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Question #40629

Difficulty: 8/10 | 🟢 Free

The Question

Let crc_r be the coefficient of xrx^r in the binomial expansion of (x2−1)5(x^2 - 1)^5. Let P(x)=c2x2+c1x+c0x3−xP(x) = \frac{c_2 x^2 + c_1 x + c_0}{x^3 - x}. Let f(x)=e∫P(x)dxf(x) = e^{\int P(x) dx} (where CC is the constant of integration) be defined on (−1,1)∖{0}(-1, 1) \setminus \{0\}. If CC is chosen such that f(12)=932f\left(\frac{1}{2}\right) = \frac{9}{32}, find the value of 192∫−1/21/2f(x)dx192 \int_{-1/2}^{1/2} f(x) dx

Options

  • A.

    37

  • B.

    0

  • C.

    61

  • D.

    74

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Coefficient of x^r in Expansion|Partial Fractions Decomposition|Integrating Factor (IF) Sign Sensitivity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to extract coefficients from a binomial expansion and use them to construct a rational function, which must then be integrated to find an integrating factor.

The core trap lies in the evaluation of e∫P(x)dxe^{\int P(x) dx}.

Students frequently forget the absolute value when integrating terms like 1/x1/x, writing ∫(1/x)dx=ln⁡(x)\int (1/x) dx = \ln(x) instead of ln⁡∣x∣\ln|x|.

If the absolute value is omitted, the resulting function f(x)=x(x2−1)2f(x) = x(x^2-1)^2 is an odd function, leading to a completely incorrect definite integral of 0 over the symmetric interval [−1/2,1/2][-1/2, 1/2].

The exponential definition f(x)=e∫P(x)dxf(x) = e^{\int P(x) dx} strictly requires the function to be positive, forcing the use of the absolute value and resulting in an even function, which correctly evaluates to 37.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterBinomial Theorem - Class 11
ConceptCoefficient of x^r in Expansion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree