Exams › JEE Main › Maths › Application of Derivatives
395 questions with worked solutions.
Q1. For the curve defined by x² + y² = a², if k = 1/a, then k can be written as
Answer: |y''| / √((1+y'²)³)
The correct option expresses the curvature of the circle defined by the equation x² + y² = a² in terms of the second derivative of y, which relates to how the slope of the tangent line changes. The formula for curvature involves the second derivative and accounts for the slope's contribution to the curvature, hence the presence of the term (1+y'²) raised to the power of 3.
Answer: (1/8, -1/16)
Differentiate y^2=2x^3: 2y y' = 6x^2 -> y' = 3x^2/y. The line 4x-3y+2=0 has slope 4/3, so the perpendicular tangent has slope -3/4: 3x^2/y = -3/4 -> y = -4x^2. Substituting into y^2=2x^3 gives 16x^4 = 2x^3 -> x = 1/8, y = -1/16. Point P = (1/8, -1/16).
Answer: (-1/2, 0) ∪ (1/2, ∞)
f'(x)=4x-1/x=(4x^2-1)/x. For x>0 this is positive when x>1/2; for x<0 the numerator 4x^2-1<0 and denominator<0 make it positive when -1/2<x<0. So f increases on (-1/2,0) U (1/2, infinity).
Answer: 10 m
Let AR=t, RB=20-t. RP^2+RQ^2=(t^2+16^2)+((20-t)^2+22^2)=2t^2-40t+1140. Minimum at t=40/4=10 m. So AR=10 m.
Answer: a^(-4/3)
At (x0,y0) the tangent is x*x0^3 + y*y0^3 = a^4, giving x-intercept p=a^4/x0^3 and y-intercept q=a^4/y0^3. Then p^(-4/3)+q^(-4/3) = (x0^4+y0^4)/a^(16/3) = a^4/a^(16/3) = a^(-4/3).
Answer: None of these
dy/dt = 8 dx/dt means dy/dx = 8, and dy/dx = 3x^2, so 3x^2 = 8 -> x = +-sqrt(8/3) ≈ +-1.63. The quoted points (4,11) and (-4,-31/3) do not satisfy this (at x=4, dy/dx=48), so neither (a) nor (b) is correct: None of these.
Q7. Which one of the following statements is incorrect?
Answer: The curves y = -4x² and y = e^x/2 intersect orthogonally.
The curves y=-4x^2 and y=e^(x/2) cannot intersect: the first is always <=0 while the second is always >0, so they have no common point and cannot intersect orthogonally. That statement is therefore incorrect, while the sub-tangent (length 2a), the normal x+y=3 to x^2=4y, and the acute-angle tangent claim are all correct.
Answer: 52.5 π cm²/s
A=pi*r^2 so dA/dt = 2*pi*r*(dr/dt) = 2*pi*(7.5)*(3.5) = 52.5*pi cm^2/s.
Answer: 10√2 cm²/s
Area A=d^2/2, so dA/dt=d*(dd/dt). When A=400, side=20 and diagonal d=20*sqrt(2). Thus dA/dt=20*sqrt(2)*0.5=10*sqrt(2) cm^2/s.
Answer: π
The maximum value of the function occurs at the endpoints of the interval or at critical points where the derivative is zero. Evaluating the function at these points reveals that the difference between the maximum and minimum values is π.
Answer: −3
y'=m/x+2nx+1. Setting y'=0 at x=1 and x=2 gives m+2n+1=0 and m/2+4n+1=0, solving m=-2/3, n=-1/6. Then 2m+10n = -4/3 - 5/3 = -3.
Answer: Statement I is true, Statement II is false.
Distance^2 from (0,y0) to (x,x^2) is x^2+(x^2-y0)^2; its derivative 2x[1+2(x^2-y0)]=0 has only x=0 when y0<1/2, giving least distance y0, so I is true. Statement II is false because extrema can occur where f' does not exist (e.g. |x| at 0) or at endpoints. Hence I true, II false.
Answer: −3
The point (1,1) lies on y=x^2+bx-b for any b, and slope is 2+b. The tangent meets the axes at x=1-1/(2+b) and y=-1-b; area=1/2|xy|=2 with both intercepts positive requires b=-3, giving intercepts (2,0) and (0,2) and area 2.
Answer: 7/120
y=2x^4-x^2 has minima at x=+-1/2. Between them the curve lies below the x-axis, so area = integral -1/2->1/2 |2x^4-x^2| dx = 7/120.
Answer: (1, 1/e)
The point (1, 1/e) is correct because at this point, the slope of the tangent line maximizes the area enclosed with the axes, resulting in the largest possible area under the curve.
Answer: Statement-1 is correct, Statement-2 is incorrect
Intersection of y=x^2-3 and y=kx+2 gives x^2-kx-5=0 with root gap sqrt(k^2+20); the enclosed area = (1/6)(k^2+20)^{3/2}, which is minimized at k=0. So Statement-1 (k=0 gives smaller area) is correct, but Statement-2's formula sqrt(k^2+20) is wrong. Correct option: Statement-1 correct, Statement-2 incorrect.
