SOLUTION: Let AB to be the vertical tower and C and D be the two points such that CD = 192 m. And let ∠ACB = θ and ∠ADB = α \[\begin{array}{*{35}{l}} tan\text{ }\theta \text{ }=\text{ }5/12 \\...
The radius of a circle is given as 15 cm and chord AB subtends an angle of 131o at the centre C of the circle. Using trigonometry, calculate: (i) the length of AB; (ii) the distance of AB from the centre C.
Since, CA = CB = 15 cm and ∠ACB = 131o Constructing a perpendicular CP from centre C to the chord AB,we get CP bisects ∠ACB as well as chord AB. =>∠ACP = 65.5o In ∆ACP, \[\begin{array}{*{35}{l}}...
Calculate AB.
SOLUTION: In ∆AMOB, \[\begin{array}{*{35}{l}} cos\text{ }{{30}^{o}}~=\text{ }AO/MO \\ \surd 3/2\text{ }=\text{ }AO/6 \\ AO\text{ }=\text{ }5.20\text{ }m \\ \end{array}\] In ∆BNO,...
Calculate BC.
SOLUTION: In ∆ADC, \[\begin{array}{*{35}{l}} CD/AD\text{ }=\text{ }tan\text{ }{{42}^{o}} \\ CD\text{ }=\text{ }20\text{ x }0.9004\text{ }=\text{ }18.008\text{ }m \\ \end{array}\] In ∆ADB,...
In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60o. Calculate the length of the board AB.
SOLUTION: In ∆PSB, \[\begin{array}{*{35}{l}} PS/PB\text{ }=\text{ }sin\text{ }{{60}^{o}} \\ PB\text{ }=\text{ }2/\text{ }\surd 3\text{ }=\text{ }1.155\text{ }m \\ \end{array}\] In ∆APQ,...
Find AD in FIG-1 and FIG-2
(i) FIG-1 SOLUTION: In ∆AEB, \[\begin{array}{*{35}{l}} AE/BE\text{ }=\text{ }tan\text{ }{{32}^{o}} \\ AE\text{ }=\text{ }20\text{ x }0.6249\text{ }=\text{ }12.50\text{ }m \\ AD\text{ }=\text{...
The angle of elevation of the top of a tower is observed to be 60o. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45o. Find: (i) the height of the tower, (ii) its horizontal distance from the points of observation.
Let AB to be the tower of height h meters and let the two points be C and D be such that CD = 30 m, ∠ADE = 45o and ∠ACB = 60o (i) In ∆ADE, \[\begin{array}{*{35}{l}} AE/DE\text{ }=\text{ }tan\text{...
From the figure, given below, calculate the length of CD.
SOLUTION: In ∆AED, \[\begin{array}{*{35}{l}} AE/\text{ }DE\text{ }=\text{ }tan\text{ }{{22}^{o}} \\ AE\text{ }=\text{ }DE\text{ }tan\text{ }{{22}^{o}}~=\text{ }15\text{ }x\text{ }0.404\text{...
Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60o and 30o; find the height of the pillars and the position of the point.
Let AB and CD be the two towers of height h m each and let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60o and ∠CPD = 30o In ∆ABP, \[\begin{array}{*{35}{l}} AB/BP\text{ }=\text{...
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48o and 36o respectively. Find the distance between the two ships (in the nearest metre) if: (i) the ships are on the same side of the light house. (ii) the ships are on the opposite sides of the light house.
Let AB to be the lighthouse and the two ships be C and D such that ∠ADB = 36o and ∠ACB = 48o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }{{48}^{o}} \\ BC\text{ }=\text{...
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30o to 45o.
Let AB to be the building of height h meters. and let the two points be C and D be such that CD = 40 m, ∠ADB = 30o and ∠ACB = 45o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{...
Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60 to 30.
Let AB to be the height of the tree, h m. and let the two points be C and D be such that CD = 20 m, ∠ADB = 30o and ∠ACB = 60o In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{...
In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, ∠ADB = 30 and ∠ACB = 45. Without using tables, find X.
In ∆ABC, \[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }{{45}^{o}}~=\text{ }1 \\ =>\text{ }BC\text{ }=\text{ }AB\text{ }=\text{ }X \\ \end{array}\] In ∆ABD,...
A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45, (ii) 60. Find the height of the tower in each case.
Let the height of the tower to be ‘h’ m. (i) Since, \[\begin{array}{*{35}{l}} \theta \text{ }=\text{ }{{45}^{o}} \\ tan\text{ }{{45}^{o}}~=\text{ }\left( h\text{ }-\text{ }1.6 \right)/\text{ }20 ...
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30o with the ground.
Let the length of the rope to be x meters. \[\begin{array}{*{35}{l}} sin\text{ }{{30}^{o}}~=\text{ }60/x \\ {\scriptscriptstyle 1\!/\!{ }_2}\text{ }=\text{ }60/x \\ x\text{ }=\text{ }120\text{ }m ...
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30o and 38o respectively. Find the distance between them, if the height of the tower is 50 m.
Let one of the persons be A , at a distance of ‘x’ meters and the second person be B at a distance of ‘y’ meters from the foot of the tower. The angle of elevation of A is 30o...
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68o with the ground. Find the height, up to which the ladder reaches.
Let the height upto which the ladder reaches as ‘h’ meters. the angle of elevation is 68o \[\begin{array}{*{35}{l}} =>\text{ }tan\text{ }{{68}^{o}}~=\text{ }h/\text{ }2.4 \\ 2.475\text{ }=\text{...
Sketch the graphs of the following functions: (i) f (x) = 2 sin x, 0 ≤ x ≤ π (ii) g (x) = 3 sin (x – π/4), 0 ≤ x ≤ 5π/4
\[\left( \mathbf{i} \right)~f\text{ }\left( x \right)\text{ }=\text{ }2\text{ }(sin\text{ }x),\text{ }0\text{ }\le \text{ }(x)\text{ }\le \text{ }\pi \] Since, \[g\text{ }\left( x \right)\text{...
The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60o. Find the height of the tower.
Let the height of the tower to be h meters. the angle of elevation is 60o => \[\begin{array}{*{35}{l}} tan\text{ }{{60}^{o}}~=\text{ }h/160 \\ \surd 3\text{ }=\text{ }h/160 \\ h\text{ }=\text{...
Prove: 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)
Let, \[P\text{ }\left( n \right)\text{ }=\text{ }1/1.2\text{ }+\text{ }1/2.3\text{ }+\text{ }1/3.4\text{ }+\text{ }\ldots \text{ }+\text{ }1/n\left( n+1 \right)\text{ }=\text{ }n/\left( n+1...
The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.
Let the length of the shadow of the tree to be x meters. Therefore, the height of the tree = √3 x meters If θ is the angle of elevation of the sun, \[\begin{array}{*{35}{l}} =>\text{ }tan\text{...
If m and n are roots of the equation: 1/x – 1/(x-2) = 3: where x ≠ 0 and x ≠ 2; find m x n.
