First-order differential equations Practice Questions

Question 1 of 10 Authentic UPSC NDA Question

What is equal to?

Step-by-Step Logic & Solution

Question 2 of 10 Authentic UPSC NDA Question

Direction for Questions (105-106): Consider the following for the two (02) items that follow: Let the curve f(x) = |x – 3|.

What is the area bounded by the curved f(x) and y = 3?

Step-by-Step Logic & Solution

Question 3 of 10 Authentic UPSC NDA Question

For the following three (03) items:

Consider the function f(x) = x|x|. What is the area bounded by the curve f(x), the x-axis and the lines x = -2 and x = 1?

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Question 4 of 10 Authentic UPSC NDA Question

For the following two (02) items:

What is u + v equal to?

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Question 5 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Which of the above is/are correct?

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Question 6 of 10 Authentic UPSC NDA Question

What is

∫ₙⁿ⁺¹ (x − [x]) dx,

where [.] is the greatest integer function and n is a natural number?

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Question 7 of 10 Authentic UPSC NDA Question

Consider the following statements:

Which of the statements given above. is/are correct?

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Question 8 of 10 Authentic UPSC NDA Question

Step-by-Step Logic & Solution

Question 9 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Let k be the area between the curve y = sin x and x-axis in the interval [0, π/4].

What is the area between the curve y = sin x and the x-axis in the interval [π/4, π/2]?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

The area \( k \) is given by the integral of \( \sin x \) from \( 0 \) to \( \frac{\pi}{4} \):

\[ k = \int_0^{\frac{\pi}{4}} \sin x \, dx \] \end{aligned} \] Now, calculate the area from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \):

\[ \text{Area} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx \end{aligned} \] By calculating the integral: \begin{aligned} \int \sin x \, dx & = -\cos x + C \\ \int_0^{\frac{\pi}{2}} \sin x \, dx & = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) \\ & = 0 + 1 = 1 \end{aligned} \] Thus, area from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \): \begin{aligned} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx & = 1 - k \end{aligned} \] Since \( k = \int_0^{\frac{\pi}{4}} \sin x \, dx \), we substitute: \begin{aligned} 1 - k & = \text{Area between } \frac{\pi}{4} \text{ and } \frac{\pi}{2} \end{aligned} \] Hence, the required area is: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx = 1 - k \] Thus, the area between the curve \( y = \sin x \) and the x-axis in the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\) is \(1 - k\).

📌 Hints / Properties Used:

Area calculation using definite integrals.
Basic properties of sine function and its integrals.
Area representation between different bounds.

📊 Visual Diagram Suggestion:

A graph of \( y = \sin x \) from \( 0 \) to \( \frac{\pi}{2} \), highlighting the areas under the curve for the intervals \([0, \frac{\pi}{4}]\) and \([\frac{\pi}{4}, \frac{\pi}{2}]\). Use shaded regions to indicate the respective areas k and \(1 - k\).

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 5 (Integrals).

Question 10 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Let k be the area between the curve y = sin x and x-axis in the interval [0, π/4].

What is the area between the curve y = cos x and the x-axis in the interval [π/4, π/2]?

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

The area \( k \) under the curve \( y = \sin x \) from \( 0 \) to \( \frac{\pi}{4} \) is calculated as:

\[ k = \int_{0}^{\frac{\pi}{4}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{4}} \]

Calculating the integral gives:

\[ k = -\cos\left(\frac{\pi}{4}\right) + \cos(0) = -\frac{\sqrt{2}}{2} + 1 = 1 - \frac{\sqrt{2}}{2} \]

Next, we need to find the area between the curve \( y = \cos x \) and the \( x \)-axis from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \):

\[ \text{Area} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]

Evaluating this integral gives:

\[ \text{Area} = \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2} \]

Thus, the area calculated is:

\[ \text{Area} = k \]

This confirms that the area is equal to \( k \). Therefore, the correct answer is option (a).

📌 Hints / Properties Used:

Integral of sine function: \(\int \sin x \, dx = -\cos x + C\)
Integral of cosine function: \(\int \cos x \, dx = \sin x + C\)
Area under a curve from \( a \) to \( b \) is given by \(\int_{a}^{b} f(x) \, dx\)

📊 Visual Diagram Suggestion:

Consider sketching the curves \( y = \sin x \) and \( y = \cos x \) over the intervals mentioned. The area can be visually represented with shaded regions under each curve, illustrating how they relate in \( [0, \frac{\pi}{4}] \) and \( [\frac{\pi}{4}, \frac{\pi}{2}] \).

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 7 (Integrals)

First-order differential equations Questions

Solution

Solution