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

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Area Under Curves

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 60A is

Answer: 16

The curve is

y = 2 x^{2} + 1

, so

\frac{d y}{d x} = 4 x

and the tangent at

(1, 3)

has slope

4

, giving

y = 4 x - 1

. The parabola meets the line

x + y = 1

(0, 1)

, the tangent meets

x + y = 1

(\frac{2}{5}, \frac{3}{5})

, and the tangent touches the curve at

(1, 3)

. Over

0 \leq x \leq 1

the parabola is the upper boundary, while the lower boundary is the line

x + y = 1

[0, \frac{2}{5}]

and the tangent on

[\frac{2}{5}, 1]

. Hence

A = \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{5}{3} - \frac{8}{25} - \frac{27}{25} = \frac{4}{15}

. Therefore

60 A = 60 \cdot \frac{4}{15} = 16

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer