Evolution of Exponential and Logarithmic Limits 1. Evaluate type of limit . Proof: . . . . 2. Evaluate type of limit . Proof: . . . = 1. 3. Evaluate type of limit . Proof: . . = 1 . Example 1: Evaluate . Solution: . . . . . Share Tweet View Read more about Evolution of Exponential and Logarithmic Limits[…]
Geometric Interpretation of Cross Product 1. . . . = 2 (Area of Triangle AOC) = Area of Parallelogram Area of the triangle OAB is . is said to be the vector area of the parallelogram with adjacent sides OA and OB. 2. If are diagonals of a parallelogram, its area = . In fig Read more about Geometric Interpretation of Cross Product[…]
Components of a Vector Let us take the points A (1, 0, 0), B (0,1,0) and C(0,0,1) on the X – axis, Y – axis and z – axis, respectively. Then, clearly , and . The vector, and each having magnitude 1, are called unit vector along the axes OX, OY and OZ, respectively and Read more about Components of a Vector[…]
Limits, Continuity and Differentiability – Method of Substitution Method of substitution: In order to evaluate we many substitute x = a + h (or) a – h, so that as x – a, h → 0. Thus (or) . This method is applied to bring the limit at zero as the most of formulate are Read more about Limits, Continuity and Differentiability – Method of Substitution[…]
Common Roots Condition for One Common Roots: Let us find the condition such that quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 may have a common root. Let α be the common root of given equations. Then, a₁α² + b₁α + c₁ = 0 and a₂α Read more about Common Roots[…]
Maxima and Minima – Problems Example 1: Show that f(x) = 3x + 7 is strictly increasing on R. f(x) = 3x + 7 Solution: Given that f(x) = 3x + 7 let x₁ x₂ ϵ R let x₁ < x₂ 3x₁ < 3x₂ 3 x₁ + 7 < 3x₂ + 7 f(x₁) < f(x₂) Read more about Maxima and Minima – Problems[…]
Functions, Limits and Continuity – Indeterminate Forms In finding the value of limits, sometimes we obtain the following forms. 0/ 0, ∞/ ∞, 0 x ∞, -∞, 0⁰, 1∞, ∞⁰ which are not determined. These forms are knowing as indeterminate forms. Example 1: Evaluate . Solution: Given that (= 0/0 form). . . . = Read more about Functions, Limits and Continuity – Indeterminate Forms[…]
Conditional Identities – II 1. sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2) Proof: Sin A + sin B + sin C = . . . . . . . . . . x. Hence proved. sinA + sinB + sinC = 4 cos(A/2) cos(B/2) cos(C/2). 2. If A + Read more about Conditional Identities – II[…]
Conditional Identities – I 1. sin2A + sin2B + sin2C = 4 sinA sinB sinC Proof: sin2A + sin2B + sin2C = 4 sinA sinB sinC L.H.S (sin2A + sin2B) + sin2C = 2sin (A + B) cos (A – B) + sin 2C A + B + C = π A + B = Read more about Conditional Identities – I[…]
General Solution of Trigonometric Function (Tan θ) General Solution of Equation tan θ = tan α: Given that, tan θ = tan α , sin θ. Cos α – sin α cos θ = 0 sin (θ – α) = 0 (θ – α) = sin⁻¹ (0) (θ – α) = sin⁻¹ (sin nπ) (θ Read more about General Solution of Trigonometric Function (Tan θ)[…]