Before You Begin

Question #40351

Difficulty: 8/10 | 🟢 Free

The Question

Let $a_r$ be the term independent of $t$ in the binomial expansion of $\left(t - \frac{1}{t}\right)^{84-r} \left(t + \frac{1}{t}\right)^r$ for $r = 0, 1, 2, \dots, 84$ . A polynomial $f(x)$ is defined as $f(x) = \sum_{r=0}^{84} \binom{84}{r} a_r x^r$ . Which of the following correctly describes the parity of $f(x)$ and the coefficient of $x^{42}$ in $f(x)$ ?

Options

A.
\text{Even function, } -\binom{84}{42}\binom{42}{21}
B.
\text{Even function, } \binom{84}{42}\binom{42}{21}
C.
\text{Odd function, } -\binom{84}{42}\binom{42}{21}
D.
\text{Neither even nor odd, } \binom{84}{42}\binom{42}{21}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Polynomial Function Parity|Coefficient of x^r in Expansion|Term Independent of x

Why This Question Is Tricky

Common Mistakes

This question tests the ability to interchange summation and coefficient extraction, a hallmark of JEE Advanced binomial problems.

The primary trap lies in attempting to evaluate $a_r$ individually; because $f(x)$ is ultimately an even function, $a_r$ vanishes for all odd $r$ , creating a deceptive sequence that breaks standard patterns.

Furthermore, the '84' trap manifests in the final step: when finding the coefficient of $x^{42}$ in $(x^2 - 1)^{42}$ , students often see the even exponent 42 and incorrectly assume a positive sign, overlooking that the binomial index $k=21$ is odd, which introduces the critical negative sign.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Polynomial Function Parity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free