jee-advanced 2004 Q11

jee-advanced · India · mains Differential equations Solving Separable DEs with Initial Conditions

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.

The value of $x$ for which $\left. \sin \left( \cot ^ { - 1 } ( 1 + x ) \right) = \cos \left( \tan ^ { - 1 } x \right) \right)$ is

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.

Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$\\
or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$\\
Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$\\
$\frac { d Y } { d X } = X + \frac { Y } { X }$\\
$\frac { d Y } { d X } - \frac { Y } { X } = X$\\
\includegraphics[max width=\textwidth, alt={}, center]{bf0af8eb-2b9c-49f0-8222-a9220127f02f-3_443_521_2175_1031}\\
I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$\\
$\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$.\\
It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$.\\
So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$\\
$\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$.\\
⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.\\

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