JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(y=\tan ^{-1}\left(\sec x^{3}-\tan x^{3}\right) \cdot \frac{\pi}{2} < x^{3} < \frac{3 \pi}{2}\), then
- A \(x y^{\prime \prime}+2 y^{\prime}=0\)
- B \(x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0\)
- C \(x^{2} y^{\prime \prime}-6 y+3 \pi=0\)
- D \(x y^{\prime \prime}-4 y^{\prime}=0\)
Answer & Solution
Correct Answer
(B) \(x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}\left(\sec x^{3}-\tan x^{3}\right)\) \(=\tan ^{-1}\left(\frac{1-\sin x^{3}}{\cos x^{3}}\right)\) \(=\tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}-x^{3}\right)}{\sin \left(\frac{\pi}{2}-x^{3}\right)}\right)\)…
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