datavorous/entrance-exam-dataset huggingface dataset:
What is the area bounded by the curves?
Consider the curves $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|-1$ and $\mathrm{g}(\mathrm{x})=\left\{\begin{array}{ll}\frac{3 \mathrm{x}}{2}, & \mathrm{x}>0 \\ 2 \mathrm{x}, & \mathrm{x} \leq 0\end{array} \quad[2016-\mathrm{I}]\right.$

Options:
A $\frac{17}{6}$ square units
B $\frac{8}{3}$ square units
C 2 square units
D $\frac{1}{3}$ square unit

The correct answer is:
$\frac{8}{3}$ square units

Solution:
$\begin{aligned} \text { Area } &=\int_{-1}^{2}[g(x)-f(x)] d x \\ &=\int_{-1}^{0}[g(x)-f(x)] d x+\int_{0}^{2}[g(x)-f(x)] d x \end{aligned}$
$\begin{aligned} &=\int_{-1}^{0}\left(2 x+x^{2}+1\right) d x+\int_{0}^{2}\left[\frac{3 x}{2}-x^{2}+1\right] d x \\ &=I_{1}+I_{2} \\ I_{1}=&\left(\frac{2 x^{2}}{2}+\frac{1}{3} x^{3}+x\right)_{-1}^{0} \\=&(0+0+0)-\left((-1)^{2}+\frac{1}{3}(-1)^{3}+(-1)\right) \\=&-\left(1-\frac{1}{3}-1\right)=\frac{1}{3} \\ I_{2}=& \int_{0}^{2}\left[\frac{3 x}{2}-x^{2}+1\right] d x \\=&\left[\frac{3}{2} \frac{x^{2}}{2}-\frac{1}{3} x^{3}+x\right]_{0}^{2} \\=&\left(\frac{3}{4} \times 4-\frac{1}{3} \times 8+2\right)-(0) \\=& 5-\frac{8}{3}=\frac{7}{3} \\ \text { Area } &=I_{1}+I_{2}=\frac{1}{3}+\frac{7}{3} \end{aligned}$
Area $=\frac{8}{3}$ square units.

Topic: Mathematics, Area Under Curves, NDA, NDA 2016 (Phase 1)
What is the area bounded by the curves $|y|=1-x^{2}$ ?

Options:
A $\frac{4}{3}$ square units
B $\frac{8}{3}$ square units
C 4 square units
D $\frac{16}{3}$ square units

The correct answer is:
$\frac{8}{3}$ square units

Solution:
Since $|y|=\left\{\begin{array}{cc}y & y>0 \\ -y & y < 0 \\ 0 & y=0\end{array}\right.$
For $y>0 \Rightarrow y=1-x^{2}$
For $y < 0 \Rightarrow y=x^{2}-1$
For $y=0 \Rightarrow x=\pm 1$
So area under the curve $=4 \times$ Area under the region OABO (symmetry)
$\begin{aligned}=& 4 \times \int_{0}^{1} 1-x^{2} \mathrm{~d} \mathrm{x} \\ & \left.=4 \times\left[x-\frac{x^{3}}{3}\right]_{0}^{1}\right] \end{aligned}$
$\quad=4\left(1-\frac{1}{3}\right)=4 \times \frac{2}{3}=\frac{8}{3}$ sq. units

Topic: Mathematics, Area Under Curves, NDA, NDA 2016 (Phase 2)
What is the area bounded by the curves $y=e^{x}$, $y=e^{-x}$ and the straight line $x=1 ?$

Options:
A $\left(e+\frac{1}{e}\right)$ sq unit
B $\left(e-\frac{1}{e}\right)$ sq unit
C $\left(e+\frac{1}{e}-2\right)$ sq unit
D $\left(e-\frac{1}{e}-2\right)$ sq unit

