Grade XI Mathematics Exam Preparation Booklet Chapter Wise - Important Questions and Solutions #GrowWithGreen
Questions Sets Q1. For two disjoint sets A and B, if n [P ( A B )] = 32 and n [P ( A B )] = 4, then find n [ P {( A B ) ( B A )}]. (2 marks) Q2. For set A = { x : x = 3 y 5, where x, y N and x 100}, find the value of n ( A ). (2 marks) Q3. Three sets have p, q and r elements. The total number of non -empty proper subsets of the first set is 120 more than the total number of non -empty proper subsets of the second set. The total number of subsets of the third set is 24 more than the total number subsets of the second set. Find the value of p + q + r. (4 marks) Q4. Let A be the set of prime numbers less than 15 and B be the set of first x odd natural numbers. If the number of relations from A to B is 1024 3, then which element of set B is the largest? Relations and Functions Q1. Draw the graph of the real function. (4 marks) Q2. Let f and g be two real functions given by f ( x ) = 9 x and g ( x ) = 2 x. If ( fg ) ( α ) = 324, then find the values of ( f g ) (2 α 1) and (3 marks) Q3. Find the range of the function =. (3 marks)
Limits and Derivatives Q1. If ( n > 0), then what is the value of n? (3 marks) Q2. Evaluate the derivative of sec 2 4 x from the first principle. (4 marks) Q3. Show that is independent of a. (3 marks) Q4. Evaluate. (3 marks) Complex Numbers and Quadratic Equations Q1. If α and β are the zeroes of the quadratic equation x 2 6 x + 13 = 0, where Im α > Im β, then write a quadratic equation whose zeroes are 5 α + 2 i and 4 β + 3 4 i. (5 marks) Q2. Find Re z, Im z, for the complex number. Also convert z into polar coordinate. (5 marks)
Q1. Solve the system of inequations: Linear Inequalities (4 marks) Q2. Solve the following system of linear inequalities graphically. 5 x + y 10, 7 x + 9 y < 63, 3 x + 4 y 12, x + y 1, x 0, y 0 (5 marks) Q3. Find the domain of the function. (2 marks) Principle of Mathematical Induction Q1. Using the principle of mathematical induction, prove that for all n N, when is divided by 3 n + 1, it always leaves 1 as remainder. (4 marks) Binomial Theorem Q1. If the 10 th and 11 th terms in the binomial expansion of ( a + 3) 49 are equal, then which two consecutive terms in the binomial expansion of ( a + 18) 99 are equal? (5 marks) Q2. Using binomial theorem, prove that. (2 marks) Q3. If the coefficients of a r 1, a r and a r +1 in the expansion of (1 + a ) 14 are in A.P., then find the value(s) of r. (5 marks)
Trigonometric Functions Q1. Find the domain and range of the trigonometric function (3 marks) Q2. If, then prove that (3 marks) Q3. Solve:. (3 marks) Q4. Find the value of the expression cos 2 x (8 cos 4 x 10 cos 2 x + 3), if cos2 x + cos 4 x + cos 6 x = 3. (3 marks) Q5. Find the values of A + B if cos A + cos B = and. (3 marks) Permutations and Combinations Q1. If there are 4 different green balls, 3 different blue balls and 2 different red balls then how many combination of balls can be chosen taking at least one green and one blue ball. (3 marks)
Q2. A round table debate is to be held between 15 students. In how many ways can they be seated if three particular students always sit together? (2 marks) Q3. A man has 7 friends. In how many ways he can invite one or more friends to his birthday party? (3 marks) Q4. If the repetition of letters is not allowed, then what is the number of 8 letter words (with or without meaning) that can be formed out of the letters of the word QUARTILE? (3 marks) Q5. If, then find the value of r. (2 marks) Sequences and Series Q1. Three positive numbers a, b and c are such that a 2 bc, b 2 ca and c 2 ab are in A.P. Show that and are in A.P. (3 marks) Q2. In an A.P., the sum of first p terms, the sum of first q terms and the sum of first r terms are in G.P. If numbers p, q and r also form a G.P., then prove that the common difference of the A.P is twice its first term. If the x th term of the A.P. is the geometric mean of the first term and the y th term, then express y in terms of x. (5 marks) Q3. Find the sum of the series 0.6 + 0.56 + 0.556 + 0.5556 +... to n terms. (4 marks)
