Mathematics Test - 28

CONCEPT:

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

CALCULATION:

Given x = at, y = a/t

\(\Rightarrow \frac{dx}{dt}=a, \frac{dy}{dt}= \frac{a}{-t^2}\)    \(\because \left(\frac{d}{dx}x^2 = nx^{n-1}\right)\)

\(\Rightarrow \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-a}{t^2}}{\frac{a}{1}} =\frac{a}{-at^2}=\frac{-1}{t^2}\)

\(\Rightarrow \frac{dy}{dx}=-\frac{1}{t^2}\)

\(\Rightarrow \frac{d^2y}{dx^2} = y'' (x) =\frac{d}{dx}\left(\frac{dy}{dx}\right)\)

\(\Rightarrow y''(x) = \frac{d}{dx}\left(-\frac{1}{t^2}\right)=\frac{d}{dt}(-t^{-2}) \frac{dt}{dx}\)

\(\Rightarrow y'' (x) = \frac{2}{t^3}\times \frac{dt}{dx}=\frac{2}{t^3}\times \frac{1}{a}\)

\(\Rightarrow y'' (x) = \frac{2}{at^3} \)

So, the correct answer is option 2.

Formula used:

• sin θ/cos θ = tan θ 
• cos θ/sin θ = cot θ 
• sin2θ + cos2θ = 1
• 2sin θ cos θ = sin 2θ

Calculation:

\(f(x) = \frac{1}{\tan x+\cot x},\)    \(0 < x < \frac{\pi}{2}?\)

⇒ \(f(x) = \frac{1}{\tan x+\cot x},\)

⇒ \(f(x) = \frac{1}{\frac{sin x}{cosx}+\frac{cosx}{sinx}}\)

⇒ \(f(x) =\frac{sinx\ cosx}{sin^2x+cos^2x}\)

⇒ f(x) = sin x cos x (\(\frac{2}{2}\))    [∵ sin²θ + cos²θ = 1]  

⇒ f(x) = \(\frac{1}{2}\)sin 2x        [∵ 2sin θ cosθ = sin 2θ]

We know that, -1 ≤ sin θ ≤ 1 

⇒  -1 ≤ sin 2x ≤ 1 

∴  f(x)_max = \(\frac{1}{2}\)(1) = 1/2

Concept:

Calculus:

• \(\rm {d\over dx}\left(a^x\right)=a^x(\log a)\).
•  

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).

Calculation:

Given expression is:

3x + 3y = 3x + y.

Differentiating w.r.t. x and using the chain rule of derivatives, we get:

⇒ 3x (log 3) + 3y (log 3) \(\rm dy\over dx\) = 3x + y (log 3) (1 + \(\rm dy\over dx\))

⇒ 3x + 3y \(\rm dy\over dx\) = 3x + y + 3x + y \(\rm dy\over dx\)

⇒ \(\rm {dy\over dx}={3^{x+y}\ -\ 3^x\over3^y\ -\ 3^{x+y}}\)

Calculation:

Given diameter of spherical balloon D = \(\rm{3\over2}(4x + 3)\)\

Radius R = \(\rm {D\over2}={3\over4}(4x+3)\)

Rate of change of radius R wrt x = \(\rm{dR\over dx}\)= \(\rm{d\over dx}\left[{3\over4}(4x + 3)\right]\)

 \(\rm{dR\over dx}\)= \(\rm{3\over4}\times{d\over dx}(4x + 3)\)

 \(\rm{dR\over dx}\) = \(\rm{3\over4}\times 4\) = 3

Volume of the spherical balloon V = \(\rm{4π\over 3}\times R^3\)

Rate of change of radius V wrt x =\(\rm{dV\over dx}\)= \(\rm{dV\over dR}\times{dR\over dx}\)

\(\rm{dV\over dx}\) = \(\rm{d\over dR}\left[{4π\over 3}\times R^3\right]\times {dR\over dx}\)​​

\(\rm{dV\over dx}\) = \(\rm{4π\over 3}\times3R^2 \times 3\) = 12πR² 

CONCEPT:

Cost and Revenue function:

• Any manufacturing company has to deal with two types of costs, the one which varies with the cost of raw material, direct labor cost, packaging, etc. is the variable cost.
• The variable cost is dependent on production output.
• As the production output increases (decreases) the variable cost will also increase (decrease).
• The other one is the fixed cost. The fixed costs are the expenses that remain the same irrespective of production output. Whether a firm makes sales or not, it must pay its fixed costs
• Cost Function: 
  • If V(x) is the variable cost of producing ‘x’ units and ‘k’ the fixed cost then, the
  • total cost C(x) is given by

⇒ C(x) = V(x) + k

• Revenue Function:
  • if R is the total revenue a company receives by selling ‘x’ units at the price ‘p’ per unit produced by it then the revenue function is given by

⇒ R(x) = p.x

Marginal cost and Marginal Revenue:

• Marginal cost and marginal revenue are the instantaneous rates of change of cost and revenue with respect to output i.e. rate of change of C(x) (or C), the cost function and R(x) (or R), the revenue function, with respect to production output ‘x’.
• Therefore, the Marginal cost (MC) and the Marginal revenue (MR) is given by

⇒  MC (Marginal cost)  = C' (x) = dC/dx

⇒  MR (Marginal revenue) =  R' (x) =dR/dx

So the correct answer is option 4.

