68 free MCQs on Application of Derivatives with worked answers and explanations. Use derivatives to study rate of change, increasing and decreasing functions, tangents and normals, and maxima and minima, with classic optimization problems.
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The tangent at a point touches the curve with slope f'(a); the normal is the line perpendicular to the tangent at that same point, with slope -1/f'(a) - together they describe the curve's local direction and the line "straight into" the curve.
Q1.
If f'(x) > 0 for all x in an interval, the function f is:
Answer: B. Strictly increasing
Why: A positive derivative throughout an interval means the function is strictly increasing on that interval.
Q2.
If f'(x) < 0 for all x in an interval, the function f is:
Answer: C. Strictly decreasing
Why: A negative derivative throughout an interval means the function is strictly decreasing on that interval.
Q3.
A critical point of a function f is a point where:
Answer: B. f'(x) = 0 or f'(x) does not exist
Why: Critical points occur where the derivative is zero or undefined; these are candidates for local maxima or minima.
Q4.
The slope of the tangent to the curve y = f(x) at point (x<sub>1</sub>, y<sub>1</sub>) is given by:
Answer: B. f'(x<sub>1</sub>)
Why: The derivative f'(x<sub>1</sub>) evaluated at the point gives the slope of the tangent line there.
Q5.
If the slope of the tangent at a point is m, the slope of the normal at that point is:
Answer: D. -1/m
Why: The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent slope, -1/m.
Q6.
At a point where the tangent is horizontal, f'(x) equals:
Answer: B. 0
Why: A horizontal tangent has slope zero, so f'(x) = 0 at that point.
Q7.
By the second derivative test, if f'(c) = 0 and f''(c) > 0, then x = c is a:
Answer: B. Local minimum
Why: A positive second derivative at a critical point indicates the curve is concave up there, giving a local minimum.
Q8.
By the second derivative test, if f'(c) = 0 and f''(c) < 0, then x = c is a:
Answer: B. Local maximum
Why: A negative second derivative at a critical point means the curve is concave down there, giving a local maximum.
Q9.
For the function f(x) = x<sup>2</sup>, the value of f'(x) is:
Answer: B. 2x
Why: Using the power rule, d/dx(x<sup>2</sup>) = 2x.
Q10.
If Area of a circle A = pi r<sup>2</sup>, then dA/dt in terms of dr/dt is:
Answer: B. 2 pi r (dr/dt)
Why: Differentiating A = pi r<sup>2</sup> with respect to t using the chain rule gives dA/dt = 2 pi r (dr/dt).
Q11.
Rolle's theorem requires which of the following conditions on [a, b]?
Answer: A. f(a) = f(b)
Why: Rolle's theorem requires f continuous on [a,b], differentiable on (a,b), and f(a) = f(b).
Q12.
The Mean Value Theorem guarantees a point c in (a, b) where f'(c) equals:
Answer: B. [f(b) - f(a)] / (b - a)
Why: MVT states f'(c) equals the average rate of change [f(b) - f(a)] / (b - a) for some c in (a,b).
Q13.
Using the approximation formula, f(x + dx) is approximately equal to:
Answer: B. f(x) + f'(x) dx
Why: The linear approximation formula states f(x + dx) approximately equals f(x) + f'(x) dx for small dx.
Q14.
The derivative dy/dx represents the ___ of a function:
Answer: A. rate of change
Why: The derivative measures how fast the output changes with the input - the rate of change.
Q15.
The slope of the tangent to a curve y = f(x) at a point is given by:
Answer: A. f′(x)
Why: The derivative f′(x) gives the slope of the tangent at each point.
Q16.
At a point of local maximum or minimum, the first derivative f′(x) is:
Answer: A. zero
Why: At a smooth turning point the tangent is horizontal, so f′(x) = 0.
Q17.
A function is increasing on an interval where its derivative f′(x) is:
Answer: A. positive
Why: A positive derivative means the function rises as x increases.
Q18.
A function is decreasing on an interval where f′(x) is:
Answer: A. negative
Why: A negative derivative means the function falls as x increases.
Q19.
The second derivative f″(x) is used to determine the ___ of a curve:
Answer: A. concavity
Why: The sign of f″(x) tells whether the curve is concave up or concave down.
Q20.
Points at which f′(x) = 0 are called ___ points:
Answer: A. critical
Why: Critical points, where the derivative is zero (or undefined), are candidates for extrema.
Q21.
Find the equation of the tangent to the curve y = x<sup>2</sup> at the point (2, 4).
Answer: A. y = 4x - 4
Why: f'(x) = 2x, so slope at x=2 is 4. Tangent: y - 4 = 4(x - 2), which simplifies to y = 4x - 4.