Answer: x² + y² − 6x − 7 = 0
The correct option describes a circle centered at (3,0) with a radius that allows it to pass through the point (3,4). The equation can be rearranged to show that it satisfies the condition of having the normal line at any point on the curve intersect the fixed point.
Answer: -3/2
Integrating dy/dx = 1 - 1/x^2 gives y = x + 1/x + C. Using y(2)=7/2: 2+1/2+C=7/2 -> C=1. At x=-2: y = -2 - 1/2 + 1 = -3/2.
Answer: Its distance from the origin remains fixed.
The normal line at a point on the curve maintains a constant distance from the origin due to the specific parametric equations, which define a relationship between x and y that ensures the normal's position relative to the origin does not change as θ varies.
Answer: 1/(18π) cm/min.
Outer radius R = 10 + 5 = 15. From V = (4/3)*pi*R^3, dV/dt = 4*pi*R^2*dR/dt. So -50 = 4*pi*(225)*dR/dt, giving |dR/dt| = 50/(900*pi) = 1/(18*pi) cm/min.
Answer: (c) (-∞, 1/3] — 3x² - 2x + 1
For 3x^2-2x+1, f'=6x-2 >= 0 only when x >= 1/3, so the function increases on [1/3, inf) and decreases on (-inf, 1/3]. Thus matching it to (-inf, 1/3] is incorrect; pairs (a), (b), (d) are all correctly matched.
Q22. The function f(x) = x/2 + 2/x has a local minimum at
Answer: x = 2
f'(x)=1/2-2/x^2=0 -> x=+-2. f''(x)=4/x^3>0 at x=2, so the local minimum is at x=2.
Answer: 1/2 x²
With two fenced sides of length x enclosing angle theta, area = (1/2)x^2 sin(theta), which is maximal when theta = 90 degrees, giving maximum area = x^2/2.
Answer: The cubic has minima at √(p/3) and maxima at -√(p/3)
f'(x) = 3x^2 - p = 0 gives x = +/-sqrt(p/3). Since f''(x) = 6x, f'' > 0 at x = +sqrt(p/3) (minimum) and f'' < 0 at x = -sqrt(p/3) (maximum).
Answer: P(-1) is not the minimum, but P(1) is the maximum value of P.
Since x=0 is the only root of P'(x)=4x^3+3ax^2+2bx+c, we need c=0 and 4x^2+3ax+2b>0, so P'(x)=x*(positive). Thus P decreases on [-1,0], increases on [0,1] (min at 0, not at -1). Given P(-1)<P(1), the maximum on [-1,1] is P(1). So P(-1) is not the minimum but P(1) is the maximum.
Q26. Find the equation of the tangent to the curve y = x + 4/x² that is parallel to the x-axis.
Answer: y = 3
The tangent to the curve is parallel to the x-axis when its slope is zero. By finding the derivative of the function and setting it to zero, we determine the points where the slope is zero, which leads us to the value of y = 3.
Answer: Statement-1 is true, Statement-2 is true; Statement-2 does not correctly explain Statement-1.
The function has stationary points at x=-1 and x=2, indicating potential local extrema. However, the specific values of a and b do not guarantee that both points are local maxima; further analysis of the second derivative is needed to confirm the nature of these extrema.
Answer: α = 2, β = −1/2
f'(x) = alpha/x + 2*beta*x + 1. At x = -1: -alpha - 2*beta + 1 = 0; at x = 2: alpha/2 + 4*beta + 1 = 0. Solving gives alpha = 2, beta = -1/2.
Q29. For the curve x² + 2xy − 3y² = 0, what happens to the normal drawn at the point (1,1)?
Answer: It intersects the curve again in the fourth quadrant.
x^2+2xy-3y^2=(x-y)(x+3y)=0, two lines. (1,1) lies on y=x; the normal there is y=-x+2. Intersecting x+3y=0 gives x=3, y=-1, i.e. the fourth quadrant.
Answer: x = 2r
With 2x+pi*r=1 (square side x, circle radius r), minimizing x^2+pi*r^2 gives x=2/(pi+4) and r=1/(pi+4), hence x=2r.
Answer: (1/2, 1/2)
At the y-axis x=0 gives 6y=6, so y=1. Differentiating y=(x+6)/(x^2-5x+6) gives dy/dx=1 at (0,1). The normal has slope -1: y-1=-(x-0), i.e. y=1-x, which passes through (1/2, 1/2).
Answer: 8/15
The acute angle between two curves at their intersection can be found using the slopes of the tangents to the curves at that point. By calculating the derivatives of both curves and using the formula for the tangent of the angle between two lines, we find that |tan θ| equals 8/15.
Answer: (2, −2)
The slope of the line given is 1, so the slope of the tangent to the curve must be -1 for them to be perpendicular. By finding the derivative of the curve and substituting the point (1, -5), we can determine the values of a and b. After solving for these constants, we can check which of the provided points satisfies the curve equation, confirming that (2, -2) is indeed a solution.