Concept:

The maximum of a function f(x) lies where,

f ' (x) = 0 and f " (x) < 0

Calculation:

Given, \(v=x^2\log{\left(\frac{1}{x}\right )}\)

⇒ \(v' = 2x \log ({1 \over x}) + x^2 {1 \over { 1\over x}}(-{1 \over x^2})\)

⇒ \(v' = -2x \log ({ x}) -x\)

⇒ \(v' = -x(2 \log x +1)\)    ----(i)

So, v' = 0 

⇒ log x = -1/2

⇒ x = e^-1/2

Now, From (i)

And v" = -2 log x - 2 - 1

⇒ v" = -2 log x - 3

⇒ v" < 0 at x = e-1/2

⇒ Velocity is maximum at x = e^-1/2

∴ The correct answer is option (2).

Concept:

Consider a function y = f(x) on a defined interval of x.

The function attains extreme values (the value can be maximum or minimum or both).

For maxima:

• Local maxima: A point is the local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby local maxima.
• Global maxima: It is the point where there is no other point has in the domain for which function has more value than global maxima.

For minima:

• Local minima: A point is the local minima of a function if there is some other point where the minimum value is less than the local minima but that point doesn’t exist nearby local minima.
• Global minima: It is the point where there is no other point has in the domain for which function has less value than global minima.

Stationary Points: Points where the derivative of the function is zero i.e., f’(x) = 0. The points can be:

• Inflection point
• Local maxima
• Local minima

Second derivative test: Let the function has a stationary point x = a

• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} < 0\)  then x = a, is a point of maxima.
• If \({\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0\)  then x = a, is a point of minima.

Application:

Given f(x) is a real -valued function such that f'(x0) = 0 for some x0 ∈ (0, 1)

Also given f"(x) > 0 for all x ∈ (0, 1)

So, Then f(x) has exactly one local minimum in (0, 1), called the point of minima.

CONCEPT:

• Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.
• Marginal revenue = R'(x) = \( \frac{dR(x)}{dx}\)
• Revenue function R(x) = x p(x) 

CALCULATION

Given: Price per unit p(x) =  30 - 2x , Marginal revenue when 5 units were produced

Revenue function, R(x) = x p(x)

⇒ R(x) = x (30 - 2x) 

⇒ R(x) = 30x - 2x²

• Marginal revenue,

⇒ R'(x) = \( \frac{dR(x)}{dx}\) 

\(⇒ R'(x) = \frac{d}{dx} [30x - 2x^2]\)

⇒ R'(x) = 30 - 4x

So, Marginal revenue at (x = 5),

⇒ R'(x = 5) = 30 - 4 × 5

⇒ R'(x = 5) = 10

• The marginal revenue when 5 commodities are in demand is Rs 10..
• So, the correct answer is option 1.

Concept:

1. As per the given information, the total cost function C(x) is given by, the sum of fixed cost and the variable const,

C(x) = x(x + 30) + 1500 = x² + 30x + 1500

2. The marginal cost MC is given by,

\(\rm MC = \frac{dC}{dx} = 2x + 30\)

Calculation:

The marginal cost of producing 20 toys,

 MC(20) = \(\rm \left. \frac{dC}{dx} \right]_{x = 20}\) = 2(20) + 30 = 70

∴ The marginal cost of producing 20 toys is Rs. 70.

Concept:

Consider a function y = f(x) on a defined interval of x.

The function attains extreme values (the value can be maximum or minimum or both).

For maxima:

• Local maxima: A point is the local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby the local maxima.
• Global maxima: It is the point where there is no other point has in the domain for which function has more value than global maxima.

Condition:

f^"(x) < 0 ⇒ maxima

f"(x) > 0 ⇒ minima

f"(x) = 0 ⇒ Point of inflection

Calculation:

Given:

f(x) = (k2 - 4)x2 + 6x3 + 8x4 

f'(x) = 2(k2 - 4)x + 18x² + 32x³ 

f''(x) = 2(k2 - 4) + 36x + 96x² 

Since, at x = 0, f(x) has local maxima

f''(0) < 0

2(k2 - 4) + 36 × 0 + 96 × 0 < 0

k2 - 4 < 0

Here, to keep the above expression less than 0, the value of k must lie in between -2 to 2.

⇒ -2 < k < 2

Mistake Points

Since the condition for maxima is inequality, don't use it as an equation, i.e. k2 - 4 = 0. This will give k = ± 2 and changes answer to K < -2 or k > 2