Saturday, 21 April 2018

Biomolecules
Class XI Chapter 9– Biomolecules Biology Question 1: What are macromolecules? Give examples. Answer Macromolecules are large complex molecules that occur in colloidal state in intercellular fluid. They are formed by the polymerization of low molecular weight micromolecules. Polysaccharides, proteins, and nucleic acids are common examples of macromolecules. Question 2: Illustrate a glycosidic, peptide and a phospho-diester bond. Answer (a) Glycosidic bond is formed normally between carbon atoms, 1 and 4, of neighbouring monosaccharide units. (b) Peptide bond is a covalent bond that joins the two amino acids by – NH – CO linkage. Page 1 of 13 (c) Phosphodiester bond is a strong covalent bond between phosphate and two sugar groups. Such bonds form the sugar phosphate backbone of nucleic acids. Question 3: What is meant by tertiary structure of proteins? Answer The helical polypeptide chain undergoes coiling and folding to form a complex threedimensional shape referred to as tertiary structure of proteins. These coils and folds are arranged to hide the non-polar amino acid chains and to expose the polar side chains. The tertiary structure is held together by the weak bonds formed between various parts of the polypeptide chain. Question 4: Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers. Answer (a) Class XI Chapter 9– Biomolecules Biology Page 2 of 13 Class XI Chapter 9– Biomolecules Biology Page 3 of 13 Molecule Structure 1. Adenosine 2. Thymidine 3. Sucrose 4. Maltose Class XI Chapter 9– Biomolecules Biology Page 4 of 13 5. Lactose 6. Ribose 7. DNA Class XI Chapter 9– Biomolecules Biology Page 5 of 13 8. RNA 9. Glycerol 10. Insulin (b) Compound Manufacturer Buyer 1. Starch products Kosha Impex (P) Ltd. Research laboratories, educational institutes, and other industries, which use biomolecules as a precursor for 2. making other products. Liquid glucose Marudhar apparels Class XI Chapter 9– Biomolecules Biology Page 6 of 13 3. Various enzymes such as amylase, protease, cellulase Map (India) Ltd Question 5: Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein? Answer Yes, if we are given a method to know the sequence of proteins, we can connect this information to the purity of a protein. It is known that an accurate sequence of a certain amino acid is very important for the functioning of a protein. If there is any change in the sequence, it would alter its structure, thereby altering the function. If we are provided with a method to know the sequence of an unknown protein, then using this information, we can determine its structure and compare it with any of the known correct protein sequence. Any change in the sequence can be linked to the purity or homogeneity of a protein. For example, any one change in the sequence of haemoglobin can alter the normal haemoglobin structure to an abnormal structure that can cause sickle cell anaemia. Question 6: Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., cosmetics, etc.) Answer Proteins used as therapeutic agents are as follows: 1. Thrombin and fibrinogen – They help in blood clotting. 2. Antigen (antibody) – It helps in blood transfusion. 3. Insulin – It helps in maintaining blood glucose level in the body. 4. Renin – It helps in osmoregulation. Class XI Chapter 9– Biomolecules Biology Page 7 of 13 Proteins are also commonly used in the manufacture of cosmetics, toxins, and as biological buffers. Question 7: Explain the composition of triglyceride. Answer Triglyceride is a glyceride, which is formed from a single molecule of glycerol, esterified with three fatty acids. It is mainly present in vegetable oils and animal fat. Structure of triglyceride The general chemical formula of triglyceride is , where R1, R2, and R3 are fatty acids. These three fatty acids can be same or different. Question 8: Can you describe what happens when milk is converted into curd or yoghurt from your understanding of proteins. Answer Proteins are macromolecules formed by the polymerization of amino acids. Structurally, proteins are divided into four levels. (a) Primary structure – It is the linear sequence of amino acids in a polypeptide chain. (b) Secondary structure – The polypeptide chain is coiled to form a threedimensional structure. (c) Tertiary structure – The helical polypeptide chain is further coiled and folded to form a complex structure. Class XI Chapter 9– Biomolecules Biology Page 8 of 13 (d) Quaternary structure – More than one polypeptide chains assemble to form the quaternary structure. Milk has many globular proteins. When milk is converted into curd or yoghurt, these complex proteins get denatured, thus converting globular proteins into fibrous proteins. Therefore, by the process of denaturation, the secondary and tertiary structures of proteins are destroyed. Question 9: Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models). Answer Ball and stick models are 3-D molecular models that can be used to describe the structure of biomolecules. In ball and stick model, the atoms are represented as balls whereas the bonds that hold the atoms are represented by the sticks. Double and triple bonds are represented by springs that form curved connections between the balls. The size and colour of various atoms are different and are depicted by the relative size of the balls. It is the most fundamental and common model of representing biomolecular structures. Class XI Chapter 9– Biomolecules Biology Page 9 of 13 In the above ball and stick model of D-glucose, the oxygen atoms are represented by red balls, hydrogen atoms by blue balls, while carbon atoms are represented by grey balls. Question 10: Attempt titrating an amino acid against a weak base and discover the number of dissociating ( ionizable ) functional groups in the amino acid. Answer Titrating a neutral or basic amino acid against a weak base will dissociate only one functional group, whereas titration between acidic amino acid and a weak acid will dissociate two or more functional groups. Question 11: Draw the structure of the amino acid, alanine. Answer Structure of alanine Question 12: What are gums made of? Is Fevicol different? Answer Gums are hetero-polysaccharides. They are made from two or more different types of monosaccharides. On the other hand, fevicol is polyvinyl alcohol (PVA) glue. It is not a polysaccharide. Class XI Chapter 9– Biomolecules Biology Page 10 of 13 Question 13: Find out a qualitative test for proteins, fats and oils, amino acids and test any fruit juice, saliva, sweat and urine for them. Answer (a) Test for protein Biuret’s test – If Biuret’s reagent is added to protein, then the colour of the reagent changes from light blue to purple. (b) Test for fats and oils Grease or solubility test (c) Test for amino acid Ninhydrin test – If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue, or purple, depending on the amino acid. Item Name of the test Procedure Result Inference 1. Fruit juice Biuret’s test Fruit juice + Biuret’s reagent Colour changes from light blue to purple Protein is present. Grease test To a brown paper, add a few drops of fruit juice. No translucent spot Fats and oils are absent or are in negligible amounts. Ninhydrin test Fruit juice + Ninhydrin reagent + boil for 5 minutes Colourless solution changes to pink, blue, or purple colour Amino acids are present. 2. Saliva Biuret’s Saliva + Biuret’s Colour changes from Proteins are Class XI Chapter 9– Biomolecules Biology Page 11 of 13 test reagent light blue to purple present. Grease test On a brown paper, add a drop of saliva. No translucent spot Fats/oils are absent. Ninhydrin test Saliva + Ninhydrin reagent + boil for 5 minutes Colourless solution changes to pink, blue, or purple colour Amino acids are present. 3. Sweat Biuret’s test Sweat + Biuret’s reagent No colour change Proteins are absent. Solubility test Sweat + Water Oily appearance Fats/oil may be present. Ninhydrin test Sweat + Ninhydrin reagent + boil for 5 minutes No colour change, solution remains colourless Amino acids are absent. 4. Urine Biuret’s test Few drops of urine + Biuret’s reagent Colour changes from light blue to purple Proteins are present. Solubility test Few drops of urine + Water Little bit of oily appearance Fats may or may not be present. Ninhydrin test Few drops of urine + Colourless solution changes to pink, Amino acids are present. Class XI Chapter 9– Biomolecules Biology Page 12 of 13 Ninhydrin reagent + boil for 5 minutes blue, or purple colour depending on the type of amino acid Question 14: Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation! Answer Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man. Question 15: Describe the important properties of enzymes. Answer Properties of enzymes (1) Enzymes are complex macromolecules with high molecular weight. (2) They catalyze biochemical reactions in a cell. They help in the breakdown of large molecules into smaller molecules or bring together two smaller molecules to form a larger molecule. (3) Enzymes do not start a reaction. However, they help in accelerating it. (4) Enzymes affect the rate of biochemical reaction and not the direction. (5) Most of the enzymes have high turnover number. Turnover number of an enzyme is the number of molecules of a substance that is acted upon by an enzyme per minute. High turnover number of enzymes increases the efficiency of reaction. Class XI Chapter 9– Biomolecules Biology Page 13 of 13 (6) Enzymes are specific in action. (7) Enzymatic activity decreases with increase in temperature. (8) They show maximum activity at an optimum pH of 6 – 8. (9) The velocity of enzyme increases with increase in substrate concentration and then, ultimately reaches maximum velocity.

