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

ધારોકે \(k\) અને \(m\) એવી ધન વાસ્તવિક સંખ્યાઓ છે કે જેથી વિધેય \(\quad f ( x )=\left\{\begin{array}{cc}3 x ^2+ k \sqrt{ x +1}, & 0< x <1 \\ mx ^2+ k ^2, & x \geq 1\end{array}\right.\) એ પ્રત્યેક \(x > 0\) માટે વિકલનીય છે, તો \(\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=........\)

  1. A \(309\)
  2. B \(310\)
  3. C \(311\)
  4. D \(312\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(309\)

Step-by-step Solution

Detailed explanation

function is differentiable \(\forall x < 0\) so \(\quad f \left(1^{-}\right)= f (1)\) \(3+\sqrt{2} k = m + k ^2......(1)\) and \(\quad f _{+}^1\left(1^{-}\right)= f _{-}^1\left(1^{+}\right)\) \(\left.2 m x\right|_{x=1}=6 x+\left.\frac{k}{2 \sqrt{x+1}}\right|_{x=1}\)…
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