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

ધારો કે \( \alpha, \beta \in \mathbb{R} \) એવા છે કે વિધેય
\( f(x)=\begin{cases}2\alpha(x^{2}-2)+2\beta x&,x<1\\ (\alpha+3)x+(\alpha-\beta)&,x\ge1\end{cases} \)
એ તમામ \( x \in \mathbb{R} \) માટે વિકલનીય છે. તો \( 34(\alpha+\beta) \) = ___ છે.

  1. A 84
  2. B 48
  3. C 36
  4. D 24
Verified Solution

Answer & Solution

Correct Answer

(B) 48

Step-by-step Solution

Detailed explanation

\(f(x)=\left\{\begin{array}{ccc}2 \alpha x^2+2 \beta x-4 \alpha & ; & x<1 \\ (\alpha+3) x+\alpha-\beta & ; & x \geq 1\end{array}\right.\) \(f\left(1^{+}\right)=2 \alpha-\beta+3, f\left(1^{-}\right)=-2 \alpha+2 \beta\)…
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