Given,
y=2x2+1
Now tangent to above equation at (1,3) is given by,
y−3=4(x−1)
⇒y=4x−1
Now plotting the diagram we get,

Required area of bounded region will be,
A=∫01(2x2+1)dx−area of△QOT−area of△PQR+area of△QRS
⇒A=(32+1)−21−89+409=4016
⇒60A=16
JEE Main 2023 — Mathematics Calculus
If A is the area in the first quadrant enclosed by the curve C:2x2−y+1=0, the tangent to C at the point (1,3) and the line x+y=1, then the value of 60A is................
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