According to ques, \[~1/x\text{ }-\text{ }1/\left( x-2 \right)\text{ }=\text{ }3\] \[\left( x\text{ }-\text{ }2\text{ }\text{ }x \right)/\text{ }\left( x\left( x\text{ }-\text{ }2 \right)...
Find the solution of the quadratic equation 2×2 – mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.
According to ques, \[m\text{ }+\text{ }5\text{ }=\text{ }0\text{ }and\text{ }n\text{ }\text{ }1\text{ }=\text{ }0\] hence, \[m\text{ }=\text{ }-5\text{ }and\text{ }n\text{ }=\text{ }1\] putting...
If p – 15 = 0 and 2×2 + px + 25 = 0: find the values of x.
According to the given question, \[p\text{ }\text{ }15\text{ }=\text{ }0\] And \[2{{x}^{2}}~+\text{ }px\text{ }+\text{ }25\text{ }=\text{ }0\] Thus, \[p\text{ }=\text{ }15\] Using\[p\]in the...
Show that one root of the quadratic equation x2 + (3 – 2a)x – 6a = 0 is -3. Hence, find its other root.
According to the given equation, \[{{x}^{2}}~+\text{ }\left( 3\text{ }\text{ }2a \right)x\text{ }\text{ }6a\text{ }=\text{ }0\] By putting \[x\text{ }=\text{ }-3\]we get \[{{\left( -3...
Glycogen is a branched-chain polymer of α-D-glucose units in which chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of C1-C6 glycosidic linkage. Structure of glycogen is similar to __________.
(i) Amylose
(ii) Amylopectin
(iii) Cellulose
(iv) Glucose
Option (ii) is the answer. Glycogen has a similar structure to amylopeptin. It's an a-D glucose unit branched chain polymer with C1-C4 glycosidic linkage for chain formation and C1-C6 glycosidic...
Which of the following polymer is stored in the liver of animals? (i) Amylose (ii) Cellulose (iii) Amylopectin (iv) Glycogen
Option (iv) is the answer. Glycogen is a type of sugar that is stored in the liver of mammals.
Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives _________.
(i) 2 molecules of glucose
(ii) 2 molecules of glucose + 1 molecule of fructose
(iii) 1 molecule of glucose + 1 molecule of fructose
(iv) 2 molecules of fructose
Option (iii) is the answer. Cane sugar (sucrose) is a disaccharide. When sucrose is hydrolyzed, one molecule of glucose and one molecule of fructose are produced.
Which of the following pairs represents anomers?
Option (C) is the answer. Anomers are isomers that differ only in the conformation of the hydroxyl group at C—1 and are known as - and -fonns.
Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure. α-helix structure of the protein is stabilised by : (i) Peptide bonds (ii) van der Waals forces (iii) Hydrogen bonds (iv) Dipole-dipole interactions
Option ( iii) is the answer. Hydrogen bonding help to keep the -helix structure of proteins stable. By twisting into a right-handed helix and hydrogen bonding the -NH group of each amino acid...
In disaccharides, if the reducing groups of monosaccharides i.e. aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar?
Option (B) is the answer. This structure represents sucrose, in which the C1—C2 glycosidic bond connects -D glucose and -D-fructose. This is a non-reducing sugar since the reducing groups of glucose...
Which of the following acids is a vitamin? (i) Aspartic acid (ii) Ascorbic acid (iii) Adipic acid (iv) Saccharic acid
Option (ii) is the answer. Vitamin C is ascorbic acid. Amino acid aspartic acid is a kind of amino acid. Dicarboxylic acids include adipic acid and saccharic acid.
Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are Are these linkages present? (i) 5′ and 3′ (ii) 1′ and 5′ (iii) 5′ and 5′ (iv) 3′ and 3′
Option (i) is the answer. Between the pentose sugars of nucleotides, there are 5′ and 3′ connections.
Nucleic acids are the polymers of ______________. (i) Nucleosides (ii) Nucleotides (iii) Bases (iv) Sugars
Option (ii) is the answer. Nucleic acids are nucleotide polymers connected together by phosphodiester linkage.
Which of the following statements is not true about glucose?
(i) It is an aldohexose.
(ii) On heating with HI, it forms n-hexane.
(iii) It is present in furanose form.
(iv) It does not give 2,4-DNP test.
Option (iii) is the answer. It's found in the pyranose structure.
Each polypeptide is a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be ____________.
(i) primary structure of proteins.
(ii) secondary structure of proteins.
(iii) the tertiary structure of proteins.
(iv) quaternary structure of proteins.
Option (i) is the answer. The main structure of proteins is the sequence of amino acids in a polypeptide chain.
DNA and RNA contain four bases each. Which of the following bases is not present in RNA? (i) Adenine (ii) Uracil (iii) Thymine (iv) Cytosine
Option (iii) is the answer. Adenine, guanine, thymine, and cytosine are the four bases found in DNA. Adenine, uracil, guanine, and cytosine are the four bases found in RNA. As a result, thymine is...
Which of the following B group vitamins can be stored in our body? (i) Vitamin B1 (ii) Vitamin B2 (iii) Vitamin B6 (iv) Vitamin B12
Option (iv) is the answer. Because vitamin B12 is water insoluble, it can be stored in the body.
Which of the following bases is not present in DNA? (i) Adenine (ii) Thymine (iii) Cytosine (iv) Uracil
Option (iv) is the answer. In DNA, uracil is absent; instead, thymine is present.
Three cyclic structures of monosaccharides are given below which of these are anomers.
(i) I and II
(ii) II and III
(iii) I and III
(iv) III is anomer of I and II
Option (i) is the answer. Anomers are cyclic configurations of monosaccharides that differ in structure at carbon-1. I and II are anomers in this case because they differ solely in carbon-1.
Which of the following reactions of glucose can be explained only by its cyclic structure?
(i) Glucose forms pentaacetate.
(ii) Glucose reacts with hydroxylamine to form an oxime.
(iii) Pentaacetate of glucose does not react with hydroxylamine.
(iv) Glucose is oxidised by nitric acid to gluconic acid
Option (iii) is the answer. The absence of a free -CHO group is indicated by the fact that glucose pentaacetate does not react with hydroxylamine. Only the cyclic nature of glucose may explain this...
Optical rotations of some compounds along with their structures are given below which of them have D configuration.
(i) I, II, III
(ii) II, III
(iii) I, II
(iv) III
Option (i) is the answer. The -OH group is on the lowest asymmetric carbon on the right side of the I, II, and III structures, which is similar to (+) glyceraldehyde.
Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.
(i) ‘a’ carbon of glucose and ‘a’ carbon of fructose.
(ii) ‘a’ carbon of glucose and ‘e’ carbon of fructose.
(iii) ‘a’ carbon of glucose and ‘b’ carbon of fructose.
(iv) ‘f’ carbon of glucose and ‘f ’ carbon of fructose.
Option (iii) is the answer. Anomeric carbon is carbon that is next to an oxygen atom in the cyclic structure of glucose or fructose. 'a' and 'b' are next to the oxygen atom, as illustrated in the...