Q4. The sum of n th term of an A.P and its corresponding H.P. is. If the first term and the common difference of the A.P are and respectively, show that the number of terms in the A.P is 27. (3 marks) Straight Lines Q1. What are the coordinates of the foot of the perpendicular drawn from the point (1, 2) to the line joining the points (1, 0) and (5, 7)? (3 marks) Q2. In a equilateral triangle ABC, the coordinates of vertex A is (4, 6) and the equation of BC is x + y 4 = 0. What are the equations of other two sides of the triangle? (3 marks) Q3. Find the point on line from where the distance to line is 10 units. (4 marks) Q4. If the line joining the points ( a b, a ) and ( b, a + b ) is perpendicular to the line joining the points ( c d, d ) and ( c, c + d ), then show that ad bc = 0, where a, b, c and d are non-zero real numbers. (2 marks) Conic Sections Q1. What are the equations of the circles touching x -axis at (4, 0) and making an intercept of 6 units on y -axis? (3 marks) Q2. Find the equation of a hyperbola with foci (0, ±7.5) and eccentricity 1.25. (2 marks)
Q3. A parabola opening upwards with vertex at origin is drawn on a coordinate plane. An isosceles, right-angled triangle is inscribed in the parabola in such a way that the vertex formed by equal sides is at the origin. If the perimeter of the triangle is then find the equation of the parabola. Also find the length of the latus rectum and the equation of the directrix. (5 marks) Q4. A parabola with vertex at origin is symmetrical about the y -axis and passes through point ( 12, 1.8). Find the equation of the parabola. Also, find the length of the latus rectum, the coordinates of foci and the equation of directrix. (3 marks) Q5. Find the centre and radius of the circle (3 marks) Q6. Find the radius of the circle with centre as (0, 3) and passing through the foci of the ellipse. (3 marks) Introduction to Three-Dimensional Geometry Q1. Show that the area of triangle whose vertices are the points A ( 2, 3, 5), B (1, 2, 3) and C (7, 0, 1) is zero. Also find the ratio in which B divides AC. (3 marks) Q2. In what ratio does the point (3, 0, 5) divide the line segment joining the points (1, 2, 3) and (6, 3, 8)? (2 marks) Q3. Find the vector and Cartesian equation of the diagonals of parallelogram ABCD if three of its vertices are A (1, 3, 0), B ( 5, 5, 2) and C ( 9, 1, 2). (6 marks)
Q4. Find the area of the quadrilateral with vertices (2, 3, 3), (0, 6, 9), ( 6, 4, 6) and ( 4, 1, 0) taken in order. (3 marks) Q5. If points ( 1, 2, 3), (4, a, 1) and ( b, 8, 5) are collinear, then find the values of a and b. Mathematical Reasoning (3 marks) Q1. Write the converse and contrapositive of the statement, If Asia is a continent, then India is a country. (2 marks) Q2. Write the negation of the statement The sum of two sides of a triangle is more than the third side. (1 mark) Q3. Which of the following sentences is a statement? Give reasons. (i) Rita is a bright student. (ii) Smoking is hazardous to health. (2 marks) Statistics Q1. The mean and variance of observations 6, x, 10, 12, y, 13, 4 and 12 are 9 and 9.25 respectively. What are the values of x and y? (3 marks) Q2. The mean and standard deviation of 16 observations was incorrectly calculated as 16 and respectively. On checking the data, it was found that the last observation was misread as 31. If this observation is excluded, then what would be the coefficient of variation of the remaining data? (5 marks) Q3. In the given frequency distribution table, the frequencies corresponding to classes 10 20 and 20 30 are kept missing.
