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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

વિધેય \(f ( x )\) \(=|2 x+1|-3|x+2|+\left|x^{2}+x-2\right|, x \in R\) જયાં વિકલનીય ન હોય તેવા બિંદુઓની સંખ્યા ......... છે.

  1. A \(6\)
  2. B \(8\)
  3. C \(2\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

\(f(x)=|2 x+1|-3|x+2|+\left|x^{2}+x-2\right|\) \(=|2 x+1|-3|x+2|+|x+2||x-1|\) \(=|2 x+1|+|x+2|(|x-1|-3)\) Critical points are \(x=\frac{-1}{2},-2,-1\) but \(x=-2\) is making a zero. twice in product so, points of non differentability are \(x =\frac{-1}{2}\) and \(x =-1\)…
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