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

અહી વિધેય  \(f(x)=\left\{\begin{array}{cc}\frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & \text { if } x \neq 0 \\ 10 & \text {; if } x=0\end{array}\right.\) એ  \(x=0\) આગળ સતત હોય તો \(\alpha\) ની કિમંત મેળવો.

  1. A \(10\)
  2. B \(-10\)
  3. C \(5\)
  4. D \(-5\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-5\)

Step-by-step Solution

Detailed explanation

\(f(x)=\left\{\begin{array}{cc}\frac{\ln (1+5 x)-\ln (1+\alpha x)}{x} & ; x \neq 0 \\ 10 & ; x=0\end{array}\right.\) \(\lim _{x \rightarrow 0} \frac{\ln (1+5 x)-\ln (1+\alpha x)}{x}=10\) Using expension…
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