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JEE Mains · Maths · STD 12 - 6. Application of derivatives

જો વક્ર  \(y = \cos \,\left( {x + f} \right),\, - 1\, - \pi \le x \le 1 + \pi ,\) નો સ્પર્શક \(x + 2y = k\) હોય તો  \(k\) મેળવો.

  1. A \(1\)
  2. B \(2\)
  3. C \(\frac{\pi }{4}\)
  4. D \(\frac{\pi }{2}\)
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Answer & Solution

Correct Answer

(D) \(\frac{\pi }{2}\)

Step-by-step Solution

Detailed explanation

Let \(y = \cos \left( {x + y} \right)\) \( \Rightarrow \frac{{dy}}{{dx}} = - \sin \left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right)\,\,\,....\left( 1 \right)\) Now, given equation of tangent is \(x + 2y = k\) \( \Rightarrow \) slope \( = \frac{{ - 1}}{2}\) So,…
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