JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Given that $y = 2x^2 + 1$ which is a parabola.

Differentiating: $\frac{dy}{dx} = 4x$

$\left. \frac{dy}{dx} \right|_{(1,3)} = 4$

$\Rightarrow$ Eqn. of tangent at $(1, 3)$ is $y - 3 = 4(x - 1)$

$\Rightarrow y = 4x - 1$ ...(i)

Solving line (i) with $x + y = 1$ ...(ii)

we get $x = \frac{2}{5}$ and $y = \frac{3}{5}$

which can be represented as

So, the required area is
$A = \int_{0}^{1} y_{Parabola} d x - \int_{0}^{2/5} y_{Line (ii)} d x - \int_{2/5}^{1} y_{Line (i)} d x$

$= \int_{0}^{1} (2 x^{2} + 1) d x - \int_{0}^{2/5} (1 - x) d x - \int_{2/5}^{1} (4 x - 1) d x$

$= [\frac{2 x ^{3}}{3} + x]_{0}^{1} - [x - \frac{x ^{2}}{2}]_{0}^{2/5} - [\frac{4 x ^{2}}{2} - x]_{2/5}^{1}$

and $15 A = 4$
or $60 A = 16$

Choose the correct answer:

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