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

ધારો કે વિધેય \(f(x)=\left(x^2+1\right)\left|x^2-a x+2\right|+\cos |x|\) બે બિંદુઓ \(x=\alpha=2\) અને \(x=\beta\) પર વિકલનીય નથી. તો બિંદુ \((\alpha, \beta)\) નું રેખા \(12 x+5 y+10=0\) થી અંતર ___ છે.

  1. A 5
  2. B 4
  3. C 3
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

\(f(x)=\left(x^2+1\right)\left|x^2-a x+2\right|+\cos |x|\) Notice that \(\cos (-x)=\cos x=\cos |x|\) which means \(\cos |x|\) is differentiable everywhere in \(x \in R\) \(\Rightarrow f(x)\) can be non differentiable where \(\left|x^2-a x+2\right|\) \(=0\)…
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