Q22.
Find the equation of the normal to the curve y = x<sup>3</sup> at the point (1, 1).
Answer: B. y - 1 = -(1/3)(x - 1)
Why: f'(x) = 3x<sup>2</sup>, slope of tangent at x=1 is 3, so normal slope is -1/3. Normal equation: y - 1 = -(1/3)(x - 1).
Q23.
Find the interval in which f(x) = x<sup>2</sup> - 4x + 3 is strictly decreasing.
Answer: B. x < 2
Why: f'(x) = 2x - 4, which is negative when x < 2. So f is strictly decreasing for x < 2.
Q24.
Find the critical points of f(x) = x<sup>3</sup> - 3x.
Answer: B. x = 1 and x = -1
Why: f'(x) = 3x<sup>2</sup> - 3 = 3(x<sup>2</sup> - 1) = 0 gives x = 1 and x = -1 as critical points.
Q25.
For f(x) = x<sup>3</sup> - 6x<sup>2</sup> + 9x + 1, the local maximum occurs at:
Answer: A. x = 1
Why: f'(x) = 3x<sup>2</sup> - 12x + 9 = 3(x-1)(x-3) = 0 at x=1,3. f''(x) = 6x - 12. f''(1) = -6 < 0, so local max at x=1.
Q26.
For f(x) = x<sup>3</sup> - 6x<sup>2</sup> + 9x + 1, the local minimum value is:
Answer: B. 1
Why: Critical points are x=1,3. f''(3) = 6(3)-12 = 6 > 0, so local min at x=3. f(3) = 27 - 54 + 27 + 1 = 1.
Q27.
Two positive numbers have sum 16. Find the maximum value of their product.
Answer: D. 64
Why: Let numbers be x and 16-x. Product P = x(16-x). dP/dx = 16-2x = 0 gives x=8, so both numbers are 8 and product = 64, which is a maximum since d<sup>2</sup>P/dx<sup>2</sup> = -2 < 0.
Q28.
Using differentials, approximate the value of the square root of 25.2.
Answer: B. 5.02
Why: Take f(x) = sqrt(x), x=25, dx=0.2. f'(x) = 1/(2 sqrt(x)) = 1/10. Approximation: 5 + (1/10)(0.2) = 5.02.
Q29.
The radius of a sphere is increasing at the rate of 2 cm/s. Find the rate of increase of its volume when the radius is 5 cm.
Answer: B. 200 pi cm<sup>3</sup>/s
Why: V = (4/3) pi r<sup>3</sup>, so dV/dt = 4 pi r<sup>2</sup> (dr/dt) = 4 pi (5)<sup>2</sup> (2) = 4 pi (25)(2) = 200 pi cm<sup>3</sup>/s.
Q30.
Find the point on the curve y = x<sup>2</sup> where the tangent is parallel to the line y = 4x - 5.
Answer: B. (2, 4)
Why: Slope of given line is 4. f'(x) = 2x = 4 gives x=2, y=4. The point is (2, 4).
Q31.
Verify Rolle's theorem for f(x) = x<sup>2</sup> - 4x + 3 on [1, 3]. The value of c satisfying f'(c) = 0 is:
Answer: C. c = 2
Why: f(1) = 0 and f(3) = 0, satisfying f(a)=f(b). f'(x) = 2x - 4 = 0 gives c = 2, which lies in (1, 3).
Q32.
Find c using the Mean Value Theorem for f(x) = x<sup>2</sup> on the interval [1, 4].
Answer: B. c = 2.5
Why: f'(c) = [f(4)-f(1)]/(4-1) = (16-1)/3 = 5. Since f'(x) = 2x, 2c = 5 gives c = 2.5, which lies in (1,4).
Q33.
A balloon's volume increases such that its radius increases at 3 cm/s. Find the rate of change of surface area when r = 4 cm. (Surface area = 4 pi r<sup>2</sup>)
Answer: C. 96 pi cm<sup>2</sup>/s
Why: dS/dt = 8 pi r (dr/dt) = 8 pi (4)(3) = 96 pi cm<sup>2</sup>/s.
Q34.
By the second-derivative test, if f′(x) = 0 and f″(x) > 0, the point is a local:
Answer: A. minimum
Why: A positive second derivative means the curve is concave up, giving a local minimum.
Q35.
If f′(x) = 0 and f″(x) < 0, the point is a local:
Answer: A. maximum
Why: A negative second derivative means the curve is concave down, giving a local maximum.
Q36.