Three structures are given below in which two glucose units are linked. Which of these linkages between glucose, units are between C1 and C4 and which linkages are between C1 and C6?
(i) (A) is between C1 and C4, (B) and (C) is between C1 and C6
(ii) (A) and (B) are between C1 and C4, (C) is between C1 and C6
(iii) (A) and (C) is between C1 and C4, (B) is between C1 and C6
(iv) (A) and (C) is between C1 and C6, (B) is between C1 and C4
Option (iii) is the answer (A) and (C) are in the Cl-C4 range, while (B) is in the Cl-C6 range.
Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a __________. (i) monosaccharide (ii) disaccharide (iii) reducing sugar (iv) non-reducing sugar
Option (ii) and (iv) are the answers. Sucrose is a non-reducing sugar and a disaccharide.
Proteins can be classified into two types on the basis of their molecular shape i.e., fibrous proteins and globular proteins. Examples of globular proteins are : (i) Insulin (ii) Keratin (iii) Albumin (iv) Myosin
Option (i) and (iii) are the answers Globulular protein is the structure that develops when a chain of polypeptides coils around to form a spherical shape. Insulin and albumin, for example, are...
Which of the following carbohydrates are branched polymer of glucose?
(i) Amylose
(ii) Amylopectin
(iii) Cellulose
(iv) Glycogen
Option (i) and (iv) are the answers. Amylopectin and glycogen are both glucose branching polymers.
Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following is acidic?
Option (ii) and (iv) are the answers. Acidic amino acids have more than one -COOH group one against the –NH2 group.
Lysine, is _______________.
(i) α-Amino acid
(ii) Basic amino acid
(iii) Amino acid synthesised in the body
(iv) β-Amino acid
Option (i), (ii) and (iii) are the answers. (a)Lysine is a kind of amino acid with the structural formula . (b) Because the number of NH2 groups (2) is more than the number of COOH groups, it is a...
Which of the following monosaccharides are present as five-membered cyclic structure (furanose structure)? (i) Ribose (ii) Glucose (iii) Fructose (iv) Galactose
Option (i) and (iii) are the answers. The five-membered cyclic structure of ribose and fructose is shown (furanose structures). They have a five-membered ring, similar to the foran compound....
In fibrous proteins, polypeptide chains are held together by ___________.
(i) van der Waals forces
(ii) disulphide linkage
(iii) electrostatic forces of attraction
(iv) hydrogen bonds
Option (ii) and (iv) are the answers. Disulphide linkage and hydrogen bonding hold polypeptide chains together in fibrous proteins.
Which of the following are purine bases?
(i) Guanine
(ii) Adenine
(iii) Thymine
(iv) Uracil
Option (i) and (ii) are the answers. Purines are made up of a six-membered nitrogen-containing ring fused together with a five-membered nitrogen-containing ring. Purine bases guanine and adenine...
Which of the following terms are correct about enzyme?
(i) Proteins
(ii) Dinucleotides
(iii) Nucleic acids
(iv) Biocatalysts
Option (i) and (iv) are the answers. Enzymes are protein molecules that act as biocatalysts in the body's chemical reactions.
Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?
Lactose is the sugar found in milk. Glucose and galactose are two monosaccharides found in lactose. Disaccharides are oligosaccharides that include two monosaccharide units.
How do you explain the presence of all the six carbon atoms in glucose in a straight chain?
When glucose is heated with HI for a long time, n-hexane develops, implying that all six carbon atoms are connected in a straight chain.
In nucleoside, a base is attached at 1C position of the sugar moiety. A nucleotide is formed by linking the phosphoric acid unit to the sugar unit of a nucleoside. At which position of sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?
When a nitrogenous base is connected to the 1' position of a five-carbon sugar, a nucleoside is produced. The 5' carbon of the sugar in a nucleoside molecule is bonded to the 5' carbon of the sugar...
Name the linkage connecting monosaccharide units in polysaccharides.
Glycosidic linkages connect the monosaccharide units of polysaccharides. When an oxide bond is created between two monosaccharide units with the loss of a water molecule, it is called a glycosidic...
Under what conditions glucose is converted to gluconic and saccharic acid?
When glucose is treated with a mild oxidising agent like Br2 water, it is transformed to gluconic acid, a six-carbon carboxylic acid. When glucose is treated with nitric acid, it is transformed to...
Monosaccharides contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule is also considered for classification. In which class of monosaccharide will you place fructose?
Carbonyl groups can be found in monosaccharides. As a result, they're categorised as either aldose or ketose. Aldose refers to monosaccharides that contain an aldehyde group. Ketose refers to...
The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicates the correlation of configuration of that particular stereoisomer. This refers to their relationship with one of the isomers of glyceraldehyde. Predict whether the following compound has ‘D’ or ‘L’ configuration.
On the left side of the C5 carbon atom, the –OH group is connected. As a result, the provided compound is in the 'L' configuration.
Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration?
D-configuration is the configuration of both aldopentoses. -D-ribose is ribose, while -D-2-deoxyribose is 2-deoxyribose.
Which sugar is called invert sugar? Why is it called so?
Invert sugar is another name for sucrose. It comes from sugarcane and sugarbeet and is a naturally occurring sugar. When sucrose is hydrolyzed, the sign of rotation changes from Dextro (+) to laevo...
Amino acids can be classified as α-, β-, -, δ- and so on depending upon the relative position of the amino group concerning the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?
The sort of amino acids that make up a polypeptide chain are -amino acids and alpha-amino acids, where the amino acid is linked to the -carbon in the molecule.
α-Helix is a secondary structure of proteins formed by twisting of the polypeptide chain into right-handed screw-like structures. Which type of interactions is responsible for making the a-helix structure stable?
The –NH group of each amino acid residue hydrogen is bound to the –C=O of an adjacent turn of the helix, forming a right-handed screw helix shape.
Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate?
Enzyme oxidoreductases is the name given to a group of enzymes that catalyse redox processes. Alcohol Dehydrogenase, for example, aids in the reduction of alcohol levels in the human body when...
During curdling of milk, what happens to sugar present in it?
The sugar found in milk, lactose, is transformed to lactic acid during curdling, which is produced by bacteria.
How do you explain the presence of five —OH groups in the glucose molecule?
When glucose is acetylated using acetic anhydride (CH3CO)2O in the presence of ZnCl2, glucose pentaacetate is formed, confirming the presence of five –OH groups.
Why does compound (A) give below not form an oxime?
The chemical in question is glucose pentaacetate. The presence of a free –C=O group in glucose indicates the presence of a free carbonyl group, as does the synthesis of oxime from glucose. Because...
Why must vitamin C be supplied regularly in diet?
Because vitamin C is a water-soluble vitamin, any excess is eliminated from the body on a regular basis. Vitamin C cannot be stored in the body, thus it must be consumed on a regular basis.
Sucrose is dextrorotatory but the mixture obtained after hydrolysis is laevorotatory. Explain.
Sucrose's aqueous solution is dextrorotatory, rotating plane-polarized light entering the solution 66.5 degrees to the right. When sucrose is hydrolyzed with dilute acids or invertase enzyme, two...
Amino acids behave like salts rather than simple amines or carboxylic acids. Explain
An amino acid has both a –NH2 and a –COOH group. The –COOH group loses a proton [H]+ in aqueous solution of the amino acid, while the –NH2 acquires a proton to create a zwitterion, which is a...
Structures of glycine and alanine are given below. Show the peptide linkage in glycylalanine.
Glycylalanine is formed when the hydroxyl group of glycine is bonded to the amine group of alanine via a peptide (-CONH) linkage.
Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to a physical change like change in temperature or a chemical change like change in pH, denaturation of protein takes place. Explain the cause.
Hydrogen bonds and other intermolecular interactions connect the amino acid residues in proteins. The hydrogen bonds are disrupted when a physical or chemical change occurs, and the native protein...
The activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol–1, while the activation energy is only 2.15 kJ mol–1 when hydrolysis is catalyzed by the enzyme sucrase. Explain.
Biocatalysts are enzymes. By providing an alternative approach, they lower the magnitude of activation energy. The enzyme sucrase lowers the activation energy of sucrose hydrolysis from 6.22 kJ...
How do you explain the presence of an aldehydic group in a glucose molecule?
Bromine water can be used to treat glucose, which results in the carboxylic acid gluconic acid, which verifies the presence of an aldehyde group.
One root of the quadratic equation 8×2 + mx + 15 = 0 is ¾. Find the value of m. Also, find the other root of the equation.
According to the given question, \[8{{x}^{2~}}+\text{ }mx\text{ }+\text{ }15\text{ }=\text{ }0\] One of the roots is\[~{\scriptscriptstyle 3\!/\!{ }_4},\]and it satisfies the given equation So,...
Solve: 2x – 3 = √(2×2 – 2x + 21)
According to the given question, \[2x\text{ }\text{ }3\text{ }=\text{ }\surd (2{{x}^{2}}~\text{ }2x\text{ }+\text{ }21)\] Squaring on both sides, we get \[{{\left( 2x\text{ }\text{ }3...
Solve: .
According to the given question, \[3\surd 2{{x}^{2}}~\text{ }5x\text{ }\text{ }\surd 2\text{ }=\text{ }0\] Or, \[3\surd 2{{x}^{2}}~\text{ }6x\text{ }+\text{ }x\text{ }\text{ }\surd 2\text{ }=\text{...
solve:
According to the given equation, \[3{{x}^{2}}~\text{ }2\surd 6x\text{ }+\text{ }2\text{ }=\text{ }0\] Or, \[3{{x}^{2}}~\text{ }\surd 6x\text{ }\text{ }\surd 6x\text{ }+\text{ }2\text{ }=\text{ }0\]...
Solve:
According to the given question, \[\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }24\] Or, \[{{x}^{2}}~\text{ }25\text{ }=\text{ }24\] Or,...
Solve: (x^2+1/x^2)-3(x-1/x)-2=0
Let \[\left( x\text{ }\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\] or,...
Solve: 2(x^2+1/x^2)-(x+1/x)=11
let \[\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
Solve:9(x^2+1/x^2)-9(x+1/x)-52=0
Let, \[~\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
Solve: x4 – 2×2 – 3 = 0
According to ques, \[{{x}^{4}}~\text{ }2{{x}^{2}}~\text{ }3\text{ }=\text{ }0\] Let’s take \[{{x}^{2}}~=\text{ }y\] Then, the equation becomes \[{{y}^{2}}~\text{ }2y\text{ }\text{ }3\text{ }=\text{...
Find the value of x, given that A^2 = B,
Solution: So, on comparison we get \[x\text{ }=\text{ }36.\]
If given matrix, find the matrix X such that: A + X = 2B + C
Solution: As per the given question,
Solve : 2×4 – 5×2 + 3 = 0
According to ques, \[~2{{x}^{4}}~\text{ }5{{x}^{2}}~+\text{ }3\text{ }=\text{ }0\] Let \[{{x}^{2}}~=\text{ }y\] The, \[2{{y}^{2}}~\text{ }5y\text{ }+\text{ }3\text{ }=\text{ }0\] \[2{{y}^{2}}~\text{...
If given matrix, find the matrix ‘X’ and matrix ‘Y’.
Solution: Now, On comparison, we get \[-28\text{ }-\text{ }3x\text{ }=\text{ }10\] \[3x\text{ }=\text{ }-38\] \[x\text{ }=\text{ }-38/3\] And, \[20\text{ }-\text{ }3y\text{ }=\text{ }-8\] \[3y\text{...
Find x and y, if:
Solution: According to the given question, On comparison, we get \[2x\text{ }+\text{ }3x\text{ }=\text{ }5\] And \[2y\text{ }+\text{ }4y\text{ }=\text{ }12\] \[5x\text{ }=\text{ }5\text{ }and\text{...
Solve: x + 4/x = -4; x ≠ 0
According to ques, \[~x\text{ }+\text{ }4/x\text{ }=\text{ }-4\] or, \[\left( {{x}^{2~}}+\text{ }4 \right)/\text{ }x\text{ }=\text{ }-4\] Or, \[{{x}^{2}}~+\text{ }4\text{ }=\text{ }-4x\] or,...
Solve: (i) A (BA) (ii) (AB) B.
Solution: \[\left( i \right)\text{ }A\text{ }\left( BA \right)\] \[\left( ii \right)\text{ }\left( AB \right)\text{ }B\]
Find the values of a, b and c.
Solution: According to the given ques, On comparison, we get \[a\text{ }+\text{ }1\text{ }=\text{ }5\Rightarrow a\text{ }=\text{ }4\] \[b\text{ }+\text{ }2\text{ }=\text{ }0\Rightarrow b\text{...
Solve: x2 – 11/4 x + 15/8 = 0
According to ques, \[~{{x}^{2}}~\text{ }11/4\text{ }x\text{ }+\text{ }15/8\text{ }=\text{ }0\] Taking L.C.M , \[\left( 8{{x}^{2}}~\text{ }22x\text{ }+\text{ }15 \right)/\text{ }8\text{ }=\text{ }0\]...
3A x M = 2B; find matrix M.
Solution: According to the given question, \[3A\text{ }x\text{ }M\text{ }=\text{ }2B\] Suppose the order of the \[matrix\text{ }M\text{ }be\text{ }\left( a\text{ }x\text{ }b \right)\] Now, we know...
Solve : a2x2 – b2 = 0
According to ques, \[{{a}^{2}}{{x}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] or, \[{{\left( ax \right)}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] Or, \[\left( ax\text{ }+\text{ }b \right)\left( ax\text{ }\text{...
Evaluate:
Solution: As per the given question,
Solve: (i) The order of the matrix X. (ii) The matrix X.
Solution: (i) Suppose, the order of the matrix be \[a\text{ }x\text{ }b\] We know that Hence, for product of matrices to be possible \[a\text{ }=\text{ }2\] And, form the order of the resultant...
