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JEE Main202225 Jul 2022Evening ShiftMathematicsLimitsActual

lim x → π 4 8 2 - cos x + sin x 7 2 - 2 sin 2 x is equal to

Options

  1. A14
  2. B7
  3. C14 2
  4. D7 2

Correct answer

A. 14

Step-by-step solution

lim x → π 4 8 2 - cos x + sin x 7 2 - 2 sin 2 x       0 0 form = lim x → π 4 - 7 cos x + sin x 6 - sin x + cos x - 2 2 cos 2 x (using L'Hospital Rule) = lim x → π 4 56 cos x - sin x 2 2 cos 2 x   0 0   form = lim x → π 4 - 56 sin x + cos x - 4 2 sin 2 x (using L'Hospital Rule) = 56 2 4 2 = 14

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