The equation of the tangent to y = f(x) at (x₁, y₁) is y − y₁ =:
Answer: A. f′(x₁)·(x − x₁)
Why: The tangent line uses the slope f′(x₁) at the point of contact.
Q37.
The rate of change of the area A = πr² of a circle with respect to its radius r is:
Answer: A. 2πr
Why: dA/dr = d(πr²)/dr = 2πr.
Q38.
For a sphere of volume V = (4/3)πr³, the rate of change of volume with respect to r is:
Answer: A. 4πr²
Why: dV/dr = 3·(4/3)πr² = 4πr² (which is the surface area).
Q39.
The slope of the normal to a curve is the ___ of the tangent’s slope:
Answer: A. the negative reciprocal
Why: The normal is perpendicular to the tangent, so its slope is the negative reciprocal.
Q40.
For a small change dx, the approximate change in y is dy =:
Answer: A. f′(x)·dx
Why: The differential dy = f′(x)·dx approximates the change in y.
Q41.
Show that among all rectangles with a given perimeter of 40 units, the one with maximum area is a square. What is the maximum area?
Answer: B. 100 sq units
Why: Let sides be x and 20-x (since 2x+2y=40 means x+y=20). Area A = x(20-x). dA/dx = 20-2x = 0 gives x=10=y, a square. Maximum area = 10 x 10 = 100 sq units.
Q42.
An open box is to be made from a square sheet of side 12 cm by cutting equal squares of side x from each corner and folding up the sides. Find x that maximizes the volume.
Answer: B. x = 2
Why: Volume V = x(12-2x)<sup>2.</sup> dV/dx = (12-2x)<sup>2</sup> - 4x(12-2x) = (12-2x)(12-6x) = 0 gives x=6 (rejected, makes V=0) or x=2. Second derivative test confirms x=2 gives maximum volume.
Q43.
Find two positive numbers x and y such that x + y = 10 and x<sup>2</sup> + y<sup>2</sup> is minimum. Find this minimum value.
Answer: C. 50
Why: Substitute y = 10−x: S = x² + (10−x)² = 2x²−20x+100. dS/dx = 4x−20 = 0 → x = 5, y = 5. d²S/dx² = 4 > 0 confirms minimum. Min S = 5² + 5² = 50.
Q44.
Among all closed cylindrical cans of a fixed volume V, the surface area is minimized when the height h and radius r satisfy:
Answer: B. h = 2r
Why: Minimizing S = 2 pi r<sup>2</sup> + 2 pi r h subject to V = pi r<sup>2</sup> h leads to the condition h = 2r, meaning the height equals the diameter.
Q45.
A man of height 2 m walks away from a lamp post of height 6 m at a speed of 1.5 m/s. Find the rate at which the length of his shadow increases.
Answer: B. 0.75 m/s
Why: Using similar triangles with distance x from pole and shadow length s: 6/(x+s) = 2/s gives 6s = 2x+2s, so 4s=2x, s=x/2. ds/dt = (1/2)(dx/dt) = (1/2)(1.5) = 0.75 m/s.
Q46.
Find the maximum value of the function f(x) = -2x<sup>3</sup> + 3x<sup>2</sup> + 12x - 5 on the interval [-2, 3].
Answer: B. At x = 2, value 15
Why: f'(x) = -6x<sup>2</sup>+6x+12 = -6(x<sup>2</sup>-x-2) = -6(x-2)(x+1) = 0 at x=2,-1. f(2)=-16+12+24-5=15. Checking endpoints f(-2)=16+12-24-5=-1, f(3)=-54+27+36-5=4, f(-1)=2+3-12-5=-12. Maximum is 15 at x=2.
Q47.
A wire of length 36 m is to be cut into two pieces. One piece (length 4x) is bent into a square of side x, and the other into a circle of radius r. To minimize the total area enclosed, what relation must hold between x and r?
Answer: B. x = 2r
Why: Total area A = x<sup>2</sup> + pi r<sup>2</sup> with 4x + 2 pi r = 36. Substituting and differentiating with respect to x, then setting dA/dx = 0, gives the condition x = 2r, meaning the side of the square equals the diameter of the circle at the minimum.
Q48.
Verify the Mean Value Theorem for f(x) = x<sup>3</sup> - 5x<sup>2</sup> - 3x on [1, 3]. Find the value of c in (1, 3) satisfying the theorem.
Answer: A. c = 7/3
Why: f(1) = 1-5-3 = -7, f(3) = 27-45-9 = -27. Average rate = [f(3)-f(1)]/(3-1) = (-27+7)/2 = -10. f'(x) = 3x<sup>2</sup>-10x-3. Solve 3c<sup>2</sup>-10c-3 = -10, giving 3c<sup>2</sup>-10c+7=0, so c = (10±4)/6, meaning c=7/3 or c=1. Only c=7/3 lies strictly inside (1,3).
