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Question #40556

Difficulty: 8/10 | 🟢 Free

The Question

Let L:y=f(x)L: y = f(x) be the line tangent to the parabola y=14x2−1y = \frac{1}{4}x^2 - 1 that is parallel to the line y=xy = x. Let RR be the feasible region in the xyxy-plane defined by the system of inequalities: y≤x,y≥f(x),andx−1x−4≥0y \le \sqrt{x}, \quad y \ge f(x), \quad \text{and} \quad \frac{x - 1}{x - 4} \ge 0 What is the area of the region RR?

Options

  • A.

    136\frac{13}{6}

  • B.

    196\frac{19}{6}

  • C.

    163\frac{16}{3}

  • D.

    116\frac{11}{6}

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

Graphical Feasible Region|Condition of Tangency|Extraneous Solution Rejection|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

The problem tests the 'Condition of Tangency' to establish a linear boundary, followed by a rigorous area calculation that embeds two major traps.

The 'Extraneous Solution Rejection' trap occurs when students blindly square x−2=xx - 2 = \sqrt{x} to find intersection points, yielding x=1x = 1 and x=4x = 4.

Assuming the region is bounded between these roots, they incorrectly integrate from 1 to 4.

Simultaneously, the 'Sign Flipping Rule' trap catches students who cross-multiply the rational inequality x−1x−4≥0\frac{x - 1}{x - 4} \ge 0 to get x≥1x \ge 1, which deceptively confirms the erroneous interval [1,4][1, 4].

The correct logical path requires analyzing the domain of the square root and the true sign intervals, revealing the actual feasible region is x∈[0,1]x \in [0, 1].

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Inequalities - Class 11
ConceptGraphical Feasible Region
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree