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

જો \(f(x)\, = {x^2} - x + 5,\,\,x > \frac{1}{2},\) અને \(g(x)\) એ તેનું વ્યસ્ત વિધેય છે તો  \(g'(7)\) મેળવો.

  1. A \(-\frac {1}{3}\)
  2. B \(\frac {1}{13}\)
  3. C \(\frac {1}{3}\)
  4. D \(-\frac {1}{13}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac {1}{3}\)

Step-by-step Solution

Detailed explanation

\(f\left( x \right) = y = {x^2} - x + 5\) \({x^2} - x + \frac{1}{4} - \frac{1}{4} + 5 = y\) \({\left( {x - \frac{1}{2}} \right)^2} + \frac{{19}}{4} = y\) \({\left( {x - \frac{1}{2}} \right)^2}\,\, = y - \frac{{19}}{4}\) \(x - \frac{1}{2} = \pm \sqrt {y - \frac{{19}}{4}} \)…
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