JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q10

← JEE Main 2022 · 27 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 90

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_1$ is twice the area $A_2$ . Then the normal to the curve perpendicular to the line $2x-12y=15$ does NOT pass through the point.

A rising curve through (4, 2): A1 is the area under the curve from x = 4 to x, A2 the area between the curve and the y-axis from y = 2 to y.

(6, 21)

(8, 9)

(10, −4)

(12, −15)

Answer and worked solution

Official answer: C

Areas. $A_1+A_2=xy-8$ and $A_1=2A_2$ , so $A_1=\frac23(xy-8)$ . Differentiating, $y=\frac23(y+xy')$ , so $y=2xy'$ and $y^2=x$ (through $(4,2)$ ).

Normal of slope $-6$ . The tangent slope $\frac1{2y}=\frac16$ gives $(9,3)$ , and the normal is $y=-6x+57$ .

Check. $(10,-4)$ gives $-3\ne-4$ . (C)

Topics:Formation and applications Tangents and normals Area under a curve

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 10