Before You Begin

Question #41328

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}(t) = (t^2 - 3)\hat{i} + k\hat{j}$ and $\vec{b}(t) = t\hat{i} + \hat{j}$ , where $k$ is a real parameter. A scalar function $f(t)$ is defined as: $f(t) = (\vec{a}(t) \cdot \vec{b}(t)) \sqrt{|\vec{a}(t)|^2 |\vec{b}(t)|^2 - |\vec{a}(t) \times \vec{b}(t)|^2}$ Let $S$ be the set of all real values of $k$ for which $f(t)$ is twice differentiable for all $t \in \mathbb{R}$ . The number of elements in $S$ is:

Options

A.
0
B.
1
C.
2
D.
3

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

Second Order Derivative Continuity|Lagrange Identity|Dot Product Projection Sign

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of vector identities and calculus, specifically the continuity of higher-order derivatives.

The primary trap lies in the condition for $f(t) = g(t)|g(t)|$ to be twice differentiable.

Students often correctly deduce that the roots of $g(t)$ must coincide with its critical points ( $g'(t)=0$ ) and find $k = \pm 2$ .

However, they falsely assume this makes the function twice differentiable everywhere.

For $k = \pm 2$ , the cubic $g(t)$ has one double root and one simple root.

At the simple root, the projection sign flips without a zero derivative, causing a jump discontinuity in the second derivative.

Recognizing that a cubic without a $t^2$ term can never be a perfect cube is essential to realizing that a simple root always exists.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Second Order Derivative Continuity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free