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JEE Mains · Maths · STD 12 - 6. Application of derivatives
If an equation of a tangent to the curve, \(y = \cos \,\left( {x + f} \right),\, - 1\, - \pi \le x \le 1 + \pi ,\) is \(x + 2y = k\) then \(k\) is equal to
- A \(1\)
- B \(2\)
- C \(\frac{\pi }{4}\)
- D \(\frac{\pi }{2}\)
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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