olympia academy
Exercise 14.1 solutions Question 1: Which of the following sentences are statements? Give reasons for your answer. (i) There are 35 days in a month. (ii) Mathematics is difficult. (iii) The sum of 5 and 7 is greater than 10. (iv) The square of a number is an even number. (v) The sides of a quadrilateral have equal length. (vi) Answer this question. (vii) The product of (–1) and 8 is 8. (viii) The sum of all interior angles of a triangle is 180°. (ix) Today is a windy day. (x) All real numbers are complex numbers. Answer : (i) This sentence is incorrect because the maximum number of days in a month is 31. Hence, it is a statement. (ii) This sentence is subjective in the sense that for some people, mathematics can be easy and for some others, it can be difficult. Hence, it is not a statement. (iii) The sum of 5 and 7 is 12, which is greater than 10. Therefore, this sentence is always correct. Hence, it is a statement. (iv) This sentence is sometimes correct and sometimes incorrect. For example, the square of 2 is an even number. However, the square of 3 is an odd number. Hence, it is not a statement. (v) This sentence is sometimes correct and sometimes incorrect. For example, squares and rhombus have sides of equal lengths. However, trapezium and rectangles have sides of unequal lengths. Hence, it is not a statement. (vi) It is an order. Therefore, it is not a statement. (vii) The product of (–1) and 8 is (–8). Therefore, the given sentence is incorrect. Hence, it is a statement. (viii) This sentence is correct and hence, it is a statement. (ix) The day that is being referred to is not evident from the sentence. Hence, it is not a statement. (x) All real numbers can be written as a × 1 + 0 × i. Therefore, the given sentence is always correct. Hence, it is a statement. Question 2: Give three examples of sentences which are not statements. Give reasons for the answers. Answer : The three examples of sentences, which are not statements, are as follows. (i) He is a doctor. It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 1 of 12 (ii) Geometry is difficult. This is not a statement because for some people, geometry can be easy and for some others, it can be difficult. (iii) Where is she going? This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement. Exercise 14.2 solutions Question 1: Write the negation of the following statements: (i) Chennai is the capital of Tamil Nadu. (ii) is not a complex number. (iii) All triangles are not equilateral triangle. (iv) The number 2 is greater than 7. (v) Every natural number is an integer. Answer : (i) Chennai is not the capital of Tamil Nadu. (ii) is a complex number. (iii) All triangles are equilateral triangles. (iv) The number 2 is not greater than 7. (v) Every natural number is not an integer. Question 2: Are the following pairs of statements negations of each other? (i) The number x is not a rational number. The number x is not an irrational number. (ii) The number x is a rational number. The number x is an irrational number. Answer : (i) The negation of the first statement is “the number x is a rational number”. This is same as the second statement. This is because if a number is not an irrational number, then it is a rational number. Therefore, the given statements are negations of each other. (ii) The negation of the first statement is “the number x is not a rational number”. This means that the number x is an irrational number, which is the same as the second statement. Therefore, the given statements are negations of each other. Question 3: Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 2 of 12 Find the component statements of the following compound statements and check whether they are true or false. (i) Number 3 is prime or it is odd. (ii) All integers are positive or negative. (iii) 100 is divisible by 3, 11 and 5. Answer : (i) The component statements are as follows. p: Number 3 is prime. q: Number 3 is odd. Both the statements are true. (ii) The component statements are as follows. p: All integers are positive. q: All integers are negative. Both the statements are false. (iii) The component statements are as follows. p: 100 is divisible by 3. q: 100 is divisible by 11. r: 100 is divisible by 5. Here, the statements, p and q, are false and statement r is true. Exercise 14.3 solutions Question 1: For each of the following compound statements first identify the connecting words and then break it into component statements. (i) All rational numbers are real and all real numbers are not complex. (ii) Square of an integer is positive or negative. (iii) The sand heats up quickly in the Sun and does not cool down fast at night. (iv) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0. Answer : (i) Here, the connecting word is ‘and’. The component statements are as follows. p: All rational numbers are real. q: All real numbers are not complex. (ii) Here, the connecting word is ‘or’. The component statements are as follows. p: Square of an integer is positive. q: Square of an integer is negative. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 3 of 12 (iii) Here, the connecting word is ‘and’. The component statements are as follows. p: The sand heats up quickly in the sun. q: The sand does not cool down fast at night. (iv) Here, the connecting word is ‘and’. The component statements are as follows. p: x = 2 is a root of the equation 3x2 – x – 10 = 0 q: x = 3 is a root of the equation 3x2 – x – 10 = 0 Question 2: Identify the quantifier in the following statements and write the negation of the statements. (i) There exists a number which is equal to its square. (ii) For every real number x, x is less than x + 1. (iii) There exists a capital for every state in India. Answer : (i) The quantifier is “There exists”. The negation of this statement is as follows. There does not exist a number which is equal to its square. (ii) The quantifier is “For every”. The negation of this statement is as follows. There exist a real number x such that x is not less than x + 1. (iii) The quantifier is “There exists”. The negation of this statement is as follows. There exists a state in India which does not have a capital. Question 3: Check whether the following pair of statements is negation of each other. Give reasons for the answer. (i) x + y = y + x is true for every real numbers x and y. (ii) There exists real number x and y for which x + y = y + x. Answer : The negation of statement (i) is as follows. There exists real number x and y for which x + y ≠ y + x. This is not the same as statement (ii). Thus, the given statements are not the negation of each other. Question 4: State whether the “Or” used in the following statements is “exclusive “or” inclusive. Give reasons for your answer. (i) Sun rises or Moon sets. (ii) To apply for a driving licence, you should have a ration card or a passport. (iii) All integers are positive or negative. Answer : Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 4 of 12 (i) Here, “or” is exclusive because it is not possible for the Sun to rise and the moon to set together. (ii) Here, “or” is inclusive since a person can have both a ration card and a passport to apply for a driving licence. (iii) Here, “or” is exclusive because all integers cannot be both positive and negative. Exercise 14.4 solutions Question 1: Rewrite the following statement with “if-then” in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd. Answer : The given statement can be written in five different ways as follows. (i) A natural number is odd implies that its square is odd. (ii) A natural number is odd only if its square is odd. (iii) For a natural number to be odd, it is necessary that its square is odd. (iv) For the square of a natural number to be odd, it is sufficient that the number is odd. (v) If the square of a natural number is not odd, then the natural number is not odd. Question 2: Write the contrapositive and converse of the following statements. (i) If x is a prime number, then x is odd. (ii) It the two lines are parallel, then they do not intersect in the same plane. (iii) Something is cold implies that it has low temperature. (iv) You cannot comprehend geometry if you do not know how to reason deductively. (v) x is an even number implies that x is divisible by 4 Answer : (i) The contrapositive is as follows. If a number x is not odd, then x is not a prime number. The converse is as follows. If a number x is odd, then it is a prime number. (ii) The contrapositive is as follows. If two lines intersect in the same plane, then they are not parallel. The converse is as follows. If two lines do not intersect in the same plane, then they are parallel. (iii) The contrapositive is as follows. If something does not have low temperature, then it is not cold. The converse is as follows. If something is at low temperature, then it is cold. (iv) The contrapositive is as follows. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 5 of 12 If you know how to reason deductively, then you can comprehend geometry. The converse is as follows. If you do not know how to reason deductively, then you cannot comprehend geometry. (v) The given statement can be written as follows. If x is an even number, then x is divisible by 4. The contrapositive is as follows. If x is not divisible by 4, then x is not an even number. The converse is as follows. If x is divisible by 4, then x is an even number. Question 3: Write each of the following statement in the form “if-then”. (i) You get a job implies that your credentials are good. (ii) The Banana trees will bloom if it stays warm for a month. (iii) A quadrilateral is a parallelogram if its diagonals bisect each other. (iv) To get A+ in the class, it is necessary that you do the exercises of the book. Answer : (i) If you get a job, then your credentials are good. (ii) If the Banana tree stays warm for a month, then it will bloom. (iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. (iv) If you want to get an A+ in the class, then you do all the exercises of the book. Question 4: Given statements in (a) and (b). Identify the statements given below as contrapositive or converse of each other. (a) If you live in Delhi, then you have winter clothes. (i) If you do not have winter clothes, then you do not live in Delhi. (ii) If you have winter clothes, then you live in Delhi. (b) If a quadrilateral is a parallelogram, then its diagonals bisect each other. (i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram. (ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Answer : (a) (i) This is the contrapositive of the given statement (a). (ii) This is the converse of the given statement (a). (b) (i) This is the contrapositive of the given statement (b). (ii) This is the converse of the given statement (b). Exercise 14.5 solutions Question 1: Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 6 of 12 Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by (i) direct method (ii) method of contradiction (iii) method of contrapositive Answer : p: “If x is a real number such that x3 + 4x = 0, then x is 0”. Let q: x is a real number such that x3 + 4x = 0 r: x is 0. (i) To show that statement p is true, we assume that q is true and then show that r is true. Therefore, let statement q be true. ∴ x3 + 4x = 0 x (x2 + 4) = 0 ⇒ x = 0 or x2 + 4 = 0 However, since x is real, it is 0. Thus, statement r is true. Therefore, the given statement is true. (ii) To show statement p to be true by contradiction, we assume that p is not true. Let x be a real number such that x3 + 4x = 0 and let x is not 0. Therefore, x3 + 4x = 0 x (x2 + 4) = 0 x = 0 or x2 + 4 = 0 x = 0 or x2 = – 4 However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0. Thus, the given statement p is true. (iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false. Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement. ∼r: x is not 0. It can be seen that (x2 + 4) will always be positive. x ≠ 0 implies that the product of any positive real number with x is not zero. Let us consider the product of x with (x2 + 4). ∴ x (x2 + 4) ≠ 0 ⇒ x3 + 4x ≠ 0 This shows that statement q is not true. Thus, it has been proved that ∼r ⇒∼q Therefore, the given statement p is true. Question 2: Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 7 of 12 Show that the statement “For any real numbers a and b, a2 = b2 implies that a = b” is not true by giving a counter-example. Answer : The given statement can be written in the form of “if-then” as follows. If a and b are real numbers such that a2 = b2 , then a = b. Let p: a and b are real numbers such that a2 = b2 . q: a = b The given statement has to be proved false. For this purpose, it has to be proved that if p, then ∼q. To show this, two real numbers, a and b, with a2 = b2 are required such that a ≠ b. Let a = 1 and b = –1 a2 = (1)2 = 1 and b2 = (– 1)2 = 1 ∴ a2 = b2 However, a ≠ b Thus, it can be concluded that the given statement is false. Question 3: Show that the following statement is true by the method of contrapositive. p: If x is an integer and x2 is even, then x is also even. Answer : p: If x is an integer and x2 is even, then x is also even. Let q: x is an integer and x2 is even. r: x is even. To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false. Let x is not even. To prove that q is false, it has to be proved that x is not an integer or x2 is not even. x is not even implies that x2 is also not even. Therefore, statement q is false. Thus, the given statement p is true. Question 4: By giving a counter example, show that the following statements are not true. (i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle. (ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2. Answer : (i) The given statement is of the form “if q then r”. q: All the angles of a triangle are equal. r: The triangle is an obtuse-angled triangle. The given statement p has to be proved false. For this purpose, it has to be proved that if q, then ∼r. To show this, angles of a triangle are required such that none of them is an obtuse angle. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 8 of 12 It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle. In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle. Thus, it can be concluded that the given statement p is false. (ii) The given statement is as follows. q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2. This statement has to be proved false. To show this, a counter example is required. Consider x2 – 1 = 0 x2 = 1 x = ± 1 One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2. Thus, the given statement is false. Question 5: Which of the following statements are true and which are false? In each case give a valid reason for saying so. (i) p: Each radius of a circle is a chord of the circle. (ii) q: The centre of a circle bisects each chord of the circle. (iii) r: Circle is a particular case of an ellipse. (iv) s: If x and y are integers such that x > y, then –x < –y. (v) t: is a rational number. Answer : (i) The given statement p is false. According to the definition of chord, it should intersect the circle at two distinct points. (ii) The given statement q is false. If the chord is not the diameter of the circle, then the centre will not bisect that chord. In other words, the centre of a circle only bisects the diameter, which is the chord of the circle. (iii) The equation of an ellipse is, If we put a = b = 1, then we obtain x2 + y2 = 1, which is an equation of a circle Therefore, circle is a particular case of an ellipse. Thus, statement r is true. (iv) x > y ⇒ –x < –y (By a rule of inequality) Thus, the given statement s is true. (v) 11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore, is an irrational number. Thus, the given statement t is false. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 9 of 12 MISCELLANEOUS of chapter 14 Question 1: Write the negation of the following statements: (i) p: For every positive real number x, the number x – 1 is also positive. (ii) q: All cats scratch. (iii) r: For every real number x, either x > 1 or x < 1. (iv) s: There exists a number x such that 0 < x < 1. Answer : (i) The negation of statement p is as follows. There exists a positive real number x, such that x – 1 is not positive. (ii) The negation of statement q is as follows. There exists a cat that does not scratch. (iii) The negation of statement r is as follows. There exists a real number x, such that neither x > 1 nor x < 1. (iv) The negation of statement s is as follows. There does not exist a number x, such that 0 < x < 1. Question 2: State the converse and contrapositive of each of the following statements: (i) p: A positive integer is prime only if it has no divisors other than 1 and itself. (ii) q: I go to a beach whenever it is a sunny day. (iii) r: If it is hot outside, then you feel thirsty. Answer : (i) Statement p can be written as follows. If a positive integer is prime, then it has no divisors other than 1 and itself. The converse of the statement is as follows. If a positive integer has no divisors other than 1 and itself, then it is prime. The contrapositive of the statement is as follows. If positive integer has divisors other than 1 and itself, then it is not prime. (ii) The given statement can be written as follows. If it is a sunny day, then I go to a beach. The converse of the statement is as follows. If I go to a beach, then it is a sunny day. The contrapositive of the statement is as follows. If I do not go to a beach, then it is not a sunny day. (iii) The converse of statement r is as follows. If you feel thirsty, then it is hot outside. The contrapositive of statement r is as follows. If you do not feel thirsty, then it is not hot outside. Question 3: Write each of the statements in the form “if p, then q”. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 10 of 12 (i) p: It is necessary to have a password to log on to the server. (ii) q: There is traffic jam whenever it rains. (iii) r: You can access the website only if you pay a subscription fee. Answer : (i) Statement p can be written as follows. If you log on to the server, then you have a password. (ii) Statement q can be written as follows. If it rains, then there is a traffic jam. (iii) Statement r can be written as follows. If you can access the website, then you pay a subscription fee. Question 4: Re write each of the following statements in the form “p if and only if q”. (i) p: If you watch television, then your mind is free and if your mind is free, then you watch television. (ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly. (iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular. Answer : (i) You watch television if and only if your mind is free. (ii) You get an A grade if and only if you do all the homework regularly. (iii) A quadrilateral is equiangular if and only if it is a rectangle. Question 5: Given below are two statements p: 25 is a multiple of 5. q: 25 is a multiple of 8. Write the compound statements connecting these two statements with “And” and “Or”. In both cases check the validity of the compound statement. Answer : The compound statement with ‘And’ is “25 is a multiple of 5 and 8”. This is a false statement, since 25 is not a multiple of 8. The compound statement with ‘Or’ is “25 is a multiple of 5 or 8”. This is a true statement, since 25 is not a multiple of 8 but it is a multiple of 5. Question 6: Check the validity of the statements given below by the method given against it. (i) p: The sum of an irrational number and a rational number is irrational (by contradiction method). (ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method). Answer : Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 11 of 12 (i) The given statement is as follows. p: the sum of an irrational number and a rational number is irrational. Let us assume that the given statement, p, is false. That is, we assume that the sum of an irrational number and a rational number is rational. Therefore, , where is irrational and b, c, d, e are integers. is a rational number and is an irrational number. This is a contradiction. Therefore, our assumption is wrong. Therefore, the sum of an irrational number and a rational number is rational. Thus, the given statement is true. (ii) The given statement, q, is as follows. If n is a real number with n > 3, then n2 > 9. Let us assume that n is a real number with n > 3, but n2 > 9 is not true. That is, n2 < 9 Then, n > 3 and n is a real number. Squaring both the sides, we obtain n2 > (3)2 ⇒ n2 > 9, which is a contradiction, since we have assumed that n2 < 9. Thus, the given statement is true. That is, if n is a real number with n > 3, then n2 > 9. Question 7: Write the following statement in five different ways, conveying the same meaning. p: If triangle is equiangular, then it is an obtuse angled triangle. Answer : The given statement can be written in five different ways as follows. (i) A triangle is equiangular implies that it is an obtuse-angled triangle. (ii) A triangle is equiangular only if it is an obtuse-angled triangle. (iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle. (iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular. (v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular. Class XI ______________________________ Chapter 14- Mathematical Reasoning ______________________________ Mathematics ______________________________ ____________________________________________________________________________________ Page 12 of 12

