M.C.Q (1 Marks)

Question 11 Mark

The area of the region bounded by the circle $x^2+y^2$ $=9$ in the first quadrant is :

Answer

(C) Area bounded by circle in first quadrant
$
=\frac{1}{4} \pi a^2=\frac{1}{4} \pi \times(3)^2=\frac{9 \pi}{4}
$
Correct option is (C) $\frac{9 \pi}{4}$.

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Question 21 Mark

Area bounded by curve $y^2=4 x, y$-axis and line $y=3$ is:

Answer

(B) $\frac{9}{4}$

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Question 31 Mark

Area of circle $x^2+y^2=4$ :

Answer

(C) $4 \pi$
Circle $x^2+y^2=4$
$
\begin{aligned}
\therefore \quad(x)^2+(y)^2 & =(2)^2 \\
\text { Area of circle } & =\pi(\text { radius })^2=\pi(2)^2 \\
& =4 \pi
\end{aligned}
$

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Question 41 Mark

The area of the region bounded by the curve $y=x^2$ and line $y=4$ is:

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Question 51 Mark

The area of the region bounded by the parabola $y=\sin ^2 x$, lines $x=\frac{\pi}{2}, x=\pi$ and $x$-axis is :

Answer

Hence the correct option is (B).

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Question 61 Mark

The area of the region between the curve $y^2=4 a x$, line $y=2 a$ and $y$-axis is :

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Question 71 Mark

The area bounded by the parabola $x^2=4 y$ and its latus rectum is:

Answer

(D)
$x^2=4 y$ and latus rectum $y=1$
The parabola is symmetrical about $y$-axis, hence required area $=2 \int_0^1 x d y$

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Question 81 Mark

The whole area of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is :

Answer

Ellipse $\frac{x^2}{25}+\frac{y^2}{16} =1$
$\Rightarrow \frac{x^2}{(5)^2}+\frac{y^2}{(4)^2} =1$

$\Rightarrow \frac{y^2}{(4)^2}=1-\frac{x^2}{(5)^2}=\frac{(5)^2-x^2}{(5)^2}$
$\Rightarrow y^2=\frac{(4)^2}{(5)^2}\left((5)^2-x^2\right)$
$\Rightarrow y=\frac{4}{5} \sqrt{(5)^2-x^2}$
$\text { Required area }= \text{ABCDA}$
$=4 \times \text { OABO }$
$\quad(\because \text { The ellipse is symmetrical about both the axes })$
$=4 \int_0^5 y d x$
$=4 \int_0^5 \frac{4}{5} \sqrt{(5)^2-x^2} d x$
$=\frac{16}{5}\left(\frac{1}{2} x \sqrt{(5)^2-x^2}+\frac{(5)^2}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right)_0^5$
$=\frac{16}{5}\left(0+\frac{(5)^2}{2} \times \frac{\pi}{2}-0-0\right)$
$=4 \times 5 \pi$
$=20 \pi \text { square units. }$
Hence the correct option is $(D)$.

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Question 91 Mark

The area of the region bounded by the curve $y=\sin x$ and $x$-axis when $0 \leq x \leq \pi$ will be :

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MATHS M.C.Q (1 Marks)

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