Sets, Relations and FunctionsMCQMTP Dec 22 - Series IQuestion 1996 of 217
All Questions

Determine f(x)\displaystyle f(x), given that f′(x)=12x2−4x\displaystyle f'(x)=12x^2-4x and f(−3)=17\displaystyle f(-3)=17

Options

Af(x)=4x3−2x2+143\displaystyle f(x)=4x^3-2x^2+143
Bf(x)=6x3−x4+137\displaystyle f(x)=6x^3-x^4+137
Cf(x)=3x4−x3−137\displaystyle f(x)=3x^4-x^3-137
DMISSING_OPTION_D
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Correct Answer

✅ Option a — f(x)=4x3−2x2+143\displaystyle f(x)=4x^3-2x^2+143

All Options:

  • Af(x)=4x3−2x2+143\displaystyle f(x)=4x^3-2x^2+143
  • Bf(x)=6x3−x4+137\displaystyle f(x)=6x^3-x^4+137
  • Cf(x)=3x4−x3−137\displaystyle f(x)=3x^4-x^3-137
  • DMISSING_OPTION_D

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Detailed Solution & Explanation

We are given the derivative of the function:
f′(x)=12x2−4xf'(x) = 12x^2 - 4x
To find f(x)\displaystyle f(x), we take the indefinite integral of f′(x)\displaystyle f'(x) with respect to x\displaystyle x:
f(x)=∫f′(x) dx=∫(12x2−4x) dxf(x) = \int f'(x) \, dx = \int (12x^2 - 4x) \, dx
Using the power rule of integration ∫xn dx=xn+1n+1+C\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C:
f(x)=12(x33)−4(x22)+Cf(x) = 12\left(\frac{x^3}{3}\right) - 4\left(\frac{x^2}{2}\right) + C
f(x)=4x3−2x2+Cf(x) = 4x^3 - 2x^2 + C

**Step 2: Find the constant of integration C\displaystyle C** using f(−3)=17\displaystyle f(-3) = 17:
17=4(−3)3−2(−3)2+C17 = 4(-3)^3 - 2(-3)^2 + C
17=4(−27)−2(9)+C17 = 4(-27) - 2(9) + C
17=−108−18+C17 = -108 - 18 + C
17=−126+C17 = -126 + C
C=17+126=143C = 17 + 126 = 143

Substituting C\displaystyle C back into the equation, we get:
f(x)=4x3−2x2+143f(x) = 4x^3 - 2x^2 + 143

Hence, **Option A** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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