Exercise 16.1 solutions Question 1: Describe the sample space for the indicated experiment: A coin is tossed three times. Answer : A coin has two faces: head (H) and tail (T). When a coin is tossed three times, the total number of possible outcomes is 23 = 8 Thus, when a coin is tossed three times, the sample space is given by: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Question 2: Describe the sample space for the indicated experiment: A die is thrown two times. Answer : When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6. When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6} The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by: S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Question 3: Describe the sample space for the indicated experiment: A coin is tossed four times. Answer : When a coin is tossed once, there are two possible outcomes: head (H) and tail (T). When a coin is tossed four times, the total number of possible outcomes is 24 = 16 Thus, when a coin is tossed four times, the sample space is given by: S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} Question 4: Describe the sample space for the indicated experiment: A coin is tossed and a die is thrown. Answer : A coin has two faces: head (H) and tail (T). A die has six faces that are numbered from 1 to 6, with one number on each face. Thus, when a coin is tossed and a die is thrown, the sample space is given by: S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Question 5: Describe the sample space for the indicated experiment: A coin is tossed and then a die is rolled only in case a head is shown on the coin. Answer : Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 1 of 27 A coin has two faces: head (H) and tail (T). A die has six faces that are numbered from 1 to 6, with one number on each face. Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by: S = {H1, H2, H3, H4, H5, H6, T} Question 6: 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person. Answer : Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy and 3 girls in room Y as B3, and G3, G4, G5 respectively. Accordingly, the required sample space is given by S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5} Question 7: One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space. Answer : A die has six faces that are numbered from 1 to 6, with one number on each face. Let us denote the red, white, and blue dices as R, W, and B respectively. Accordingly, when a die is selected and then rolled, the sample space is given by S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6} Question 8: An experiment consists of recording boy-girl composition of families with 2 children. (i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births? (ii) What is the sample space if we are interested in the number of girls in the family? Answer : (i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB} (ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl. Hence, the required sample space is S = {0, 1, 2} Question 9: A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment. Answer : It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 2 of 27 When two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW} Question 10: An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space. Answer : A coin has two faces: head (H) and tail (T). A die has six faces that are numbered from 1 to 6, with one number on each face. Thus, in the given experiment, the sample space is given by S = {HH, HT, T1, T2, T3, T4, T5, T6} Question 11: Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment? Answer : 3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-defective (N). The sample space of this experiment is given by S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN} Question 12: A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment? Answer : When a coin is tossed, the possible outcomes are head (H) and tail (T). When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6. Thus, the sample space of this experiment is given by: S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66} Question 13: The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment. Answer : If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too. Thus, the sample space of this experiment is given by S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)} Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 3 of 27 Question 14: An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment. Answer : A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers. A coin has two faces: head (H) and tail (T). Hence, the sample space of this experiment is given by: S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT} Question 15: A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment. Answer : The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R1, R2 and the 3 black balls as B1, B2, and B3. The sample space of this experiment is given by S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6} Question 16: A die is thrown repeatedly until a six comes up. What is the sample space for this experiment? Answer : In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained. Hence, the sample space of this experiment is given by S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …} Exercise 16.2 solutions Question 1: A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive? Answer : When a die is rolled, the sample space is given by S = {1, 2, 3, 4, 5, 6} Accordingly, E = {4} and F = {2, 4, 6} Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 4 of 27 It is observed that E ∩ F = {4} ≠ Φ Therefore, E and F are not mutually exclusive events. Question 2: A die is thrown. Describe the following events: (i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3 (iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3 Also find Answer : When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}. Accordingly: (i) A = {1, 2, 3, 4, 5, 6} (ii) B = Φ (iii) C = {3, 6} (iv) D = {1, 2, 3} (v) E = {6} (vi) F = {3, 4, 5, 6} A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ B ∪ C = {3, 6}, E ∩ F = {6} D ∩ E =Φ, A – C = {1, 2, 4, 5} D – E = {1, 2, 3}, Question 3: An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die C: The sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive? Answer : When a pair of dice is rolled, the sample space is given by Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 5 of 27 It is observed that A ∩ B =Φ B ∩ C =Φ Hence, events A and B and events B and C are mutually exclusive. Question 4: Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”. C denote the event “three tails show” and D denote the event ‘a head shows on the first coin”. Which events are (i) mutually exclusive? (ii) simple? (iii) compound? Answer : When three coins are tossed, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Accordingly, A = {HHH} B = {HHT, HTH, THH} C = {TTT} D = {HHH, HHT, HTH, HTT} We now observe that A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ C ∩ D = Φ (i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive. (ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 6 of 27 (iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events. Question 5: Three coins are tossed. Describe (i) Two events which are mutually exclusive. (ii) Three events which are mutually exclusive and exhaustive. (iii) Two events, which are not mutually exclusive. (iv) Two events which are mutually exclusive but not exhaustive. (v) Three events which are mutually exclusive but not exhaustive. Answer : When three coins are tossed, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (i) Two events that are mutually exclusive can be A: getting no heads and B: getting no tails This is because sets A = {TTT} and B = {HHH} are disjoint. (ii) Three events that are mutually exclusive and exhaustive can be A: getting no heads B: getting exactly one head C: getting at least two heads i.e., A = {TTT} B = {HTT, THT, TTH} C = {HHH, HHT, HTH, THH} This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S (iii) Two events that are not mutually exclusive can be A: getting three heads B: getting at least 2 heads i.e., A = {HHH} B = {HHH, HHT, HTH, THH} This is because A ∩ B = {HHH} ≠ Φ (iv) Two events which are mutually exclusive but not exhaustive can be A: getting exactly one head B: getting exactly one tail That is A = {HTT, THT, TTH} B = {HHT, HTH, THH} It is because, A ∩ B =Φ, but A ∪ B ≠ S (v) Three events that are mutually exclusive but not exhaustive can be A: getting exactly three heads B: getting one head and two tails Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 7 of 27 C: getting one tail and two heads i.e., A = {HHH} B = {HTT, THT, TTH} C = {HHT, HTH, THH} This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ S Question 6: Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤ 5 Describe the events (i) (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) Answer : When two dice are thrown, the sample space is given by Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 8 of 27 (vii) B and C = B ∩ C Question 7: Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤ 5 State true or false: (give reason for your answer) (i) A and B are mutually exclusive (ii) A and B are mutually exclusive and exhaustive (iii) (iv) A and C are mutually exclusive (v) A and are mutually exclusive (vi) are mutually exclusive and exhaustive. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 9 of 27 Answer : (i) It is observed that A ∩ B = Φ ∴ A and B are mutually exclusive. Thus, the given statement is true. (ii) It is observed that A ∩ B = Φ and A ∪ B = S ∴ A and B are mutually exclusive and exhaustive. Thus, the given statement is true. (iii) It is observed that Thus, the given statement is true. (iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠Φ ∴A and C are not mutually exclusive. Thus, the given statement is false. (v) ∴A and are not mutually exclusive. Thus, the given statement is false. (vi) It is observed that ; However, Therefore, events are not mutually exclusive and exhaustive. Thus, the given statement is false. Exercise 16.3 solutions Question 1: Which of the following can not be valid assignment of probabilities for outcomes of sample space S = Assignment ω1 ω2 ω3 ω4 ω5 ω6 ω7 (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 10 of 27 (b) (c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (d) –0.1 0.2 0.3 0.4 –0.2 0.1 0.3 (e) Answer : (a) ω1 ω2 ω3 ω4 ω5 ω6 ω7 0.1 0.01 0.05 0.03 0.01 0.2 0.6 Here, each of the numbers p(ωi ) is positive and less than 1. Sum of probabilities Thus, the assignment is valid. (b) ω1 ω2 ω3 ω4 ω5 ω6 ω7 Here, each of the numbers p(ωi ) is positive and less than 1. Sum of probabilities Thus, the assignment is valid. (c) ω1 ω2 ω3 ω4 ω5 ω6 ω7 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Here, each of the numbers p(ωi ) is positive and less than 1. Sum of probabilities Thus, the assignment is not valid. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 11 of 27 (d) ω1 ω2 ω3 ω4 ω5 ω6 ω7 –0.1 0.2 0.3 0.4 –0.2 0.1 0.3 Here, p(ω1) and p(ω5) are negative. Hence, the assignment is not valid. (e) ω1 ω2 ω3 ω4 ω5 ω6 ω7 Here, Hence, the assignment is not valid. Question 2: A coin is tossed twice, what is the probability that at least one tail occurs? Answer : When a coin is tossed twice, the sample space is given by S = {HH, HT, TH, TT} Let A be the event of the occurrence of at least one tail. Accordingly, A = {HT, TH, TT} Question 3: A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number greater than or equal to 3 will appear, (iii) A number less than or equal to one will appear, (iv) A number more than 6 will appear, (v) A number less than 6 will appear. Answer : The sample space of the given experiment is given by S = {1, 2, 3, 4, 5, 6} (i) Let A be the event of the occurrence of a prime number. Accordingly, A = {2, 3, 5} Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 12 of 27 (ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6} (iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C = {1} (iv) Let D be the event of the occurrence of a number greater than 6. Accordingly, D = Φ (v) Let E be the event of the occurrence of a number less than 6. Accordingly, E = {1, 2, 3, 4, 5} Question 4: A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) black card. Answer : (a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements. Therefore, there are 52 points in the sample space. (b) Let A be the event in which the card drawn is an ace of spades. Accordingly, n(A) = 1 (c) (i)Let E be the event in which the card drawn is an ace. Since there are 4 aces in a pack of 52 cards, n(E) = 4 (ii)Let F be the event in which the card drawn is black. Since there are 26 black cards in a pack of 52 cards, n(F) = 26 Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 13 of 27 Question 5: A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12 Answer : Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Accordingly, n(S) = 12 (i) Let A be the event in which the sum of numbers that turn up is 3. Accordingly, A = {(1, 2)} (ii) Let B be the event in which the sum of numbers that turn up is 12. Accordingly, B = {(6, 6)} Question 6: There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? Answer : There are four men and six women on the city council. As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements. Let A be the event in which the selected council member is a woman. Accordingly, n(A) = 6 Question 7: A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. Answer : Since the coin is tossed four times, there can be a maximum of 4 heads or tails. When 4 heads turns up, is the gain. When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain. When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 14 of 27 When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss. When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss. There are 24 = 16 elements in the sample space S, which is given by: S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT} ∴n(S) = 16 The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs. ∴Probability (of winning Rs 4.00) = The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs. ∴Probability (of winning Rs 1.50) = The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs. ∴Probability (of losing Re 1.00) The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs. Probability (of losing Rs 3.50) = The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs. Probability (of losing Rs 6.00) = Question 8: Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails. Answer : When three coins are tossed once, the sample space is given by S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} ∴Accordingly, n(S) = 8 It is known that the probability of an event A is given by (i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH} Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 15 of 27 ∴P(B) = (ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH} ∴P(C) = (iii) Let D be the event of the occurrence of at least 2 heads. Accordingly, D = {HHH, HHT, HTH, THH} ∴P(D) = (iv) LetE be the event of the occurrence of at most 2 heads. Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT} ∴P(E) = (v) Let F be the event of the occurrence of no head. Accordingly, F = {TTT} ∴P(F) = (vi) Let G be the event of the occurrence of 3 tails. Accordingly, G = {TTT} ∴P(G) = (vii) Let H be the event of the occurrence of exactly 2 tails. Accordingly, H = {HTT, THT, TTH} ∴P(H) = (viii) Let I be the event of the occurrence of no tail. Accordingly, I = {HHH} ∴P(I) = (ix) Let J be the event of the occurrence of at most 2 tails. Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH} ∴P(J) = Question 9: If is the probability of an event, what is the probability of the event ‘not A’. Answer : Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 16 of 27 It is given that P(A) = . Accordingly, P(not A) = 1 – P(A) Question 10: A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) an consonant Answer : There are 13 letters in the word ASSASSINATION. ∴Hence, n(S) = 13 (i) There are 6 vowels in the given word. ∴Probability (vowel) = (ii) There are 7 consonants in the given word. ∴Probability (consonant) = Question 11: In a lottery, person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint: order of the numbers is not important.] Answer : Total number of ways in which one can choose six different numbers from 1 to 20 Hence, there are 38760 combinations of 6 numbers. Out of these combinations, one combination is already fixed by the lottery committee. ∴Required probability of winning the prize in the game = Question 12: Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 Answer : (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F). However, here, P(A ∩ B) > P(A). Hence, P(A) and P(B) are not consistently defined. (ii)P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 17 of 27 It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F). Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B). Hence, P(A) and P(B) are consistently defined. Question 13: Fill in the blanks in following table: P(A) P(B) P(A ∩ B) P(A ∪ B) (i) … (ii) 0.35 … 0.25 0.6 (iii) 0.5 0.35 … 0.7 Answer : (i) Here, We know that (ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6 We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ∴0.6 = 0.35 + P(B) – 0.25 ⇒P(B) = 0.6 – 0.35 + 0.25 ⇒P(B) = 0.5 (iii)Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7 We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ∴0.7 = 0.5 + 0.35 – P(A ∩ B) ⇒P(A ∩ B) = 0.5 + 0.35 – 0.7 ⇒P(A ∩ B) = 0.15 Question 14: Given P(A) = and P(B) = . Find P(A or B), if A and B are mutually exclusive events. Answer : Here, P(A) = , P(B) = For mutually exclusive events A and B, P(A or B) = P(A) + P(B) ∴P(A or B) Question 15: Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 18 of 27 If E and F are events such that P(E) = , P(F) = and P(E and F) = , find:(i) P(E or F), (ii) P(not E and not F). Answer : Here, P(E) = , P(F) = , and P(E and F) = (i) We know that P(E or F) = P(E) + P(F) – P(E and F) ∴P(E or F) = (ii) From (i), P(E or F) = P (E ∪ F) = Question 16: Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive. Answer : It is given that P (not E or not F) = 0.25 Thus, E and F are not mutually exclusive. Question 17: A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P (not B) and (iii) P(A or B). Answer : It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16 Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 19 of 27 (i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58 (ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52 (iii) We know that P(A or B) = P(A) + P(B) – P(A and B) ∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74 Question 18: In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology. Answer : Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology. Accordingly, P(A) = 40% = = P(B) = 30% P(A and B) = 10% We know that P(A or B) = P(A) + P(B) – P(A and B) Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6. Question 19: In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both? Answer : Let A and B be the events of passing first and second examinations respectively. Accordingly, P(A) = 0.8, P(B) = 0.7 and P(A or B) = 0.95 We know that P(A or B) = P(A) + P(B) – P(A and B) ∴0.95 = 0.8 + 0.7 – P(A and B) ⇒ P(A and B) = 0.8 + 0.7 – 0.95 = 0.55 Thus, the probability of passing both the examinations is 0.55. Question 20: The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination? Answer : Let A and B be the events of passing English and Hindi examinations respectively. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 20 of 27 Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e., P(A) = 0.75 We know that P(A or B) = P(A) + P(B) – P(A and B) ∴0.9 = 0.75 + P(B) – 0.5 ⇒ P(B) = 0.9 – 0.75 + 0.5 ⇒ P(B) = 0.65 Thus, the probability of passing the Hindi examination is 0.65. Question 21: In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS. (iii) The student has opted NSS but not NCC. Answer : Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS. Total number of students = 60 Number of students who have opted for NCC = 30 ∴P(A) = Number of students who have opted for NSS = 32 Number of students who have opted for both NCC and NSS = 24 (i) We know that P(A or B) = P(A) + P(B) – P(A and B) Thus, the probability that the selected student has opted for NCC or NSS is . (ii) Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 21 of 27 Thus, the probability that the selected students has neither opted for NCC nor NSS is . (iii) The given information can be represented by a Venn diagram as It is clear that Number of students who have opted for NSS but not NCC = n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8 Thus, the probability that the selected student has opted for NSS but not for NCC = Miscellaneous of chapter 16 Question 1: A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? Answer : Total number of marbles = 10 + 20 + 30 = 60 Number of ways of drawing 5 marbles from 60 marbles = (i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles. 5 blue marbles can be drawn from 20 blue marbles in ways. ∴Probability that all marbles will be blue = Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 22 of 27 (ii) Number of ways in which the drawn marble is not green = ∴Probability that no marble is green = ∴Probability that at least one marble is green = Question 2: 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade? Answer : Number of ways of drawing 4 cards from 52 cards = In a deck of 52 cards, there are 13 diamonds and 13 spades. ∴Number of ways of drawing 3 diamonds and one spade = Thus, the probability of obtaining 3 diamonds and one spade = . Question 2: 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade? Answer : Number of ways of drawing 4 cards from 52 cards = In a deck of 52 cards, there are 13 diamonds and 13 spades. ∴Number of ways of drawing 3 diamonds and one spade = Thus, the probability of obtaining 3 diamonds and one spade = Question 3: A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3) Answer : Total number of faces = 6 (i) Number faces with number ‘2’ = 3 (ii) P (1 or 3) = P (not 2) = 1 − P (2) (iii) Number of faces with number ‘3’ = 1 Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 23 of 27 Question 4: In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets? Answer : Total number of tickets sold = 10,000 Number prizes awarded = 10 (i) If we buy one ticket, then P (getting a prize) = ∴P (not getting a prize) = (ii) If we buy two tickets, then Number of tickets not awarded = 10,000 − 10 = 9990 P (not getting a prize) = (iii) If we buy 10 tickets, then P (not getting a prize) = Question 5: Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same sections? (b) you both enter the different sections? Answer : My friend and I are among the 100 students. Total number of ways of selecting 2 students out of 100 students = (a) The two of us will enter the same section if both of us are among 40 students or among 60 students. ∴ Number of ways in which both of us enter the same section = ∴ Probability that both of us enter the same section (b) P(we enter different sections) Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 24 of 27 = 1 − P(we enter the same section) = Question 6: Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. Answer : Let L1, L2, L3 be three letters and E1, E2, and E3 be their corresponding envelops respectively. There are 6 ways of inserting 3 letters in 3 envelops. These are as follows: There are 4 ways in which at least one letter is inserted in a proper envelope. Thus, the required probability is . Question 7: A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∩ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′) Answer : It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35 (i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B) ∴P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88 (ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law] ∴P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12 (iii) P(A ∩ B′) = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19 (iv) We know that Question 8: Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 25 of 27 From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: S. No. Name Sex Age in years 1. Harish M 30 2. Rohan M 33 3. Sheetal F 46 4. Alis F 28 5. Salim M 41 A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years? Answer : Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age. Accordingly, P(E) = and P(F) = Since there is only one male who is over 35 years of age, We know that Thus, the probability that the spokesperson will either be a male or over 35 years of age is . Question 9: If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed? Answer : (i)When the digits are repeated Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5. The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed. ∴Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 − 1 = 250 − 1 = 249 [In this case, 5000 can not be counted; so 1 is subtracted] A number is divisible by 5 if the digit at its units place is either 0 or 5. Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 26 of 27 ∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 − 1 = 100 − 1 = 99 Thus, the probability of forming a number divisible by 5 when the digits are repeated is . (ii)When repetition of digits is not allowed The thousands place can be filled with either of the two digits 5 or 7. The remaining 3 places can be filled with any of the remaining 4 digits. ∴Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2 = 48 When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits. ∴Here, number of 4-digit numbers starting with 5 and divisible by 5 = 3 × 2 = 6 When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits. ∴Here, number of 4-digit numbers starting with 7 and divisible by 5 = 1 × 2 × 3 × 2 = 12 ∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18 Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is . Question 10: The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase? Answer : The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9. Number of ways of selecting 4 different digits out of the 10 digits = Now, each combination of 4 different digits can be arranged in ways. ∴Number of four digits with no repetitions = There is only one number that can open the suitcase. Thus, the required probability is . Class XI ______________________________ Chapter 16- Probability ______________________________ Mathematics ______________________________ __________________________________________________________________________________________ Page 27 of 27

