Application of Derivatives
Mathematics · Class 12 · All
If f(x) = x^2, then the equation of the normal to the curve at the point (1,1) is
All 15 questions in this quiz
The primary objective of optimization problems in Application of Derivatives is to
- Find the derivative of a function.
- Find the maximum or minimum value of a quantity.
- Solve systems of linear equations.
- Calculate determinants.
Which theorem states that if a function is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0?
- Mean Value Theorem
- Rolle's Theorem
- Intermediate Value Theorem
- L'Hospital's Rule
If a function is increasing on an interval, then its first derivative on that interval is
- Negative
- Zero
- Positive
- Undefined
For the function f(x) = -x^2 + 6x - 5, the maximum value is
- 4
- 5
- 6
- 8
If f(x) = x^3 - 6x^2 + 9x + 1, then the value of x at which the function has a local maximum is
- x = 0
- x = 1
- x = 3
- x = 2
Which of the following is the necessary condition for a function to have a local maximum or minimum at x = a, provided the derivative exists?
- f(a) = 0
- f''(a) = 0
- f'(a) = 0
- f(a) > 0
If the second derivative of a function at x = a is positive and f'(a) = 0, then x = a is
- Point of maximum
- Point of minimum
- Point of inflection
- Point of discontinuity
If f(x) = x^2, then the equation of the normal to the curve at the point (1,1) is
- y = 2x - 1
- y = -x + 2
- y = x - 2
- y = -2x + 3
If f(x) = x^3 - 3x^2, then the critical points of the function are
- x = 0 only
- x = 2 only
- x = 0 and x = 2
- x = -2 and x = 2
If the slope of the tangent to a curve at a point is 2, then the slope of the normal at the same point is
- 2
- -2
- 1/2
- -1/2
The equation of the tangent to the curve y = x^2 at the point (1,1) is
- y = 2x - 1
- y = x + 1
- y = 2x + 1
- y = x - 1
If f'(x) = 0 at x = 5, then x = 5 is necessarily
- A point of maximum
- A point of minimum
- A stationary point
- A point of discontinuity
The slope of the tangent to the curve y = x^3 at x = 2 is
- 4
- 8
- 12
- 16
If the derivative of a function changes from positive to negative at x = a, then x = a is
- Point of minimum
- Point of maximum
- Point of inflection
- Point of discontinuity
If f(x) = x^2 - 4x + 3, then the interval on which the function is increasing is
- x < 2
- x > 2
- x < 0
- x > 4