JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 18)
If A is the area in the first quadrant enclosed by the curve $$\mathrm{C: 2 x^{2}-y+1=0}$$, the tangent to $$\mathrm{C}$$ at the point $$(1,3)$$ and the line $$\mathrm{x}+\mathrm{y}=1$$, then the value of $$60 \mathrm{~A}$$ is _________.
Answer
16
Explanation
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$$ y=2 x^2+1 $$
Tangent at (1, 3) : $y-3=4(x-1)$
$$ y=4 x-1 $$
$\mathrm{A}=\int\limits_0^1\left(2 \mathrm{x}^2+1\right) \mathrm{dx}-$ area of $(\Delta \mathrm{QOT})-$ area of $(\Delta \mathrm{PQR})+$ area of $(\Delta \mathrm{QRS})$
$$ \mathrm{A}=\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60} $$
$$ \therefore $$ $$ 60 A=16 $$
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