Q591: A function $f(x)=|1-x+|x||$ is:...

Class CBSE Class 12 Mathematics Continuity and Differentiability Q #591

KNOWLEDGE BASED

UNDERSTAND

1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE

A function $f(x)=|1-x+|x||$ is:

(A) discontinuous at $x=1$ only

(B) discontinuous at $x=0$ only

(D) continuous everywhere

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AI Tutor Explanation

Detailed Solution

Step 1: Analyze the inner absolute value function

We have $f(x) = |1 - x + |x|| $. First, let's analyze the inner absolute value function $|x|$. $|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$

Step 2: Rewrite the function based on the inner absolute value

Now, we can rewrite the function $f(x)$ based on the definition of $|x|$: If $x \geq 0$, then $f(x) = |1 - x + x| = |1| = 1$. If $x < 0$, then $f(x) = |1 - x - x| = |1 - 2x|$.

Step 3: Analyze the second absolute value function

Now, let's analyze the absolute value function $|1 - 2x|$. $|1 - 2x| = \begin{cases} 1 - 2x, & \text{if } 1 - 2x \geq 0 \Rightarrow x \leq \frac{1}{2} \\ -(1 - 2x) = 2x - 1, & \text{if } 1 - 2x < 0 \Rightarrow x > \frac{1}{2} \end{cases}$

Step 4: Combine the cases to define the piecewise function

Combining these cases, we can define $f(x)$ as a piecewise function: $f(x) = \begin{cases} 1, & \text{if } x \geq 0 \\ |1 - 2x|, & \text{if } x < 0 \end{cases}$ $f(x) = \begin{cases} 1, & \text{if } x \geq 0 \\ 1 - 2x, & \text{if } x < 0 \text{ and } x \leq \frac{1}{2} \\ 2x - 1, & \text{if } x < 0 \text{ and } x > \frac{1}{2} \end{cases}$ Since $x < 0$ implies $x \leq \frac{1}{2}$ and $x > \frac{1}{2}$ is impossible, we have: $f(x) = \begin{cases} 1, & \text{if } x \geq 0 \\ 1 - 2x, & \text{if } x < 0 \end{cases}$

Step 5: Check for continuity

Now, we need to check for continuity. For $x > 0$, $f(x) = 1$, which is continuous. For $x < 0$, $f(x) = 1 - 2x$, which is continuous. We need to check the continuity at $x = 0$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 1 = 1$ $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 - 2x) = 1 - 2(0) = 1$ $f(0) = 1$ Since $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) = 1$, the function is continuous at $x = 0$.

Step 6: Conclusion

Since the function is continuous for $x > 0$, $x < 0$, and at $x = 0$, the function is continuous everywhere.

Final Answer: continuous everywhere

AI Suggestion: Option D

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the properties of absolute value functions and how they affect continuity. The student must decompose the function into piecewise components and then evaluate the limits to determine continuity.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concept of continuity and absolute value functions, rather than just recalling facts or applying a specific algorithm. The student needs to understand how the absolute value function changes the behavior of the overall function and how to analyze continuity at a point.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of continuity and differentiability, which is a core topic in the syllabus. The question involves absolute value functions, which are commonly encountered in calculus problems.