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Here are the NCERT solutions for Class 12 Science Maths Chapter 7 on Probability, featuring clear step-by-step explanations. Widely favored by Class 12 Science students, these solutions prove valuable for efficiently completing homework assignments and exam preparation. You can access all questions and answers from Chapter 7 of the NCERT Book for Class 12 Science Maths here at no cost. Page No 538: Question 1: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E). ANSWER: It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2 Page No 538: Question 2: Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32 ANSWER: It is given that P(B) = 0.5 and P(A ∩ B) = 0.32 Page No 538: Question 3: If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B) ANSWER: It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4 (i) P (B|A) = 0.4 (ii)  (iii) P(A∪B) = P(A) + P(B) − P(A∩B)⇒P(A∪B)=0.8 + 0.5 − 0.32 = 0.98PA∪B = PA + PB – PA∩B⇒PA∪B=0.8 + 0.5 – 0.32 = 0.98 Page No 538: Question 4: Evaluate P (A ∪ B), if 2P (A) = P (B) =and P(A|B) = ANSWER: It is given that, It is known that,  Page No 538: Question 5: If P(A), P(B) =and P(A ∪ B) =, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A) ANSWER: It is given that  (i)  (ii) It is known that,  (iii) It is known that,  Page No 538: Question 6: A coin is tossed three times, where (i) E: head on third toss, F: heads on first two tosses (ii) E: at least two heads, F: at most two heads (iii) E: at most two tails, F: at least one tail ANSWER: If a coin is tossed three times, then the sample space S is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} It can be seen that the sample space has 8 elements. (i) E = {HHH, HTH, THH, TTH} F = {HHH, HHT} E ∩ F = {HHH} (ii) E = {HHH, HHT, HTH, THH} F = {HHT, HTH, HTT, THH, THT, TTH, TTT} E ∩ F = {HHT, HTH, THH} Clearly,  (iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH} F = {HHT, HTT, HTH, THH, THT, TTH, TTT} Page No 539: Question 7: Two coins are tossed once, where (i) E: tail appears on one coin, F: one coin shows head (ii) E: not tail appears, F: no head appears ANSWER: If two coins are tossed once, then the sample space S is (ii) E = {HH} F = {TT} ∴ E ∩ F = Φ P (F) = 1 and P (E ∩ F) = 0 ∴ P(E|F) = Page No 539: Question 8: A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses ANSWER: If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216 Page No 539: Question 9: Mother, father and son line up at random for a family picture E: son on one end, F: father in middle ANSWER: If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be S = {MFS, MSF, FMS, FSM, SMF, SFM} ⇒ E = {MFS, FMS, SMF, SFM} F = {MFS, SFM} ∴ E ∩ F = {MFS, SFM} Page No 539: Question 10: A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. ANSWER: Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements. A: Obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} B: Black die results in a 5. = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} ∴ A ∩ B = {(5, 5), (5, 6)} The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B). P(A|B) = P(A∩B)P(B) = 236636 = 26 = 13PA|B = PA∩BPB = 236636 = 26 = 13 (b) E: Sum of the observations is 8. = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} F: Red die resulted in a number less than 4. The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F). Page No 539: Question 11: A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} Find (i) P (E|F) and P (F|E) (ii) P (E|G) and P (G|E) (ii) P ((E ∪ F)|G) and P ((E ∩ G)|G) ANSWER: When a fair die is rolled, the sample space S will be S = {1, 2, 3, 4, 5, 6} It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5} (i) E ∩ F = {3} (ii) E ∩ G = {3, 5} (iii) E ∪ F = {1, 2, 3, 5} (E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5} E ∩ F = {3} (E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3} Page No 539: Question 12: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional …

NCERT Solutions for Class 12 Science Maths Chapter 7 – Probability Read More »

Explore comprehensive NCERT Solutions for Class 12 Science Maths Chapter 6 on Linear Programming, featuring straightforward step-by-step explanations. These meticulously crafted solutions have gained immense popularity among class 12 Science students, proving invaluable for completing homework assignments efficiently and preparing for exams. All questions and answers from Chapter 6 of the NCERT Book for class 12 Science Maths are available here at your disposal, allowing you to access them free of charge. Page No 513: Question 1: Maximise Z = 3x + 4y Subject to the constraints: ANSWER: The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows. The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points are as follows. Corner point Z = 3x + 4y O(0, 0) 0 A(4, 0) 12 