CBSE Class 12 Mathematics Application Of Derivatives MCQs Set B

Question. A particle moves along the curve 6y = x³ + 2. The point ‘P’ on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate, are (4, 11) and (-4 , -31/3) .
(a) x-coordinates at the point P are ± 4
(b) y-coordinates at the point P are 11 and -31/3
(c) Both (a) and (b)
(d) None of the above

Answer : C

Question. The difference between the greatest and least values of the function f(x) = sin²x – x, on [-π/2 , π/2] is
(a) π/2
(b) π
(c) 3π
(d) π/4

Answer : B

Question. The slope of the normal to the curve
(a) x = a cos³θ, y = a sin³θ at θ = π/4 is 0
(b) x = 1 – a sinθ, y = b cos²θ at θ = π/2 is a/2b
(c) Both (a) and (b) are true
(d) Both (a) and (b) are not true

Answer : D

Question. Test to examine local maxima and local minima of a given function is/are
(a) first derivative test
(b) second derivative test
(c) Both (a) and (b)
(d) None of these

Answer : C

Question. If at x = 1, the function x4 – 62x² + ax + 9 attains its maximum value on the interval [0, 2], then the value of a is
(a) 110
(b) 10
(c) 55
(d) None of these

Answer : D

Question. The maximum value of In x/x in (2 , ∞) is
(a) 1
(b) e
(c) 2/e
(d) 1/e

Answer : D

Question. The maximum value of (1/x)x is
(a) e
(b) e^e
(c) e^1/e
(d) (1/e)1/e

Answer : C

Question. The radius of a cylinder is increasing at the rate of 3 m/s and its altitude is decreasing at the rate of 4 m/s. The rate of change of volume when radius is 4 m and altitude is 6m, is
(a) 20 π m³/s
(b) 40 π m³/s
(c) 60 π m³/s
(d) None of these

Answer : D

Question. The straight line x/a + y/b = 2 touches the curve (x/a)n + (y/b)n = 2 at the point (a, b) for
(a) n = 1, 2
(b) n = 3, 4, –5
(c) n = 1, 2, 3
(d) any value of n

Answer : D

Question. Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius a.
(a) 2a/3
(b) 2a/√3
(c) a/3
(d) a/5

Answer : B

Question. A right circular cylinder which is open at the top and has a given surface area, will have the greatest volume if its height h and radius r are related by
(a) 2h = r
(b) h = 4r
(c) h = 2r
(d) h = r

Answer : D

Question. The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of
(a) π/4
(b) π/3
(c) π/2
(d) π/6

Answer : C

Question. The maximum area of rectangle inscribed in a circle of diameter R is
(a) R²
(b) R²/2
(c) R²/4
(d) R²/8

Answer : B

Question. If I be an open interval contained in the domain of a real valued function f and if x₁ < x₂ in I, then which of the following statements is true?
(a) f is said to be increasing on I, if f(x₁) ≤ f(x₂) for all x₁, x₂ < I
(b) f is said to be strictly increasing on I, if f(x₁) < f(x₂) for all x₁, x₂ < I
(c) Both (a) and (b) are true
(d) Both (a) and (b) are false

Answer : C

Question. If f be a function defined on an interval I and there exists a point c in I such that f(c) > f(x), for all x ∈ I, then
(a) function ‘f’ is said to have a maximum value in I
(b) the number f(c) is called the maximum value of f in I
(c) the point c is called a point of maximum value of f in I
(d) All the above are true

Answer : D
 

Case Based Questions

Megha wants to prepare ahandmade gift box for her friend’s birthday at home. For making lower part of box, she takes a square piece of cardboard of side 20 cm.