The correct answer is:
$\left(e+\frac{1}{e}-2\right)$ sq unit

Solution:
Given equations of curves are $y=e^{x}$ and $y=e^{-x}$. $\Rightarrow e^{x}=\frac{1}{e^{x}} \quad \Rightarrow e^{2 x}=e^{0} \Rightarrow x=0$
Also, equation of straight line gives $x=1$
$\begin{aligned} \text { Required area } &=\int_{0}^{1}\left(e^{x}-e^{-x}\right) d x \\ &=\left[e^{x}+e^{-x}\right]_{0}^{1}=e+e^{-1}-\left(e^{0}+e^{-0}\right) \\ &=\left(e+\frac{1}{e}-2\right) \text { sq unit } \end{aligned}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2011 (Phase 1)
What is the area bounded by the lines $x=0, y=0$ and $x+$ ] $2=0 ?$

Options:
A $\frac{1}{2}$ square unit
B 1 square unit
C 2 square units
D 4 square units

The correct answer is:
2 square units

Solution:
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \times 2 \times 2=2$ square units

Topic: Mathematics, Area Under Curves, NDA, NDA 2013 (Phase 2)
What is the area bounded by $y=\tan x, y=0$ and $x=\frac{\pi}{4}$ ?

Options:
A $\ln 2$ square units
B $\frac{\ln 2}{2}$ square units
C 2 ( en 2 )square units
D Nonc of these

The correct answer is:
$\frac{\ln 2}{2}$ square units

Solution:
Required area $=\int_{0}^{\frac{\pi}{4}} \tan x \mathrm{dx}$
$\quad=\ln |\sec x|_{0}^{\frac{\pi}{4}}=\ln \sqrt{2}=\frac{\ln 2}{2}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2013 (Phase 2)
What is the area bounded hy the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$
$(x, y \geq 0)$ and the coordinate axes?

Options:
A $\frac{5 a^{2}}{6}$
B $\frac{a^{2}}{3}$
C $\frac{a^{2}}{2}$
D $\frac{a^{2}}{6}$

The correct answer is:
$\frac{a^{2}}{6}$

Solution:
Area bounded by curve $\sqrt{x}+\sqrt{y}=\sqrt{a}(\mathrm{x}, \mathrm{y} \geq 0$ ) and coordinate axes is
$=\int_{0}^{a} y d x=\int_{0}^{a} a+x-2 \sqrt{a} \sqrt{x} d x$
$\quad(\because \sqrt{y}=\sqrt{a}-\sqrt{x} \Rightarrow y=a+x-2 \sqrt{a} \sqrt{x})$
$=a x+\frac{x^{2}}{2}=\left.\frac{2 \sqrt{a} x^{3 / 2}}{3 / 2}\right|_{0} ^{a}$
$=a^{2}+\frac{a^{2}}{2}-\frac{4}{3} a^{2}=\frac{3 a^{2}}{2}-\frac{4}{3} a^{2}$
$=\frac{9 a^{2}-8 a^{2}}{6}=\frac{a^{2}}{6}$ sq. unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2011 (Phase 2)
What is the area enclosed by the curve $2 \mathrm{x}^{2}+\mathrm{y}^{2}=1$ ?

Options:
A $2 \pi$
B $\pi$
C $\frac{\pi}{2}$
D $\frac{\pi}{\sqrt{2}} \quad$

The correct answer is:
$\frac{\pi}{\sqrt{2}} \quad$

Solution:
Given equation of curve $2 x^{2}+y^{2}=1$ is an ellipse which can be written as $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{1}=1$
Area of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $A=\pi$ ab sq unit
Here, $\mathrm{a}=\frac{1}{\sqrt{2}}, \mathrm{~b}=1$.
: Required area $=\pi \cdot \frac{1}{\sqrt{2}} \cdot 1=\frac{\pi}{\sqrt{2}}$ sq unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2017 (Phase 2)
What is the area enclosed by the equation
$x^{2}+y^{2}=2 ?$

Options:
A $4 \pi$ square units
B $2 \pi$ square units
C $4 \pi^{2}$ square units
D 4 square units