B(0, 4) 16 → Maximum Therefore, the maximum value of Z is 16 at the point B (0, 4). Page No 514: Question 2: Minimise Z = −3x + 4y subject to. ANSWER: The feasible region determined by the system of constraints,x ≥ 0, and y ≥ 0, is as follows. The corner points of the feasible region are O (0, 0), A (4, 0), B (2, 3), and C (0, 4). The values of Z at these corner points are as follows. Corner point Z = −3x + 4y 0(0, 0) 0 A(4, 0) −12 → Minimum B(2, 3) 6 C(0, 4) 16 Therefore, the minimum value of Z is −12 at the point (4, 0). Page No 514: Question 3: Maximise Z = 5x + 3y subject to. ANSWER: The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows. The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and . The values of Z at these corner points are as follows. Corner point Z = 5x + 3y 0(0, 0) 0 A(2, 0) 10 B(0, 3) 9 → Maximum Therefore, the maximum value of Z is Page No 514: Question 4: Minimise Z = 3x + 5y such that. ANSWER: The feasible region determined by the system of constraints, , and x, y ≥ 0, is as follows. It can be seen that the feasible region is unbounded. The corner points of the feasible region are A (3, 0), , and C (0, 2). The values of Z at these corner points are as follows. Corner point Z = 3x + 5y A(3, 0) 9 7 → Smallest C(0, 2) 10 As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z. For this, we draw the graph of the inequality, 3x + 5y < 7, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 3x + 5y < 7 Therefore, the minimum value of Z is 7 at. Page No 514: Question 5: Maximise Z = 3x + 2y subject to. ANSWER: The feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, and y ≥ 0, is as follows. The corner points of the feasible region are A (5, 0), B (4, 3), and C (0, 5). The values of Z at these corner points are as follows. Corner point Z = 3x + 2y A(5, 0) 15 B(4, 3) 18 → Maximum C(0, 5) 10 Therefore, the maximum value of Z is 18 at the point (4, 3). Page No 514: Question 6: Minimise Z = x + 2y subject to. ANSWER: The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y ≥ 0, is as follows. The corner points of the feasible region are A (6, 0) and B (0, 3). The values of Z at these corner points are as follows. Corner point Z = x + 2y A(6, 0) 6 B(0, 3) 6 It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6 Thus, the minimum value of Z occurs for more than 2 points. Therefore, the value of Z is minimum at every point on the line, x + 2y = 6 Page No 514: Question 7: Minimise and Maximise Z = 5x + 10y subject to. ANSWER: The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and y ≥ 0, is as follows. The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30), and D (40, 20). The values of Z at these corner points are as follows. Corner point Z = 5x + 10y A(60, 0) 300 → Minimum B(120, 0) 600 → Maximum C(60, 30) 600 → Maximum D(40, 20) 400 The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30). Page No 514: Question 8: Minimise and Maximise Z = x + 2y subject to. ANSWER: The feasible region determined by the constraints, x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, and y ≥ 0, is as follows. The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200). The values of Z at these corner points are as follows. Corner point Z = x + 2y A(0, 50) 100 → Minimum B(20, 40) 100 → Minimum C(50, 100) 250 D(0, 200) 400 → Maximum The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40). Page No 514: Question 9: Maximise Z = − x + 2y, subject to the constraints: . ANSWER: The feasible region determined by the constraints,  is as follows. It can be seen that the feasible region is unbounded. The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows. Corner …

NCERT Solutions for Class 12 Science Maths Chapter 6 – Linear Programming Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Maths Chapter 5 on Three Dimensional Geometry, featuring clear, step-by-step explanations. Widely favored by class 12 Science students, these solutions prove invaluable for completing homework efficiently and preparing for exams. The Three Dimensional Geometry Solutions offer a user-friendly resource to grasp key concepts quickly. Access free answers to all questions from the NCERT Book of class 12 Science Maths Chapter 5, enhancing your understanding and aiding in academic success. Page No 467: Question 1: If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines. ANSWER: Let direction cosines of the line be l, m, and n. Therefore, the direction cosines of the line are Page No 467: Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes. ANSWER: Let the direction cosines of the line make an angle α with each of the coordinate axes. ∴ l = cos α, m = cos α, n = cos α Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are  Page No 467: Question 3: If a line has the direction ratios −18, 12, −4, then what are its direction cosines? ANSWER: If a line has direction ratios of −18, 12, and −4, then its direction cosines are Thus, the direction cosines are. Page No 467: Question 4: Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear. ANSWER: The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7). It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2), are given by, x2 − x1, y2 − y1, and z2 − z1. The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3. The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6. It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional. Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear. Page No 467: Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2) ANSWER: The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2). The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6. Therefore, the direction cosines of AB are The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4. Therefore, the direction cosines of BC are −217√, −317√, −217√-217, -317, -217The direction ratios of CA are 3−(−5), 5−(−5) and −4−(−2) i.e. 8, 10 and -2. Therefore the direction cosines of CA are 8(8)2 + (10)2 + (−2)2√, 10(8)2 + (10)2 + (−2)2√, −2(8)2 + (10)2 + (−2)2√8242√, 10242√, −2242√442√, 542√, −142√882 + 102 + -22, 1082 + 102 + -22, -282 + 102 + -228242, 10242, -2242442, 542, -142 Page No 477: Question 1: Show that the three lines with direction cosines  are mutually perpendicular. ANSWER: Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0 (i) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. (ii) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. (iii) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. Thus, all the lines are mutually perpendicular. Page No 477: Question 2: Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). ANSWER: Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). The direction ratios, a1, b1, c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4. The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4. AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0 a1a2 + b1b2+ c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4 = 6 + 10 − 16 = 0 Therefore, AB and CD are perpendicular to each other. Page No 477: Question 3: Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5). ANSWER: Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5). The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4. The direction ratios, a2, b2, c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4. AB will be parallel to CD, if  Thus, AB is parallel to CD. Page No 477: Question 4: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector. ANSWER: It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is  It is known that the line which passes through point A and parallel to is given by is a constant. This is the required equation of the line. Page No 477: Question 5: Find the equation of the line in vector and in Cartesian form that passes through the point with position vector  and is in the direction . ANSWER: It is given that the line passes through the point with position vector It is known that a line through a point with position vector and parallel to is given by the equation,  This is the required equation of the line in vector form. Eliminating λ, we obtain the Cartesian form equation as This is the required equation of the given line in Cartesian form. Page No …

NCERT Solutions for Class 12 Science Maths Chapter 5 – Three Dimensional Geometry Read More »

Explore detailed solutions for Class 12 Science Maths Chapter 4 on Vector Algebra, featuring clear, step-by-step explanations. Widely embraced by Class 12 Science students, these Maths Vector Algebra Solutions are invaluable for efficiently finishing homework assignments and gearing up for examinations. Free access to all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Maths is available here. Page No 428: Question 1: Represent graphically a displacement of 40 km, 30° east of north. ANSWER: Here, vector represents the displacement of 40 km, 30° East of North. Page No 428: Question 2: Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 metres north-west (iii) 40° (iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2 ANSWER: (i) 10 kg is a scalar quantity because it involves only magnitude. (ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction. (iii) 40° is a scalar quantity as it involves only magnitude. (iv) 40 watts is a scalar quantity as it involves only magnitude. (v) 10–19 coulomb is a scalar quantity as it involves only magnitude. (vi) 20 m/s2 is a vector quantity as it involves magnitude as well as direction. Page No 428: Question 3: Classify the following as scalar and vector quantities. (i) time period (ii) distance (iii) force (iv) velocity (v) work done ANSWER: (i) Time period is a scalar quantity as it