Q49.
The sum of the surface areas of a cube and a sphere is constant. Show that the sum of their volumes is minimum when the edge of the cube equals the diameter of the sphere. If the cube's edge is a and sphere's radius is r at the minimum, what is the relationship?
Answer: B. a = 2r
Why: Differentiating the volume sum subject to the fixed surface area constraint and setting the derivative to zero leads to a = 2r, meaning the edge of the cube equals the diameter of the sphere.
Q50.
Find the maximum area of an isosceles triangle inscribed in a circle of radius r, with the triangle's apex at the top of a vertical diameter.
Answer: A. (3 sqrt(3)/4) r<sup>2</sup>
Why: Setting up the area as a function of the half-angle and differentiating shows the maximum area inscribed isosceles triangle (which turns out equilateral) is (3 sqrt(3)/4) r<sup>2</sup>, achieved when the triangle is equilateral.
Q51.
A point moves along the curve y = x<sup>3</sup> - 3x. Find the points where the tangent to the curve is parallel to the x-axis, and classify them.
Answer: A. (1,-2) is local min, (-1,2) is local max
Why: y' = 3x<sup>2</sup>-3 = 0 gives x=1,-1. y''=6x. At x=1, y''=6>0 so local minimum at (1,-2). At x=-1, y''=-6<0 so local maximum at (-1,2).
Q52.
The function f(x) = −x² + 4x attains its maximum at x =:
Answer: A. 2
Why: f′(x) = −2x + 4 = 0 gives x = 2.
Q53.
The maximum value of f(x) = −x² + 4x is:
Answer: A. 4
Why: f(2) = −4 + 8 = 4.
Q54.
Among all rectangles of a fixed perimeter, the area is greatest when the rectangle is a:
Answer: A. square
Why: For a fixed perimeter, the square encloses the greatest area.
Q55.
For f(x) = x³ − 3x, the critical points occur at x =:
Answer: A. ±1
Why: f′(x) = 3x² − 3 = 0 gives x² = 1, so x = ±1.
Q56.
The point where a curve changes its concavity is called a point of:
Answer: A. inflection
Why: A point of inflection is where the second derivative changes sign.
Q57.
If f′(x) > 0 for every x in an interval, the function is:
Answer: A. strictly increasing
Why: A positive derivative throughout means the function strictly increases.
Q58.
The derivative of the position of a particle with respect to time gives its:
Answer: A. velocity
Why: Velocity is the rate of change of position with time.
Q59.
The second derivative of position with respect to time gives the:
Answer: A. acceleration
Why: Acceleration is the rate of change of velocity, i.e. the second derivative of position.
Q60.
A rectangle of fixed perimeter 20 has maximum area when each side equals:
Answer: A. 5
Why: Maximum area is a square; perimeter 20 gives side 20/4 = 5.
Q61.
The value of x > 0 at which x<sup>1/x</sup> attains its maximum is:
Answer: B. x = e
Why: Setting the derivative of (1/x)lnx to zero gives lnx = 1, so x = e.
Q62.
The function f(x) = x³ − 3x is decreasing on the interval:
Answer: C. (−1, 1)
Why: f'(x) = 3x² − 3 < 0 when −1 < x < 1.
Q63.
The equation of the tangent to y = x² at the point (1, 1) is:
Answer: A. y = 2x − 1
Why: The slope is 2, so y − 1 = 2(x − 1), i.e. y = 2x − 1.
Q64.
Two numbers have a sum of 20. The maximum value of their product is:
Answer: C. 100
Why: The product x(20 − x) is maximised at x = 10, giving 10·10 = 100.
Q65.
A spherical balloon's volume increases at 100 cm³/s. When the radius is 5 cm, the radius increases at:
Answer: A. 1/π cm/s
Why: dV/dt = 4πr²·dr/dt gives dr/dt = 100/(4π·25) = 1/π cm/s.
Q66.
Using differentials, the approximate value of √26 is:
Answer: B. 5.1
Why: √26 ≈ √25 + (1/(2·5))·1 = 5 + 0.1 = 5.1.
Q67.
The minimum value of f(x) = x + 1/x for x > 0 is:
Answer: C. 2
Why: By AM–GM x + 1/x ≥ 2, with equality at x = 1.
Q68.
The point on the parabola y² = 4x nearest to (2, 1) is:
Answer: A. (1, 2)
Why: Parametrising as (t², 2t) and minimising the squared distance gives t³ = 1, so t = 1 and the point is (1, 2).