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Question #41861

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be the set of all integer solutions to the system of inequalities: $\log_2(x^2 - 4x + 4) \le 4$ $2^{x+2} + 2^{4-x} \le 65$ Let $N$ be the number of elements in set $A$ . If $S_n$ denotes the sum of the first $n$ terms of a sequence $\{a_n\}$ and is given by $S_n = 4n^2 - 3n + 7$ , then the value of $\sum_{k=1}^{N} a_{2k-1}$ is:

Options

A.
246
B.
253
C.
343
D.
350

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

System of Inequalities|Sigma Notation Properties|Sum to n terms vs. General Term

Why This Question Is Tricky

Common Mistakes

This question tests two distinct, high-frequency traps in JEE Advanced.

First, in the system of inequalities, students often solve the logarithmic inequality by exponentiating to get $(x-2)^2 \le 16$ , yielding $-2 \le x \le 6$ , but forget the implicit domain constraint of the logarithm ( $x-2 \neq 0 \implies x \neq 2$ ).

This leads to an incorrect upper limit $N=7$ instead of $N=6$ .

Second, when deriving the general term $a_n$ from the sum $S_n$ , students blindly apply $a_n = S_n - S_{n-1} = 8n - 7$ for all $n$ , ignoring that $S_0 \neq 0$ implies the formula fails for $a_1$ .

They incorrectly use $a_1 = 1$ instead of the true $a_1 = S_1 = 8$ .

The options are specifically designed to catch any combination of these two errors.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	System of Inequalities
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free