JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q76

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MathematicsSingle correct+4 / -1Question 76 of 90

The area (in sq. units) of the region described by $\{(x,y):y^2\le2x\text{ and }y\ge4x-1\}$ is :

\dfrac7{32}

\dfrac5{64}

\dfrac{15}{64}

\dfrac9{32}

Answer and worked solution

Official answer: D

The parabola $x=\frac{y^2}2$ and the line $x=\frac{y+1}4$ meet at $y=1$ and $y=-\frac12$ . The region lies between them:
$\int_{-1/2}^1\left(\frac{y+1}4-\frac{y^2}2\right)dy=\left[\frac{y^2}8+\frac y4-\frac{y^3}6\right]_{-1/2}^1=\frac9{32}.$

Topics:Area between curves

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 76