Software Testing Question Bank (20-05-2023)

  QUESTION BANK   Unit 1 part1 Explain different Phases of software project. Define Quality assurance and quality control. Differentiate between Quality assurance and quality control. Differentiate between Verification and validation. Explain different Principles of Testing. What Is Bug Life Cycle In Software Testing, Different...

SOftware Testing Question Bank

Question Bank 1:A. Explain the "Pesticide paradox" in software testing and how to avoid it.B. Differentiate between verification and validation in software testing.C. Identify common challenges in integration testing and propose solutions for successful testing and project outcomes.D. Determine the boundary values for an input field accepting numbers between 1 and 100, inclusive....

Solution 09

 Given: y = sin⁻¹(2x√(1-x²)) Using the chain rule with d/dx[sin⁻¹(u)] = 1/√(1-u²) · du/dx Let u = 2x√(1-x²) Step 1: Find du/dx using the product rule. u = 2x · √(1-x²) du/dx = 2x · d/dx[√(1-x²)] + √(1-x²) · d/dx[2x] For d/dx[√(1-x²)] = d/dx[(1-x²)^(1/2)] = (1/2)(1-x²)^(-1/2) · (-2x) = -x/√(1-x²) So: du/dx = 2x · (-x/√(1-x²)) + √(1-x²) · 2 du/dx = -2x²/√(1-x²) + 2√(1-x²) du/dx...

solution 6 ,7 and 8

 6. If y = ln((1+x)/(1-x)), find dy/dx and simplify Using the chain rule with ln(u) where u = (1+x)/(1-x): dy/dx = 1/u · du/dx = (1-x)/(1+x) · du/dx Find du/dx using the quotient rule: du/dx = [(1-x)(1) - (1+x)(-1)]/(1-x)² du/dx = [1-x + 1+x]/(1-x)² du/dx = 2/(1-x)² Therefore: dy/dx = (1-x)/(1+x) · 2/(1-x)² dy/dx = 2/[(1+x)(1-x)] dy/dx = 2/(1-x²) 7. A function is given implicitly...

Solution 3, 4 and 5

 3. If y = tan⁻¹((1+x)/(1-x)), find dy/dx Using the chain rule with the derivative of tan⁻¹(u) = 1/(1+u²) · du/dx Let u = (1+x)/(1-x) First, find du/dx using the quotient rule: du/dx = [(1-x)(1) - (1+x)(-1)]/(1-x)² du/dx = [1-x + 1+x]/(1-x)² du/dx = 2/(1-x)² Now apply the chain rule: dy/dx = 1/(1 + ((1+x)/(1-x))²) · 2/(1-x)² Simplify the denominator: 1 + ((1+x)/(1-x))² = [(1-x)²...

solution 2

 To differentiate y = x^x, I need to use logarithmic differentiation since both the base and exponent contain x. Given: y = x^x Step 1: Take the natural logarithm of both sides. ln y = ln(x^x) ln y = x ln x Step 2: Differentiate both sides with respect to x. d/dx(ln y) = d/dx(x ln x) Step 3: Apply the chain rule on the left side and product rule on the right side. Left side:...

Solution 1

 To find the derivative of y = ln(sin x), I'll use the chain rule. Given: y = ln(sin x) Using the chain rule: dy/dx = d/dx[ln(sin x)] The derivative of ln(u) is 1/u · du/dx, where u = sin x. So: dy/dx = 1/(sin x) · d/dx(sin x) Since d/dx(sin x) = cos x: dy/dx = 1/(sin x) · cos x = cos x/sin x Therefore: dy/dx = cot x We can also write this as cos x/sin x, but cot x is the...

Solution Q. 10

 Given the equation x=etan⁡−1(y−x2x2)x=etan−1(x2y−x2​), we need to find (dydx)(dxdy​) at x=1x=1.First, we take the natural logarithm of both sides:ln⁡x=tan⁡−1(y−x2x2)lnx=tan−1(x2y−x2​)Let u=y−x2x2u=x2y−x2​. Then the equation becomes:ln⁡x=tan⁡−1(u)lnx=tan−1(u)Taking the tangent of both sides, we get:u=tan⁡(ln⁡x)u=tan(lnx)Substituting back u=y−x2x2u=x2y−x2​,...