If the total number of observations of the given data is 50 and the mean is 25.8, then find the mean deviation about the median of the data. (6 marks) Q4. A data set containing 70 observations has variance and coefficient of variation as 25 and 12.5 respectively. Another data set containing 80 observations has variance and coefficient of variation as 30.25 and 10 respectively. If the two data sets are combined, then what would be the mean? (5 marks) Probability Q1. Four cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that (i) 2 will be red and 2 will be black (ii) among the 4 cards, 2 will be queen (2 marks) Q2. A carton consists of bulbs in which some are good and others are defective. Shahid will buy a bulb, if it is good, but will not buy if it is defective. The shopkeeper draws one bulb at random to give it to him. If the probability that Shahid will not buy a bulb is 0.048, then find: (a) (b) Probability that he will buy it (1 mark) Number of good bulbs in the carton, if there are 12 defective bulbs (2 marks) (3 marks)
Q3. There are 42 balls of colours blue, green, and red in a bag. The probability of selecting a blue and red ball is and respectively. One ball of each colour is added to the bag. What is the probability of selecting a green ball? (3 marks) Q4. For any two events A and B, if and P( A B ) = 0.1, then show that (a) A and B are mutually exclusive events (b) A and B are exhaustive events (4 marks)
Solutions Sets Ans 1. It is given that n [P ( A B )] = 32. n [P( A B ) = 2 5 n ( A B ) = 5 It is also given that n [P ( A B ) = 4 = 2 2. n ( A B ) = 2 For two disjoint sets A and B, it is known that A B and B A are disjoint sets. n (( A B ) ( B A )) = n ( A B ) + n ( B A ) (1) It is known that n ( A B ) + n ( B A ) + n ( A B ) = n ( A B ) n ( A B ) + n ( B A ) = n ( A B ) n ( A B ) n {( A B ) ( B A )} = 5 2 [Using equation (1)] n {( A B ) ( B A )} = 3 n [P{( A B ) ( B A )}] = 2 3 = 8 Ans 2. A = { x : x = 3 y 5, where x, y N and x 100} It can be observed that for y = 1, x = 2 N. For y = 2, 3, 4 35, the respective values of x are 1, 4, 7, 10, 100, which are natural numbers. For y > 35, x > 100 Thus, A = {1, 4, 5, 7, 10, 100} There are 35 1 = 34 natural numbers from 2 to 35. Thus, set A contains 34 elements. Hence, n ( A ) = 34. Ans 3. The total number of subsets of a set A containing n elements is 2 n. Here, set A is not the proper subset of set A. Also, empty Φ is a subset of A. Total number of non -empty proper subsets of a set containing n elements = 2 n 2 Total number of non -empty proper subsets of first set = 2 p 2 Total number of non -empty proper subsets of second set = 2 q 2 Total number of subsets of second set = 2 q Total number of subsets of third set = 2 r 2 According to the given condition, (2 p 2) = 120 + (2 q 2) (2 p 2) (2 q 2) = 120 2 p 2 q = 120 2 q (2 p q 1) = 120 2 q (2 p q 1) = 8 15 2 q (2 p q 1) = 2 3 (2 4 1)
q = 3 and p q = 4 q = 3 and p = 7 (1) 2 r = 24 + 2 q [Given] 2 r 2 q = 24 2 r 8 = 24 [Using (1)] 2 r = 32 r = 5 Thus, p + q + r = 7 + 3 + 5 = 15 Ans 4. A is the set of prime numbers less than 15. A = {2, 3, 5, 7, 11, 13} n ( A ) = 6 B is the set of first x odd natural numbers. n ( B ) = x Hence, n ( A B ) = 6 x Number of relations from A to B = 2 6 x According to the given information, 2 6 x = 1024 3 2 6 x = (2 10 ) 3 2 6 x = 2 30 6 x = 30 x = 5 Thus, set B contains first five odd natural numbers. B = {1, 3, 5, 7, 9} Hence, the largest element in set B is 9. Relations and Functions Ans 1. The domain of the given function (which is real line) can be divided into three parts. They are (, 1), [1, 3] and (3, ). For x (, 1), and In this case, f ( x ) = (1 x ) + (3 x ) = 4 2 x For x [1, 3], In this case, f ( x ) = x 1 + 3 x = 2 For x (3, ), In this case, f ( x ) = x 1 + x 3 = 2 x 4 Thus, the graph of the given function can be drawn as
Ans 2. It is given that ( fg ) ( α ) = 324 f ( α ). g ( α ) = 324 [( fg ) ( x ) = f ( x ). g ( x )] 9 α.2 α = 324 [ f ( x ) = 9 x, g ( x ) = 2 x ] (9 2) α = 324 18 α = 18 2 α = 2 ( f g ) (2 α 1) = ( f g ) (2 2 1) = ( f g ) (3) = f (3) g (3) [( f g ) ( x ) = f ( x ) g ( x )] = 9 3 2 3 = 729 8 = 721 Ans 3. 1 sin 5 x 1 for all x R 1 sin 5 x 1 for all x R (1) 3 4 sin 5 x 5 for all x R
4 sin 5 x 0 for all x R is defined for all x R Thus, the domain of f ( x ) = R From equation (1): Hence, range of Limits and derivatives Ans 1. Since n > 0, the required value of n is 49.
Ans 2. Let f ( x ) = sec 2 4 x
Thus, derivative of sec 2 4 x is 8 tan4 x sec 2 4 x. Ans 3., which is independent of a Thus, is independent of a. Ans 4. The given function is.
At, the value of the function takes the form. Put When.
Complex Numbers and Quadratic Equations Ans 1. The given quadratic equation i.e., x 2 6 x + 13 = 0 is of the form ax 2 + bx + c = 0, where a = 1, b = 6 and c = 13. Discriminant, D = b 2 4 ac = ( 6) 2 4.1.13 = 16 < 0 The zeroes of the given quadratic equation are given by Therefore, zeroes of the given quadratic equation are. Here, Thus, the required quadratic equation whose zeroes are 5 α + 2 i and 4 β + 3 4 i is Ans 2.
Now We have, Let and
and Thus, polar form of, where Linear Inequalities Ans 1.
The solution set of the given inequations is the intersection of (1) and (2). Thus, the solution set of the given inequations is Ans 2. The given system of linear inequalities is 5 x + y 10 (1) 7 x + 9 y < 63 (2) 3 x + 4 y 12 (3) x + y 1 (4) x 0 (5) y 0 (6) The graphs of lines 5 x + y = 10, 7 x + 9 y = 63, 3 x + 4 y = 12 and x + y = 1 are drawn in the figure below. Point (0, 0) satisfies inequalities (1), (2) and (3) but point (0, 0) does not satisfy inequality (4). Therefore, inequality (1) represents the region below line 5 x + y = 10 (including line 5 x + y = 10); inequality (2) represents the region below line 7 x + 9 y = 63 (excluding line 7 x + 9 y = 63); inequality (3) represents the region below line 3 x + 4 y = 12 (including line 3 x + 4 y = 12) and inequality (4) represents the region above line x + y = 1 ((including line x + y = 1). Since x 0 and y 0, every point in the common shaded region in the first quadrant, including the points on the respective lines and axes, represents the solution of the system of linear inequalities.
Ans 3. For f ( x ) to be defined, Now, Also, Domain of f ( x ) = Principle of Mathematical Induction Ans 1. Let the given statement be P ( n ): when is divided by 3 n + 1, it always leaves 1 as remainder.