involves only magnitude. (ii) Distance is a scalar quantity as it involves only magnitude. (iii) Force is a vector quantity as it involves both magnitude and direction. (iv) Velocity is a vector quantity as it involves both magnitude as well as direction. (v) Work done is a scalar quantity as it involves only magnitude. Page No 428: Question 4: In Figure, identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal ANSWER: (i) Vectors and are coinitial because they have the same initial point. (ii) Vectorsandare equal because they have the same magnitude and direction. (iii) Vectorsand are collinear but not equal. This is because although they are parallel, their directions are not the same. Page No 428: Question 5: Answer the following as true or false. (i)  andare collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal. ANSWER: (i) True. Vectors  andare parallel to the same line. (ii) False. Collinear vectors are those vectors that are parallel to the same line. (iii) False. It is not necessary for two vectors having the same magnitude to be parallel to the same line. (iv) False. Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points. Page No 440: Question 1: Compute the magnitude of the following vectors: ANSWER: The given vectors are: Page No 440: Question 2: Write two different vectors having same magnitude. ANSWER: Hence, are two different vectors having the same magnitude. The vectors are different because they have different directions. Page No 440: Question 3: Write two different vectors having same direction. ANSWER: The direction cosines of are the same. Hence, the two vectors have the same direction. Page No 440: Question 4: Find the values of x and y so that the vectors are equal ANSWER: The two vectors will be equal if their corresponding components are equal. Hence, the required values of x and y are 2 and 3 respectively. Page No 440: Question 5: Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7). ANSWER: The vector with the initial point P (2, 1) and terminal point Q (–5, 7) can be given by, Hence, the required scalar components are –7 and 6 while the vector components are  Page No 440: Question 6: Find the sum of the vectors. ANSWER: The given vectors are. Page No 440: Question 7: Find the unit vector in the direction of the vector. ANSWER: The unit vector in the direction of vector is given by. Page No 440: Question 8: Find the unit vector in the direction of vector, where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively. ANSWER: The given points are P (1, 2, 3) and Q (4, 5, 6). Hence, the unit vector in the direction of  is . Page No 440: Question 9: For given vectors, and , find the unit vector in the direction of the vector  ANSWER: The given vectors are and. Page No 440: Question 10: Find a vector in the direction of vector which has magnitude 8 units. ANSWER: Hence, the vector in the direction of vector which has magnitude 8 units is given by, Page No 440: Question 11: Show that the vectorsare collinear. ANSWER: . Hence, the given vectors are collinear. Page No 440: Question 12: Find the direction cosines of the vector  ANSWER: Hence, the direction cosines of  Page No 440: Question 13: Find the direction cosines of the vector joining the points A (1, 2, –3) and B (–1, –2, 1) directed from A to B. ANSWER: The given points are A (1, 2, –3) and B (–1, –2, 1). Hence, the direction cosines of are  Page No 440: Question 14: Show that the vector is equally inclined to the axes OX, OY, and OZ. ANSWER: Therefore, the direction cosines of  Now, let α, β, and γbe the angles formed by with the positive directions of x, y, and z axes. Then, we have Hence, the given vector is equally inclined to axes OX, OY, and OZ. Page No 440: Question 15: Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  respectively, in the ration 2:1 (i) internally (ii) externally ANSWER: The position vector of point R dividing the line segment joining two points P and Q in the ratio m: n is given by: Position vectors of P and Q are given as: (i) The position vector of point R which divides the line joining two points P and Q internally in the ratio 2:1 is given by, (ii) The …

NCERT Solutions for Class 12 Science Maths Chapter 4 – Vector Algebra Read More »

Explore comprehensive NCERT Solutions for Class 12 Science Maths Chapter 3 on Differential Equations, featuring clear and straightforward step-by-step explanations. Widely acclaimed among class 12 Science students, these Maths Differential Equations Solutions are a valuable resource for efficiently completing homework assignments and preparing for exams. Free access to all questions and answers from Chapter 3 of the NCERT Book for class 12 Science Maths is available here, ensuring a convenient and effective study experience. Page No 382: Question 1: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is four. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined. Page No 382: Question 2: Determine order and degree(if defined) of differential equation  ANSWER: The given differential equation is: The highest order derivative present in the differential equation is. Therefore, its order is one. It is a polynomial equation in. The highest power raised tois 1. Hence, its degree is one. Page No 382: Question 3: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the given differential equation is. Therefore, its order is two. It is a polynomial equation inand. The power raised tois 1. Hence, its degree is one. Page No 382: Question 4: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the given differential equation is. Therefore, its order is 2. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined. Page No 382: Question 5: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is two. It is a polynomial equation inand the power raised tois 1. Hence, its degree is one. Page No 382: Question 6: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is three. The given differential equation is a polynomial equation in. The highest power raised tois 2. Hence, its degree is 2. Page No 382: Question 7: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is three. It is a polynomial equation in. The highest power raised tois 1. Hence, its degree is 1. Page No 383: Question 8: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is one. The given differential equation is a polynomial equation inand the highest power raised tois one. Hence, its degree is one. Page No 383: Question 9: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is two. The given differential equation is a polynomial equation inandand the highest power raised tois one. Hence, its degree is one. Page No 383: Question 10: Determine order and degree(if defined) of differential equation  ANSWER: The highest order derivative present in the differential equation is. Therefore, its order is two. This is a polynomial equation inandand the highest power raised tois one. Hence, its degree is one. Page No 383: Question 11: The degree of the differential equation is (A) 3 (B) 2 (C) 1 (D) not defined ANSWER: The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined. Hence, the correct answer is D. Page No 383: Question 12: The order of the differential equation is (A) 2 (B) 1 (C) 0 (D) not defined ANSWER: The highest order derivative present in the given differential equation is. Therefore, its order is two. Hence, the correct answer is A. Page No 385: Question 1: ANSWER: Differentiating both sides of this equation with respect to x, we get: Now, differentiating equation (1) with respect to x, we get: Substituting the values ofin the given differential equation, we get the L.H.S. as: Thus, the given function is the solution of the corresponding differential equation. Page No 385: Question 2: ANSWER: Differentiating both sides of this equation with respect to x, we get: Substituting the value ofin the given differential equation, we get: L.H.S. == R.H.S. Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 3: ANSWER: Differentiating both sides of this equation with respect to x, we get: Substituting the value ofin the given differential equation, we get: L.H.S. == R.H.S. Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 4: ANSWER: Differentiating both sides of the equation with respect to x, we get: L.H.S. = R.H.S. Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 5: ANSWER: Differentiating both sides with respect to x, we get: Substituting the value ofin the given differential equation, we get: Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 6: ANSWER: Differentiating both sides of this equation with respect to x, we get: Substituting the value ofin the given differential equation, we get: Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 7: ANSWER: Differentiating both sides of this equation with respect to x, we get:  L.H.S. = R.H.S. Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 8: ANSWER: Differentiating both sides of the equation with respect to x, we get: Substituting the value ofin equation (1), we get: Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 9: ANSWER: Differentiating both sides of this equation with respect to x, we get: Substituting the value ofin the given differential equation, we get: Hence, the given function is the solution of the corresponding differential equation. Page No 385: Question 10: ANSWER: Differentiating both sides of this equation with respect to x, we get: Substituting the value ofin the given …

NCERT Solutions for Class 12 Science Maths Chapter 3 – Differential Equations Read More »