Thursday, 19 April 2018

Olympia Academy

NEET/JEE Main/IIT Foundation Courses Tirupattur, Vellore-D.t

Olympia Academy

II-50% - ONE MARK - REVISION EXAM - 2016 A

MATHEMATICS

CLASS : XII MARKS : 100

[ CHAP – 5,6,7,8,10 B.B & C.B ] TIME : 1.00 hrs

I. Choose the correct answer : 100 x 1 = 100

1. The gradient of the curve y = -2x³ +3x + 5 at x = 2 is

1) -20 2) 27 3) -16 4) -21

2. The velocity v of a particle moving along a straight line when at a distance x from the origin is

given by a + bv²= x² where a and b are constants. Then the acceleration is

3. The slope of the normal to the curve y = 3x² at the point whose x coordinate is 2 is

4. The equation of the normal to the curve

at the point

1) 3

= 27t - 80 2) 5

= 27 t -80 3) 3

= 27 t + 80 4)

5. The parametric equation of the curve x^2/3 + y^2/3 = a^2/3 are

1) x = a sin³ q , y = a cos³ q 2) x = a cos³q , y = a sin³ q

3) x = a³ sin q , y = a³ cos q 4) x = a³ cos q , y = a³sin q

6. What is the surface area of a sphere when the volume is increasing at the same rate as its radius?

1) 1 2)

3) 4

7. If y = 6x - x³and x increases at the rate of 5 units per second, the rate of change of slope when x = 3 is

1) -90 units/sec 2)90 units/sec 3)180 units/sec 4)-180 units/sec

8. The angle between the parabolas y² = x and x² = y at the origin is

9. The value of ‘a’ so that the curves y = 3e^x and y =

intersect orthogonally is

1) -1 2) 1 3)

4) 3

10. The Rolle’s constant for the function y = x² on [ -2, 2] is

2) 0 3) 2 4) -2

11. The value of ‘c’ of Lagranges Mean Value Theorem for f (x) =

when a = 1 and b = 4 is

12. If f (a) = 2; f ' (a) = 1 ; g (a) = -1 ; g ' (a) = 2, then the value of

1) 5 2) -5 3) 3 4) -3

13. The function f (x) = x² is decreasing in

1) ( -

) 2) ( -

, 0 ) 3) ( 0,

) 4) ( -2,

)

14. The least possible perimeter of a rectangle of area 100 m² is

1. 10 2. 20 3. 40 4. 60

15. Which of the following curves is concave down?

1) y = - x ² 2) y = x ² 3) y = e ^x 4) y = x² + 2x - 3

16. If u = x ^y then

is equal to

1) yx^y-1 2) u log x 3) u log y 4) xy^x-1

17. The curve y²(x-2) = x²(1+x) has

1) an asymptote parallel to x-axis 2) an asymptote parallel to y-axis

3) asymptotes parallel to both axes 4) no asymptotes

18. If u = log

, then

1) 0 2) u 3) 2 u 4) u^-1

19. An asymptote to the curve y² (a + 2x) = x² (3a - x) is

1) x = 3a 2) x = -

3) x =

4) x = 0

20. If u = f

then

is equal to

1) 0 2) 1 3) 2u 4) u

21. If u = sin ^{- 1}

and f = sinu, then f is a homogeneous function of degree

1) 0 2) 1 3) 2 4) 4

22. If u =

, then

2) u 3)

4) - u

23. Identify the true statements in the following:

(i) If a curve is symmetrical about the origin, then it is symmetrical about both axes.

(ii) If a curve is symmetrical about both the axes, then it is symmetrical about the origin.

(iii) A curve f (x, y) = 0 is symmetrical about the line y = x if f (x, y) = f (y, x).

(iv) For the curve f (x, y) = 0, if f (x, y) = f (-y, - x), then it is symmetrical about the origin.

1. (ii), (iii) 2. (i), (iv) 3. (i), (iii) 4. (ii), (iv)

24. The percentage error in the 11^th root of the number 28 is approximately _________ times the

percentage error in 28.

3) 11 4) 28

25. The curve 9y² = x²(4-x²) is symmetrical about

1. y-axis 2. x-axis 3. y = x 4. both the axes

26. The value of

is 1)

3) 0 4)

27. The value of

is 1)

2) 0 3)

28. The value of

29. The value of

3. 0 4.

