Let S = (0, 2p) – ${\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}} .$ Let y = y(x), $x \in S,$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}, y (\frac{π}{4}) = \frac{1}{2} .$ If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $y = \sqrt[]{2} s i n x i s \frac{k π}{12},$ then k is equal to………..

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Raj Pandey | Contributor-Level 9

2 weeks ago

$\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}$

$d y = \frac{s e c^{2} x d x}{{(1 + t a n x)}^{2}}$

$\Rightarrow y = - \frac{1}{1 + t a n x} + c$

when $x = \frac{π}{4}, y = \frac{1}{2}$ gives c = 1

so $x + \frac{π}{4} = \frac{5 π}{6} o r \frac{13 π}{6} \Rightarrow x = \frac{7 π}{12} o r \frac{23 π}{12}$

sum of all solutions =

$π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{42 π}{12}$

Hence k = 42

<math> <mrow> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>x</mi> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mi>d</mi> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mi>s</mi> <mi>e</mi> <msup> <mrow> <mi>c</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>x</mi> <mi>d</mi> <mi>x</mi> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>+</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>y</mi> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>x</mi> </mrow> </mfrac> <mo>+</mo> <mi>c</mi> </mrow> </math> when   <math> <mrow> <mi>x</mi> <mo>=</mo> <mfrac> <mrow> <mi>π</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>, </mo> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> gives c = 1so <math> <mrow> <mi>x</mi> <mo>+</mo> <mfrac> <mrow> <mi>π</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <mi>π</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mfrac> <mrow> <mn>1</mn> <mn>3</mn> <mi>π</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>⇒</mo> <mi>x</mi> <mo>=</mo> <mfrac> <mrow> <mn>7</mn> <mi>π</mi> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mfrac> <mrow> <mn>2</mn> <mn>3</mn> <mi>π</mi> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> sum of all solutions = <math> <mrow> <mi>π</mi> <mo>+</mo> <mfrac> <mrow> <mn>7</mn> <mi>π</mi> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>2</mn> <mn>3</mn> <mi>π</mi> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>2</mn> <mi>π</mi> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> Hence k = 42

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