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Question #41321

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = f(x)$ be a differentiable function defined for $x > 0$ that satisfies the equation $\begin{vmatrix} x^2 & y & 1 \\ 2x & y' & \frac{1}{x} \\ a & b & 0 \end{vmatrix} + \begin{vmatrix} y & x^2 & 1 \\ y' & 2x & \frac{1}{x} \\ -a & -b & 0 \end{vmatrix} = 0$ where $a$ and $b$ are distinct positive integers. If the curve passes through the points $(1, a+b-2)$ and $(2, ab)$ , find the value of $f(3)$ .

Options

A.
21
B.
3
C.
15
D.
24

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

Implicit Differentiation Implicit Domain|Arbitrary Constants vs Parameters|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply determinant properties during implicit differentiation.

The primary trap is the 'Row/Column Exchange Parity'.

When simplifying $\det(B)$ to match the structure of $\det(A)$ , students must swap the first two columns, which requires multiplying the determinant by $-1$ .

Failing to do so leads to the incorrect differential equation $y' - y/x = -x$ .

The problem further tests 'Arbitrary Constants vs Parameters' by forcing students to use the boundary conditions to form a Diophantine equation $(a-2)(b-2) = 2$ .

The trap path also yields a valid Diophantine equation $(a-2)(b-2) = -2$ , leading to a false sense of security and the incorrect answer 3.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Implicit Differentiation Implicit Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free