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QJEEM23Ma98fc: JEE Main 2023 Mathematics - Class 12 JEEM23Ma98fc Solution: If A is the area in the first quadrant enclosed by the curve ,…

If A is the area in the first quadrant enclosed by the curve C:2x2−y+1=0\mathrm{C: 2 x^{2}-y+1=0} , the tangent to C\mathrm{C} at the point (1,3)(1,3) and the line x+y=1\mathrm{x}+\mathrm{y}=1 , then the value of 60 A60 \mathrm{~A} is _________.

Step-by-Step Verified Solution

JEE Main 2023 (Online) 11th April Evening Shift Mathematics - Area Under The Curves Question 40 English Explanation

y=2x2+1y=2 x^2+1

Tangent at (1, 3) : y−3=4(x−1)y-3=4(x-1)

y=4x−1y=4 x-1

A=∫01(2x2+1)dx−\mathrm{A}=\int\limits_0^1\left(2 \mathrm{x}^2+1\right) \mathrm{dx}- area of (ΔQOT)−(\Delta \mathrm{QOT})- area of (ΔPQR)+(\Delta \mathrm{PQR})+ area of (ΔQRS)(\Delta \mathrm{QRS})

A=(23+1)−12−98+940=1660\mathrm{A}=\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60}

∴\therefore 60A=1660 A=16

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