JEE Main & Advanced
JEE Main Paper (Held On 11-Jan-2019 Evening)
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= \[=\int\limits_{0}^{2}{({{x}^{2}}+1)}dx-\frac{1}{2}\left( \frac{5}{4} \right)5\] \[=\left[ \frac{{{x}^{3}}}{3}+x \right]_{0}^{2}-\frac{25}{8}=\frac{37}{24}\]