A) \[\frac{14}{3}\]
B) \[\frac{56}{3}\]
C) \[\frac{8}{3}\]
D) \[\frac{32}{3}\]
Correct Answer: C
Solution :
\[y\,=\,{{x}^{2}}\,-\,1\,\to \,(1)\] |
Tangent at P (2, 3) to (1) is \[y\,-\,3\,=\,4\,(x-2)\,as\,{{\left( \frac{dy}{dx} \right)}_{\text{p}}}\,=\,4\] |
Required area \[=\,\Delta \text{PTN}\,-\,\text{Area}\,\text{of}\,\text{curve}\,\text{A}\text{.P}\] on v-axis \[=\,\frac{1}{2}\,(\text{NT}\times \text{NP})\,-\,\int_{1}^{3}{\sqrt{y+1}dy}\] |
\[=\frac{1}{2}(8\times 2)-\frac{2}{3}\left[ {{(y+1)}^{3/2}} \right]_{-1}^{3}\] |
\[=\,8\,-\,\frac{16}{3}=\,\frac{8}{5}\]sq. units. |
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