question_answer

The area (in sq. units) in the first quadrant bounded by the parabola\[y={{x}^{2}}+1,\] the tangent to it at the point (2, 5) and the coordinate axes is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

A) 8/3                                           

B) 187/24

C) 14/3                             

D)   37/24

Correct Answer: D

Solution :

Given parabola is \[y={{x}^{2}}+1\] Tangent to the parabola at (2, 5) is\[\frac{1}{2}(y+5)=2x+1\]\[\Rightarrow \]\[y+5=4x+2\]\[\Rightarrow \]\[4x-y=3\]                                      ?(i) Eq. (i) cuts the x-axis at\[\left( \frac{3}{4},0 \right).\] Now, the required area= \[=\int\limits_{0}^{2}{({{x}^{2}}+1)}dx-\frac{1}{2}\left( \frac{5}{4} \right)5\] \[=\left[ \frac{{{x}^{3}}}{3}+x \right]_{0}^{2}-\frac{25}{8}=\frac{37}{24}\]