question_answer

The area (in sq. units) bounded by the parabola \[y={{x}^{2}}-1\], the tangent at the point (2, 3) to it and the y-axis is: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) \[\frac{56}{3}\]

B) \[\frac{32}{3}\]

C) \[\frac{8}{3}\]

D) \[\frac{14}{3}\]

Correct Answer: C

Solution :

  \[y={{x}^{2}}-1\] \[y+1)={{x}^{2}}\,\,\,\,\,\,\,\,\,\frac{dy}{dx}\,\,=\,\,2x\] \[{{\left( \frac{dy}{dx} \right)}_{(2,\,\,3)}}\,\,\,\,\,\,\,=\,\,\,\,\,4\] \[(y-3)=4(x-2)\] \[y-\text{ }3=4x\text{ }-\text{ }8\] \[y=\text{ }4x\text{ }-\text{ }5\] \[\int\limits_{-5}^{3}{\left( \frac{y+5}{4} \right)}\,dy\,\,-\,\,\,\,\,\int\limits_{-1}^{3}{\sqrt{y+1\,}dy}\] \[=\,\,\frac{1}{4}\,\left[ \frac{{{y}^{2}}}{2}+5y \right]{{\,}^{3}}_{-5}\,\,-\,{{\left[ 2\frac{{{(y+1)}^{3/2}}}{3} \right]}^{3}}_{-1}\] \[=\,\,\frac{1}{4}\,\left[ \left( \frac{9}{2}+15 \right)-\,\left( \frac{25}{2}\,-\,25 \right) \right]\,\,-\,\frac{2}{3}\,\,[{{(4)}^{3/2}}]\] \[=\,\,\,\,\,\,\frac{1}{4}\,\left[ \frac{39}{2}\,\,+\,\,\frac{25}{2} \right]\,\,-\,\,\frac{2}{3}\,[8]\] \[=\,\,\,\,\,\frac{1}{4}\,\,\times \,\frac{64}{2}\,-\,\,\frac{16}{3}\,\] \[=\,\,\,\,8\,\,-\,\,\frac{16}{3}\,\,=\,\,\frac{8}{3}\]