Based on the above information, answer the following questions

Question. Volume of the open box formed by folding up the cutting corner can be expressed as
(a) V = x(20 – 2x)²
(b) V = x/2(20 + x)(20 + x)
(c) V = x/3(20 – 2x) (20 + 2x)
(d) V = x(20 – 2x)(20 – x)

Answer : A

Question. If x cm be the length of each side of the square cardboard which is to be cut off from corners of the square piece of side 20 cm, then possible value of x will be given by the interval
(a) [0, 20]
(b) [0, 10]
(c) [0, 3]
(d) None of these

Answer : B

Question. Megha is interested in maximising the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum?
(a) 12 cm
(b) 8 cm
(c) (10/3) cm
(d) 2 cm

Answer : C

Question. The values of x for which dV/dx = 0, are

Answer : D

Question. The maximum value of the volume is

Answer : D
 

An owner of an electric bike rental company have determined that if they charge customers ₹x per day to rent a bike, where 50 ≤ x ≤ 200, then number of bikes (n), they rent per day can be shown by linear function n(x) = 2000 – 10x. If they charge ₹50 per day or less, they will rent all their bikes. If they charge ₹200 or more per day, they will not rent any bike.

Based on the above information, answer the following questions:

Question. If R(x) denote the revenue, then maximum value of R(x) occur when x equals
(a) 10
(b) 100
(c) 1000
(d) 50

Answer : B

Question. Total revenue R as a function of x can be represented as
(a) 2000x – 10x²
(b) 2000x + 10x²
(c) 2000 – 10x
(d) 2000 – 5x²

Answer : A

Question. The number of bikes rented per day, if x = 105 is
(a) 850
(b) 900
(c) 950
(d) 1000

Answer : C

Question. At x = 260, the revenue collected by the company is
(a) ₹10
(b) ₹500
(c) ₹0
(d) ₹1000

Answer : C

Question. Maximum revenue collected by company is
(a) ₹40,000
(b) ₹50,000
(c) ₹75,000
(d) ₹1,00,000

Answer : D
 

Mr. Sahil is the owner of a high rise residential society having 50 apartments. When he set rent at ₹10,000/month, all apartments are rented. If he increases rent by ₹250/ month, one fewer apartment is rented. The maintenance cost for each occupied units is ₹500 month.

Based on the above information answer the following questions:

Question. If x represents the number of apartments which are not rented, then the profit expressed as a function of x is
(a) (50 – x) (38 + x)
(b) (50 + x) (38 – x)
(c) 250(50 – x)(38 + x)
(d) 250(50 + x) (38 – x)

Answer : C

Question. If P is the rent price per apartment and N is the number of rented apartment, then profit is given by
(a) NP
(b) (N – 500)P
(c) N(P – 500)
(d) none of these

Answer : C

Question. If P = 11,000 then the profit is
(a) ₹4,83,000
(b) ₹5,00,000
(c) ₹5,05,000
(d) ₹6,50,000

Answer : A

Question. If P = 10,500 then N =
(a) 47
(b) 48
(c) 49
(d) 50

Answer : B

Question. The rent that maximises the total amount of profit is
(a) ₹11,000
(b) ₹11,500
(c) ₹15,800
(d) ₹16,500

Answer : B
 

P(x) = –5x² + 125x + 37500 is the total profit function of a company, where x is the production of the company

Question. What will be the maximum profit?
(a) ₹38,28,125
(b) ₹38,281.25
(c) ₹39,000
(d) None

Answer : B

Question. What will be the production when the profit is maximum?
(a) 37,500
(b) 12.5
(c) – 12.5
(d) – 37,500

Answer : B

Question. When the production is 2 units what will be the profit of the company?
(a) 37,500
(b) 37,730
(c) 37,770
(d) None

Answer : B

Question. Check in which interval the profit is strictly increasing.
(a) (12.5, ∞)
(b) for all real numbers
(c) for all positive real numbers
(d) (0, 12.5)

Answer : D

Question. What will be production of the company when the profit is ₹38,250?
(a) 15
(b) 30
(c) 2
(d) data is not sufficient to find

Answer : A