Applications of Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Applications of Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have to find the second derivative.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

Calculation:

Let f(x) = x³ - 12x² + kx – 8

Differentiating with respect to x, we get

⇒ f’(x) = 3x² – 24x + k

It is given that function attains its maximum value of the interval [0, 3] at x = 2

∴ f’(2) = 0

⇒ 3 × 2² – (24 × 2) + k = 0

∴ k = 36

Concept:  

If f′(x) >  0 at each point in an interval, then the function is said to be strictly increasing.

Calculations:

Given , f(x) = x2 - 4x 

Differentiating, we get

f'(x) = 2x - 4

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 2x - 4 > 0

⇒ x > 2

∴ x ∈ (2, ∞)

Mistake Points

If a, b ∈ R and a

• Open interval is indicated by (a, b) = {x : a
• Closed interval is indicated by [a, b] = {x : a ≤ x ≤ b}, i.e. 'a' and 'b' are included.
• [a, b) = {x : a ≤ x
• (a, b ] = { x : a

Since '2' is not included here, Option (1) will not be correct.

Concept:

The equation of tangent at point (x1 , y1) with slope m is given by 

Calculations:

Given curve is y = 2x sin x 

Taking derivative on both side, we get

Put x =  to find the equation of tangent at the point .

The equation of tangent at point (x₁ , y₁) with slope m is given by 

y = 2x

Hence, the equation of the tangent line to the curve y = 2x sin x at the point  is 2x.

From the question, the derivative given is:

Now,

∵

From the question, let’s assume,

From the question, we need to find .

Now,

Now,

Concept:

Differentiation formula

Quotient rule:

Calculation:

f(x) = 

⇒ f^'(x) = 

⇒ f′(x) = 0 at x = e

⇒ f′(x) > 0 when x

⇒ f′(x) e

So, f(x) is maxima at x = e

⇒  f(e) = 

∴ The maximum value of  is 

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) decreasing.

Calculation:

Given:

f(x) = x - e^x

Differentiating with respect to x, we get

⇒ f’(x) = 1 - e^x

For increasing function,

f'(x) > 0

⇒ 1 – e^x > 0

⇒ e^x

⇒ e^x 0

∴ x

So, x ∈ (-∞, 0) 

CONCEPT:

• If y = f(x), then dy/dx denotes the rate of change of y with respect to x its value at x = a is denoted by: 
• Decreasing rate is represented by negative sign whereas increasing rate is represented by positive sign.
• Volume of cylinder is given by: πr²h where r is the radius and h is the altitude.

CALCULATION:

Given: dr/dt = 3m/s and dh/dt = - 4m/s

As we know that, volume of cylinder is given by: πr2h where r is the radius and h is the altitude.

Let V = πr2h

Now by differentiating v with respect to t we get,

Now by substituting r = 4m, h = 6m, dr/dt = 3m/s and dh/dt = - 4m/s we get,

Hence, option D is the correct answer.

Concept:

Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as monotonic function.

First derivative test:

• If  for all x in (a, b) then, f(x) is an increasing function in (a, b).
• If  for all x in (a, b) then, f(x) is an decreasing function in (a, b).

For strictly increasing or decreasing:

• 0\) for strictly increasing.
•  for strictly decreasing.

Calculation:

Given: 

Differentiating w.r.t x

So, for x ∈ (0, 1), f’(x)

Hence, f(x) is decreases for x ∈ (0, 1).

Concept:

Circumference of circle = 2πr, where r is the radius of the circle.

Calculation:

Given:  .

Circumference of circle (C) = 2πr

Differentiating both the sides with respect to t, we get:

⇒          ----(1)

By, substituting the values of  the equation(1), we get:

⇒ 

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given:f(x) = 3x2 +  3

Let x + Δx = 3.01 = 3 + 0.01

Therefore, x = 3 and Δx = 0.01

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(3.01) = 3x2 +  3 + (6x)Δx

= f(3.01) = 3(3)² + 3 + (6⋅3)(0.01)

= f(3.01) = 30 + 0.18

= f(3.01) = 30.18