Applications of Derivatives MCQs

Ans: (d) 5√3 cm²/s

Explanation: Assume that x be the side of an equilateral triangle and A be its area.

• Hence, A = \((\frac{\sqrt3}{4})\)x² square units
• Differentiate both sides with respect to t, we get
• dA/dt = \((\frac{\sqrt3}{4})\)2x(\(\frac{dx}{dt}\)) …(1)
• Given that x = 20 cm and \(\frac{dx}{dt}\) = 2 cm/s
• Now, substitute the values in (1), we get
• dA/dt = \((\frac{\sqrt3}{4})\)2(20)(2)
• Therefore, \(\frac{dA}{dt}\) = 5√3 cm²/s

Ans: (d) 12x-2

Explanation: Given, f(x) = 6x² – 2x + 6

• Now taking the derivative of f(x),
• d/dx f(x) = d/dx (6x² – 2x + 6)
• Let us split the terms of the function is:
• d/dx f(x) = d/dx (6x²) – d/dx (2x) + d/dx (6)
• Using the formulas:
• d/dx (kx) = k and d/dx (xn) = nxn – 1
• ⇒ d/dx f(x) = 6(2x) – 2(1) + 0 = 12x – 2

Ans: (b) Always increasing

Explanation: As it is given f(x) is equal to x+cos x

• Derivative of required function is given as
• f’(x) = 1 – sinx
• f’(x)>0 for all values of x.
• Since sin x is lying between -1 and +1
• So f(x) is always increasing

Ans: (a) 6.9

Explanation: Given: f(3) = 5, and f'(3) = 19

• To find: f(3.1)
• Using formula for approximation:
• L(x) = f(a) + f'(a)(x−a)
• L(x) = 5 + 19 (3.1-3)
• L(x) = 5 + 19(0.1)
• L(x) = 6.9

Ans: (b) 0, 8/9

Explanation: Given function: y = 3x³ - 4x² + 6

• Using the second order derivative test we can find the maxima and minima of a function:
• Taking first order derivative:
• y = 3x³ - 4x² + 6 -------------(eq 1)
• Differentiate both sides (eq 1), w.r.t - x.
• ⇒ dy/dx = d/dx (3x³) - d/dx (4x²) + d/dx (6)
• ⇒ dy/dx = 9x² - 8x + 0
• ⇒ dy/dx = 9x² - 8x ----------------(eq 2)
• Putting dy/dx = 0 to find critical points.
• 9x² - 8x = 0
• x (9x - 8) = 0
•  x = 0,8/9
• The critical points are 0 & 8/9.

Ans: (a) dy/dx = \(\lim _{x→0}\)(f(x+h)-f(x))/h

Explanation: A derivatives of a functions at point x = h is given as follows:

dy/dx = \(\lim _{x→0}\)(f(x+h)-f(x))/h

Ans: (c) 2x-1

Explanation: Given, f(x) = x² – x + 6

• Now taking the derivative of f(x),
• d/dx f(x) = d/dx (x² – x + 6)
• Let us split the terms of the function is:
• d/dx f(x) = d/dx (x²) – d/dx (x) + d/dx (6)
• Using the formulas:
• d/dx (kx) = k and d/dx (xn) = nxn – 1
• ⇒ d/dx f(x) = (2x) – (1) + 0 = 2x – 1

Ans: (d) All of the above

Explanation: Derivatives have a wide range of applications,which includes determining the rate of change of a variable or quantity, obtaining the approximation value of a function, determining the equation of tangent and normal to the required curve, and calculating the minimum and maximum values of algebraic expressions. 

Ans: (a) e^x

Explanation: A function is said to be monotonic if the value of the function is either increasing or decreasing in the entire domain of a function. From the above options it is also possible in option (a)

Ans: (a) 45√3 cm²/s

Explanation: Assume that x be the side of an equilateral triangle and A be its area.

• Hence, A = (√3/4)x² square units
• Differentiate both sides with respect to t, we get
• dA/dt = (√3/4)2x(dx/dt) …(1)
• Given that x = 30 cm and dx/dt = 3 cm/s
• Now, substitute the values in (1), we get
• dA/dt = (√3/4)2(30)(3)
• Therefore, dA/dt = 45√3 cm²/s

Ans: (c) (2,8)

Explanation: As known : dy/dt = (4x)(dx/dt)

• 8. (dx/dt) = 4x.(dx/dt)
• x = 2
• Now, substitute x = 2 in y=2x², we get
• y = 2(2)² = 8
• Therefore, y = 8
• Hence, the coordinate is (2, 8)

Ans: (b) Always increasing

Explanation: As it is given f(x) is equal to 2x + 3cos x

• Derivative of required function is given as
• f’(x) = 2 – 3sinx
• f’(x)>0 for all values of x.
• Since sin x is lying between -3/2 and +3/2
• So f(x) is always increasing

Ans: (d) 17

Explanation: Given: f(2) = 15, and f'(2) = 10

• To find: f(2.2)
• Using formula for approximation:
• L(x) = f(a) + f'(a)(x−a)
• L(x) = 15 + 19 (2.2-2)
• L(x) = 15 + 10(0.2)
• L(x) = 17

Ans: (d) is an increasing function

Explanation: As given in the question: f(x) = 3x + cos x

• f’(x) = 3 – sin x
• Now, f’(x) will be positive for all values of x
• It means, f’ (x)>0 ∀ x∈R

Ans: (a) (3, 4)

Explanation: As given: y = x+3…(1)

• y²= 12x …(2)
• Substitute (1) in (2), we get
• (x+3)² = 12x
• x²+9+6x = 12x
• x²-6x+9 = 0, which is equal to (x-3)²=0
• ⇒ x = 3
• Now, substitute x = 3 in y=x+3, we get
• y = 3+1 = 4.
• Hence, the line y = x+1 is a tangent to the curve (3,4).

Ans: (b) sin x

Explanation: A function is said to be non-monotonic if the value of the function is either increasing or decreasing in the particular domain of a function. From the above options it is also possible in option (b)

Ans: (b) 2%

Explanation: As the formula for the time period of a pendulum is given as T = 2π × √(l/g)

• dT/dl = π/√gl
• Given that the percentage error in measuring the length “l” = 4%
• Δl/l = 4/100
• Δl = 4l/100
• Therefore, the approximate error in T = dT = (dT/dl)Δl
• dT = T/100 = 2% of T.
• Hence, the percentage error in T is 2%.