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

Calculation: 

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

f'(x) = 0 

∴ \(x=\frac{\sqrt{6}-2}{2}\)

Now

⇒ \(f(x)=\left(x^{2}-2 x-\frac{9}{2}\right)\left(2 x^{2}+4 x-1\right)+24 x+7.5\)

∴ \(f\left(\frac{\sqrt{6}-2}{2}\right)=M=12 \sqrt{6}-\frac{33}{2} \)

Hence, the correct answer is Option 1.

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,

\( f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1 \)

First derivative,

\( f'(x) = 6x^2 - 18ax + 12a^2 \)

Set \( f'(x) = 0 \) to find critical points:

\( 6x^2 - 18ax + 12a^2 = 0 \)

Dividing by 6:

\( x^2 - 3ax + 2a^2 = 0 \)

Let the roots be \( p \) and \( q \). Then, by Vieta's formulas,

• \( p + q = 3a \)
• \( pq = 2a^2 \)

Given condition,

\( p^2 = q \)

Substitute \( q = p^2 \):

\( p + p^2 = 3a \)

\( p \cdot p^2 = p^3 = 2a^2 \)

From sum,

\( a = \frac{p + p^2}{3} \)

From product,

\( a^2 = \frac{p^3}{2} \)

Squaring the sum and equating to product:

\( \left( \frac{p + p^2}{3} \right)^2 = \frac{p^3}{2} \)

Solving for \( p \) gives \( p = 2 \).

Then,

\( a = \frac{2 + 4}{3} = 2 \)

Finally, evaluate \( f(3) \):

\( f(3) = 2(3)^3 - 9(2)(3)^2 + 12(2)^2 (3) + 1 \)

\( = 54 - 162 + 144 + 1 = 37 \)

∴ The correct answer is Option 4.

Calculation:

Given, S = {x ∈ ℝ/x² + 30 ≤ 11x}

∴ x² + 30 ≤ 11x

⇒ x² - 11x + 30 ≤ 0

⇒ (x - 5) (x - 6) ≤ 0

⇒ x ∈ [5, 6]

Now, f(x) = 3x³ - 18x² + 27x - 40

⇒ f'(x) = 9x² - 36x + 27

⇒ f'(x) = 9(x² - 4x + 3)

= 9 [(x² - 4x + 4) − 1]

= 9(x - 2)² - 9

∴ f'(x) > 0 ∀x ∈ [5, 6]

∴ f(x) is strictly increasing in the interval [5, 6]

∴ Maximum value of f(x) when x ∈ [5, 6] is f(6)

= 122

∴ Maximum value is 122.

The correct answer is Option 1.

Concept:

Second Derivative Test: Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then

• x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0. The value f (c) is the local maximum value of f .
• x = c is a point of local minima if f c ′( ) 0 = and f ″(c) > 0. In this case, f (c) is the local minimum value of f .
• The test fails if f ′(c) = 0 and f ″(c) = 0. 

Calculation:

Given that 

f(x) = - \(\frac{3}{4}\)x4 - 8x3 - \(\frac{45}{2}\)x2 + 105

Differentiating wrt x

⇒ f'(x) = -3x³ -  24x² - 45x 

Again differentiating wrt x

⇒ f''(x) = -9x²  - 48x - 45

Now, for local minima and maxima  f'(x) = 0 gives

⇒ -3x³ - 24x² - 45x = 0 

⇒ -3x(x² + 8x + 15) = 0 ⇒ 3x(x + 3)(x + 5) = 0

⇒ x = 0, -3, -5

At x = 0 we have

f''(0) = -45 < 0

So, x = 0 is a point of local maxima

At x = -3 , f''(-3) = 18 > 0

So, x = -3 is a point of local minima

At x  = -5, f"(-5) = -30 < 0

So, x = -5 is a point of local maxima

∴ Option 3 is correctddxxn=nxn−1" role="presentation" style="display: inline; position: relative;" tabindex="0">=nxn−1

Concept:

The points where the f'(x) = 0 are known as critical values.

At critical point, if f ''(x) > 0 then function has minima.

If f ''(x) < 0 then function has maxima.

Calculation:

Given, f (x) = \(\left\{\begin{array}{cc} x^3+x^2+10 x, & x<0 \\ -3 \sin x, & x ≥ 0 \end{array}\right.\)

For x ≥ 0

f(x) = − 3sin x

⇒ f '(x) = − 3cos x

⇒ f ′(0) = − 3

For x < 0, f (x) = x³ + x² + 10x

⇒ f ′(x) = 3x² + 2x + 10

⇒ f ′(0) = 10

⇒ f ′(x) > 0 for x < 0 and f ′(x) < 0 for x ≥ 0

⇒ f (x) has maxima at x = 0.

∴ At x = 0, there is a point of maximum.

The correct answer is Option 1.

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 = \(\frac12\)

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

So, we get minimum value at x = \(\frac12\)

f(\(\frac12\)) = (\(\frac12\))² - \(\frac12\) + 2 = \(\frac74\)

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) < 0.
• 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, the local maximum value of the function occurs at x = 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 < 0

∴ 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 < 0 for all value of x

∴ 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 \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) 

⇒ f'(x) = \(\rm p\;cos \; x + \dfrac {3\;cos \;3x}{3}\)

⇒ f'(x) = \(\rm p\;cos \; x + cos \;3x\)

⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \;3(\dfrac {\pi}{3})\)

⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\)

The function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\)

Therefore,  \(\rm f'(\dfrac {\pi}{3}) = 0\)

⇒ \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\) = 0

⇒ \(\rm \dfrac p 2 -1 = 0\)

⇒ \(\rm \dfrac p 2 =1\)

⇒ \(\rm p = 2\)

 Hence, the value of p for which the function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\) 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) < 0.
• 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) = \(\rm \dfrac{d}{dx}f(x)=\dfrac{d}{dx}(2x^3-21x^2+36x-20)\) = 6x2 - 42x + 36.

And, f''(x) = \(\rm \dfrac{d^2}{dx^2}f(x)=\dfrac{d}{dx}\left[\dfrac{d}{dx}f(x)\right]=\dfrac{d}{dx}(6x^2-42x+36)\) = 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).