Let y = y(x) satisfies the equation $\frac{dy}{dx} - | A | = 0$ , for all x > 0, where $A = [\begin{matrix} y & sinx & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \frac{1}{x} \end{matrix}]$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is

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$\frac{π}{2} - \frac{4}{π}$

$\frac{3 π}{2} - \frac{1}{π}$

$\frac{π}{2} + \frac{4}{π}$

$\frac{π}{2} - \frac{1}{π}$

answer is A.

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Detailed Solution

$\begin{matrix} | A | = |\begin{array}{ccc} y & \sin x & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \frac{1}{x} \end{array}| \\ = y (- \frac{1}{x}) + \sin x (2) + 1 (+ 2) = - \frac{y}{x} + 2 \sin x + 2 \end{matrix}$

Now, $\frac{dy}{dx} - | A | = 0$ gives

$\frac{dy}{dx} + \frac{y}{x} - 2 \sin x - 2 = 0$

or $\frac{dy}{dx} + \frac{y}{x} = 2 \sin x + 2$

$\begin{array}{l} IF = e^{\int \frac{1}{x} dx} = e^{\log x} = x \\ ∴ y \cdot IF = \int (2 \sin x + 2) \cdot IFdx \\ y \cdot x = \int (2 xsin x + 2 x) dx \\ yx = 2 [- xcos x + \sin x] + x^{2} + c . . . (i) \end{array}$

Given, $y (π) = π + 2$ , putting in Eq. (i)

$\begin{array}{l} (π + 2) π = 2 (- πcos π + \sin π) + π^{2} + c \\ \Rightarrow π^{2} + 2 π = π^{2} + 2 π + c \Rightarrow c = 0 \end{array}$

From Eq. (i),

$yx = x^{2} - 2 xcos x + 2 \sin x$

Put $x = \frac{π}{2}$

$\begin{array}{l} y \cdot (\frac{π}{2}) = {(\frac{π}{2})}^{2} - 2 (\frac{π}{2}) \cos \frac{π}{2} + 2 \sin (\frac{π}{2}) \\ \Rightarrow y \cdot (\frac{π}{2}) = \frac{π}{2} - 2 (0) + \frac{2}{π / 2} \Rightarrow y \cdot (\frac{π}{2}) = \frac{π}{2} + \frac{4}{π} \end{array}$