30. The value of

4) 0

31. The area of the region bounded by the graph of y = sinx and y = cosx between x = 0 and

x =

3) 2

4) 2

32. The area bounded by the parabola y² = x and its latus rectum is

33. The volume, when the curve y =

from x = 0 to x = 4 is rotated about x- axis is

1) 100

34. Volume of solid obtained by revolving the area of the ellipse

about major and minor axes are in the ratio

1) b² : a² 2) a² : b² 3) a : b 4) b : a

35. The length of the arc of the curve x^2/3 + y^2/3 = 4 is

1) 48 2) 24 3) 12 4) 96

36. The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from the centre is

1) 20

2) 40

3)10

4) 30

37. The value of

1) 0 2) 2 3) log 2 4) log 4

38. The value of

3. 0 4.

39. The area bounded by the line y = x, the x-axis, the ordinates x = 1, x = 2 is

40. The area between the ellipse

and its auxillary circle is

b(a - b) 2) 2

a (a - b) 3)

a (a - b) 4) 2

b ( a - b)

41. The integrating factor of

1) log x 2) x² 3) e^x 4) x

42. The integrating factor of dx + xdy = e^-y sec² y dy is

1) e^x 2) e^-x 3) e^y 4) e^-y

43. The solution of

+ mx = 0 , where m < 0 is

1) x = ce^my 2) x = ce^-my 3) x = my + c 4) x = c

44. The differential equation

1. of order 2 and degree 1 2. of order 1 and degree 2

3. of order 1 and degree 6 4. of order 1 and degree 3

45. The differential equation of all circles with centre at the origin is

1) x dy + y dx = 0 2) x dy - y dx = 0 3) x dx + y dy = 0 4) x dx - y dy = 0

46. The complementary function of (D² + 1 ) y = e^2x is

1) (Ax + B)e^x 2) A cos x +B sinx 3) (Ax + B)e^{2 x} 4) (Ax + B)e^-x

47. The differential equation of the family of lines y = mx is

= m 2) y dx - xdy = 0 3)

4) ydx + xdy = 0

48. The degree of the differential equation c=

, where c is a constant is

1) 1 2) 3 3) -2 4) 2

49. The differential equation satisfied by all the straight lines in xy-plane is

= a constant 2.

3. y+

= 0 4.

50. The differential equation obtained by eliminating a and b from y = ae^3x + be^-3x is

51. If f ^¢(x) =

and f (1) = 2, then f (x) is

52. The particular integral of (3D² + D-14)y = 13e^2x is

1) 26xe^2x 2)13xe^2x 3) xe^2x 4)

53. The particular integral of the differential equation f (D) y = e^ax , where f (D) = (D-a) g(D),

g(a) ¹0 is

1) me^ax 2)

3) g(a)e^ax 4)

54. If cos x is an integrating factor of the differential equation

+ Py = Q , then P =

1) -cot x 2) cot x 3) tan x 4) -tan x

55. The integrating factor of

1) e^x 2) log x 3)

4) e^-x

56. A random variable X has the following probability distribution

X	0	1	2	3	4	5
P(X=x)		2a	3a	4a	5a

Then P(1

4) is

57. X is a discrete random variable which takes the values 0, 1, 2 and P(X=0) =

P(X=1) =

, then the value of P(X =2 ) is

58. Given E(X+ c) = 8 and E(X-c) = 12 then the value of c is

1) -2 2) 4 3) -4 4) 2

59. Variance of the random variable X is 4. Its mean is 2. Then E(X²) is

1) 2 2) 4 3) 6 4) 8

60. Var (4X + 3) is

1) 7 2) 16 Var(X) 3) 19 4) 0

61. The mean of a binomial distribution is 5 and its its standard deviation is 2. Then the value

of n and p are

62. In 16 throws of a die getting an even number is considered a success. Then the variance of

the successes is

1. 4 2. 6 3. 2 4. 256

63. If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that

they are of the same colours without replacement, is

64. If a random variable X follows Poisson distribution such that E(X² ) = 30, then the variance of the distribution is

1. 6 2. 5 3. 30 4. 25

65. For a Poisson distribution with parameter

=0.25 the value of the 2^nd moment about the origin is

1)0.25 2)0.3125 3)0.0625 4)0.025

66. If

is a p.d. f. of a normal distribution with mean

, then

1. 1 2. 0.5 3. 0 4. 0.25

67. If

is a p.d. f. of a normal variate X and X ~ N(

), then

1. undefined 2. 1 3. 0.5 4. -0.5

68. The marks secured by 400 students in a Mathematics test were normally distributed with mean 65. If 120 students got more marks above 85, the number of students securing marks between 45 and 65 is

1)120 2)20 3)80 4)160

69. The random variable X follows normal distribution f(x) = c

. Then the value of ‘c’ is

3) 5

70. In a Poisson distribution if P(X = 2) =P(X=3) then the value of its parameter

1) 6 2) 2 3) 3 4) 0

71. Let ‘h' be the height of the tank. Then the rate of change of pressure ‘p’ of the tank with respect to height is

72. Food pockets were dropped from an helicopter during the flood and distance fallen in ‘t’ seconds is given by

(g = 9.8 m/

). Then the speed of the food pocket after it has fallen for ‘2’ seconds is

1. 19.6 m/sec 2. 9.8 m/sec 3. -19.6 m/sec 4. -9.8 m/sec

73. A continuous graph y = f (x) is such that

. Then

has a

1. vertical tangent y =

2. horizontal tangent x =

3. vertical tangent x =

4. horizontal tangent y =

74. The point that separates the convex part of a continuous curve from the concave part is

1. the maximum point 2. the minimum point

3. the inflection point 4. critical point

75. Which of the following statement is incorrect?

1. Initial velocity means velocity at t = 0

2. Initial acceleration means acceleration at t = 0

3. If the motion is upward, at the maximum height, the velocity is not zero

4. If the motion is horizontal, v = 0 when the particle comes to rest

76. In the law of mean, the value

satisfies the condition

77. If

is a differentiable function of x and y ; x and y are differentiable functions of ‘t’ then

78. The differential on y of the function

4. 0

79. The differential of

80. The curve

has

1. only one loop between x=0 and x=1

2. two loops between x=-1 and x=0

3. two loops between x=-1 and 0; 0 and 1

4. no loop

81. The x-intercept of the curve

2. 6, 0 3.

82. The curve

is symmetrical about

1. x-axis only 2. y-axis only 3. both the axes 4. both the axes and origin

83. If f(x) is even function

1. 0 2. 2

4. -2

84.

85. The surface area obtained by revolving the area bounded by the curve y= f(x), the two ordinates x=a, x=b and x-axis, about x-axis is

3. 2

4.2

86. If

, then

87. The arc length of the curve y = f(x) from x=a to x=b is

3. 2

4.2

88.

1. -

3. -

4. 2

89. The order and degree of the differential equation

are

1. 2, 1 2. 1, 2 3. 1, 1 4. 2,2

90. The order and degree of the differential equation

are

1. 1, 1 2. 1, 2 3. 2, 1 4. 0, 1

91. The order and degree of the differential equation

are

1. 2, 1 2. 1, 2 3. 2,

4. 2,2

92. The order and degree of the differential equation

are

1. 1, 1 2. 1, 2 3. 2, 1 4. 2,2

93. The solution of a linear differential equation

where P and Q are functions of x is

94. The solution of a linear differential equation

where P and Q are functions of y is

95. A continuous random variable takes

1. only a finite number of values

2. all possible values between certain given limits

3. infinite number of values

4. a finite or countable number of values

96. A continuous random variable X has probability density function ‘f(x)’ then

97. Which of the following is or are correct regarding normal distribution curve ?

a. symmetrical about the line X =

(mean)

b. Mean = median = mode

c. unimodal

d. Points of inflection are at X =

1. a, b only 2. b,d only 3. a.b,c only 4. all

98. For a standard normal distribution the mean and variance are

3. 0, 1 4. 1, 1

99. If X is a continuous random variable then which of the following is incorrect ?

100. Which of the following is not true regarding the normal distribution?

1. skewness is zero. 2. mean = median = mode

3. the points of inflection are at X =

4. maximum height of the curve is

RKV MATRIC HIGHER SECONDARY SCHOOL – JEDARPALAYAM

II-50% - ONE MARK - REVISION EXAM - 2016

MATHEMATICS

CLASS : XII MARKS : 100

DATE : 05.11.16 [ CHAP – 5,6,7,8 & 10 B.B.& C.B ] TIME : 1.00 hrs

I. Choose the correct answer : 100 x 1 = 100

1. x = 2 ,y; y = -2x³ +3x + 5 vd;w tistiuapd; rha;t[

1) -20 2) 27 3) -16 4) -21

2. MjpapypUe;J xU neh;f;nfhl;oy; x bjhiytpy; efUk; g[s;spapd; jpirntfk; v vdt[k; a +bv² = x²vdt[k; bfhLf;fg;gl;Ls;sJ/ ,';F a kw;Wk; b khwpypfs;/ mjd; KLf;fk; MdJ

3. y = 3x² vd;w tistiuf;F x ,d; Maj;bjhiyt[ 2 vdf; bfhz;Ls;s g[s;spapy; br';nfhl;od; rha;thdJ 1)

vDk; tistiuf;F g[s;sp

vd;w g[s;spapy; br';nfhl;od; rkd;ghL

1) 3

= 27t - 80 2) 5

= 27 t -80 3) 3

= 27 t + 80 4)

vDk; tistiuapd; Jiz myFr; rkd;ghLfs;

a) x = a sin³ q , y = a cos³ q c) x = a cos³q , y = a sin³ q

b) x = a³ sin q , y = a³ cos q d) x = a³ cos q , y = a³sin q

6. xU nfhsj;jpd; fd mst[ kw;Wk; Muj;jpy; Vw;gLk; khWtPj';fs; vz;zstpy; rkkhf ,Uf;Fk; nghJ nfhsj;jpd; tisgug;g[

