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Question #41139

Difficulty: 8/10 | 🟢 Free

The Question

Let L1L_1 and L2L_2 be two lines in R3\mathbb{R}^3 given by: L1:r⃗=λ(i^+2j^)L_1: \vec{r} = \lambda(\hat{i} + 2\hat{j}) L2:r⃗=(i^−2j^+k^)+μ(2j^+k^)L_2: \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu(2\hat{j} + \hat{k}) Let dd be the shortest distance between L1L_1 and L2L_2. Let f:R→Rf: \mathbb{R} \to \mathbb{R} be defined as f(x)=x3−3dx2+3(3d−2)xf(x) = x^3 - 3dx^2 + 3(3d-2)x. Consider the vectors u⃗=xi^+j^+k^\vec{u} = x\hat{i} + \hat{j} + \hat{k}, v⃗=i^+xj^+k^\vec{v} = \hat{i} + x\hat{j} + \hat{k}, and w⃗=i^+j^+xk^\vec{w} = \hat{i} + \hat{j} + x\hat{k}. A function g:R→Rg: \mathbb{R} \to \mathbb{R} is defined as: g(x)={[u⃗ v⃗ w⃗]∣u⃗×v⃗∣2if u⃗ and v⃗ are non-collinearkif u⃗ and v⃗ are collinearg(x) = \begin{cases} \frac{[\vec{u} \ \vec{v} \ \vec{w}]}{|\vec{u} \times \vec{v}|^2} & \text{if } \vec{u} \text{ and } \vec{v} \text{ are non-collinear} \\ k & \text{if } \vec{u} \text{ and } \vec{v} \text{ are collinear} \end{cases} If g(x)g(x) is continuous for all x∈Rx \in \mathbb{R}, which of the following is true?

Options

  • A.

    f is bijective and k=12f \text{ is bijective and } k = \frac{1}{2}

  • B.

    f is many-one and k=12f \text{ is many-one and } k = \frac{1}{2}

  • C.

    f is bijective and k=0f \text{ is bijective and } k = 0

  • D.

    f is many-one and k=0f \text{ is many-one and } k = 0

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Correct Answer

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Explanation

How the Examiner Thinks

Monster

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

Concepts Behind This Question

Composite Injectivity/Surjectivity|Shortest distance between skew lines|Collinearity vs Coplanarity|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question tests the integration of 3D geometry with functional properties, specifically targeting two common traps.

First, the 'Cross Multiplication Trap': students frequently make a sign error in the j^\hat{j} component when computing the cross product b1⃗×b2⃗\vec{b_1} \times \vec{b_2}.

This leads to an incorrect shortest distance d=2/3d=2/3, which alters the derivative of f(x)f(x) to have distinct real roots, falsely making f(x)f(x) many-one.

Second, the 'Collinearity vs Coplanarity Trap': students often confuse the condition for two vectors being collinear (u⃗×v⃗=0⃗\vec{u} \times \vec{v} = \vec{0}) with three vectors being coplanar ([u⃗ v⃗ w⃗]=0[\vec{u} \ \vec{v} \ \vec{w}] = 0).

This confusion leads them to evaluate the limit at the coplanar point x=−2x=-2 instead of the collinear point x=1x=1, resulting in an incorrect continuous extension value of k=0k=0.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptComposite Injectivity/Surjectivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree