-ordinates for the points on the ellipse x² + 3y² = 37 at which the normal is parallel to the line 6x – 5y = 2.
Solution:
Equation of the ellipse is x² + 3y² = 37
⇒ \(\frac{x^2}{37}+\frac{y^2}{\left(\frac{37}{3}\right)}\) = 1
a² = 37, b² = \(\frac{37}{3}\)
Slope of the normal =

The normal is parallel to 6x – 5y = 2

Case i) The co-ordinates of P are
(a cos θ, b sin θ)

Case ii) The co-ordinates of P are (a cos θ, b sin θ)

Question 5.
Find the value of k if 4x+y + k = 0isa tangent to the ellipse x² + 3y² = 3.
Solution:
Equation of the ellipse is x² + 3y² = 3
\(\frac{x^2}{3}+\frac{y^2}{1}\) = 1
a² = 3, b² = 1
Equation of the line is 4x + y + k = 0
y = -4x – k
m = -4, c = -k.
Condition for tangency is c² = a² m² + b²
(-k)² = 3(-4)² + 1
k²= 48 + 1 = 49
k = ±7.

Question 6.
Find the condition for the line x cos α + y sin α = p to be a tangent to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.
Solution:
Equation of the ellipse is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) ……….. (1)
Equation of the line is x cos α + y sin α = p
y sin α = – x cos α + p cos α p

Condition for tangency is c² = a² m² + b²

II.

Question 1.
Find the equations of tangent and normal to the ellipse 2x² + 3y² = 11 at the point whose ordinate is 1.
Solution:
Equation of the ellipse is 2x² + 3y² = 11
Given y = 1
2×2 + 3 = 11 ⇒ 2x² = 8
x² = 4
x = ±2
Points on the ellipse are P (2, 1) and
Q(-2, 1)

Case i)P (2, 1)
Equation of the tangent is 2x.2 + 3y.1 = 11
4x + 3y = 11
The normal is perpendicular to the tangent Equation of the normal at P can be taken as 3x – 4y = k.
The normal passes through P (2, 1)
6 – 4 = k ⇒ k = 2
Equation of the normal at P is 3x – 4y = 2.

Case ii) Q (-2, 1)
Equation of the tangent at Q is
2x(-2) + 3y.1 = 11
– 4x + 3y = 11
4x – 3y + 11 = 0
Equation of the normal can be taken as
3x + 4y = k
The normal passes through Q (-2, 1)
-6 + 4 = k ⇒ k = -2
Equation of the normal at Q is 3x + 4y = -2
or 3x + 4y + 2 = 0.

Question 2.
Find the equations to the tangents to the ellipse, x² + 2y² = 3 drawn from the point (1, 2) and also find the angle between these tangents.
Solution:
Equations of the ellipse is x² + 2y² = 3
⇒ \(\frac{x^2}{3}+\frac{y^2}{\left(\frac{3}{2}\right)}\) = 1
a² = 3, b² = \(\frac{3}{2}\)
Suppose m is the slope of the tangent. It passes through P (1, 2)
Equation of the tangent is

y – 2 = m(x – 1)
= mx – m
y = mx + (2 – m)
Condition for tangency is c² = a²m² + b²
(2 – m)² = 3(m²) + \(\frac{3}{2}\)
4 + m² – 4m = 3m² + \(\frac{3}{2}\)
2m² + 4m – \(\frac{5}{2}\)
4m² + 8m – 5 = 0
(2m – 1) (2m + 5) = 0
m = \(\frac{1}{2}\) or –\(\frac{5}{2}\)

Case i) m = \(\frac{1}{2}\)
Equation of the tangent is y = \(\frac{1}{2}\) x + 2 – \(\frac{1}{2}\)
= \(\frac{x}{2}+\frac{3}{2}\)
2y = x + 3
x- 2y + 3 = 0

Case ii) m = –\(\frac{5}{2}\)
Equation of the tangent is
y = – \(\frac{5}{2}\)x + (2 + \(\frac{5}{2}\))
= –\(\frac{5}{2}\)x + \(\frac{9}{2}\)
2y = – 5x + 9
or 5x + 2y – 9 = 0
Angle between the tangents is given by

Question 3.
Fincbthe equation of tangents to the ellipse 2x² + y² = 8 which are
i) Parallel to x – 2y – 4 = 0
Solution:
Slope will be : \(\frac{1}{2}\)
Equation of tangent y = mx ± \(\sqrt{a^2m^2+b^2}\)

2y – x + 6 = 0 required equation of tangents,
x – 2y ± 6 = 0.

ii) Perpendicular to x + y + 2 = 0
Solution:
Slope of tangent be ‘1’ as it is perpendicular to above line
y = mx ± \(\sqrt{a^2m^2+b^2}\)
y = x ± \(\sqrt{4+8}\)
y = x ± 2√3
⇒ x – y ± 2√3 = 0.

iii) Which makes an angle \(\frac{\pi}{4}\) with x-axis.
Solution:
Equation of tangent y = x ± 2√3.
⇒ x – y ± 2√3 = 0.

Question 4.
A circle of radius 4, is concentric with the ellipse 3x² + 13y² = 78. Prove that a common tangent is inclined to the major axis at an angle \(\frac{\pi}{4}\).
Solution:
Equation of the ellipse is 3x² + 13y² = 78
\(\frac{x^2}{26}+\frac{y^2}{6}\) = 1 ……… (1)
Centre of the ellipse is (0, 0)
Equation of the circle is x² + y² = 16 ………. (2)
Equation of tangent at P(θ) to the circle is
x cos θ + y sin θ = 16 ……. (3)

(3) is a tangent to the ellipse.

16 = 26 cos² θ + 6 sin² θ
= 26 cos² θ + 6(1 – cos² θ)
= 26 cos² θ + 6-6 cos² θ
20 cos² θ = 10
cos² θ = \(\frac{10}{20}=\frac{1}{2}\)
cos θ = ± \(\frac{1}{\sqrt{2}}\)
θ = \(\frac{\pi}{4}\)

III.

Question 1.
Show that the foot of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve
(x² + y²)² = a²x² + b²y².
Solution:
Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Equation of the tangent at P (θ) is
\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1

CN is the perpendicular from C on the tangent slope of CN = \(\frac{y_1}{x_1}\)
∴ Slope of PT × slope of CN = -1


Locus of N (xv y,) is (x² + y²)² = a²x² + b²y²

Question 2.
Show that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.

Solution:
Hint : Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)
Equation of the tangent to the ellipse is
y = mx ± \(\sqrt{a^2m^2+b^2}\)
⇒ y – mx = ± \(\sqrt{a^2m^2+b^2}\) ………… (1)
Equation to the perpendicular from either focus (+ae, 0) on this tangent is
y = – \(\frac{1}{m}\) (x ± ae)
my = -(x ± ae)
my + x = ± ae ……….. (2)
Squaring and adding (1) and (2)
(y – mx)² + (my + x)² = a²m² + b² + a²e²
y² + m²x² – 2mxy + m²y² + x² + 2mxy = a²m² + a² – a²e² + a²e²
(x² + y²) (1 + m²) = a² (1 + m²)
⇒ x² + y² = a²
The locus is the auxiliary circle concentric with the ellipse.

Question 3.
The tangent and normal to the ellipse x² + 4y² = 4 at a point P(θ) on it meets the major axis in Q and R respectively. If 0 < θ < \(\frac{x}{2}\) and QR = 2, then show that θ = cos^–\(\frac{2}{3}\)
Solution:

Equation of the ellipse is x² + 4y² = 4
\(\frac{x^2}{4}+\frac{y^2}{1}\) = 1
The equation of the tangent at P (θ) is
\(\frac{x}{2}\).cos θ + \(\frac{y}{1}\) sin θ = 1
Equation of x-axis is y = 0
\(\frac{x}{2}\) cos θ = 1 ⇒ x = \(\frac{2}{\cos\theta}\)
Co-ordinates of Q are (\(\frac{2}{\cos\theta}\), θ)
Equation of the normal at
P(θ) is \(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = a² – b²
\(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = 3
Substituting y = 0 we get \(\frac{2x}{\cos\theta}\) = 3
x = \(\frac{3}{2}\). cos θ
Co-ordinates of R are (\(\frac{3}{2}\). cos θ, o)

Given QR = 2
\(\frac{-3 \cos ^2 \theta+4}{2 \cos \theta}\) = 2
-3 cos² θ+ 4 = 4 cos θ
3 cos² θ + 4 cos θ – 4 = 0
(3 cos θ – 2) ( cos θ + 2) = 0
3 cos θ – 2 = 0 or cos θ + 2 = 0
cos θ = \(\frac{2}{3}\) or cos θ = – 2
cos θ always lies between -1 and 1
∴ cos θ = \(\frac{2}{3}\)
i.e., θ = cos^-1(\(\frac{2}{3}\))