Assertion (A) & Reason (B) MCQ

Question 11 Mark

Assertion $(A) :$ The area of the region bounded by the curve $y^2=4 x$ and the line $x=3$ is $8 \sqrt{3}$ sq. units.
Reason $(R):$ The area is symmetric about $x$ and $y$ axes.

Answer

$(c) :$ We have, $y^2=4 x$ and $x=3$.

$\therefore $ Required area
$=2 \int_0^3|y| d x=2 \int_0^3 2 \sqrt{x} d x$
$=4\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^3$
$=\frac{8}{3}(3 \sqrt{3})=8 \sqrt{3} \text { sq. units }$
$\therefore \quad$ Assertion is true.
Clearly, reason is false.

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Question 21 Mark

Assertion $(A) :$ The area bounded by the curve $y=2 \cos x$ and the $x-$axis from $x=0$ to $x=2 \pi$ is $8$ sq. units.
Reason $(R) :$ Maximum value of the curve $y=2 \cos x$ is $2 .$

Answer

$(b):$ We have, $y=2 \cos x$
Let us draw the graph of $2 \cos x$ between $0$ to $2 \pi$.

$\therefore $ Required area
$=\int_0^{\pi / 2} 2 \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} 2 \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} 2 \cos x d x$
$=2[\sin x]_0^{\pi / 2}+\left|[2 \sin x]_{\pi / 2}^{3 \pi / 2}\right|+[2 \sin x]_{3 \pi / 2}^{2 \pi}$
$=2\left[\sin \frac{\pi}{2}-0\right]+\left|2\left[\sin \frac{3 \pi}{2}-\sin \frac{\pi}{2}\right]\right|+2\left[\sin 2 \pi-\sin \frac{3 \pi}{2}\right]$
$=2+2 \times 2+2=2+4+2=8 \text { sq. units }$

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Question 31 Mark

Assertion (A) : The area of the smaller region bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is $\frac{3}{2}(\pi-2)$ sq. units.
Reason (R) : Formula to calculate the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is $\frac{a b}{4}(\pi-2)$ sq. units.

Answer

(a): Clearly, reason is true.
We have, equation of ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and line $\frac{x}{3}+\frac{y}{2}=1$
$\therefore \quad$ Here, $a=3, b=2$
$\therefore \quad$ Required area $=\frac{a b}{4}(\pi-2)$
$=\frac{3 \times 2}{4}(\pi-2)$ sq. units $=\frac{3}{2}(\pi-2)$ sq. units

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Question 41 Mark

Assertion $(A):$ The area bounded by the curves $y^2=4 a^2(x-1)$ and lines $x=1$ and $y=4 a$ is $\frac{8 a}{3}$ sq. units.
Reason $(R) :$ The area enclosed between the parabola $y^2=49 x$ and its latus rectum $\frac{8 a^2}{3}$ sq. units.

Answer

$(b) :$ On solving $y^2=4 a^2(x-1)$ and $y=4 a$, we get $x=5$

$\therefore \text { Required area }=\int_1^5(4 a-2 a \sqrt{x-1}) d x$
$=\left[4 a x-2 a \cdot \frac{(x-1)^{3 / 2}}{3 / 2}\right]_1^5=\frac{16 a}{3} \text { sq. units }$
Clearly, reason is true.

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Question 51 Mark

Assertion (A): The area bounded by the parabola $y^2=4 a x$ and the line $x=a$ and $x=4 a$ is $\frac{56 a^2}{3}$ sq. units.
Reason (R) : The area bounded by the parabola $y^2=49 x$ and $y=m x$ is $8 a^2 / 3 m^3$ sq. units.

Answer

Required area $=2 \int_a^{4 a} \sqrt{4 a x} d x=4 \sqrt{a}\left[\frac{x^{3 / 2}}{3 / 2}\right]_a^{4 a}$
$=\frac{8}{3} \sqrt{a}\left(8 a^{3 / 2}-a^{3 / 2}\right)=\frac{56 a^2}{3}$ sq. units
Clearly, reason is true.

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MATHS Assertion (A) & Reason (B) MCQ

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