KCET · Maths · Differentiation
\(\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]\) is
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(-\frac{1}{2}\)
- D \(-\frac{3}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right\}\) put \(x=2 \cos \theta\)
\(\Rightarrow \frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\left.\frac{2+\cos \theta \cdot 2}{2-\cos \theta \cdot 2}\right)}\right\}\right.\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right\}\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}}\right)\right\}\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1}(\cot \theta / 2)\right)\right\}\) \(=\frac{d}{d x}\left\{\cos ^{2} \frac{\theta}{2}\right\}=\frac{1}{2} \cdot \frac{d}{d x}(1+\cos \theta)\) \(=\frac{d}{d x}\left(\frac{1}{2}+\frac{1}{2} \cdot \frac{x}{2}\right)\) \(=\frac{d}{d x}\left(\frac{1}{2}+\frac{x}{4}\right)\) \(=(0+1 / 4)=1 / 4\)
\(\Rightarrow \frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\left.\frac{2+\cos \theta \cdot 2}{2-\cos \theta \cdot 2}\right)}\right\}\right.\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right\}\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}}\right)\right\}\)
\(=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1}(\cot \theta / 2)\right)\right\}\) \(=\frac{d}{d x}\left\{\cos ^{2} \frac{\theta}{2}\right\}=\frac{1}{2} \cdot \frac{d}{d x}(1+\cos \theta)\) \(=\frac{d}{d x}\left(\frac{1}{2}+\frac{1}{2} \cdot \frac{x}{2}\right)\) \(=\frac{d}{d x}\left(\frac{1}{2}+\frac{x}{4}\right)\) \(=(0+1 / 4)=1 / 4\)
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