Chapter 6: Application of Derivatives

Given,

f (x) = √3 sin x – cos x – 2ax + b, a ³ 1

On differentiating both sides w.r.t. x, we get

f’ (x) = √3 cos x + sin x – 2a

For increasing function, f’ (x) < 0

Therefore, the given function is decreasing in R.

Given, f(x) = tan^–1(sin x + cos x) in (0, π/4).

Differentiating both sides w.r.t. x, we got

Therefore, the given function f(x) is an increasing function in (0, π/4).

Given, curve y = – x³ + 3x² + 9x – 27

Differentiating both sides w.r.t. x, we get

dy/dx = -3x² + 6x + 9

Let slope of the curve dy/dx = z

So, z = -3x² + 6x + 9

Differentiating both sides w.r.t. x, we get

dz/dx = -6x + 6

For local maxima and local minima,

dz/dx = 0

-6x + 6 = 0 ⇒ x = 1

d²z/dx² = -6 < 0 Maxima

Putting x = 1 in equation of the curve y = (-1)³ + 3(1)² + 9(1) – 27

= -1 + 3 + 9 – 27 = -16

Maximum slope = -3(1)² + 6(1) + 9 = 12

Therefore, (1, -16) is the point at which the slope of the given curve is maximum and maximum slope = 12.

Therefore, the given function has maximum value at x = π/6 and the maximum value is 2.

Long Answer (L.A.)

Let ∆ABC be the right-angled triangle in which ∠B = 90^o

Let AC = x, BC = y

So, AB = √(x² + y²)

∠ACB = θ

Let z = x + y (given)

Now, the area of ∆ABC = ½ x AB x BC

Therefore, the area of the given triangle is maximum when the angle between its hypotenuse and a side is π/3.

Given, f (x) = x⁵ – 5x⁴ + 5x³ – 1

Differentiating the function,

f’ (x) = 5x⁴ – 20x³ + 15x²

For local maxima and local minima, f’ (x) = 0

The maximum value of the function at x = 1

f(x) = (1)⁵ – 5(1)⁴ + 5(1)³ – 1

= 1 – 5 + 5 – 1 = 0

The minimum value at x = 3 is

f(x) = (3)⁵ – 5(3)⁴ + 5(3)³ – 1

= 243 – 405 + 135 – 1

= 378 – 406 = -28

Therefore, the function has its maxima at x = 1 and the maximum value = 0 and its has minimum value at x = 3 and its minimum value is -28.

Let’s consider that the company increases the annual subscription by Rs x.

So, x is the number of subscribers who discontinue the services.

Total revenue, R(x) = (500 – x) (300 + x)

= 150000 + 500x – 300x – x²

= -x² + 200x + 150000

Differentiating both sides w.r.t. x, we get R’(x) = -2x + 200

For local maxima and local minima, R’(x) = 0

-2x + 200 = 0 ⇒ x = 100

R’’(x) = -2 < 0 Maxima

So, R(x) is maximum at x = 100

Therefore, in order to get maximum profit, the company should increase its annual subscription by Rs 100.

The given curve is x²/a² + y²/b² = 1 and the straight-line x cos α + y sin α = p

Differentiating equation (i) w.r.t. x, we get

Therefore, a² cos² α + b² sin² α = p².

Let x be the length of the side of the square base of the cubical open box and y be its height.

So, the surface area of the open box

Let’s consider x and y to be the length and breadth of given rectangle ABCD.

According to the question, the rectangle will be resolved about side AD which making a cylinder with radius x and height y.

So, the volume of the cylinder V = πr²h = πx²y …. (1)

Now, perimeter of rectangle P = 2(x + y)

36 = 2(x + y)

18 = x + y

y = 18 – x ….. (ii)

Putting the value of y in the equation (i), we get

V = πx²(18 – x) = π(18x² – x³)

Differentiating both sides w.r.t. x, we get

dV/dx = π(36x – 3x²) …. (iii)

For local maxima and local minima dV/dx = 0

π(36x – 3x²) = 0

36x – 3x² = 0

3x(12 – x) = 0

x ≠ 0 and 12 – x = 0 ⇒ x = 12

From equation (ii), we have

y = 18 – 12 = 6

Differentiating equation (iii) w.r.t. x, we get

d²V/dx² = π(36 – 6x)

At x = 12,

d²V/dx² = π(36 – 6 x 12) = π(36 – 72)

= -36 π < 0 maxima

Now, volume of the cylinder so formed = πx²y

= π x (12)² x (6) = π(144)² x 6 = 864π cm³

Therefore, the required dimension are 12 cm and 6 cm and the maximum volume is 864π cm³.