Class CBSE Class 12 Mathematics Differential Equations Q #653
KNOWLEDGE BASED
APPLY
1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE
Let \(f^{\prime}(x)=3(x^{2}+2x)-\frac{4}{x^{3}}+5,\) and \(f(1)=0\). Then, \(f(x)\) is:
(A) \(x^{3}+3x^{2}+\frac{2}{x^{2}}+5x+11\)
(B) \(x^{3}+3x^{2}+\frac{2}{x^{2}}+5x-11\)
(C) \(x^{3}+3x^{2}-\frac{2}{x^{2}}+5x-11\)
(D) \(x^{3}-3x^{2}-\frac{2}{x^{2}}+5x-11\)
Correct Answer: B

AI Tutor Explanation

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Step-by-Step Solution

Given \(f^{\prime}(x)=3(x^{2}+2x)-\frac{4}{x^{3}}+5\), we need to find \(f(x)\) such that \(f(1)=0\).

First, integrate \(f^{\prime}(x)\) with respect to \(x\):

\(f(x) = \int f^{\prime}(x) \, dx = \int \left(3x^{2}+6x-\frac{4}{x^{3}}+5\right) \, dx\)

\(f(x) = \int (3x^2 + 6x - 4x^{-3} + 5) \, dx\)

\(f(x) = 3\int x^2 \, dx + 6\int x \, dx - 4\int x^{-3} \, dx + 5\int dx\)

\(f(x) = 3\left(\frac{x^3}{3}\right) + 6\left(\frac{x^2}{2}\right) - 4\left(\frac{x^{-2}}{-2}\right) + 5x + C\)

\(f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x + C\)

Now, use the initial condition \(f(1) = 0\) to find the constant \(C\):

\(0 = (1)^3 + 3(1)^2 + \frac{2}{(1)^2} + 5(1) + C\)

\(0 = 1 + 3 + 2 + 5 + C\)

\(0 = 11 + C\)

\(C = -11\)

Substitute the value of \(C\) back into the expression for \(f(x)\):

\(f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11\)

Correct Answer: \(x^{3}+3x^{2}+\frac{2}{x^{2}}+5x-11\)

AI Suggestion: Option B

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concept of integration to find the function f(x) given its derivative and an initial condition.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure (integration) to arrive at the solution. The student must know how to integrate different types of functions and how to use the initial condition to find the constant of integration.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration techniques and their ability to apply them.