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

જો વિધેય \(f(x)\, = \frac{1}{x} - \frac{{k - 1}}{{{e^{2x}} - 1}}\), \(x\, \ne \,0\) એ \(x = 0\) આગળ સતત હોય તો જોડ \((k,f(0))\) = . . . 

  1. A \((3, 1 )\)
  2. B \((3, 2)\)
  3. C \(\left( {\frac{1}{3},2} \right)\)
  4. D \((2, 1)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((3, 1 )\)

Step-by-step Solution

Detailed explanation

if the funtion is continuous at \(x=0\), then \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) will exist and \(f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right)\) Now,…
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