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

જો વિધેય \(f(x)=\frac{1}{x} \log _{e}(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}) , \quad x<0\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad k \quad, \quad x=0\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} ,\,\,\, x>0\) એ  \(x=0\) આગળ સતત હોય તો \(\frac{1}{a}+\frac{1}{b}+\frac{4}{k}\) ની કિમંત મેળવો.

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

Answer & Solution

Correct Answer

(A) \(-5\)

Step-by-step Solution

Detailed explanation

If \(f(\mathrm{x})\) is continuous at \(\mathrm{x}=0, \mathrm{RHL}=\mathrm{LHL}=f(0)\) \(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} \cdot \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}+1}\) (Rationalisation)…
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