JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}\) where \(a>b>0,\) then \(\frac{d x}{d y}\) at \(\left(\frac{\pi}{4}, \frac{\pi}{4}\right)\) is
- A \(\frac{a-b}{a+b}\)
- B \(\frac{a+b}{a-b}\)
- C \(\frac{2 a+b}{2 a-b}\)
- D \(\frac{a-2 b}{a+2 b}\)
Answer & Solution
Correct Answer
(B) \(\frac{a+b}{a-b}\)
Step-by-step Solution
Detailed explanation
\((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}\) \(\Rightarrow a^{2}-\sqrt{2} a b \cos y+\sqrt{2} a b \cos x\) \(-2 b^{2} \cos x \cos y=a^{2}-b^{2}\) Differentiating both sides: \(0-\sqrt{2} ab \left(-\sin y \frac{ dy }{ dx }\right)+\sqrt{2} ab (-\sin x )\)…
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