Maxima and minima Practice Questions - NDA Mathematics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

How many extreme values does sin4x + 2x, where 0 < x < π/2 have?

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Question 2 of 10 Authentic UPSC NDA Question

The function y has a relative maxima at x = 0 for

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Question 3 of 10 Authentic UPSC NDA Question

Direction: Consider the following for the next two items
that follow:

What is the greatest value of f(x)?

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Question 4 of 10 Authentic UPSC NDA Question

What is the least value of f(x)?

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Question 5 of 10 Authentic UPSC NDA Question

Direction for Questions (63-64): Consider the following for the two (02) items that follow: Let the function f(x) = x² + 9.

Consider the following statements:

I. f(x) is an increasing function.

II. f(x) has local maximum at x = 0

Which of the statements given above is/are correct?

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Question 6 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Consider the function

f(x) = 1 − ∛(x − 1)².

The function has

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Question 7 of 10 Authentic UPSC NDA Question

A wire of length 20 cm is to be bent into a rectangle.

Which of the following statements is/are correct?

I. The rectangle of the largest area is the square.

II. It is possible to form a rectangle of an area of 27 cm2.

Select the answer using the code given below.

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Question 8 of 10 Authentic UPSC NDA Question

Consider the following statements:

Statement I: The function

f(x) = (x³ + 128)/x

has a minimum value 48 at x = 4.

Statement II: As x increases through 4, f′(x) changes sign from positive to negative.

Which one of the following is correct in respect of the above statements?

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Question 9 of 10 Authentic UPSC NDA Question

Consider the following statements in respect of the function

f(x) = x

in the interval (−1, 1):

I. The function attains maximum value.

II. The function attains minimum value.

Which of the statements given above is/are correct?

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

The function \( f(x) = x \) is defined on the open interval \((-1, 1)\). A function attains its maximum and minimum values at its endpoints in a closed interval. However, since \((-1, 1)\) is open, it does not include its endpoints. Thus, the function does not attain maximum or minimum values within this interval. \[ \text{Maximum value} = \lim_{x \to 1^-} f(x) = 1 \] \[ \text{Minimum value} = \lim_{x \to -1^+} f(x) = -1 \] Since the function approaches these values asymptotically but does not reach them within the interval, neither a maximum nor a minimum value is attained in \((-1, 1)\).

📌 Hints / Properties Used:

Open intervals do not include endpoints.
A function attains extrema in closed intervals.

📊 Visual Diagram Suggestion:

A coordinate chart showing the line \( y = x \) within the interval \((-1, 1)\) where the y-values approach but never reach -1 and 1, illustrating that maximum and minimum values are not attained.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Mathematics, Chapter on Functions.

Question 10 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Let S and T be the sets where

f(x) = x³/3 − 5x²/2 + 6x + 7

decreases and increases respectively.

What is T equal to?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

To find the intervals of increase and decrease of the function

\[ f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7 \]

calculate its derivative:

\[ f'(x) = x^2 - 5x + 6 \]

Set the derivative equal to zero to find critical points:

\[ x^2 - 5x + 6 = 0 \]

Factoring, we get:

\[ (x - 2)(x - 3) = 0 \]

Thus, critical points are \(x = 2\) and \(x = 3\). To determine where \(f\) is increasing or decreasing, examine intervals: 1. Interval \((- \infty, 2)\): Choose \(x = 0\) - \(f'(0) = 0^2 - 5(0) + 6 = 6 > 0\) (increasing) 2. Interval \((2, 3)\): Choose \(x = 2.5\) - \(f'(2.5) = (2.5)^2 - 5(2.5) + 6 = -0.75 < 0\) (decreasing) 3. Interval \((3, \infty)\): Choose \(x = 4\) - \(f'(4) = 16 - 20 + 6 = 2 > 0\) (increasing) Hence, \(f(x)\) is increasing in the intervals \((- \infty, 2)\) and \((3, \infty)\) and decreasing in the interval \((2, 3)\). Therefore, set \(T\), where the function increases, is:

\[ T = (-\infty, 2) \cup (3, \infty) \]

This corresponds to answer choice B: \(\{x < 2\} \cup \{x > 3\}\).

📌 Hints / Properties Used:

Calculus: Derivative test for increasing/decreasing functions
Finding critical points by setting the derivative to zero

📊 Visual Diagram Suggestion:

Graph of the function \(f(x)\) showing critical points at \(x = 2\) and \(x = 3\) with clearly marked intervals of increase and decrease.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 5 (Continuity and Differentiability).

Maxima and minima Questions

Solution

Solution