This statement can be rewritten as: when is exactly divisible by 3 n + 1. Therefore, the equivalent form of the given statement can be written as: P ( n ): is exactly divisible by 3 n + 1, for all n N For n = 1,, which is divisible by 3 2 (= 3 1+1 ). Thus, P ( n ) is true for n = 1. Let P ( n ) be true for n = k. i.e., is exactly divisible by 3 k + 1. Let, where α N (1) We shall prove that P ( k + 1) is true. = 3 k + 2. β, where is natural number = 3 ( k + 1) + 1. β Thus, is divisible by 3 ( k + 1) + 1 Thus, P ( n ) is true for n = k + 1 whenever P ( n ) is true for n = k. Hence, by principle of mathematical induction, P ( n ) is true for all n N. Binomial Theorem Ans 1. The ( r + 1) th term, in the binomial expansion of ( a + 3) 49 is given by It is given that in the binomial expansion of ( a + 3) 49, T 10 = T 11.
In the binomial expansion of ( a + 18) 99 i.e., (12 + 18) 99, the ( R + 1) th term is given by Let the ( x + 1) th and ( x + 2) th terms in the binomial expansion of (12 + 18) 99 be equal. T x + 1 = T x + 2
Thus, the 60 th and 61 st terms in the binomial expansion of (9 + 18) 99 are equal. Ans 2. On using binomial theorem, we obtain Ans 3. The ( r + 1) th term in the expansion of (1 + a ) 14 is. Therefore, a r occurs in ( r + 1) th term and its coefficient is. Coefficient of a r 1 = Coefficient of a r + 1 = The coefficients of a r 1, a r and a r + 1 are in A.P.
r = 5 Or r = 9 Thus, the value of r is 5 or 9. Trigonometric Functions Ans 1. is not defined for where n Z. is not defined for.
is not defined for. Thus, the domain of The range of is R ( 1, 1). The range of 2 is R ( 1 2, 1 2). Thus, the range of f ( x ) is R ( 3, 1). Ans 2. On adding 1 to both the sides, we obtain Hence, the result is proved. Ans 3. Consider the L.H.S. of the given equation.
Ans 4. Ans 5. On dividing equation (2) by equation (1), we obtain
Thus, Permutations and Combinations Ans 1. Out of 4 different green balls, at least one can be selected in ( 4 C 1, 4 C 2 + 4 C 3 + 4 C 4 ) ways Similarly, out of 3 different blue balls, at least one blue balls can be selected in (2 3 1) = 7 ways. After selecting at least one green and at least one blue ball, red ball can be selected in 2 2 = 4 ways Thus, the total number of combinations of balls = 15 7 4 = 420 Ans 2. Considering the three particular students as one student. So, the number become 13. These can be arranged in a circle in (13 1)! = 12! ways. The three boys can be seated in 3! ways. Required number of ways = 12! 3! Ans 3. The man has 7 friends. He can invite 1 friend to his birthday party in ways. He can invite 2, 3, 4, 5, 6 and 7 friends in and ways respectively. Thus, total number of ways in which he can invite one or more friends to his birthday party
Ans 4. There are 8 different letters in the word QUARTILE. There will be as many ways as there are ways of filling 8 vacant places in succession by the 8 different letters. The first place can be filled in 8 different ways by any of the letters Q, U, A, R, T, I, L and E, following which, the second place can be filled in 7 different ways by any of the remaining 7 letters. Similarly, the third, fourth, fifth, sixth, seventh and eighth places can be filled in 6, 5, 4, 3, 2 and 1 different ways respectively. Therefore, by multiplication principle, the required number of 8-letter words (with or without meaning) that can be formed out of the letters of the word QUARTILE is 8 7 6 5 4 3 2 1 = 40320 Ans 5. Thus, the value of r is 4. Sequences and Series Ans 1. It is given that a 2 bc, b 2 ca and c 2 ab are in A.P.
