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

Calculation: 

f(x) = 2x³ + (2p – 7)x² + 3(2p – 9)x – 6

⇒ f'(x) = 6x² + 2(2p – 7)x + 3(2p – 9)

⇒ f'(0)

∴ 3(2p – 9)

⇒ 

⇒ 

Hence, the correct answer is Option 3.

Concept:

Local Extrema (Maxima and Minima):

• The local maxima of a function occur where the derivative changes from positive to negative.
• The local minima of a function occur where the derivative changes from negative to positive.
• To find the points of local maxima or minima, we calculate the first derivative of the function and set it equal to zero to find critical points.
• The second derivative test can further confirm whether the critical points correspond to maxima or minima:
  • If 0 \), the function has a local minimum at that point.
  • If , the function has a local maximum at that point.
  • If , the test is inconclusive.

Given Function:

• The function is defined as follows:
  • For ,
  • For ,

Graph Behavior:

• The function is a rational function, and its behavior at critical points and over intervals needs to be analyzed.
• We analyze the function in the interval and for local minima and maxima.

Calculation:

Step 1: Finding the first derivative of the function.

The given function is , and we need to find its first derivative.

Step 2: Solving for critical points.

To find the critical points, we solve . This will give us the points where the function has potential maxima or minima.

Step 3: Second derivative test for confirmation.

The second derivative is used to confirm the nature of the critical points (whether they are maxima or minima).

Step 4: Local Maxima and Minima.

The results of the calculations reveal that:

• Statement A: The point is a point of local maxima of .
• Statement B: The point is a point of local minima of .
• Statement C: The number of points of local maxima of in the interval is 3.
• Statement D: The number of points of local minima of in the interval is 1.

Conclusion:

Hence, the correct answers are:

• Statement B: The point is a point of local minima of .
• Statement C: The number of points of local maxima of in the interval is 3.
• Statement D: The number of points of local minima of in the interval is 1.

Calculation: 

⇒ f'(x) = 8x³ - 36x + 8 = 4(2x³ - 9x + 2) 

f'(x) = 0 

∴ 

Now

⇒ 

∴ 

Hence, the correct answer is Option 1.

Concept:

• The question involves testing continuity of rational functions, minima of polynomial functions, and differentiability of composite trigonometric and floor functions.
• Continuity requires the numerator and denominator to cancel out the points where the denominator becomes zero.
• For finding the minima, the first derivative is used to locate critical points, and the second derivative test confirms whether the point is a minima.
• Non-differentiability occurs where absolute functions like sin |x - k| cause sharp turns (cusps).

Calculation:

P) Let k(x) = 10x³ - 45x² + 60x + 35

⇒ k′(x) = 30x² - 90x + 60 = 30(x - 1)(x - 2)

⇒ k(x) is continuous in [1, 2]

⇒ [k(1), k(2)] are same integer for all x ∈ [1, 2]

⇒ Minimum value of n = 9     

(P → 2)

Q) g(x) = (2n² - 13n - 15) / (n² + 3x)

⇒ (2n² - 13n - 15) / (n² + 3x) ≥ 0 for g(x) ≥ 0 for g(x) = mix of n & x

⇒ Minimum value of n is 5     

(Q → 1)

R) h(x) = (x² - 9)²(x² + 2x + 3)

⇒ h(x) has a local minima at x = 3 for n = 6

⇒ (3 + δ) h(3 + δ) (δ is a small positive real number)

⇒ has local minimum at x = 3 for n = 6

⇒ (R → 4)

S) g(x) = sin |x - k| + cos |x - k - 1/2| + sin |x - k - 1| + cos |x - k - 3/2| + ⋯ + sin |x - 4| + cos |x - 9/2|

as sin |x - k| is non-differentiable at x = k    

but, cos |x - λ| is differentiable at x = λ

⇒ g(x) is non-differentiable at x₀ = 0, 1, 2, 3, 4 (5 points)

⇒ (S → 3)

∴ The correct matching is P → 2, Q → 1, R → 4, S → 3.

Hence, Option 2 is the correct answer.

Concept:

Finding local maxima and minima of a cubic function:

• To find local maxima and minima of a cubic function, first differentiate the function and find the critical points by setting the derivative equal to zero.
• The critical points correspond to values of x" id="MathJax-Element-124-Frame" role="presentation" style="position: relative;" tabindex="0">xx $x$ where the function attains local maxima or minima.
• Use given conditions on the critical points to find the parameters and evaluate the function at required points.

Calculation:

Given,

First derivative,

Set to find critical points:

Dividing by 6:

Let the roots be and . Then, by Vieta's formulas,

Given condition,

Substitute :

From sum,

From product,

Squaring the sum and equating to product:

Solving for gives .

Then,

Finally, evaluate :

∴ The correct answer is Option 4.

Concept:

Following steps to finding 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: If f"(x) Is greater than 0 then the function is said to be minima

Calculation:

f(x) = x2 - x + 2

f'(x) = 2x - 1

Set the derivative equal to 0, we get

f'(x) = 2x - 1 = 0

⇒ x = 

Now, f''(x) = 2 > 0

So, we get minimum value at x =

f() = ()² - + 2 =

Hence, option (3) is correct. 

Concept:

|x| ≥ 0 for every x ∈ R

Calculation:

Let f(x) = |x + 3| - 2

As we know that |x| ≥ 0 for every x ∈ R

∴ |x + 3| ≥ 0

The minimum value of function is attained when |x + 3| = 0

Hence, Minimum value of f(x) = 0 – 2 = -2 

Concept:

For a function y = f(x):

• Relative (local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x) .
• At the points of relative (local) minima, f''(x) > 0.

Calculation:

For the given function f(x) = 3x4 + 4x3 - 12x2 + 12, first let's find the points of local maxima or minima:

f'(x) = 12x³ + 12x² - 24x = 0

⇒ 12x(x² + x - 2) = 0

⇒ x(x + 2)(x - 1) = 0

⇒ x = 0 OR x = -2 OR x = 1.

f''(x) = 36x² + 24x - 24.

f''(0) = 36(0)2 + 24(0) - 24 = -24.

f''(-2) = 36(-2)2 + 24(-2) - 24 = 144 - 48 - 24 = 72.

f''(1) = 36(1)2 + 24(1) - 24 = 36 + 24 - 24 = 36.

Since, at x = 0 the value f''(0) = -24 0.

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 find 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:

Given:

f(x) = e^x

Differentiating with respect to x, we get

⇒ f’(x) = e^x

For maximum value f’(x) = 0

∴ f’(x) = e^x = 0

Exponential function can never assume zero for any value of x, therefore function does not have local maxima or minima.

Concept:

Following steps to finding maxima 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.
• f"(x) is less than 0 then the given function is said to be maxima

Calculation:

Here, f(x) = x3 + 2x2 - 4x + 6 

f'(x) = 3x2 + 4x - 4

Set f'(x) = 0

3x2 + 4x - 4 = 0 

⇒3x2 + 6x - 2x - 4 = 0

⇒ 3x(x + 2) - 2(x + 2) = 0

⇒ (3x - 2)(x + 2) = 0

So, x = -2 OR x = 2/3

Now, f''(x) = 6x + 4

f''(-2) = -12 + 4 = -8 

∴ At x = -2, Maximum value of f(x) exists.

Hence, option (1) is correct.

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:

Slope m of the curve is given by dy/dx = 0

And, condition for slope to be maximum: d2y/dx2 = 0

(dy/dx)x = a gives the value of maximum slope.

Calculation:

y = – x3 + 3x2 + 9x – 27

dy/dx = – 3x2 + 6x + 9 = slope of the curve

Now, double differentiation:

d2y/dx2 = – 6x + 6 = – 6 (x – 1)

d2y/dx2 = 0

⇒ – 6 (x – 1) = 0

⇒ x = 1

clearly, d3y/dx3 = – 6

∴ The slope is maximum when x = 1.

(dy/dx)x = 1 = – 3 (1)2 + 6 × 1 + 9 = 12

Concept:

If the function f(x) has an extremum at x = a then f'(a) = 0

Calculations:

Given, the function is  

⇒ f'(x) = 

⇒ f'(x) = 

⇒ f'() = 

⇒ f'() = 

The function  has an extremum at 

Therefore,  

⇒  = 0

⇒ 

⇒ 

⇒ 

 Hence, the value of p for which the function  has an extremum at  is 2.

Concept:

For a function y = f(x):

• Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x)
• At the points of relative (local) minima, f''(x) > 0.

Calculation:

Let's say that the function is y = f(x) = 2x3 - 21x2 + 36x - 20.

∴ f'(x) =  = 6x2 - 42x + 36.

And, f''(x) =  = 12x - 42.

For maxima/minima points, f'(x) = 0.

⇒ 6x² - 42x + 36 = 0

⇒ x2 - 7x + 6 = 0

⇒ x2 - 6x - x + 6 = 0

⇒ x(x - 6) - (x - 6) = 0

⇒ (x - 6)(x - 1) = 0

⇒ x - 6 = 0 OR x - 1 = 0

⇒ x = 6 OR x = 1.

Now, let's check these points for maxima/minima by inspecting the values of f''(x) at these points.

f''(6) = 12(6) - 42 = 72 - 42 = 30.

f''(1) = 12(1) - 42 = 12 - 42 = -30.

Since, f''(6) = 30 > 0, it is the point of minimum value.

And the minimum value is f(6):

= 2(6)3 - 21(6)2 + 36(6) - 20

= 432 - 756 + 216 - 20

= -128.

Concept:

The centre of the circle inscribed in a square is the same as the centre of the square.

Calculation:

The given lines forming the square are:

x2 - 7x + 12 = 0

⇒ (x - 4)(x - 3) = 0

⇒ x = 4 and x = 3

And, y2 - 13y + 42 = 0

⇒ (y - 7)(y - 6) = 0

⇒ y = 7 and y = 6

The co-ordinates of the centre will be the mid-point of sides: (3.5, 6.5).