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JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities

Let  \({f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)\) where \(x \in R\;\) and \(k \ge 1\). Then \({f_4}\left( x \right) - {f_6}\left( x \right) \) is equals

  1. A \(\frac{1}{4}\)
  2. B \(\frac{1}{{12}}\)
  3. C \(\frac{1}{6}\)
  4. D \(\frac{1}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{{12}}\)

Step-by-step Solution

Detailed explanation

\({f_4}(x)\, - \,{f_6}(x)\, = \frac{1}{4}({\sin ^4} + {\cos ^4}x) - \frac{1}{6}({\sin ^6} + {\cos ^6}x)\) \( = \frac{1}{4}(1 - 2{\sin ^2}{\cos ^2}x) - \frac{1}{6}(1 - 3{\sin ^2}{\cos ^2}x)\) \( = \frac{1}{4} - \frac{1}{6} = \frac{{3 - 2}}{{12}} = \frac{1}{{12}}\)