1) 1 2)

3) 4

7. y = 6x - x³ nkYk; x MdJ tpdhof;F 5 myFfs; tPjj;jpy; mjpfhpf;fpd;wJ/

x = 3 vDk; nghJ mjd; rha;tpd; khWtPjk;

1) -90 myFfs;/tpdho 3) 90 myFfs;/tpdho

3) 180 myFfs;/tpdho 4) -180 myFfs;/tpdho

8. y² = x kw;Wk; x² = y vd;w gutisa';fSf;fpilna Mjpapy; mika[k; nfhzk;

9. y = 3e^x kw;Wk; y =

vd;Dk; tistiufs; br';Fj;jhf btl;of; bfhs;fpd;wd vdpy; ‘a’ ,d; kjpg;g[ 1) -1 2)1 3)

4) 3

10. y = x² vd;w rhh;gpw;F [ -2, 2] ,y; nuhypd; khwpyp

2) 0 3) 2 4) -2

11. a = 1 kw;Wk; b = 4 vdf; bfhz;L. f (x) =

vd;w rhh;gpw;F byf;uh";rpapd; ,ilkjpg;g[j; njw;wj;jpd;go mika[k; ‘c’ ,d; kjpg;g[

12. f (a) = 2; f ' (a) = 1 ; g (a) = -1 ; g ' (a) = 2 vdpy;

,d; kjpg;g[

1) 5 2) -5 3) 3 4) -3

13/ f (x) = x² vd;w rhh;g[ ,w';Fk; ,ilbtsp

1) ( -

) 2) ( -

, 0 ) 3) ( 0,

) 4) ( -2,

)

14. 100 kP ² gug;g[ bfhz;Ls;s brt;tfj;jpd; kPr;rpW Rw;wst[

1. 10 2. 20 3. 40 4. 60

15. gpd;tUk; tistiufSs; vJ fPH;nehf;fp FHpt[ bgw;Ws;sJ>

1) y = - x ² 2) y = x ² 3) y = e ^x 4) y = x² + 2x - 3

16. u = x ^y vdpy;

f;Fr; rkkhdJ

1) yx ^y-1 2) u log x 3) u log y 4) xy ^x-1

17. y ² ( x - 2 ) = x² ( 1 + x ) vd;w tistiuf;F

1) x - mr;Rf;F ,izahd xU bjhiyj; bjhLnfhL cz;L

2) y - mr;Rf;F ,izahd xU bjhiyj; bjhLnfhL cz;L

3) ,U mr;RfSf;Fk; ,izahd bjhiyj; bjhLnfhLfs; cz;L

4) bjhiyj; bjhLnfhLfs; ,y;iy

18. u = log

vdpy;

vd;gJ

1) 0 2) u 3) 2u 4) u^-1

19. y² (a + 2x) = x² (3a - x) vd;w tistiuapd; bjhiyj; bjhLnfhL

1) x = 3a 2) x = -a/2 3) x = a/2 4) x = 0

20. u = f

vdpy;.

,d; kjpg;g[

1) 0 2)1 3) 2u 4) u

21. u = sin ^{- 1}

kw;Wk; f = sin u vdpy;. rkgoj;jhd rhh;g[ f ,d;go

1) 0 2) 1 3) 2 4) 4

22. u =

vdpy;

2) u 3)

4) - u

23/ gpd;tUtdtw;Ws; rhpahd Tw;Wfs;:

1) xU tistiu Mjpia bghWj;J rkr;rPh; bgw;wpUg;gpd; mJ ,U

mr;Rfisg; bghWj;JK; rkr;rPh; bgw;wpUf;Fk;/

2) xU tistiu ,U mr;Rfisg; bghWj;J rkr;rPh; bgw;wpUg;gpd; mJ

Mjpiag; bghWj;Jk; rkr;rPh; bgw;wpUf;Fk;/

3) f (x , y) = 0 vd;w tistiu y = x vd;w nfhl;ilg; bghWj;J rkr;rPh;

bgw;Ws;sJ vdpy; f (x , y) = f ( y , x)

4) f (x , y) = 0 vd;w tistiuf;F f (x , y) = f (-y , -x ) cz;ikahapd; mJ

Mjpiag; bghWj;J rkr;rPh; bgw;wpUf;Fk;/

1) (ii), (iii) 2) (i), (iv) 3) (i), (iii) 4)(ii), (iv)

24. 28 ,d; 11 Mk; go rjtpfpjg; gpiH njhuhakhf 28 ,d; rjtpfpjg; gpiHiag; nghy; _____ kl';fhFk;/

3) 11 4) 28

25. 9y² = x²(4-x²) vd;w tistiu vjw;F rkr;rPh;>

1) y -mr;R 2) x -mr;R 3) y = x 4) ,U mr;Rfs;

26.

,d; kjpg;g[ 1)

3)0 4)

27.

,d; kjpg;g[ 1)

2) 0 3)

28.

,d; kjpg;g[ 1)

29.

,d; kjpg;g[ 1.

3. 0 4.

30.

,d; kjpg;g[ 1)

4) 0

31. x = 0 ,ypUe;J x =

tiuapyhd y = sinx kw;Wk; y = cos x vd;w tistiufspd; ,ilg;gl;l gug;g[

3) 2

4) 2

32. gutis y² = x f;Fk; mjd; brt;tfyj;jpw;Fk; ,ilg;gl;l gug;g[

33. y =

vd;w tistiu x = 0 tpypUe;J x = 4 tiu x- mr;ir mr;rhf

itj;Jr; RHw;wg;gLk; jplg;bghUspd; fd mst[

1) 100

34.

vd;w ePs;tl;lj;jpd; gug;ig bel;lr;R. Fw;wr;R ,tw;iw bghWj;Jr; RHw;wg;gLk; jplg;bghUspd; fd mst[fspd; tpfpjk;

1) b² : a² 2) a² : b² 3) a : b 4) b : a

35. x^2/3 + y^2/3 = 4 vd;w tistiuapd; tpy;ypd; ePsk;

1) 48 2) 24 3) 12 4) 96

36/ Muk; 5 cs;s nfhsj;ij js';fs; ikaj;jpypUe;J 2 kw;Wk; 4 J}uj;jpy; btl;Lk; ,U ,izahd js';fSf;F ,ilg;gl;l gFjpapd; tisg;gug;g[

1) 20

2) 40

3)10

4) 30

37.

,d; kjpg;g[

1)0 2) 2 3) log 2 4) log 4

38.

,d; kjpg;g[

3. 0 4.

39. y = x vd;w nfhl;ow;Fk; x- mr;R. nfhLfs; x = 1 kw;Wk; x = 2 Mfpatw;wpw;Fk; ,ilg;gl;l mu';fj;jpd; gug;g[

40.

vd;w ePs; tl;lj;jpw;Fk; mjd; Jiz tl;lj;jpw;Fk; ,ilg;gl;l gug;g[

b(a - b) 2) 2

a (a - b) 3)

a (a - b) 4) 2

b ( a - b)

41.

vd;w tiff;bfGr; rkd;ghl;od; bjhiff; fhuzp

1) log x 2) x² 3) e^x 4) x

42. dx + xdy = e^-y sec² y dy ,d; bjhiff; fhuzp

1) e^x 2) e^-x 3) e^y 4) e^-y

43. m < 0, Mf ,Ug;gpd;

+ mx = 0 ,d; jPh;t[

1) x = ce^my 2) x = ce^-my 3) x = my + c 4) x = c

44.

vd;w tiff;bfGtpd;

1) thpir 2 kw;Wk; go 1 2) thpir 1 kw;Wk; go 2

3) thpir 1 kw;Wk; go 6 4) thpir 1 kw;Wk; go 3

45/ Mjpg;g[s;spia ikakhff; bfhz;l tl;l';fspd; bjhFg;gpd; tiff;bfGr;

rkd;ghL

1) x dy + y dx = 0 2) x dy - y dx = 0 3) x dx + y dy = 0 4) x dx - y dy = 0

46. (D² + 1 ) y = e^2x ,d; epug;g[r; rhh;g[

1) (Ax + B)e^x 2) A cos x +B sinx 3) (Ax + B)e^{2 x} 4) (Ax + B)e^-x

47. y = mx vd;w neh;f;nfhLfspd; bjhFg;gpd; tiff;bfGr; rkd;ghL

= m 2) y dx - xdy = 0 3)

4) ydx + xdy = 0

48. c =

vd;w tiff;bfGr; rkd;ghl;od; go

1) 1 2) 3 3) -2 4) 2

49. xy- jsj;jpYs;s vy;yh neh;f;nfhLfspd; bjhFg;gpd; tiff; bfGr; rkd;ghL

= xU khwpyp 2.

3. y+

= 0 4.

50. y = ae^3x + be^-3x vd;w rkd;ghl;oy; a iaa[k; b iaa[k; ePf;fpf; fpilf;Fk;

tiff;bfGr; rkd;ghL

51. f ^¢(x) =

kw;Wk; f (1) = 2 vdpy; f (x) vd;gJ

52. (3D² + D-14)y = 13e^2x ,d; rpwg;g[r; jPh;t[

1) 26xe^2x 2)13xe^2x 3) xe^2x 4) x²/2e^2x

53. f (D) = (D-a) g(D) , g(a) ¹0 vdpy; tiff;bfGr; rkd;ghL f (D) y = e^ax ,d;

rpwg;g[j; jPh;t[

1) me^ax 2)

3) g(a)e^ax 4)

54.

+Py = Q vd;w tiff;bfGr; rkd;ghl;od; bjhiff; fhuzp cos x

vdpy;. P ,d; kjpg;g[

1) -cot x 2) cot x 3) tan x 4) -tan x

55.

,d; bjhiff; fhuzp

1) e^x 2) log x 3)

4) e^-x

56. X vd;w rktha;g;g[ khwpapd; epfH;jft[g; guty; gpd;tUkhW:

X	0	1	2	3	4	5
P(x = X)	1/4	2a	3a	4a	5a	1/4

P(1£ x £ 4) ,d; kjpg;g[

57. X vd;w xU jdpepiy rktha;g;g[ khwp 0 , 1 , 2 vd;w kjpg;g[fisf; bfhs;fpwJ/ nkYk; P (X= 0) =

, vdpy; P (X = 1) =

, vdpy; P(X = 2 ) ,d; kjpg;g[

58. E(X+ c) = 8 kw;Wk; E (x-c) = 12 vdpy; c ,d; kjpg;g[

1) -2 2) 4 3) -4 4) 2

59. X vd;w rktha;g;g[ khwpapd; gutw;go 4 nkYk; ruhrhp 2 vdpy; E(X²) ,d;

kjpg;g[

1) 2 2) 4 3) 6 4) 8

60. Var (4X + 3) ,d; kjpg;g[

1) 7 2) 16Var(X) 3) 19 4) 0

61. xU <UWg;g[g; gutypd; ruhrhp 5 nkYk; jpl;ltpyf;fk; 2 vdpy; n kw;Wk;

p ,d; kjp;gg[fs;

62. xU gfilia 16 Kiwfs; tPRk; nghJ. ,ul;ilg;gil vz; fpilg;gJ btw;wpahFk; vdpy; btw;wpapd; gutw;go

1)4 2)6 3)2 4)256

63. ed;F fiyf;fg;gl;l 52 rPl;Lfs; bfhz;l rPl;Lf;fl;oypUe;J 2 rPl;Lfs; vLf;fg;gLfpd;wd/ ,uz;Lk; xnu epwj;jpy; ,Uf;f epfH;jft[

64. xU rktha;g;g[ khwp X gha;!hd; gutiyg; gpd;gw;WfpwJ/ nkYk; E(X²) = 30 vdpy; gutypd; gutw;go

1)6 2)5 3)30 4)25

65. gha;!hd; gutypd; gz;gsit

=0.25 vdpy; ,uz;lhtJ tpyf;fg; bgUf;Fj; bjhif

1)0.25 2)0.3125 3)0.0625 4)0.025

66. xU ,ay;epiyg; gutypd; epfH;jft[ mlh;j;jpr; rhh;g[ f (x) ,d; ruhrhp

m vdpy;

,d; kjpg;g[

1)1 2)0.5 3)0 4)0.25

67. xU ,ay;epiy khwp X ,d; epfH;jft[ mlh;j;jpr; rhh;g[ f(x) kw;Wk;

X~N(m , s²) vdpy;

1) tiuaWf;f KoahjJ 2)1 3) 0.5 4) -0.5

68. 400 khzth;fs; vGjpa fzpjj; njh;tpd; kjpg;bgz;fs; ,ay;epiyg; gutiy xj;jpUf;fpwJ/ ,jd; ruhrhp 65. nkYk; 120 khzth;fs; 85 kjpg;bgz;fSf;F nky; bgw;wpUg;gpd;. kjpg;bgz;fs; 45 ,ypUe;J 65 f;Fs; bgWk; khzth;fspd; vz;zpf;if

1)120 2)20 3)80 4)160

69. xU rktha;g;g[ khwp X , ,ay;epiyg; guty; f(x)=c

I gpd;gw;WfpwJ vdpy; c ,d; kjpg;g[

3) 5

70. xU gha;!hd; gutypy; P(X = 2) = P(X =3) vdpy;. gz;gsit

,d; kjpg;g[

1)6 2)2 3)3 4)0

71/ xU ePu;j; bjhl;oapd; cauk;

vd;f/ mj;bjhl;oapd; mGj;jk;

MdJ cauj;ijg; bghWj;J khWk; tPjk;

72/ bts;sg; bgUf;fj;jpd; nghJ K:yk; ,lg;gl;l czt[g; bghUl;fs;

tpdhoapy; fle;j J}uk;

tpdho²) vdpy; mJ nghlg;gl;l 2?tpdhofSf;Fg; gpd; mg;bghUspd; ntfk;/

1/ 19/6 kP

tpdho 2/ 9/8 kP

tpdho

3/ ?19/6 kP

tpdho 4/ ?9/8 kP

tpdho

73/ bjhlu;r;rpahd tistiu

MdJ

vd;w g[s;spapy;

vDk;nghJ

vdpy;

f;F

vd;w epiyf;Fj;jhd bjhLnfhL cz;L

vd;w fpilkl;l bjhLnfhL cz;L

vd;w epiyf;Fj;jhd bjhLnfhL cz;L

vd;w fpilkl;l bjhLnfhL cz;L

74/ xU bjhlu;r;rpahd tistiuapy; FHpt[ gFjpapypUe;J Ftpt[ gFjpahf khw;wk; bgWk; g[s;sp

1/ bgUk g[s;sp 2/ rpWk g[s;sp

3/ tist[ khw;Wg; g[s;sp 4/ khWepiyg;g[s;sp

75. fPH;f;fhQqk; Tw;wpy; vJ rupay;y>

1 bjhlf;f jpirntfk; vd;gJ

tpYs;s jpirntfk;

2 bjhlf;f KLf;fk; vd;gJ

tpYs;s KLf;fk;

3 xU Jfs; br';Fj;jhfr; brd;W mjpfgl;r cauk; mila[k; nghJ mjd;

jpirntfk; g{r;rpaky;y

4/ xU JfshdJ fpilkl;l ,af;fj;jpy; njf;f epiyf;F tUk; neuj;jpy;

76/ ,ilkjpg;g[ tpjpapd;go

tpd; kjpg;g[ ve;j epge;jidia epiwt[ bra;a ntz;Lk;/

77/

vd;gJ x kw;Wk; y ,y; tifaplj;jf;f rhu;g[/ nkYk;

kw;Wk;

vd;git

My; Md tifaplj;jf;f rhu;g[fs; vdpy;

78.

vdpy;

,d; tifaPL

4. 0

79.

,d; tifaPL

80.

vd;w tistiu

kw;Wk;

f;fpilna xU fz;zp bgw;Ws;sJ

kw;Wk;

f;fpilna ,U fz;zpfis bgw;Ws;sJ

kw;Wk;

kw;Wk; 1 fSf;fpilna ,U fz;zpfs; bgw;Ws;sJ/

4/ fz;zpfs; VJk; bgwtpy;iy/

81.

vd;w tistiuapy;

btl;Lj;Jz;L.

2. 0, 6 3/

82.

vd;w tistiu vjidg; bghWj;J rkr;rPu; bgw;Ws;sJ/

-mr;ir kl;Lk; 2/

mr;ir kl;Lk;

3. ,U mr;Rf;fis 4/ ,U mr;Rf;fs; kw;Wk; Mjpia

83/

Xu; ,ul;ilg;gilr; rhu;bgdpy;

1/ 0 2/

84/

85/

vd;w tistiu

Mfpa nfhLfs;

mr;R Mfpatw;why; milg;gLk; gug;gpid

-mr;irg; bghWj;J RHw;wpdhy; Vw;gLk; jplg;bghUspd; tisgug;g[

86/

vdpy;

87/

vd;w tistiuf;F

apypUe;J

tiu cs;s tpy;ypd; ePsk;/

88/

1. -

3. -

4) 2

89.

vd;w tiff;bfGr; rkd;ghl;od; tupir kw;Wk; go

90/

vd;w tiff;bfGr; rkd;ghl;od; tupir kw;Wk; go

91/

vd;w tiff;bfGr; rkd;ghl;od; tupir kw;Wk; go

92/

vd;w tiff;bfGr; rkd;ghl;od; tupir kw;Wk; go

93/

vd;w neupa tiff;bfGr; rkd;ghl;oy;

kw;Wk;

Mfpait

,d; rhu;g[fshf ,Ug;gpd;/ jPu;t[

94/

vd;w neupa tiff;bfGr; rkd;ghl;oy; P kw;Wk;

Mfpait

,d; rhu;g[fshf ,Ug;gpd;. jPu;t[

95. xU bjhlh; rktha;g;g[ khwp

1. Kot[w;w fzj;jpd; kjpg;g[fisg; bgWfpwJ.

2. Fwpg;gpl;l xU ,ilbtspapYs;s vy;yh kjpg;g[fisa[k; bgWfpwJ

3. vz;zpyl';fh kjpg;g[fisg; bgWfpwJ/

4. xU Kot[w;w my;yJ vz;zplj;jf;f kjpg;g[fisg; bgWfpwJ/

96. xU bjhlh; rktha;g;g[ khwp X ,d; epfH;jft[ mlh;j;jpr; rhh;g[ ‘f(x)’ vdpy;

97. ,ay;epiyg; gutiyg; bghWj;J gpd;tUtdtw;Ws; vit my;yJ vJ rhp ?

a. X =

(ruhrhp) vd;w nfhl;ow;Fr; rkr;rPuhdJ b. ruhrhp = ,ilepiy mst[ = KfL

c. xU Kfl;Lg; guty; d. X =

tpy; tist[ khw;Wg;g[s;spfs; cs;sd/

1. a, b kl;Lk; 2. b,d kl;Lk; 3. a.b,c kl;Lk; 4.midj;Jk;

98. jpl;l ,ay;epiyg; gutypd; ruhrhpa[k; gutw;goa[k;

3. 0, 1 4. 1, 1

99. X xU bjhlh; rktha;g;g[ khwp vdpy; vJ jtW>

100. ,ay;epiyg; gutypd; nghJ fPnH bfhLf;fg;gl;l Tw;wpy; vJ rhpahdjy;y>

1. nfhl;lf;bfG g[{r;rpakhFk;/ 2. ruhrhp = ,ilepiy mst[ = KfL

3. tist[ khw;Wg;g[s;spfs; X =

4. tistiuapd; kPg;bgU cauk;

MATHS HUNTERS

Saturday, 21 April 2018

Thursday, 19 April 2018

Olympia Academy

NEET/JEE Main/IIT Foundation Courses Tirupattur, Vellore-D.t

Olympia Academy

Blog Archive

About Me