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

ધારો કે \(f(x)\) એ બહુપદ્દી વિધેય છે કે જેથી  \(f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64\). તો ,  \(\lim _{x \rightarrow 1} \frac{f(x)}{x-1}\) ની કિમત ....... છે.

  1. A \(-15\)
  2. B \(-60\)
  3. C \(60\)
  4. D \(15\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-15\)

Step-by-step Solution

Detailed explanation

\(\operatorname{Lt}_{x \rightarrow 1} \frac{f(x)}{x-1}=f^{\prime}(1)(\) and \(f(1)=0)\) \(f(x)+f^{\prime}(x)+t^{\prime \prime}(x)=x^{5}+64\) \(f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)=5 x^{4}\)…
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