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JEE Mains · Maths · STD 12 - 9. differential equations

माना अवकल समीकरण \(x \tan \left(\frac{y}{x}\right) d y=\left(y \tan \left(\frac{y}{x}\right)-x\right)\) \(dx ,-1 \leq x \leq 1, y \left(\frac{1}{2}\right)=\frac{\pi}{6}\) का हल \(y = y ( x )\) है। तो वक्रों \(x =0, x =\frac{1}{\sqrt{2}}\) तथा \(y = y ( x )\) द्वारा ऊपरी आधे निर्देशांक तल में घिरे क्षेत्र का क्षेत्रफल है

  1. A \(\frac{1}{12}(\pi-3)\)
  2. B \(\frac{1}{6}(\pi-1)\)
  3. C \(\frac{1}{8}(\pi-1)\)
  4. D \(\frac{1}{4}(\pi-2)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{8}(\pi-1)\)

Step-by-step Solution

Detailed explanation

We have \(\tan \left(\frac{y}{x}\right)(x d y-y d x)=-x d x\) \(\tan \left(\frac{y}{x}\right)\left(\frac{x d y-y d x}{x^{2}}\right)=-\frac{x}{x^{2}} d x\) \(\int \tan \left(\frac{y}{x}\right)\left(d\left(\frac{y}{x}\right)\right)=\int-\frac{1}{x} d x\)…
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