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

વિધેય \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11\) કે જ્યાં \(\mathrm{x} \in[0,1]\) માં મ્ધયકમાન પ્રમેય અનુસાર \(c\) ની કિમંત મેળવો.

  1. A \(\frac{2}{3}\)
  2. B \(\frac{\sqrt{7}-2}{3}\)
  3. C \(\frac{4-\sqrt{5}}{3}\)
  4. D \(\frac{4-\sqrt{7}}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{4-\sqrt{7}}{3}\)

Step-by-step Solution

Detailed explanation

\(f(0)=11\) \(f(1)=16\) \(\frac{\mathrm{f}(1)-\mathrm{f}(0)}{1-0}=3 \mathrm{c}^{2}-8 \mathrm{c}+8\) \(\Rightarrow 3 c^{2}-8 c+8=5\) \(\Rightarrow 3 c^{2}-8 c+3=0\) \(c \in[0,1] \Rightarrow c=\frac{4-\sqrt{7}}{3}\)
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