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Question #40343

Difficulty: 8/10 | 🟢 Free

The Question

The coordinates $(x, y, z)$ of a point on a line $L$ are given by the relation: $\frac{3x - 6}{2} = \frac{2 - y}{3} = \frac{2z - 4}{4} = t$ where $t \in \mathbb{R}$ is a parameter. A real-valued function $f$ is defined as: $f(t) = \sin^{-1}\left( \frac{x + y + z - 6}{2} \right) + \sqrt{4 - t} + \sqrt{t + 10}$ If the domain of $f(t)$ is the interval $[a, b]$ , find the value of $a^2 + b^2$ .

Options

A.
52
B.
8
C.
116
D.
72

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

Real-Valued Function Domain|Parametric form pitfalls|Symmetric vs Non-symmetric form

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle parametric equations of a line in non-symmetric form and map them to a function's domain.

The primary trap lies in the 'normalization' of the line equation.

Students frequently convert the given equation to the standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$ with coprime integer direction ratios (e.g., multiplying by 3 to get direction ratios 2, -9, 6), mistakenly assuming the new parameter $\lambda$ is equivalent to the original parameter $t$ .

This leads to an incorrect domain interval of $[-2, 2]$ and the trap answer 8.

Additionally, the problem requires careful intersection of multiple domain constraints (inverse trigonometric and square roots), punishing students who only check partial conditions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Real-Valued Function Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free