Thus, and are in A.P. Ans 2. Since p, q and r are in G.P., q 2 = pr...(1) Let a be the first term and d be the common difference of the A.P. It is also given that S p, S q and S r are in G.P.
Thus, the given result is verified. The A.P. is given by a, a + d, a + 2 d, a + 3 d... i.e., a, 3 a, 5 a, 7 a... Here, n th term = (2 n 1) a x th term = (2 x 1) a ; y th term = (2 y 1) a Since the x th term is the geometric mean of a and the y th term, (2 x 1) 2 a 2 = a (2 y 1) a (2 x 1) 2 = (2 y 1) 4 x 2 4 x + 1 = 2 y 1 4 x 2 4 x + 2 = 2 y y = 2 x 2 2 x + 1 Ans 3. The given series is 0.6 + 0.56 + 0.556 + 0.5556 +... Here, the terms are not in G.P.
Thus, sum of the given series to n term =. Ans 4. Here, a = and d = The sum of n th term of an A.P and its corresponding H.P. is.
Hence, the number of terms in the A.P. is 27. Straight Lines Ans 1. Let A, B, C be the points (1, 2), (1, 0), (5, 7) respectively. Let the foot D of perpendicular AD to BC divide BC in the ratio λ : 1 λ. Coordinates of point D are (5λ + (1 λ), 7λ + (1 λ) 0) = (4λ + 1, 6λ + 1). Since AD BC, Thus, the required coordinates are.
Ans 2. Slope of BC, m 1 = 1 Let the slope of the line which makes an angle of 60 with the given line be m. Equation of other two sides are: Ans 3.
Hence, any point on line is of the form, where. Let be a point on line from where the distance to line is 10 units. It is known that the distance of any line from a point is. Thus, the distance of the line from the point is given by According to the given information, When k = 4, the required point on line is
When the required point on line is Thus, (4, 9) and are the points on line from where the distance to the line is 10 units. Ans 4. Slope ( m 1 ) of the line joining the points ( a b, a ) and ( b, a + b ) Slope ( m 2 ) of the line joining the points ( c d, d ) and ( c, c + d ) Since the two lines are perpendicular, the product of their slopes is 1. Hence, the relation is proved. Ans 1. Conic Sections Let the circle whose centre is C intersect y -axis at A and B. Also, let CL be the perpendicular on AB i.e. CL AB.
Intercept on y -axis AB = 6 units (Given) AL = AB = 3 units And LC = 4 Thus, x coordinate of centre C is 4 and y coordinate is ±5 i.e. C (4, ±5). Therefore, required equations of circles are: ( x 4) 2 + ( y ± 5) 2 = 25 x 2 + 16 8 x + y 2 + 25 ± 10 y = 25 x 2 + y 2 8 x ± 10 y + 16 = 0 Ans 2. Since the foci is on y -axis, the equation of the hyperbola is of the form. Foci are (0, ±7.5), which are of the form (0, ± c ). Hence, c = 7.5 The eccentricity of the hyperbola is given as 1.25. It is known that. Thus, equation of the required hyperbola is. Ans 3. The equation of a parabola opening upwards with vertex at origin is of the form. The isosceles, right-angled triangle inscribed in the parabola can be drawn as:
Let OM = l Since ΔAOB is isosceles, AOM = = 45 and AM OM In right-angled triangle AOM, AB = AM + BM = l + l = 2 l Hence, the coordinates of A are ( l, l ) and the coordinates of B are ( l, l ). It is given that the perimeter of ΔOAB is The coordinates of A and B are ( 24, 24) and (24, 24) respectively. Point A ( 24, 24) is a point on parabola.
Thus, the equation of the parabola is. Length of the latus rectum of the parabola = 4 a = 4 6 units = 24 units Equation of directrix of parabola x 2 = 4 ay is given by y = a. Thus, the equation of the directrix of the parabola is y = 6. Ans 4. It is given that the parabola passes through point ( 12, 1.8). This means that the parabola passes through the third quadrant. It is also given that the vertex of the parabola is at origin and is symmetrical about the y -axis. Thus, the equation of the parabola is of the form Parabola (i) passes through point ( 12, 1.8). Therefore, this point satisfies equation (i). Thus, the equation of the parabola is. Length of the latus rectum = 4 a = 4 20 = 80 units Coordinates of the foci = (0, a ) = (0, 20) Equation of directrix = y = a i.e., y = 20 Ans 5. The equation of the circle can be rewritten as:
Comparing this equation with ( x h ) 2 + ( y k ) 2 = r 2 : Thus, the centre of the circle is and its radius is Ans 6. Since the denominator of is larger than the denominator of, the major axis is along the x -axis. Comparing the given equation with : a 2 = 16 and b 2 = 9 The coordinate of the foci are Thus, radius of the circle = distance between (0, 3) and Introduction to Three-Dimensional Geometry Ans 1. For the area of triangle ABC to be zero, the points A, B and C need to be collinear. For the same we need to show that AB + BC = AC. It can be seen that AB + BC = AC, thus A, B and C are collinear. So, the area of the triangle ABC is zero. Let B divides AC in the ratio λ : 1.
Coordinates of point B = (1, 2, 3) = 7λ 2 = λ + 1 Thus, the required ratio is 1 : 2. Ans 2. Let the point (3, 0, 5) divide the line segment joining the points (1, 2, 3) and (6, 3, 8) in the ratio k : 1. By section formula, we obtain Equating x, y, and z coordinates, we obtain Solving these equations, we obtain Thus, the required ratio is 2:3. Ans 3. Let the coordinates of vertex D of parallelogram ABCD be ( x, y, z ). Since ABCD is a parallelogram, diagonals AC and BD bisect each other. The coordinates of the mid-point of AC are The coordinates of the mid-point of BD are
Therefore, the coordinates of vertex D are ( 3, 3, 0). Diagonal AC passes through points A (1, 3, 0) and C ( 9, 1, 2). The position vectors of points A and C are It is known that the vector equation of a line passing through two points whose position vectors are is given by Therefore, the vector equation of diagonal AC is given as: It is known that the Cartesian equation of a line passing through two points ( x 1, y 1, z 1 ) and ( x 2, y 2, z 2 ) is given by. Since diagonal AC passes through points A (1, 3, 0) and C ( 9, 1, 2), its Cartesian equation is given as Thus, the vector and Cartesian equation of diagonal AC of parallelogram ABCD are constant. respectively, where λ is an arbitrary Diagonal BD passes through points B ( 5, 5, 2) and D ( 3, 3, 0).
The position vectors of vertices B and D of parallelogram ABCD are respectively. The vector equation of diagonal BD is given as: The Cartesian equation of diagonal BD is given as: Thus, the vector and Cartesian equation of diagonal BD of parallelogram ABCD are arbitrary constant. respectively, where μ is an Ans 4. Let ABCD be the quadrilateral formed by vertices A (2, 3, 3), B (0, 6, 9), C ( 6, 4, 6) and D ( 4, 1, 0). Using distance formula, ABCD is a parallelogram such that AB = BC = CD = AD. Hence, ABCD is a rhombus.
Area of rhombus ABCD Now, AC BD Thus, area of quadrilateral ABCD units square Ans 5. Points P ( 1, 2, 3), Q (4, a, 1) and R ( b, 8, 5) are given to be collinear. Let point Q (4, a, 1) divide the line segment joining P ( 1, 2, 3) and R ( b, 8, 5) in the ratio k :1. Using section formula, Coordinates of Q (4, a, 1) From (3): Q externally divides PR in the ratio 1:2. From (1):
From (2): Hence, the values of a and b are 4 and 6 respectively. Mathematical Reasoning Ans 1. The given compound statement can be broken into component statements as p : Asia is a continent q : India is a country The converse of the statement p q is q p. Therefore, the converse of the given statement is If India is a country, then Asia is a continent. The contrapositive of the statement p q is q p. Therefore, the contra positive of the given statement is If India is not a country, then Asia is not a continent. Ans 2. The negation of the given statement can be written as It is not the case that the sum of two sides of a triangle is more than the third side. Ans 3. (i) This sentence is subjective as some people may agree that Rita is bright but others may not, i.e., this sentence will not always be true. Hence, this is not a statement. (ii) This sentence is a scientifically established fact that smoking is hazardous to health. Hence, this sentence is always true. Hence, this is a statement. Ans 1. Mean = 9 Statistics Variance = 9.25
Then from x + y = 15 and x y = 1 2 x = 16 x = 8 y = 7 If x y = 1 Then from x y = 1 and x + y = 15 x = 7 and y = 8 Ans 2. For the incorrect data, and However,
When the last observation (that was misread as 31) is excluded, 15 observations are left. Then, for the remaining data, x i = 256 31 = 225, x i 2 = 5456 961 = 4495 Standard deviation Thus, coefficient of variation Ans 3. Let the missing frequencies corresponding to classes 10 20 and 20 30 be x and y respectively. Accordingly, the given data can be written as
It is given that mean, Now, the median ( M ) of the given data can be calculated as: Class Frequency Cumulative ( f ) frequency 0 10 4 4 10 20 15 19 20 30 10 29 30 40 15 44 40 50 6 50 Total N = 50 N = 50, which is even Median = Mean of
It is clear that the median class is 20 30. l = Lower limit of the median class = 20 N = 50 c = Cumulative frequency of class preceding the median class = 19 f = Frequency of the median class = 10 h = Width of each class = 10 Now, the mean deviation about the median can be calculated as: Class Frequency Mid-point ( f i ) ( x i ) 0 10 4 5 21 84 10 2 15 15 11 165 0 20 3 10 25 1 10 0 30 4 15 35 9 135 0 40 5 6 45 19 114 0 Total 50 508 Thus, mean deviation about median M.D. ( M ) is given by Ans 4. For the first data set: Number of observations, n 1 = 70 Variance, = 25 Coefficient of variation, (C.V. 1 ) = 12.5
Mean of the first data set = 40 For the second data: Number of observations, n 2 = 80 Variance, = 30.25 Coefficient of variation, (C.V. 2 ) = 10 Mean of the second data set = 55 Mean of the combined data sets Probability Ans 1. (i) In a pack of cards, 26 cards are red and 26 are black. (ii) Required probability In a pack of 52 cards, there are 4 queens. Number of ways to choose 2 queens = 4 C 2 48 C 2
Required probability Ans 2. (a) Let E be the event that Shahid will buy a bulb. Then, will be the event that he will not buy a bulb. We know, P (E) + P ( ) = 1 P (E) + 0.048 = 1 P (E) = 1 0.048 = 0.952 (b) Let there are x numbers of good bulbs in the carton. Since there are 12 defective bulbs, there are ( x + 12) bulbs in the carton. Now, probability that Shahid will not buy a bulb Therefore, there are 238 good bulbs in the carton. Ans 3. Let the number of green-coloured balls in the bag be x. Initial number of balls in the bag = 42 P (blue ball) P (red ball) P (green ball) = 1 [P (blue ball) + P (red ball)]
One ball of each colour is added to the bag. Total number of balls in the bag = 42 + 3 = 45 Number of green balls in the bag = 8 + 1 = 9 P (green ball) Thus, the required probability is. Ans 4. (a) P( A B ) = 0.1 P[( A B ) ] = 0.1 1 P ( A B ) = 0.1 P( A B ) = 0.9 P( A ) + P( B ) P( A B ) = Thus, A and B are mutually exclusive events. (b) P( A B ) = P( A ) + P( B ) P( A B )
Thus, A and B are exhaustive events.