Maxima & Minima MCQ Quiz - Objective Question with Answer for Maxima & Minima - Download Free PDF

Explanation:

f(x) = sin x + cos 2x in [0, 2π]

f’(x) = cos x – 2 sin 2x

Since sin2x = 2 sinx cosx, the above can be written as:

f'(x) = cos x [1 – 4 sinx]

Now,

Equating this to zero, we solve for x as:

f'(x) = cos x [1 – 4 sinx] = 0

⇒ cos x = 0; 1 – 4sinx = 0

For cosx = 0,

⇒ x = π/2, 3π/2 in [0, 2π]

Now,

Also, for sin x = 1/4, f'(x) = 0.

Also f’(x) exists for all x in [0, 2π]

Checking for all the points in the defined range, we get:

Now,

f(0) = 1; f(2π) = 1; f(π/2) = 0; f(3π/2) = -2

At sinx = 1/4 , f(x) = 1/2 + (1 – 2(1/4)²) = 9/8

∴ The maximum value of f(x) are 9/8.

Hence Option (2) is the answer.

Concept:

A point c in the domain of a function f at which either f'(c) = 0 or if f is nondifferentiable is called the critical point.

The slope of tangents at critical points is zero

Let f be a real-valued function and c be an interior point in the domain of f such that f'(c)=0.

• If f'(x) > 0 at every point close to the left of c and f'(x) < 0 at every point right of c, then is a local maximum.
• If f'(x) > 0 at every point close to the right of c and f'(x) < 0 at every point left of c, then is a local minimum.
• If f'(x) does not change sign as x increases or decreases, then such a point is called the point of inflection.

Let \(\rm F=(x^3+y^3+z^3)+\lambda(x^2+y^2+z^2-3)\) 

\(\begin{array}{l} {F_x} = 3{x^2} +2 \lambda x = 0 \Rightarrow {{ 2 \lambda }}{} = {-3x}\\ {F_y} = 3{y^2} +2 \lambda y = 0 \Rightarrow {{ 2 \lambda }}{} = {-3y}\\ {F_z} = 3{z^2} +2 \lambda z = 0 \Rightarrow {{ 2 \lambda }}{} = {-3z}\\ \therefore {x^2} + {y^2} + {z^2} = 3\\ \therefore \lambda =\pm \ \frac {3}{2} \Rightarrow x = \pm 1,y = \pm 1,z = \pm 1 \end{array}\)

∴ stationary point is (-1, -1, -1)

∴ Minimal value \(= {\left( { 1} \right)^3} + {\left( { 1} \right)^3} + {\left( { 1} \right)^3} = 3\)

Hence Option(1) is the answer.

Answer : 1

Solution :

To determine the local minimum of the function \(f(x)=\frac{x}{2}+\frac{2}{x}\), we start by finding the derivative of the function. This will help identify the critical points where potential minima or maxima might occur.

The derivative of f(x) is found using the power rule and the chain rule:

\(f^{\prime}(x)=\frac{d}{d x}\left(\frac{x}{2}\right)+\frac{d}{d x}\left(\frac{2}{x}\right)=\frac{1}{2}-\frac{2}{x^{2}} .\)

Set the derivative equal to zero to find the critical points:

\(\frac{1}{2}-\frac{2}{z^{2}}=0 .\)

Rearranging, we get:

\(\frac{1}{2}-\frac{2}{z^{2}}=0 .\)

Multiplying both sides by a2 to clear the fraction:

\(2=\frac{1}{2} x^{2}\)

Multiplying both sides by 2:

4 = x².

Thus, z = ±2.

Now, we evaluate whether these critical points are minima or maxima using the second derivative test.

\(f^{\prime \prime}(x)=\frac{d}{d x}\left(\frac{1}{2}-\frac{2}{x^{2}}\right)=0+\frac{d}{d x}\left(-\frac{2}{x^{2}}\right)=\frac{4}{x^{2}}\)

At x = 2:

\(f^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{4}{8}=\frac{1}{2}>0 \text {. }\)

This indicates that x = -2, the function f has a local minimum.

At x = -2:

\(f^{\prime \prime}(-2)=\frac{4}{(-2)^{3}}=\frac{4}{-8}=-\frac{1}{2}<0 .\)

This indicates that at a = -2, the function f has a local maximum.

Option 3 and Option 4 (i.e., for a 0 and x = 1) are not even candidates given by solving the critical point equation, and moreover, f(x) is not defined for x = 0 due to division by zero.

Therefore, the function f(x) = \(\frac{x}{2}+\frac{2}{x}\) has a local minimum at x = 2.

So, the correct answer is:

Option 1 : x = 2

Explanation:

As peer given data:

f(x) = x3 - 27x + 4

​f'(x) = 3x2 - 27

​f'(x) = 0

3x2 - 27 = 0

x = ±3

x = 3 ∈ (1, 6)

So at x = 3 function has point at local minima.

