NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

If the area bounded by $y = x^2$ and $y = x$ is $A$ sq units, then the area bounded by $y = x^2$ and $y = 1$ is:

Choose the correct answer:

Explanation

1. Finding $A$ (Area between $y = x^2$ and $y = x$ ):

Points of intersection: $x^2 = x \implies x(x-1) = 0 \implies x = 0, 1$ .

A = \int_{0}^{1} (x - x^2) dx

A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}

A = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

2. Finding the area bounded by $y = x^2$ and $y = 1$ :

Points of intersection: $x^2 = 1 \implies x = -1, 1$ .

Since the curve is symmetric about the y-axis:

\text{Area} = 2 \int_{0}^{1} (1 - x^2) dx

\text{Area} = 2 \left[ x - \frac{x^3}{3} \right]_{0}^{1}

\text{Area} = 2 \left[ 1 - \frac{1}{3} \right] = 2 \left( \frac{2}{3} \right) = \frac{4}{3}

3. Expressing the result in terms of $A$ :

We know $A = \frac{1}{6}$ , so $8A = 8 \times \frac{1}{6} = \frac{4}{3}$ .

Alternatively, let's check the options using $A = \frac{1}{6}$ :

(a) $2(\frac{1}{6}) + 1 = \frac{1}{3} + 1 = \frac{4}{3}$
(b) $2(\frac{1}{6}) = \frac{1}{3}$
(c) $2(\frac{1}{6}) + 2 = \frac{7}{3}$
(d) $\frac{1}{6} + 2 = \frac{13}{6}$

Correct Option: (a) $(2A + 1)$ sq units