Category: N-Scheme
Unit – III – 3.1 TRIGONOMETRY
Quadrant Sign of ratios Remember I All are +ve ( 90 – θ, 360 + θ ) All II Sin and cosec are +ve, Other ratios are –ve ( 90 + θ, 180 – θ ) Silver III tan θ and cot θ are +ve, Other ratios are –ve ( 180 + θ, 270 – θ ) Tea IV cos θ and sec θ are +ve, Other ratios are –ve ( 180 + θ, 360 – θ ) Cups Ratio Falls in Quadrant Change of ratio Sin…
Read MoreUnit – I – ALGEBRA
Matrix and its applications are very important part of Mathematics. Also it is one of the most powerful tools of Mathematics. Matrix notation and operation are used in electronic spread sheet programmes for personal computer, business budgeting, sales projection, cost estimation, analyzing the results of an experiment etc. Also many physical operations such as magnification, rotation and reflection through a plane can be represented mathematically by matrices. Also matrix used in Cryptography. A matrix is a rectangular array of numbers arranged in rows and columns enclosed by brackets. If there…
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N-SCHEME-ENGINEERING MATHEMATICS-II MODEL EXAM QUESTION PAPER/2021
MODEL EXAM QUESTION PAPER 40022 ENGINEERING MATHEMATICS – II Time : 3.00 Hours Date: 22-05-2021 Max.Marks: 100 Note: 1. Answer all question in PART A. Each question carries one mark. 2. Answer any ten questions in PART B. Each question carries two marks. 3. Answer all question by selecting either A or B. Each question carries fifteen marks. 4. Clarkes Table and programmable calculators are not permitted. PART – A (5×1=5) 1. Find the value of ‘p’if the pair of lines px2-5xy + 7y2 = 0 …
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N – 2.2 – Product of two vectors – Exercise Problems with solutions
Part – A Soln: = 1(1) + 1(1) + 0 = 1 + 1 = 2 Soln: = 2(3) + 3(2) – 2 (6) = 6 + 6 -12 = 0 Soln: 2(p) + 1(3) – 5 (-2) = 0 2p + 3 +10 = 0 2p + 13 = 0 Soln: Part –B Soln: Soln: Part –C Soln: = 1(1) + 2(1) + 1(-3) = 0 = 1(7) + 1(-4) – 3(1) = 7 – 4 – 3 = 0 = 7(1) -4 (2) + 1 (1) = 7…
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N – 2.2 – Product of two vectors – Exercise Problems
Part – A Part –B Part –C
Read MoreN – 2.1 – Vector Introduction – Exercise Problems with Solutions
Part – A Soln: Given Sol.: Given Soln: Part –B 1. Show that the points whose position vectors Soln: Given Part –C Prove that the points Soln: Given The given triangle is an equilateral triangle. 2. Prove that the points whose position vectors are Soln: Given The given triangle is an right angled triangle.
Read More2.1 – N – Vector Introduction – Exercise Problems
Part – A Part –B 1. Show that the points whose position vectors Part –C Prove that the points 2. Prove that the points whose position vectors are
Read More1.3 – N – CONICS – Exercise Problems with solutions
Part – A Prove that the equation x2 – 2xy + y2 – 16x – 12y + 22 = 0 is a parabola. Soln: x2 – 2xy + y2 – 16x – 12y + 22 = 0 —————– ( 1 ) Condition for ( 1 ) to represent parabola is h2 = ab From ( 1 ) a = 1, b = 1 2h = -2 ⇒ h = -1 h2 = ab (-1)2 = 1 ( 1) 1 = 1 1 = 1. ∴ ( 1 ) represents a…
Read More1.3 – N – CONICS Exercise Problems
Part – A Prove that the equation x2 – 2xy + y2 – 16x – 12y + 22 = 0 is a parabola. 2. Show that the equation 7×2 + 3xy + 2y2 – x + 2y – 1 = 0 represents an ellipse. Part – C Find axis, vertex, focus and equation of directrix for y2 + 4x + 6y + 17 = 0.