Explanation:

f(x, y) = (x2 + y2)e2−x−y 

f_x = 2xe2−x−y - (x2 + y2)e2−x−y = (2x - (x2 + y2))e2−x−y

fy = 2ye2−x−y - (x2 + y2)e2−x−y = (2y - (x2 + y2))e2−x−y

Critical points are given by

fx = fy = 0  

⇒ (2x - (x2 + y2))e2−x−y  = 0

⇒ 2x - x² - y² = 0

⇒ 2x = (x2 + y2)...(i)

Similarly, 2y = (x2 + y2)

⇒ 2y = 2x ⇒ x = y

Then by (i)

2x = 2x² ⇒ x = 0, 1

Then y = 0 when x = 0 and y = 1 when x = 1

So critical points are (0, 0), (1, 1)

So, f(x, y) will be maximum at critical points only.

Now, f(0, 0) = 0 and f(1, 1) = 2

Answer is 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 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 

Concept:

Intermediate Value Theorem:

It states that if 'f' is a continuous function whose domain contains the interval [a, b], then it takes any given value between f(a) and f(b) at some point within the interval.

Considerd an interval of I = [a, b], and a continuous function f(x) then,

If 'u' is the number between f(a) and, f(b), then

\(\min \left[ {f\left( a \right),\;f\left( b \right)} \right] < u < \max \left[ {f\left( a \right),f\left( b \right)} \right]\)

\(c \in \left( {a,b} \right)\;and,\;f\left( c \right) = u\)

Calculation:

Given polynomial,

p(x) = x4 - 4x3 + 2

Using intermediate value theorem,

p(0) = 0 - 0 + 2 = 2

p(1) = 1 - 4 + 2 = -1

Since,

p(0) > 0 .... (1)

p(1) < 0 .... (2)

From equation (1) and (2), there exist a value 'x' in between 0 and 1 where,

p(x) = 0

Hence, in the open interval (0, 1), the polynomial p(x) has one real root.

Concept:

The box has a square base and has a open top. Let the side of the square base be x  and the height of the box be y.
Now the surface area (S) can be given as:

S = (Area of square base + Area of Sides)

S = x² + 4xy
And the volume (V) is given as:

V = x²y

Calculation:

Given:
S = 1200 m²

As, S = x2 + 4xy

1200 = x² + 4xy
\(y = \frac{{1200 - {x^2}}}{{4x}} = \frac{{300}}{x} - \frac{x}{4}\)

Now,

Volume is

V = x2y

putting value of y

\(V = {x^2} \times \left( {\frac{{300}}{x} - \frac{x}{4}} \right) = 300x - \frac{{{x^3}}}{4}\)

Now for maximum volume

\(\frac{{dV}}{{dx}} = 0\)

\(\frac{d}{{dx}}\left( {300x - \frac{{{x^3}}}{4}} \right) = 300 - \frac{{3{x^2}}}{4} = 0\)

\({x^2} = \frac{{300 \times 4}}{3} = 400\)

\(x = \sqrt {400} = 20\)

Now,

\(\frac{{{d^2}V}}{{d{x^2}}} = - \frac{{6x}}{4}\)

\({\left( {\frac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 20}} = - \frac{{6\left( {20} \right)}}{4} = - 30 < 0\)

∴ At x = 20 volume will be maximum.

\(y = \frac{{300}}{x} - \frac{x}{4} = \frac{{300}}{{20}} - \frac{{20}}{4} = 15 - 5 = 10\)

Maximum volume is

V = x²y

(V)_max = (20)² × 10 = 4000 m²

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:

Let f(x , y) have continuous second-order partial derivatives in some disc centred at a critical point (a, b), and let

\(D = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left( {{f_{xy}}\left( {a,b} \right)} \right)^2}\)

• If D > 0 and f_xx(a, b) > 0, then f has a relative minimum at (a, b)
• If D > 0 and f_xx(a, b) < 0, then f has a relative maximum at (a, b)
• If D < 0, then f has a saddle point at (a, b)
• If D = 0, then no conclusion can be drawn

Calculation:

f(x, y) = 2 + 2x + 2y – x² – y²

f_x = 2 – 2x, f_y = 2 – 2y

f_xx = -2, f_yy = -2

f_xy = 0

f_x = 0 2 – 2x = 0 x = 1

f_y = 0 2 – 2y = 0 y = 1

Now, the critical point (a, b) = (1, 1)

At (a, b) = (1, 1)

f_xx = -2, f_yy = -2, f_xy = 0

D = 4 > 0 and f_xy < 0

So, the function f(x, y) has a relative maximum at the point (1, 1).

f(x, y) at a point (a, b) = (1, 1) is,

f(1, 1) = 4

The given region is triangular plate in the first quadrant, bounded by the lines x = 0, y = 0 and y = 9 – x. It is represented as shown below.

From the above figure, the critical points are: (0, 0), (0, 9), (9, 0)

Now let us check for the absolute maximum and minimum of f(x, y) at all the critical points.

At (0, 0), f(0, 0) = 2

At (0, 9), f(0, 9) = -61

At (9, 0), f(9, 0) = -61

From the above values,

The absolute minimum value of f(x, y) = -61

Concept:

The method of finding Maxima and Minima of Y = f(x)

1) Find f’(x) and f”(x) for given function Y = f(x)

2) Equate f’(x) to zero to obtain stationary points x = a

3) calculate f”(x) at each stationary points x = a (i.e f”(a))

4) we obtain following three conditions

(i) If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)

(ii) If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)

(iii) If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a 

Calculation:

Given function f(x) = x2 – 4x + 2

1) f’(x) = 2x - 4, f”(x) = 2

2) f’(x) = 0, ∴ 2x - 4 = 0, x = 2(stationary point)

3) f"(x) at stationary point x = 2, f”(x) = 2

4) f"(x) > 0

∴ f(x) is minimum at x = 2
i.e.,f(2)= (2)² – 4(2) + 2 = – 2 

∴ The optimum value of f(x) is – 2 (minimum)

Concept:

Maxima and Minima of a function:

Let f(x) is supposed to be continuous for all values of x in the neighbourhood of x = a. Then f(a) is called as maximum or minimum of f(x) according to as f(a+h) is less than or greater than for all sufficiently small independent value of h, positive or negative, provided both of them are not equal to zero.

Necessary conditions for a point to be maxima or minima: The necessary conditions for f(x) to have maxima or minima at x = a is that

\({\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} = 0{\rm{\;}}\)

Sufficient conditions for a point to be maxima and minima:

\({\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} > 0 \to {\rm{minima}}\)

\({\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}_{\left( {{\rm{x}} = {\rm{a}}} \right)}} < 0 \to {\rm{maxima}}\)

Calculation:

Given, 

\(f\left( x \right) = x{e^{ - x}}\)

On differentiating both sides,

\(f'\left( x \right) = - x{e^{ - x}} + {e^{ - x}}\)

The stationary points are calculated as,

For \(f'\left( x \right) = 0,\;x = 1\)

At x = 1, f" (x) <

∴ Maximum exists at \(x = 1\) and is equal to \(f\left( 1 \right) = {e^{ - 1}}\)

Concept:

A point c in the domain of a function f at which either f’(c) = 0 or if f is no differentiable is called the critical point.

The slope of tangents at critical points is zero.

Let f be a real-valued function and c be an interior point in the domain of f such that f’(c) = 0

a) If f’(x) > 0 at every point close to the left of c and f’(x) < 0 at every point right of c, then it is a local maximum

b) If f’(x) > 0 at every point close to the right of c and f’(x) < 0 at every point left of c, then it is a local minimum

c) Also, if f’’(x) < 0 then it is maxima and if f’’(x) > 0 then it is minima.

d) If f’(x) does not change sign as x increases or decreases then such a point is called inflection.

Calculation:

f(x) = (x² - 4)²

f’(x) = 2(x² - 4)(2x) = 4x³ – 16x

Let f’(x) = 0

Then, x = 0, +2, -2

f’’(x) = 12x² – 16

f’’(0) = -16 < 0 (maxima)

f’’(2) = 82 > 0 (minima)

f’’(-2) = 32 > 0 (minima)

Concept:

The method of finding Maxima and Minima of y = f(x)

• Find f’(x) and f”(x) for given function y = f(x)
• Equate f’(x) to zero to obtain stationary points x = a
• Calculate f”(x) at each stationary points x = a (i.e f”(a))

The following three conditions are obtained:

1. If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)
2. If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)
3. If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a 

Calculation:

Given:

The function f(x) = 2x3 - 3x2

f’(x) = 6x² - 6x, f”(x) = 12x - 6

For maximum and minimum value of f(x)

f’(x) = 0, 

∴ 6x2 - 6x = 0,

x = 0, 1 (stationary point)

Checking for the global minimum in the domain [- 1, 2],

f''(1) = 12 × 1 - 6 = 6 > 0 ⇒ At 'x = 1' f(x) is having local minima. 

f''(0) = 12 × 0 - 6 = - 6 < 0 ⇒ At 'x = 0' f(x) is having local maxima. 

@x = 1: f(1) = 2 × (1)3 - 3 × (1)2 = -1

@x = 0: f(0) = 2 × (0)3 - 3 × (0)2 = 0

Since the interval given is [-1, 2]

 ⇒  f(-1) = 2 × (-1)3 - 3 × (-1)2 = -5

 ⇒ f (2) = 2 × (2)3 - 3 × (2)2 = 4

The global minimum value of f(x) will be at 'x = -1' and its magnitude will be f(-1) = -5

\(f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\)

At x = 0, f(x) = 0

At any value of x between 0 and 1, f(x) is positive.

At x = 1, f(x) = 0

At any value of x between 1 and 2, f(x) is negative.

At x = 2, f(x) = 0

The curve for \(f\left( x \right) = x\;\left( {x-1} \right)\;\left( {x-2} \right)\) will approximately be

Therefore, in the given interval [1, 2], the maximum value of f(x) is 0.