Question:
If \( f(x) = |x^2 + x - 6| \) is not differentiable at \( x = a \) and \( x = b \), then \( a^2 + b^2 = \)
If \( f(x) = |x^2 + x - 6| \) is not differentiable at \( x = a \) and \( x = b \), then \( a^2 + b^2 = \)
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Modulus functions are non-differentiable at points where inside expression = 0.
Updated On: Apr 21, 2026
- \(11 \)
- \(14 \)
- \(12 \)
- \(13 \)
- \(16 \)
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The Correct Option is D
Solution and Explanation
Concept:
Non-differentiable where expression inside modulus = 0.
Step 1: Solve quadratic.
\[ x^2 + x - 6 = 0 \] \[ (x+3)(x-2)=0 \Rightarrow x = -3, 2 \]
Step 2: Compute sum.
\[ a^2 + b^2 = (-3)^2 + 2^2 = 9 + 4 = 13 \]
Step 1: Solve quadratic.
\[ x^2 + x - 6 = 0 \] \[ (x+3)(x-2)=0 \Rightarrow x = -3, 2 \]
Step 2: Compute sum.
\[ a^2 + b^2 = (-3)^2 + 2^2 = 9 + 4 = 13 \]
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