Solve: (2x + 3)2 = 81
According to ques, \[{{\left( 2x\text{ }+\text{ }3 \right)}^{2}}~=\text{ }81\] Taking square root both sides, \[2x\text{ }+\text{ }3\text{ }=\text{ }\pm \text{ }9\] Or, \[2x\text{ }=\text{ }\pm...
If given matrix , find x and y, if: (i) x, y ∈ W (whole numbers) (ii) x, y ∈ Z (integers)
Solution: According to the given question, \[{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }25\] And, \[-2{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }-2\] \[\left( i \right)\text{ }x,\text{ }y~\in ~W\text{ }\left(...
Find x and y, if:
Solution: On comparison, we get \[3x\text{ }+\text{ }18\text{ }=\text{ }15\] And \[12x\text{ }+\text{ }77\text{ }=\text{ }10y\] \[3x\text{ }=\text{ }-3\] And \[y\text{ }=\text{ }\left( 12x\text{...
Solve each of the following equations: 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0; x 3,x -3/2
According to ques, \[4{{x}^{2}}~+\text{ }6x\text{ }+\text{ }x\text{ }\text{ }3\text{ }+\text{ }3x\text{ }+\text{ }9\text{ }=\text{ }0\] Or, \[4{{x}^{2}}~+\text{ }10x\text{ }+\text{ }6\text{ }=\text{...
Find x and y, if:
Solution: On comparison, we get \[6x\text{ }-\text{ }10\text{ }=\text{ }8\] And \[-2x\text{ }+\text{ }14\text{ }=\text{ }4y\] \[6x\text{ }=\text{ }18\] And \[y\text{ }=\text{ }\left( 14\text{...
If the given matrix. simplify: A2 + BC.
Solution: \[{{A}^{2}}~+\text{ }BC\]
If given matrix. Then show that: (i) A(B + C) = AB + AC (ii) (B – A)C = BC – AC.
Solution: \[\left( i \right)\text{ }A\left( B\text{ }+\text{ }C \right)\] \[AB\text{ }+\text{ }AC\] So, \[A\left( B\text{ }+\text{ }C \right)\text{ }=\text{ }AB\text{ }+\text{ }AC\] \[\left( ii...
If given matrix and A2 = I, find a and b.
Solution: \[{{A}^{2}}\] Given, \[~{{A}^{2~}}=\text{ }I\] On comparison, we get \[1\text{ }+\text{ }a\text{ }=\text{ }1\] \[a\text{ }=\text{ }0\] And, \[-1\text{ }+\text{ }b\text{ }=\text{ }0\]...
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
let the present ages of father and his son be \[~x\text{ }years\text{ }and\text{ }\left( 45\text{ }\text{ }x \right)\text{ }years\] hence five years ago, Father’s age \[=\text{ }\left( x\text{...
Find the matrix A, if B =given matrix and B2 = B + ½A.
Solution: \[{{B}^{2}}\] \[{{B}^{2}}~=\text{ }B\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}A\] \[{\scriptscriptstyle 1\!/\!{ }_2}A\text{ }=\text{ }{{B}^{2}}-\text{ }B\] \[A\text{ }=\text{...
Solve : (i) (A + B)^2 (ii) A2 + B2
Solution: According to the given ques, \[\left( i \right)\text{ }\left( A\text{ }+\text{ }B \right)\] \[So,\text{ }{{\left( A\text{ }+\text{ }B \right)}^{2}}~=\text{ }\left( A\text{ }+\text{ }B...
The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.
According to ques, \[S\text{ }=\text{ }n\left( n\text{ }+\text{ }1 \right)\] Also, \[S\text{ }=\text{ }420\] So, \[n\left( n\text{ }+\text{ }1 \right)\text{ }=\text{ }420\] Or, \[{{n}^{2}}~+\text{...
Solve: (i) AB (ii) A^2 – AB + 2B
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{AB}\text{ }\] \[\left( ii \right)\text{ }{{\mathbf{A}}^{\mathbf{2}}}-\text{ }\mathbf{AB}\text{ }+\text{ }\mathbf{2B}\]
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
Let the speed of the second train be \[~x\text{ }km/hr.\] Then, the speed of the first train is \[\left( x\text{ }+\text{ }5 \right)\text{ }km/hr\] Let O be the position of the railway station,...
Solve: (i) A – B (ii) A^2
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{A}\text{ }-\text{ }\mathbf{B}\text{ }\] \[\text{ }\left( \mathbf{ii} \right)\text{ }{{\mathbf{A}}^{\mathbf{2}~}}\]
BA = M^2, find the values of a and b.
Solution: $BA$ \[{{M}^{2}}\] So, \[BA\text{ }={{M}^{2}}\] On comparison, we get \[-2b\text{ }=\text{ }-2\] \[b\text{ }=\text{ }1\] And, \[a\text{ }=\text{ }2\]
If the given matrix and I is a unit matrix of the same order as that of M; show that: M2 = 2M + 3I
Solution: \[{{M}^{2}}\] \[2M\text{ }+\text{ }3I\] Hence, \[{{M}^{2}}~=\text{ }2M\text{ }+\text{ }3I\]
A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Let the usual speed of the plane to be \[x\text{ }km/hr\] The distance to travel \[=\text{ }1500km\] since, Time = Distance/ Speed As the ques suggests, \[{{x}^{2}}~+\text{ }250x\text{ }\text{...
Find A2 + AC – 5B
Solution: \[{{A}^{2}}\] $AC$ $5B$ \[{{A}^{2}}~+\text{ }AC\text{ }-\text{ }5B\text{ }=\]
Is the following possible: A^2
Solution: \[{{A}^{2}}~=\text{ }A\text{ }x\text{ }A,\text{ }\]isn’t possible because the number of columns isn’t equal to its number of rows in matrix A.
Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.
let the original number of persons to be x. According to ques, Total money which was divided is \[=\text{ }Rs\text{ }6500\] Each person’s share is \[=\text{ }Rs\text{ }6500/x\] Then, as the question...
Is the following possible: (i) AB (ii) BA
Solution: \[\left( i \right)\text{ }AB\] \[\left( ii \right)\text{ }BA\]
Solve: (i) (AB) C (ii) A (BC)
Solution: According to the given ques, \[\left( i \right)\text{ }\left( AB \right)\] \[\left( AB \right)\text{ }C\] \[\left( ii \right)\text{ }BC\] \[A\text{ }\left( BC \right)\] So, \[\text{...
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for: (i) the onward journey; (ii) the return journey. If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
According to ques, Distance \[=\text{ }400\text{ }km\] Average speed of the airplane \[=\text{ }x\text{ }km/hr\] Also, speed while returning \[=\text{ }\left( x\text{ }+\text{ }40 \right)\text{...
Find x and y, if:
Solution: (i) On comparison, we get \[5x\text{ }-\text{ }2\text{ }=\text{ }8\] \[5x\text{ }=\text{ }10\] \[x\text{ }=\text{ }2\] And, \[20\text{ }+\text{ }3x\text{ }=\text{ }y\] \[20\text{ }+\text{...
A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight.
Let the number of people staying overnight as x. According to ques, total hotel bill \[~=\text{ }Rs\text{ }4800\] Now,hotel bill for each person \[=\text{ }Rs\text{ }4800/x\] therefore,...
If find x and y when x and y when A2 = B.
Solution: \[{{A}^{2}}~\] \[{{A}^{2}}~=\text{ }B\] On comparison, we get \[4x\text{ }=\text{ }16\] \[x\text{ }=\text{ }4\] And, \[1\text{ }=\text{ }-y\] \[y\text{ }=\text{ }-1\]
If the given matrix and I is a unit matrix of order 2×2, find: (i) A^2 (ii) B^2A
Solution: (i) \[{{A}^{2}}\] (ii) \[~{{B}^{2}}\] \[{{B}^{2}}A\]
A trader buys x articles for a total cost of Rs 600. (i) Write down the cost of one article in terms of x. If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less. (ii) Write down the equation in x for the above situation and solve it for x.
According to ques, Number of articles \[=\text{ }x\] And, the total cost of articles \[=\text{ }Rs\text{ }600\] Again, (i) Cost of one article \[=\text{ }Rs\text{ }600/x\] (ii) also,...
If the given matrix and I is a unit matrix of order 2×2, find: (i) AI (ii) IB
Solution: (i) AI = (ii) IB=
If given matrix and I is a unit matrix of order 2×2, find: (i) AB (ii) BA
Solution: According to the given question (i) (ii)
The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it. (iv) Hence, find the speed of the train.
(iii) According to the question, \[4x\text{ }+\text{ }1728\text{ }=\text{ }{{x}^{2}}~+\text{ }16x\] Or, \[{{x}^{2}}~+\text{ }12x\text{ }\text{ }1728\text{ }=\text{ }0\] Or, \[{{x}^{2}}~+\text{...
Evaluate: if possible: If not possible, give reason.
Solution: The product of the given matrices isn’t possible as per the rule the number of columns in the first matrix isn’t equal to the number of rows in the second matrix.
Evaluate: if possible: If not possible, give reason.
Solution: \[=\text{ }\left[ 6\text{ }+\text{ }0 \right]\text{ }=\text{ }\left[ 6 \right]\] \[=\text{ }\left[ -2+2\text{ }3-8 \right]\text{ }=\text{ }\left[ 0\text{ }-5 \right]\]
The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (i) the time taken by the car to reach town B from A, in terms of x; (ii) the time taken by the train to reach town B from A, in terms of x.
According to ques, Speed of car = \[x\text{ }km/hr\] Speed of train = \[\left( x\text{ }+\text{ }16 \right)\text{ }km/hr\] Time = \[Distance/\text{ }Speed\] (i)Time taken by the car to reach town B...
(i) find the matrix 2A + B. (ii) find a matrix C such that:
(ii) Solution: (i) \[2A\text{ }+\text{ }B\] (ii)
Solve:
Solution: According to the given question, the matrix is
From given data below find (i) 2A – 3B + C (ii) A + 2C – B
Solution: \[\left( i \right)\text{ }2A\text{ }-\text{ }3B\text{ }+\text{ }C\] \[\left( ii \right)\text{ }A\text{ }+\text{ }2C\text{ }-\text{ }B\]
Find x and y if: (i) 3[4 x] + 2[y -3] = [10 0]
(ii) Solution: From L.H.S, we have \[3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]\] \[=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~\]...
Evaluate:
(I) (ii) Solution: According to the given ques, (i) (ii)
Evaluate: (i) 3[5 -2]
(ii) Solution: (i) \[3\left[ 5\text{ }-2 \right]\text{ }=\text{ }\left[ 3\times 5\text{ }3x-2 \right]\text{ }=\text{ }\left[ 15\text{ }-6 \right]\] (ii)
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. Find its common ratio.
According to the given ques, Hence, the common ratio is \[3/5.\]
How many terms of the series 2 + 6 + 18 + ….. must be taken to make the sum equal to 728?
. According to the given question, G.P: \[2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..\] Here, \[a\text{ }=\text{ }2\] And \[r\text{ }=\text{ }6/2\text{ }=\text{ }3\] Also given,...
Find the sum of G.P.: 3, 6, 12, …., 1536.
According to the given question, G.P: \[3,\text{ }6,\text{ }12,\text{ }\ldots .,\text{ }1536\] So, \[a\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }1536\] and \[r\text{ }=\text{ }6/3\text{ }=\text{...
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
According to the given question, For a G.P., \[r\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }486\] and \[{{S}_{n}}~=\text{ }728\] \[1458\text{ }-\text{ }a\text{ }=\text{ }728\text{ }x\text{ }2\text{...
The 4th term and the 7th term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P.
According to the given question, \[{{t}_{4}}~=\text{ }1/27\] and \[{{t}_{7~}}=\text{ }1/729\] We know that, \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4~}}=\text{...
A boy spends Rs.10 on first day, Rs.20 on second day, Rs.40 on third day and so on. Find how much, in all, will he spend in 12 days?
Amount spent on \[{{1}^{st}}~day\text{ }=\text{ }Rs\text{ }10\] Amount spent on \[{{2}^{nd}}~day\text{ }=\text{ }Rs\text{ }20\] Amount spent on \[{{3}^{rd}}~day\text{ }=\text{ }Rs\text{ }40\]...
The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.
\[First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27\] And, \[{{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81\]...
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
According to the given question, G.P: \[1\text{ }+\text{ }4\text{ }+\text{ }16\text{ }+\text{ }64\text{ }+\text{ }\ldots \ldots ..\] Here, \[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{...
Find the sum of G.P.:
(i) (ii) Solution: (i) According to the given question Here, \[a\text{ }=\text{ }\left( x\text{ }+\text{ }y \right)/\text{ }\left( x\text{ }-\text{ }y \right)\] And \[~r\text{ }=\text{ }1/\left[...
Find the sum of G.P.: (i) 1 – 1/2 + 1/4 – 1/8 + …….. to 9 terms (ii) 1 – 1/3 + 1/32 – 1/33 + ……… to n terms
(i) According to the given question G.P: \[1\text{ }-\text{ }1/2\text{ }+\text{ }1/4\text{ }-\text{ }1/8\text{ }+\text{ }\ldots \ldots ..\text{ }to\text{ }9\text{ }terms\] Here, \[a\text{ }=\text{...
Find the sum of G.P.: (i) 1 + 3 + 9 + 27 + ………. to 12 terms (ii) 0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 terms.
(i) According to the given question G.P: \[1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }+\text{ }27\text{ }+\text{ }\ldots \ldots \ldots .\text{ }to\text{ }12\text{ }terms\] Here, \[a\text{ }=\text{...
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that: (q – r) log a + (r – p) log b + (p – q) log c = 0
The first term of the G.P. be A and its common ratio be R. Hence, \[{{p}^{th}}~term\text{ }=\text{ }a\text{ }\Rightarrow \text{ }A{{R}^{p\text{ }-\text{ }1}}~=\text{ }a\] \[{{q}^{th}}~term\text{...
Find the G.P. 1/27, 1/9, 1/3, ……, 81; find the product of fourth term from the beginning and the fourth term from the end.
According to the given question, G.P. \[1/27,\text{ }1/9,\text{ }1/3,\text{ }\ldots \ldots ,\text{ }81\] Here, \[a\text{ }=\text{ }1/27,\] \[common\text{ }ratio\text{ }\left( r \right)\text{...
Find the third term from the end of the G.P. 2/27, 2/9, 2/3, ……., 162
Given series: \[2/27,\text{ }2/9,\text{ }2/3,\text{ }\ldots \ldots .,\text{ }162\] Here, \[a\text{ }=\text{ }2/27\] \[r\text{ }=\text{ }\left( 2/9 \right)\text{ }/\text{ }\left( 2/27 \right)\]...
Prove that : (v)
Prove that : (iii) (iv)
(iii) $\sin \left(28^{\circ}+\mathrm{A}\right)=\sin \left[90^{\circ}-\left(62^{\circ}-\mathrm{A}\right)\right]=\cos \left(62^{\circ}-\mathrm{A}\right)$ (iv)
Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32
Given series: \[\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32\] Here, \[a\text{ }=\text{ }\surd 2\] \[r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2\]...
Prove that: (i) (ii)
(i) $\tan \left(55^{\circ}+x\right)=\tan \left[90^{\circ}-\left(35^{\circ}-x\right)\right]=\cot \left(35^{\circ}-x\right)$ (ii) $\sec \left(70^{\circ}-\theta\right)=\sec...
Evaluate: (vii) (viii)
(vii) $3 \cos 80^{\circ} \operatorname{cosec} 10^{\circ}+2 \cos 59^{\circ} \operatorname{cosec} 31^{\circ}$ $=3 \cos (90-10)^{0} \operatorname{cosec} 10^{\circ}+2 \cos (90-31)^{0}...
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term
According to the given question, Product of \[{{3}^{rd}}~and\text{ }{{8}^{th}}\] terms of a G.P. is \[243\] The general term of a G.P. First term \[a\] And Common ratio \[r\]is given by,...
Evaluate : (v) (vi)
(v) $\sin 27^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \cos 27^{\circ}$ $=\sin (90-63)^{0} \sin 63^{\circ}-\cos 63^{\circ} \cos (90-63)^{\circ}$ $=\cos 63^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \sin...
Evaluate : (iii) (iv)
(iii) (iv) $\cos 40^{\circ} \operatorname{cosec} 50^{\circ}+\sin 50^{\circ}$ sec $40^{\circ}$ $=\cos (90-50)^{0} \operatorname{cosec} 50^{\circ}+\sin (90-50)^{0} \sec 40^{\circ}$ $=\sin 50^{\circ}...
Evaluate:(i) (ii)
(i) (ii) 1+1=2
If the first and the third terms of a G.P are 2 and 8 respectively, find its second term.
According to the given question, \[{{t}_{1}}~=\text{ }2\text{ }and\text{ }{{t}_{3}}~=\text{ }8\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{1}}~=\text{...
Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P.
Given, \[{{t}_{4}}~=\text{ }1/18\text{ }and\text{ }{{t}_{7}}~=\text{ }-1/486\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4}}~=\text{ }a{{r}^{4\text{ }-\text{...
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
According to the given question, \[{{t}_{5}}~=\text{ }81\text{ }and\text{ }{{t}_{2}}~=\text{ }24\] We know that, General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So,...
(i) If , show that: (ii) If , show that:
(i) Since, $2 \sin \mathrm{A}-1=0$ Therefore, $\sin A=1 / 2$ since, $\sin 30^{\circ}=1 / 2$ Hence, $\mathrm{A}=30^{\circ}$ LHS = $\sin 3 A=\sin 3\left(30^{\circ}\right)=\sin 30^{\circ}=1$...
If tan A = n tan B and sin A = m sin B, prove that: cos^2 A = m^2 – 1/ n^2 – 1
$\tan A=n \tan B$ $n=\tan A / \tan B$ And, $\sin A=m \sin B$ $\mathrm{m}=\sin \mathrm{A} / \sin \mathrm{B}$ Substituting RHS in $\mathrm{m}$ and $\mathrm{n}$ $m^{2}-1 / n^{2}-1$...
If , show that:
RHS, $\left(p^{2}-1\right) /\left(p^{2}+1\right)$ $=\frac{(\sec A+\tan A)^{2}-1}{(\sec A+\tan A)^{2}+1}$ $=\frac{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec A-1}{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec...
If and , show that:
LHS = $a^{2} / x^{2}-b^{2} / y^{2}$ $=\frac{a^{2}}{a^{2} \cos ^{2} \theta}-\frac{b^{2}}{b^{2} \cot ^{2} \theta}$ $=\frac{1}{\cos ^{2} \theta}-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$...
If sin A + cos A = p and sec A + cosec A = q, then prove that: q(p^2 – 1) = 2p
LHS = $q\left(p^{2}-1\right)=(\sec A+\operatorname{cosec} A)\left[(\sin A+\cos A)^{2}-1\right]$ $=(\sec A+\operatorname{cosec} A)\left[\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1\right]$ $=(\sec...
Which term of the G.P. :
Solution: In the given G.P. First term, \[a\text{ }=\text{ }-10\] Common ratio, \[r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)\]...
Prove: (xv) (xvi)
(xv) LHS = $\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}$ $=\sec ^{4} \mathrm{~A}\left(1-\sin ^{2} \mathrm{~A}\right)\left(1+\sin ^{2} \mathrm{~A}\right)-2 \tan...
Prove : (xvii)
$(1+\tan A+\sec A)(1+\cot A-\operatorname{cosec} A)$ $=1+\cot A-\operatorname{cosec} A+\tan A+1-\sec A+\sec A+\operatorname{cosec} A-\operatorname{cosec} A \sec A$ $=2+\cos \mathrm{A} / \sin...
Prove : (xiii) (xiv)
LHS = = RHS (xiv) LHS = = RHS
The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
According to the given question, The mid-point of \[\left( 2a,\text{ }4 \right)\text{ }and\text{ }\left( -2,\text{ }2b \right)\text{ }is\text{ }\left( 1,\text{ }2a\text{ }+\text{ }1 \right)\] By...
Prove : (xi) (xii)
LHS = = RHS (xii) LHS = = RHS
Prove (ix) (x)
LHS= = RHS (x) LHS =
Prove : (vii) (viii)
(vii) LHS =$=(\sin A /(1-\cos A))-\cot A$ Since, $\cot A=\cos A / \sin A$ $=\left(\sin ^{2} A-\cos A+\cos ^{2} A\right) /(1-\cos A) \sin A$ $=(1-\cos A) /(1-\cos A) \sin A$ $=1 / \sin \mathrm{A}$...
Prove: (v) (vi)
(v) LHS= cot A/ (1 – tan A) + tan A/ (1 – cot A) = RHS (vi) LHS= cos A/ (1 + sin A) + tan A = RHS
The mid-point of the line-segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
According to the given question, The mid-point of the line-segment joining \[\left( 4a,\text{ }2b\text{ }-\text{ }3 \right)\text{ }and\text{ }\left( -4,\text{ }3b \right)\text{ }is\text{ }\left(...
Prove: (iii) (iv)
(iii) LHS= 1 – sin2 A/ (1 + cos A) = RHS (iv) LHS= (1 – cos A)/ sin A + sin A/ (1 – cos A) = RHS
Find the co-ordinates of the centroid of a triangle ABC whose vertices are: A (-1, 3), B (1, -1) and C (5, 1)
By the centroid of a triangle formula, we get the co- ordinates of the centroid of triangle \[ABC\]as \[=\text{ }\left( 5/3,\text{ }1 \right)\]
Use tables to find sine of: (i) 21° (ii) 34° 42′
(i) Taking LHS, $1 /(\cos A+\sin A)+1 /(\cos A-\sin A)$ (ii) Taking LHS, $\operatorname{cosec} A-\cot A$ $=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin...
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find (i) the length of radius AC (ii) the coordinates of B.
(i) \[Radius\text{ }AC\text{ }=\text{ }\surd [\text{ }{{\left( 3\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( -7\text{ }-\text{ }5 \right)}^{2}}~]\] \[=\text{ }\surd [\text{ }({{5}^{2}}~+\text{...
Use trigonometrical tables to find tangent of: (iii) 17° 27′
(iii) $\tan 17^{\circ} 27^{\prime}=\tan \left(17^{\circ} 24^{\prime}+3^{\prime}\right)=0.3134+0.0010=0.3144$
Use trigonometrical tables to find tangent of: (i) 37° (ii) 42° 18′
(i) $\tan 37^{\circ}=0.7536$ (ii) $\tan 42^{\circ} 18^{\prime}=0.9099$
The mid-point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B.
Solution: As, \[point\text{ }A\text{ }lies\text{ }on\]\[x-axis\], we can assume the co-ordinates of \[point\text{ }A\text{ }to\text{ }be\text{ }\left( x,\text{ }0 \right).\] As, \[point\text{...
Use tables to find cosine of: (v) 9° 23’ + 15° 54’
$(\mathrm{v}) \cos \left(9^{\circ} 23^{\circ}+15^{\circ} 54^{\circ}\right)=\cos 24^{\circ} 77^{\circ}=\cos 25^{\circ} 17^{\circ}=\cos \left(25^{\circ}...
Use tables to find cosine of: (iii) 26° 32’ (iv) 65° 41’
(iii) $\cos 26^{\circ} 32^{\prime}=\cos \left(26^{\circ} 30^{\prime}+2^{\prime}\right)=0.8949-0.0003=0.8946$ (iv) $\cos 65^{\circ} 41^{\prime}=\cos \left(65^{\circ}...
In what ratio is the line joining A (0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates of the point where AB intersects the x-axis.
Suppose, that the line segment \[AB\]intersects the \[x-axis\]by point \[P\text{ }\left( x,\text{ }0 \right)\]in the ratio \[k:\text{ }1.\] \[0\text{ }=\text{ }\left( -k\text{ }+\text{ }3...
Use tables to find cosine of: (i) 2° 4’ (ii) 8° 12’
(i) $\cos 2^{\circ} 4^{\prime}=0.9994-0.0001=0.9993$ (ii) $\cos 8^{\circ} 12^{\prime}=\cos 0.9898$
Use tables to find sine of: (v) 10° 20′ + 20° 45′
(v) $\sin \left(10^{\circ} 20^{\prime}+20^{\circ} 45^{\prime}\right)=\sin 30^{\circ} 65^{\prime}=\sin 31^{\circ} 5^{\prime}=0.5150+0.0012=0.5162$
A line segment joining A (-1, 5/3) and B (a, 5) is divided in the ratio 1: 3 at P, point where the line segment AB intersects the y-axis. (i) Calculate the value of ‘a’. (ii) Calculate the co-ordinates of ‘P’.
As, the line segment \[AB\] intersects the \[y-axis\text{ }at\text{ }point\text{ }P\], suppose the co-ordinates of \[point\text{ }P\] be \[\left( 0,\text{ }y \right).\] And, \[P\text{...
Use tables to find sine of: (iii) 47° 32′ (iv) 62° 57′
(iii) $\sin 47^{\circ} 32^{\prime}=\sin \left(47^{\circ} 30^{\prime}+2^{\prime}\right)=0.7373+0.0004=0.7377$ (iv) $\sin 62^{\circ} 57^{\prime}=\sin \left(62^{\circ}...
Use tables to find sine of: (i) 21° (ii) 34° 42′
(i) $\sin 21^{\circ}=0.3584$ (ii) $\sin 34^{\circ} 42^{\prime}=0.5693$
Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.
Suppose \[P\text{ }and\text{ }Q\] to be the points of trisection of the line segment joining \[A\text{ }\left( 6,\text{ }-9 \right)\text{ }and\text{ }B\text{ }\left( 0,\text{ }0 \right).\] So,...
A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP: PB = 3: 5 and AQ: QC = 3: 5. Show that: PQ = 3/8 BC.
According to the given question, Point \[P\text{ }lies\text{ }on\text{ }AB\] such that \[AP:\text{ }PB\text{ }=\text{ }3:\text{ }5.\] So, the co-ordinates of point P are \[=\text{ }\left(...
A triangle ABC is right angled at B; find the value of (sec A. cosec C – tan A. cot C)/ sin B
Since, ABC is a right angled triangle right angled at B Therefore, \[\begin{array}{*{35}{l}} A\text{ }+\text{ }C\text{ }=\text{ }{{90}^{o}} \\ \left( sec\text{ }A.\text{ }cosec\text{ }C\text{...
A (20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of a point P in AB such that: 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
According to the given question, \[3PB\text{ }=\text{ }AB\] So, \[AB/PB\text{ }=\text{ }3/1\] \[\left( AB\text{ }-\text{ }PB \right)/\text{ }PB\text{ }=\text{ }\left( 3\text{ }-\text{ }1...
Evaluate: (ix)
(ix) \[14\text{ }sin\text{ }{{30}^{o}}~+\text{ }6\text{ }cos\text{ }{{60}^{o}}~-\text{ }5\text{ }tan\text{ }{{45}^{o}}\] \[=\text{ }14\text{ }\left( 1/2 \right)\text{ }+\text{ }6\text{ }\left( 1/2...
Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP: PC = 3: 2. Find the length of line segment AP.
According to the given question, \[BP:\text{ }PC\text{ }=\text{ }3:\text{ }2\] Then by section formula, the co-ordinates of \[point\text{ }P\]are: \[=\text{ }\left( 15/5,\text{ }40/5 \right)\]...