# Application of Derivatives Class 12 Mathematics Exam Questions

Please refer to Application of Derivatives Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

## Class 12 MathematicsExam Questions Application of Derivatives

Class 12 Mathematics students should read and understand the important questions and answers provided below for Application of Derivatives which will help them to understand all important and difficult topics.

Question. The amount of pollution content added in air in a city due to x-diesel vehicles is given by P(x) = 0.005×3 + 0.02x2 + 30x. Find the marginal increase in pollution content when 3 diesel vehicles are added.
Answer. P(x) = 0.005x3 + 0.02x2 + 30x
Differentiating w.r.t. x, we get
The marginal increase in pollution content when x = 3
P'(x) = 0.015x2 + 0.04x + 30
P'(3) = 0.015(3)2 + 0.04(3) + 30
= 0.135 + 0.12 + 30 = 30.255

Question. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5, find the marginal revenue, when x = 5.
Answer. R(x) = 3x2 + 36x + 5
Differentiating w.r.t. x
MR = R'(x) = 6x + 36
When x = 5, R'(5) = 6(5) + 36 = 30 + 36 = 66

Question. The total expenditure (in ₹) required for providing the cheap edition of a book for poor and deserving students is given by R(x) = 3x2 + 36x where x is the number of sets of books If the marginal expenditure is defined as dR/dx write the marginal expenditure required for 1200 such sets.
Answer. R(x) = 3x2 + 36x
Differentiating w.r.t. x, we have

Question. The volume of a sphere is increasing at the rate of 3 cubic centimeters per second. Find the rate of increase of its surface area, when the radius is 2 cm.
Answer. Let r be the radius, v be the volume, and s be the surface area of the sphere.
Then v = 4/3 πr3 and s = 4πr2, where r is a function of time t.

Question. The volume of a cube is increasing at the rate of 9 cm3/s. How fast is its surface area increasing when the length of an edge is 10 cm?
Answer. Let the volume of a cube and length of an edge be V and x respectively.

Question. The length x, of a rectangle is decreasing at the rate of 5 cm/minute and the width y, is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle.
Answer. Let Area of rectangle, A = lb = xy

= 8(4) + 6(–5) = 32 – 30 = 2 cm2/min
∴ Increase in area = 2 cm2/min

Question. The volume of a sphere is increasing at the rate of 8 cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12 cm.
Answer. Let v be the volume

Question. The radius r of a right circular cone is decreasing at the rate of 3 cm/minute and the height h is increasing at the rate of 2 cm/ minute. When r = 9 cm and h = 6 cm, find the rate of change of its volume.

∴ Change in volume of cone = – 54π cm3/min.
∴ Volume decreases at the rate of 54π cm3/minute

Question. A balloon, which always remains spherical, has a variable diameter 2/3 (3x + 1). Find the rate of change of its volume with respect to x

Question. Find the intervals in which the function f given

Hence f (x) is increasing on (– ∞, – 1] and [1, ∞) f (x) is decreasing on (– 1, 1).

Question. Find the intervals in which the following function is (a) increasing (a) decreasing :
f (x) = x3 – 12x2 + 36x + 17
f (x) = x3 – 12x2 + 36x + 17
∴f (x) = 3x2 – 24x + 36
= 3(x2 – 8x + 12)
= 3(x2 – 2x – 6x + 12)
= 3[x(x – 2) – 6(x – 2)]
= 3(x – 2) (x – 6)
f ’(x) = 0                     ⇒ x = 2, 6

Question. Find the equation of the tangent to the curve

which is parallel to the line 4x – 2y + 5 = 0.

Question. Find the equation of the tangent to the curve x2 + 3y = 3, which is parallel to the line y – 4x + 5 = 0.
Answer. x2 + 3y = 3 …(i)
Differentiating both sides w.r.t. x, we get

Question. Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.
Answer. Let surface area of open cylinder = πK
… where [K is a constant

∴ Volume is maximum at r = h
∴ Volume of open cylinder is maximum,
when its height is equal to the radius of the base.

Question. If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is π/3
Sol. In rt. ‘ ABC,
Let base, AB = x units
Hypotenuse, AC = y units
Let ! be the angle between hypotenuse and base

Let x + y = k
y = k – x                                                                    …(i)
BC2 = AC2 – AB2                                                 (Pythagoras’ theorem)
= y2 – x2
= (k – x)2 – x2 [from                                                  (i)]
BC2 = k2 + x2 – 2kx – x2 = k2 – 2kx                        …(ii)

Question. Prove that the radius of the right circular cylinder of greatest curved surface area, which can be inscribed in a given right circular cone is half that of the cone.

Question. Show that a cylinder of given volume, open at the top, has minimum total surface area if its height is equal to radius of the base.
Sol.
Let r be the radius h be the height and kπ be the volume of cylinder open at top Volume of cylinder, πr2h = kπ …(a constant)

Question. Find the values of x for which f (x) = [x(x – 2)]2 is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis.
Answer. Part I : f (x) = [x(x – 2)]2                                       …(i)
∴ f ‘(x) = 2[x(x – 2)]1.[x.1 + (x – 2).1]
= 2x(x – 2) (2x – 2) = 4x(x –1) (x– 2)
For stationary points, f ‘(x) = 0
∴ 4x(x –1)(x– 2) = 0
∴ x = 0, 1, 2

Given function is increasing in (0, 1) and (2, ∞)
Part II : For required points
f(x) = 0,
x = 0, 1, 2
Putting the values of x in (i),
When x = 0,
f (0) = [0(0 – 2)]2 = 0           ∴ Pt. (0, 0)
When x = 1,
f (1) = [1(1 – 2)]2 = 1           ∴ Pt. (1, 1)
When x = 2,
f (2) = [2(2 – 2)]2 = 0           ∴ Pt. (2, 0)
∴ Required points are (0, 0), (1, 1) and (2, 0).