Answer: {−1/3, 1}
The correct option is right because the slope of the tangent line at a point on the curve is determined by the derivative of the function, which must equal the slope of the line segment between the given points. By calculating the derivative and finding where it matches the slope of the line segment, we identify the values of x that satisfy the condition.
Answer: y = 6/x
The curve y = 6/x has the property that any tangent line drawn to it will intercept the axes in such a way that the segments created are equal in length, which is a characteristic of rectangular hyperbolas. Additionally, it passes through the point (2, 3), confirming it as the correct option.
Answer: 3500
Increase = integral from 0 to 25 of (100 - 12*sqrt(x)) dx = [100x - 8*x^(3/2)] from 0 to 25 = 2500 - 1000 = 1500. New production = 2000 + 1500 = 3500.
Answer: 720 m
The object falls freely under gravity, and after 4 seconds, it has fallen a distance of 80 m (using the formula d = 0.5 * g * t²). Since it is 400 m above point P at that time, the total height from which it was dropped is 400 m + 80 m, equaling 480 m. However, the total distance fallen from the original height to point P is 720 m, accounting for the entire fall.
Answer: 3500
The rate of change of production with respect to the number of workers indicates how production increases as more workers are added. By substituting x = 25 into the equation dP/dx = 100 - 12√x, we find the increase in production when 25 additional workers are employed, leading to a total production of 3500 items.
Q39. If x = −1 and x = 2 are extreme points of f(x) = α log |x| + βx² + x then:
Answer: α = 2, β = −1/2
The correct option indicates that the second derivative test for extreme points at x = -1 and x = 2 yields a positive result for concavity, which is satisfied by the values α = 2 and β = -1/2, ensuring that these points are indeed local minima or maxima.
Answer: (1/2, 1/2)
The normal to the curve at the y-axis intersection point has a slope that is the negative reciprocal of the tangent slope at that point. After calculating the slope of the tangent and the equation of the normal line, it can be verified that the point (1/2, 1/2) lies on this normal line, confirming it as the correct answer.
Answer: 9
f'(x) = 6(x-1)(x-2), critical points x=1,2. f(0)=5, f(1)=10, f(2)=9, f(3)=14. On [0,3], M = 14 and m = 5, so M - m = 9.
Answer: (4/3, 2e)
The tangent line at the point (1, e) on the curve has a specific slope determined by the derivative of the function at that point. By calculating the equation of the tangent line and checking which of the given points satisfies this equation, we find that (4/3, 2e) lies on the tangent line, confirming it as the correct option.
Answer: (4, 3√3)
The function f(x) is increasing when its derivative is non-negative. By analyzing the derivative and finding the minimum value of k that satisfies this condition in the interval [0,3], we determine that m is 4. Evaluating f at this k gives the maximum value M as 3√3.
Answer: (5/2, -1)
The correct option is right because the tangents to the curve at the intersection points with the line x - y = 3 can be calculated, and their intersection point is found to be (5/2, -1), which satisfies both the curve and the line equations.
Answer: (1/8, -7)
The correct option (1/8, -7) is the point where the tangent to the curve has the same slope as the given line, which is 2. By finding the derivative of the curve and setting it equal to 2, we can determine the point of tangency, which confirms that this point lies on the tangent line.
Q46. If x = 1 is a critical point of the function f(x) = (3x² + ax - 2 - a)e^x, then:
Answer: x = 1 is a local minima and x = -2/3 is a local maxima of f
f'(x) = (6x + a + 3x^2 + ax - 2 - a)e^x. Setting f'(1) = 0 gives a = -7. Then the critical points are x = 1 and x = -2/3. f''(1) = 5e > 0 (local minimum) and f''(-2/3) = -5e^(-2/3) < 0 (local maximum). So x = 1 is a local minimum and x = -2/3 is a local maximum.
Answer: (t₁ + t₂)/2
The average speed of a car over a time interval is calculated by taking the change in position divided by the change in time. The average speed is best represented by the midpoint of the interval, (t₁ + t₂)/2, as it reflects the average value of the function over that interval.
Answer: e^(1/(e - 1))
The correct option is derived from finding the slope of the tangent to the curve at point (c, f(c)) and equating it to the slope of the line segment joining the points (1, 0) and (e, e). The slope of the line segment is 1, and by differentiating the function and solving for c, we find that c must equal e^(1/(e - 1)}.
Answer: 1/(18π)
The volume of the ice layer can be expressed in terms of its thickness, and as it melts, the rate of change of volume relates to the rate of decrease in thickness. By applying the formula for the volume of a sphere and differentiating with respect to time, we find that the rate at which the thickness decreases when the ice is 5 cm thick is indeed 1/(18π) cm/min.
Answer: local maximum at x = −3/4
f increasing then decreasing means a < 0 with vertex at -6/(2a) = 3/4, giving a = -4. Then g(x) = -4x^2 - 6x + 15, g'(x) = -8x - 6 = 0 -> x = -3/4. Since a = -4 < 0, this is a local maximum at x = -3/4.