Question:
Let $y = \dfrac{3x^3 - 2x^2 + x}{|x|}, \; x \ne 0$. Then $\dfrac{dy}{dx}$ at $x=-2$ is equal to:
Let $y = \dfrac{3x^3 - 2x^2 + x}{|x|}, \; x \ne 0$. Then $\dfrac{dy}{dx}$ at $x=-2$ is equal to:
Show Hint
Break modulus functions into cases before differentiating.
Updated On: Apr 24, 2026
- $14$
- $-12$
- $-14$
- $12$
- $10$
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
• $|x| = -x$ for $x<0$
Step 1: For $x=-2$ (negative)
\[ |x| = -x \] \[ y = \frac{3x^3 - 2x^2 + x}{-x} = -3x^2 + 2x - 1 \]
Step 2: Differentiate
\[ \frac{dy}{dx} = -6x + 2 \]
Step 3: Substitute $x=-2$
\[ = -6(-2) + 2 = 12 + 2 = 14 \] Final Conclusion:
\[ = 14 \]
• $|x| = -x$ for $x<0$
Step 1: For $x=-2$ (negative)
\[ |x| = -x \] \[ y = \frac{3x^3 - 2x^2 + x}{-x} = -3x^2 + 2x - 1 \]
Step 2: Differentiate
\[ \frac{dy}{dx} = -6x + 2 \]
Step 3: Substitute $x=-2$
\[ = -6(-2) + 2 = 12 + 2 = 14 \] Final Conclusion:
\[ = 14 \]
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