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if f(x)=(x^2-4)|(x^3-6x^2+11x-6)|+x/(1+|...

if `f(x)=(x^2-4)|(x^3-6x^2+11x-6)|+x/(1+|x|)`then set of points at which the function if non differentiable is

A

{-2,2,1,3}

B

{-2,0,3}

C

{-2,2,0}

D

{1,3}

Text Solution

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The correct Answer is:
To determine the set of points at which the function \( f(x) = (x^2 - 4) |x^3 - 6x^2 + 11x - 6| + \frac{x}{1 + |x|} \) is non-differentiable, we will follow these steps: ### Step 1: Identify the components of the function The function consists of two main parts: 1. \( (x^2 - 4) |x^3 - 6x^2 + 11x - 6| \) 2. \( \frac{x}{1 + |x|} \) ### Step 2: Analyze the absolute value function The expression \( |x^3 - 6x^2 + 11x - 6| \) is non-differentiable at points where \( x^3 - 6x^2 + 11x - 6 = 0 \). We need to find the roots of the polynomial: \[ x^3 - 6x^2 + 11x - 6 = 0 \] ### Step 3: Factor the polynomial Using the Rational Root Theorem or synthetic division, we can find the roots. Testing \( x = 1 \): \[ 1^3 - 6(1^2) + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \] Thus, \( x = 1 \) is a root. We can factor the polynomial as follows: \[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \] Now, we can factor \( x^2 - 5x + 6 \): \[ x^2 - 5x + 6 = (x - 2)(x - 3) \] Thus, we have: \[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \] ### Step 4: Identify critical points The critical points where \( |x^3 - 6x^2 + 11x - 6| \) is non-differentiable are \( x = 1, 2, 3 \). ### Step 5: Check the effect of \( (x^2 - 4) \) The term \( (x^2 - 4) \) becomes zero at \( x = 2 \) and \( x = -2 \). At \( x = 2 \), since \( (x^2 - 4) \) is multiplied by \( |x^3 - 6x^2 + 11x - 6| \), it neutralizes the non-differentiability at this point. ### Step 6: Analyze \( \frac{x}{1 + |x|} \) The term \( \frac{x}{1 + |x|} \) is non-differentiable at \( x = 0 \) due to the absolute value in the denominator. ### Step 7: Compile the non-differentiable points From our analysis, the points where \( f(x) \) is non-differentiable are: - \( x = 1 \) (due to the absolute value) - \( x = 3 \) (due to the absolute value) - \( x = 0 \) (due to the absolute value in the denominator) ### Final Answer The set of points at which the function is non-differentiable is: \[ \{0, 1, 3\} \]
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