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Question #60434

Difficulty: 8/10 | 🟢 Free

The Question

Let $f, g, h$ be thrice differentiable functions on $\mathbb{R}$ . For a constant $a \in \mathbb{R}$ , define the determinants: $W(a) = \begin{vmatrix} f(a) & g(a) & h(a) \\ f'(a) & g'(a) & h'(a) \\ f''(a) & g''(a) & h''(a) \end{vmatrix}, \quad V(a) = \begin{vmatrix} f(a) & g(a) & h(a) \\ f'(a) & g'(a) & h'(a) \\ f'''(a) & g'''(a) & h'''(a) \end{vmatrix}$ Let the limits $L_1$ and $L_2$ be defined as follows: $L_1 = \lim_{x \to a} \frac{1}{(x-a)^2} \begin{vmatrix} f(a) & g(a) & h(a) \\ f(x) & g(x) & h(x) \\ f'(a) & g'(a) & h'(a) \end{vmatrix}$ $L_2 = \lim_{x \to a} \frac{1}{(x-a)^3} \left( \begin{vmatrix} f(x) & g(x) & h(x) \\ f(a) & g(a) & h(a) \\ f'(a) & g'(a) & h'(a) \end{vmatrix} + L_1 (x-a)^2 \right)$ If $W(a) \neq 0$ , then the value of $L_2$ is:

Options

A.
$\frac{1}{6} V(a)$
B.
$-\frac{1}{6} V(a)$
C.
$\frac{1}{3} V(a)$
D.
\text{Does not exist}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

L'Hopital's Pitfall|Determinant Product Rule|Singular vs. Non-Singular Threshold|Row/Column Exchange Parity

Why This Question Is Tricky

Examiner Levers

Logical Assertion

Tests the causal relationship between two true statements.

Difficulty Boost : +4

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the intersection of L'Hopital's rule, Taylor expansions, and determinant properties, specifically targeting the 'Row Exchange Parity' trap.

When students differentiate the determinant for $L_1$ , they often fail to track the row swaps required to match the standard Wronskian-like form $W(a)$ .

Missing the single row swap in $P''(a)$ leads to the incorrect value $L_1 = W(a)/2$ .

If this erroneous value is substituted into the recursive limit $L_2$ , the $(x-a)^2$ terms in the numerator add up instead of canceling out, leaving a dominant $W(a)(x-a)^2$ term over $(x-a)^3$ .

Since $W(a) \neq 0$ , this causes the limit to diverge, baiting the student into incorrectly choosing 'Does not exist'.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	L'Hopital's Pitfall
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free