Explore comprehensive NCERT Solutions for Class 12 Science Maths Chapter 2: Application Of Integrals. These solutions offer straightforward, step-by-step explanations, making them a favored resource among Class 12 Science students. Whether you’re tackling homework assignments or gearing up for exams, these Maths Application Of Integrals Solutions prove invaluable for efficient preparation. Access free answers to all questions from Chapter 2 of the NCERT Book for Class 12 Science Maths, ensuring a reliable aid in your academic journey. Page No 365: Question 1: Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis. ANSWER: The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD. Page No 365: Question 2: Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant. ANSWER: The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD. Page No 366: Question 3: Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant. ANSWER: The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD. Page No 366: Question 4: Find the area of the region bounded by the ellipse  ANSWER: The given equation of the ellipse, , can be represented as It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area of OAB Therefore, area bounded by the ellipse = 4 × 3π = 12π units Page No 366: Question 5: Find the area of the region bounded by the ellipse  ANSWER: The given equation of the ellipse can be represented as It can be observed that the ellipse is symmetrical about x-axis and y-axis. ∴ Area bounded by ellipse = 4 × Area OAB Therefore, area bounded by the ellipse =  Page No 366: Question 6: Find the area of the region in the first quadrant enclosed by x-axis, line and the circle  ANSWER: The area of the region bounded by the circle, , and the x-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is . Area OAB = Area ΔOCA + Area ACB Area of OAC  Area of ABC  Therefore, required area enclosed = 3√2 + π3 − 3√2 = π3 square units32 + π3 – 32 = π3 square units Page No 366: Question 7: Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line  ANSWER: The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the area ABCDA. It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, , is  units. Page No 366: Question 8: The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a. ANSWER: The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts. ∴ Area OAD = Area ABCD It can be observed that the given area is symmetrical about x-axis. ⇒ Area OED = Area EFCD From (1) and (2), we obtain Therefore, the value of a is . Page No 366: Question 9: Find the area of the region bounded by the parabola y = x2 and  ANSWER: The area bounded by the parabola, x2 = y,and the line,, can be represented as The given area is symmetrical about y-axis. ∴ Area OACO = Area ODBO The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1). Area of OACO = Area ΔOAM – Area OMACO Area of ΔOAM    Area of OMACO  ⇒ Area of OACO = Area of ΔOAM – Area of OMACO Therefore, required area = units Page No 366: Question 10: Find the area bounded by the curve x2 = 4y and the line x = 4y – 2 ANSWER: The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the shaded area OBAO. Let A and B be the points of intersection of the line and parabola. Coordinates of point . Coordinates of point B are (2, 1). We draw AL and BM perpendicular to x-axis. It can be observed that, Area OBAO = Area OBCO + Area OACO … (1) Then, Area OBCO = Area OMBC – Area OMBO Similarly, Area OACO = Area OLAC – Area OLAO Therefore, required area =  Page No 366: Question 11: Find the area of the region bounded by the curve y2 = 4x and the line x = 3 ANSWER: The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO. The area OACO is symmetrical about x-axis. ∴ Area of OACO = 2 (Area of OAB) Therefore, the required area is units. Page No 366: Question 12: Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is A. π B.  C.  ANSWER: The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as Thus, the correct answer is A. Page No 366: Question 13: Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is A. 2 B.  C.  ANSWER: The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as Thus, the correct answer is B. Page No 371: Question 1: Find the area of the circle 4×2 + 4y2 = 9 which is interior to the parabola x2 = 4y ANSWER: The required area is represented by the shaded area OBCDO. Solving the given equation of circle, 4×2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as. It can be observed that the required area is symmetrical about y-axis. ∴ Area OBCDO = 2 × Area OBCO We draw BM perpendicular to OA. Therefore, the coordinates of M are. Therefore, the required area OBCDO is units Page No 371: Question 2: Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1 ANSWER: The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by the shaded area as On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, …

NCERT Solutions for Class 12 Science Maths Chapter 2 – Application Of Integrals Read More »