Read More1.2 – N – Analytical Geometry Exercise Problems With Solutions
Part – A 1. Find the equation of the circle with centre (1, -2) and radius 5 units. Soln: We know that the equation of circle is (x – h )2 + (y – k )2 = r2 Here h = 1, k = – 2 (given) and r = 5 (x – 1 )2 + (y + 2 )2 = 52 x2 – 2x + 1+ y2 + 4y + 4= 25 x2 + y2 – 2x + 4y + 5 -25 = 0 x2 + y2 –…
Read More1.2 – N – Analytical Geometry – II Exercise Problems
Part – A 1. Find the equation of the circle with centre (1, -2) and radius 5 units. 2. Find the centre and radius of the circle x2 + y2 + 10x + 8y + 5 = 0 . 3. Find the equation of the circle passing through the point (1 , 1 ) and concentric to the circle x2 + y2 + 4x + 6y – 15 = 0. Part – B 1. Find the equation of the circle passing through the point A (2, 3) and having its centre at…
Read More1.1 – N – Analytical Geometry Exercise Problems With Solutions
Part – A 1. Find the perpendicular distance from the point (3, -5) to the straight line 3x-4y-26=0. Soln: W.K.T The length of the perpendicular distance from (x1,y1) to the line ax + by + c = 0 is ± (ax1 + by1 + c)/ √(a2 + b2 ) Given straight line is 3x-4y-26 =0 Given point (x1,y1) = (3, -5) i.e ± 3(3) – 4(-5) – 26/ √(32 + (-4)2 ) = 3 / 5 2. Find the distance between the line 3x+4y = 9 and 6x+8y = 15. Soln: W.K.T …
Read More1.1 – N – Analytical Geometry Exercise Problems
Part – A 1. Find the perpendicular distance from the point (3, -5) to the straight line 3x-4y-26=0. 2. Find the distance between the line 3x+4y = 9 and 6x+ 8y = 15. 3. Show that the lines 3x+2y+9=0 and 12x+8y-15 =0 are parallel. 4. Find ‘p’ such that the lines 3x+4y = 8 and px + 2y = 7 are parallel. 5. Show that the lines 27x-18y+25 =0 and 2x + 3y+7 =0 are perpendicular. 6. Find the value of k if the lines 2x + ky -11 =0 and 5x…
Read MoreDiploma Continuous Assessment Test -2/ April – 2021
Course: First year Diploma course in Engineering & Technology Subject & Code : Engineering Mathematics – II (40022) Time : 2 Hours Date: 27-04-2021 Max. Marks: 50 PART – A Answer all questions ( 6×1=6 marks) Soln: = 1 ( 1 – 0 ) – 1 ( 0 – 1 ) + 0 ( 0 – 1 ) = 1 ( 1 ) -1 (- 1 ) + 0 = 1 + 1 = 2 2. Define vector Differential operator. Soln: Soln: = yz + 3×2 ( 1 ) …
Read MoreDEFINITE INTEGRALS
Definition of Definite Integrals: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Example : Soln: du = cos x dx Example : Soln: Adding ( 1 ) & ( 2 )
Read MoreBERNOULLLI’S FORMULA
If u and v are functions x, then Bernoulli’s form of integration by parts formula is Where u΄, u΄΄,u΄΄΄….. are successive differentiation of the function u and v, v1, v2, v3, …………. the successive integration of the function dv. Note: The function ‘u’ is differentiated up to constant. Example : Soln: ILATE Example : Soln: Example : Soln: ILATE Example : Soln: ILATE Example : Soln: ILATE
Read More5.1 INTEGRATION BY PARTS
Introduction: When the integrand is a product of two functions and the method of decomposition or substitution can not be applied, then the method of by parts is used. Integraiton by parts formula: The above formula is used by taking proper choice of ‘u’ and ‘dv’. ‘u’ should be chosen based on thefollowing order of Preference. Simply remember ILATE 1. Inverse trigonometric functions: 2. Logarithmic functions: log x 3. Algebraic functions: 4. Trigonometric functions: sin x, cos x, tan x, etc. 5. Exponential functions: Example: Soln: ILATE u = x …
Read More4.3 STANDARD INTEGRALS
List of Formulae: Example : Soln: Example : Soln: Example : Soln: Example : Soln: Put u = 3x + 2 Example : Soln: Example : Soln: Put u = 3x – 2 Example : Soln: Example : Soln: Amazon.in Widgets Example : Soln: Example : Soln:
Read More4.2 INTEGRATION BY SUBSTITUTION
So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences. which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using basic integration formula. When the integrand (the function to be integrated) is either in multiplication or in division form and if the derivative of one full or meaningful part of the function is equal to…
Read More4.1 INTEGRATION – DECOMPOSITION METHOD
Sir Sardar Vallabhai Patel, called the Iron Man of India integrated several princely states together while forming our country Indian Nation after independence. Like that in Maths while finding area under a curve through integration, the area under the curve is divided into smaller rectangles and then integrating (i.e) summing of all the area of rectangles together. So, integration means of summation of very minute things of the same kind. Integration as the reverse of differentiation: Integration can also be introduced in another way, called integration as the reverse…
Read MoreAPPLICATION OF VECTOR DIFFERENTIATION
is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of (abbreviated as ‘Div ‘) is defined as, Basic properties of Divergence: If A, B are vector functions and ‘f’ is a scalar function, then Example: Soln: = yz + 3×2 ( 1 ) + ( 0 – y3 ) = yz + 3×2 – y3 Example: Soln: At ( 1, -1, 2) = -2 – 4 – 1 = -7 Example: Soln:…
Read MoreVECTOR DIFFERENTIATION
Vector point function and Vector field: Let P be any point in a region ‘D’ of space. Let r be the position vector of P. If there exists a vector function F corresponding to each P, then such a function F is called a vector function and the region D is called a vector field. Example: consider the vector function Let P be a point whose position vector is At P , the value of F is obtained by putting x = 2, Y = I, z = 3 in…
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