Question:medium

The function \( f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos|x| \) is non-differentiable at

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If modulus is multiplied by zero at a point, cusp may vanish — check carefully.
Updated On: Jun 17, 2026
  • \( -1 \)
  • \( 0 \)
  • \( 1 \)
  • \( 2 \)
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The Correct Option is D

Solution and Explanation

To determine where the function \( f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos|x| \) is non-differentiable, we need to check the points where:

  1. The term \( |x^2 - 3x + 2| \) changes its form, which is typically at the roots or critical points of the expression inside the absolute value.
  2. The derivative of \( \cos|x| \) is discontinuous, which happens at the points where the absolute value function changes its behavior.

Let's analyze each part separately:

1. Analysis of \( |x^2 - 3x + 2| \):

Factorizing the quadratic expression:

\(x^2 - 3x + 2 = (x - 1)(x - 2)\)

The expression changes sign at \( x = 1 \) and \( x = 2 \). Hence, these are potential points of non-differentiability.

2. Analysis of \( \cos|x| \):

The derivative of \( \cos|x| \) with respect to \( x \) will involve the derivative of the inner function, which is discontinuous at \( x = 0 \).

Points of Interest:

  • \( x = 1 \): Here, \( (x - 1)(x - 2) \) changes from negative to positive, indicating a possible non-differentiable point.
  • \( x = 2 \): Another point where \( (x - 1)(x - 2) \) changes from positive to negative, indicating another possible point of non-differentiability.
  • \( x = 0 \): The derivative of \( \cos|x| \) is discontinuous at this point, but checking the full expression will confirm its differentiability as \( |x^2 - 3x + 2| \) is continuous at \( x = 0 \).

Conclusion:

Upon examining these points in the expression \( f(x) \), we find:

  • The function is continuous across \( x = 0 \) due to the behavior of \( |x^2 - 3x + 2| \).
  • Discontinuity at \( x = 2 \) due to the change in \( (x - 1)(x - 2) \) influencing \( |x^2 - 3x + 2| \).

Therefore, the function \( f(x) \) is non-differentiable at \( x = 2 \).

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