Unlock the power of NCERT Solutions for Class 12 Science Maths Chapter 1: Integrals! Dive into step-by-step explanations that make tackling homework and exam prep a breeze. Access all the questions and answers from the NCERT Book for Class 12 Science Maths Chapter 1, absolutely free by DD Target PMT Page No 299: Question 1: sin 2x ANSWER: The anti derivative of sin 2x is a function of x whose derivative is sin 2x. It is known that, Therefore, the anti derivative of Page No 299: Question 2: Cos 3x ANSWER: The anti derivative of cos 3x is a function of x whose derivative is cos 3x. It is known that, Therefore, the anti derivative of . Page No 299: Question 3: e2x ANSWER: The anti derivative of e2x is the function of x whose derivative is e2x. It is known that, Therefore, the anti derivative of . Page No 299: Question 4: ANSWER: The anti derivative of is the function of x whose derivative is . It is known that, Therefore, the anti derivative of . Page No 299: Question 5: ANSWER: The anti derivative of  is the function of x whose derivative is . It is known that, Therefore, the anti derivative of  is . Page No 299: Question 6: ANSWER: Page No 299: Question 7: ANSWER: Page No 299: Question 8: ANSWER: Page No 299: Question 9: ANSWER: Page No 299: Question 10: ANSWER: Page No 299: Question 11: ANSWER: Page No 299: Question 12: ANSWER: Page No 299: Question 13: ANSWER: On dividing, we obtain Page No 299: Question 14: ANSWER: Page No 299: Question 15: ANSWER: Page No 299: Question 16: ANSWER: Page No 299: Question 17: ANSWER: Page No 299: Question 18: ANSWER: Page No 299: Question 19: ANSWER: Page No 299: Question 20: ANSWER: Page No 299: Question 21: The anti derivative of equals ANSWER: Hence, the correct answer is C. Page No 299: Question 22: If such that f(2) = 0, then f(x) is (A) (B)  (C)  (D)  ANSWER: It is given that, ∴Anti derivative of  ∴ Also, Hence, the correct answer is A. Page No 304: Question 1: ANSWER: Let = t ∴2x dx = dt Page No 304: Question 2: ANSWER: Let log |x| = t ∴  Page No 304: Question 3: ANSWER: Let 1 + log x = t ∴  Page No 304: Question 4: sin x ⋅ sin (cos x) ANSWER: sin x ⋅ sin (cos x) Let cos x = t ∴ −sin x dx = dt Page No 304: Question 5: ANSWER: Let  ∴ 2adx = dt Page No 304: Question 6: ANSWER: Let ax + b = t ⇒ adx = dt Page No 304: Question 7: ANSWER: Let  ∴ dx = dt Page No 304: Question 8: ANSWER: Let 1 + 2×2 = t ∴ 4xdx = dt Page No 304: Question 9: ANSWER: Let  ∴ (2x + 1)dx = dt Page No 304: Question 10: ANSWER: Let  ∴ Page No 304: Question 11: ANSWER: Page No 304: Question 12: ANSWER: Let  ∴  Page No 304: Question 13: ANSWER: Let  ∴ 9x2dx = dt Page No 304: Question 14: ANSWER: Let log x = t ∴  Page No 304: Question 15: ANSWER: Let  ∴ −8x dx = dt Page No 304: Question 16: ANSWER: Let  ∴ 2dx = dt Page No 304: Question 17: ANSWER: Let  ∴ 2xdx = dt Page No 305: Question 18: ANSWER: Let  ∴  Page No 305: Question 19: ANSWER: Dividing numerator and denominator by ex, we obtain Let  ∴  Page No 305: Question 20: ANSWER: Let  ∴  Page No 305: Question 21: ANSWER: Let 2x − 3 = t ∴ 2dx = dt Page No 305: Question 22: ANSWER: Let 7 − 4x = t ∴ −4dx = dt Page No 305: Question 23: ANSWER: Let  ∴  Page No 305: Question 24: ANSWER: Let  ∴  Page No 305: Question 25: ANSWER: Let  ∴  Page No 305: Question 26: ANSWER: Let  ∴  Page No 305: Question 27: ANSWER: Let sin 2x = t ∴  Page No 305: Question 28: ANSWER: Let  ∴ cos x dx = dt Page No 305: Question 29: cot x log sin x ANSWER: Let log sin x = t Page No 305: Question 30: ANSWER: Let 1 + cos x = t ∴ −sin x dx = dt Page No 305: Question 31: ANSWER: Let 1 + cos x = t ∴ −sin x dx = dt Page No 305: Question 32: ANSWER: Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt Page No 305: Question 33: ANSWER: Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt Page No 305: Question 34: ANSWER: Page No 305: Question 35: ANSWER: Let 1 + log x = t ∴  Page No 305: Question 36: ANSWER: Let  ∴  Page No 305: Question 37: ANSWER: Let x4 = t ∴ 4×3 dx = dt Let  ∴ From (1), we obtain Page No 305: Question 38: equals ANSWER: Let  ∴  Hence, the correct answer is D. Page No 305: Question 39: equals A.  B.  C.  ANSWER: Hence, the correct answer is B. Page No 307: Question 1: ANSWER: Page No 307: Question 2: ANSWER: It is known that,  Page No 307: Question 3: cos 2x cos 4x cos 6x ANSWER: It is known that, Page No 307: Question 4: sin3 (2x + 1) ANSWER: Let  Page No 307: Question 5: sin3x cos3x ANSWER: Page No 307: Question 6: sin x sin 2x sin 3x ANSWER: It is known that,  Page No 307: Question 7: sin 4x sin 8x ANSWER: Page No 307: Question 8: ANSWER: Page No 307: Question 9: ANSWER: Page No 307: Question 10: sin4x ANSWER: Page No 307: Question 11: cos4 2x ANSWER: Page No 307: Question 12: ANSWER: Page No 307: Question 13: ANSWER: Page No 307: Question 14: ANSWER: Page No 307: Question 15: ANSWER: Page No 307: Question 16: tan4x ANSWER: From equation (1), we obtain Page No 307: Question 17: ANSWER: Page No 307: Question 18: ANSWER: Page No 307: Question 19: ANSWER: Page No 307: Question 20: ANSWER: Page No 307: Question 21: sin−1 (cos x) ANSWER: It is known that, Substituting in equation (1), we obtain Page No 307: Question 22: ANSWER: Page No 307: Question 23:  is equal to A. tan x + cot x + C B. tan x + cosec x + C C. − tan x + cot x + C D. tan x + sec x + C ANSWER: Hence, the correct answer is A. Page No 307: Question 24:  equals A. − cot (exx) + C B. tan (xex) + C C. tan (ex) + C D. cot (ex) + C ANSWER: Let exx = t Hence, the correct answer is B. Page No 315: Question 1: ANSWER: Let x3 = t ∴ 3x2dx = dt …

NCERT Solutions for Class 12 Science Maths Chapter 1 – Integrals Read More »

Discover comprehensive NCERT solutions for Class 12 Science Mathematics Chapter 6: Application of Derivatives, featuring straightforward, step-by-step explanations. Highly sought after by Class 12 Science students, these solutions are a valuable resource for efficiently finishing homework assignments and preparing for exams. All questions and answers from Chapter 6 of the NCERT Mathematics textbook for Class 12 Science are available here at no cost, serving as a convenient aid for students. Page No 197: Question 1: Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm ANSWER: The area of a circle (A)with radius (r) is given by, Now, the rate of change of the area with respect to its radius is given by,  Hence, the area of the circle is changing at the rate of 6Ï€ cm when its radius is 3 cm. Hence, the area of the circle is changing at the rate of 8Ï€ cm when its radius is 4 cm. Page No 197: Question 2: The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm? ANSWER: Let x be the length of a side, V be the volume, and s be the surface area of the cube. Then, V = x3 and S = 6×2 where x is a function of time t. It is given that. Then, by using the chain rule, we have: ∴ ⇒ Thus, when x = 12 cm,  Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of cm2/s. Page No 197: Question 3: The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm. ANSWER: The area of a circle (A) with radius (r) is given by, Now, the rate of change of area (A) with respect to time (t) is given by, It is given that, ∴ Thus, when r = 10 cm, Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s. Page No 197: Question 4: An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? ANSWER: Let x be the length of a side and V be the volume of the cube. Then, V = x3. ∴ (By chain rule) It is given that, ∴ Thus, when x = 10 cm, Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long. Page No 197: Question 5: A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? ANSWER: The area of a circle (A) with radius (r) is given by. Therefore, the rate of change of area (A) with respect to time (t) is given by, [By chain rule] It is given that. Thus, when r = 8 cm, Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s. Page No 198: Question 6: The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? ANSWER: The circumference of a circle (C) with radius (r) is given by C = 2πr. Therefore, the rate of change of circumference (C) with respect to time (t) is given by, (By chain rule) It is given that. Hence, the rate of increase of the circumference Page No 198: Question 7: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle. ANSWER: Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have: and  (a) The perimeter (P) of a rectangle is given by, P = 2(x + y) Hence, the perimeter is decreasing at the rate of 2 cm/min. (b) The area (A) of a rectangle is given by, A = x⋅ y ∴ When x = 8 cm and y = 6 cm,  Hence, the area of the rectangle is increasing at the rate of 2 cm2/min. Page No 198: Question 8: A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. ANSWER: The volume of a sphere (V) with radius (r) is given by, ∴Rate of change of volume (V) with respect to time (t) is given by,  [By chain rule] It is given that. Therefore, when radius = 15 cm, Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is Page No 198: Question 9: A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm. ANSWER: The volume of a sphere (V) with radius (r) is given by. Rate of change of volume (V) with respect to its radius (r) is given by, Therefore, when radius = 10 cm, Hence, the volume of the balloon is increasing at the rate of 400π cm2. Page No 198: Question 10: A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall? ANSWER: Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall. Then, by Pythagoras theorem, we have: x2 + y2 = 25 [Length of the ladder = 5 m] Then, the rate …

NCERT Solutions for Class 12 Science Maths Chapter 6 – Application Of Derivatives Read More »