The correct answer is:
$2 \pi$ square units

Solution:
Given equation of circle is $x^{2}+y^{2}=2$ $\Rightarrow y=\sqrt{2-x^{2}}$
Required area $=4 \times$ Area of shaded portion $-4 \int_{0}^{\sqrt{2}} \sqrt{2-x^{2}} d x$
$\sqrt{2-x^{2}} d x$
Let $x=\sqrt{2} \sin t \Rightarrow t=\sin ^{-1}\left(\frac{x}{\sqrt{2}}\right)$
$d x=\sqrt{2} \cos t \cdot d t$
$\therefore \int \sqrt{2-x^{2}} \cdot d x=\int \sqrt{2-2 \sin ^{2} t} \cdot \sqrt{2} \cos t \cdot d t$
$=\int \sqrt{2 \cos ^{2} t} \cdot \sqrt{2} \cos t \cdot d t$
$=2 \int \cos ^{2} t \cdot d t$
We know,
$\int \cos ^{n} x \cdot d x=\frac{n-1}{n} \int \cos ^{n-2}(x) d x+\frac{\cos ^{n-1} x \cdot \sin x}{n}$
$=2\left[\frac{\cos t \cdot \sin t}{2}+\frac{1}{2} \int 1 \cdot d t\right]=2\left[\frac{\cos t \cdot \sin t}{2}+\frac{t}{2}\right]$
$=\cos t \sin t+t$
$=\cos \left(\sin ^{-1} \frac{x}{\sqrt{2}}\right) \times \sin \left(\sin ^{-1} \frac{x}{\sqrt{2}}\right)+\sin ^{-1} \frac{x}{\sqrt{2}}$
$=\sqrt{1-\frac{x^{2}}{2}} \cdot \frac{x}{\sqrt{2}}+\sin ^{-1} \frac{x}{\sqrt{2}}$
$=\frac{x}{2} \sqrt{2-x^{2}}+\sin ^{-1} \frac{x}{\sqrt{2}}$
$\therefore 4 \cdot \int_{0}^{\sqrt{2}} \sqrt{2-x^{2}} \cdot d x=4\left[\frac{x}{2} \sqrt{2-x^{2}}+\sin ^{-1} \frac{x}{\sqrt{2}}\right]_{0}^{\sqrt{2}}$
$=4\left[0+\sin ^{-1} \frac{\sqrt{2}}{\sqrt{2}}-0-\sin ^{-1} \frac{0}{\sqrt{2}}\right]$
$=4\left[\sin ^{-1} 1-\sin ^{-1} 0\right]=4\left(\frac{\pi}{2}-0\right)=2 \pi$ sq. units.

Topic: Mathematics, Area Under Curves, NDA, NDA 2011 (Phase 2)
What is the area of a triangle whose vertices are at $(3,-1,2)$, $(1,-1,-3)$ and $(4,-3,1)$ ?

Options:
A $\frac{\sqrt{165}}{2}$
B $\frac{\sqrt{135}}{2}$
C 4
D 2

The correct answer is:
$\frac{\sqrt{165}}{2}$

Solution:
Let the vertices of the $\Delta \mathrm{ABC}$ are $\mathrm{A}(3,-1,2)$, $\mathrm{B}(1,-1,-3)$ and $\mathrm{C}(4,-3,1)$
Let $\overrightarrow{\mathrm{OA}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and
$$
\overrightarrow{\mathrm{OC}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\mathrm{k}
$$
Area of $\Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
Now, $\overrightarrow{\mathrm{AB}}=\overline{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\therefore \quad$ Required Area $=\frac{1}{2}\left\|\begin{array}{|ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 0 & 5 \\ -1 & 2 & 1\end{array}\right\|$
$=\frac{1}{2}|\hat{i}(-10)-\hat{j}(2+5)+\hat{k}(4)|$
$=\frac{1}{2}|-10 \hat{i}-7 \hat{j}+4 \hat{k}|$
$=\frac{1}{2} \sqrt{100+49+16}=\frac{1}{2} \sqrt{165}$ squnit

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
What is the area of one of the loops between the curve $y=$ $\mathrm{c} \sin \mathrm{x}$ and $\mathrm{x}$ -axis ?

Options:
A $\mathrm{c}$
B $2 \mathrm{c}$
C $3 \mathrm{c}$
D $4 \mathrm{c}$

The correct answer is:
$2 \mathrm{c}$

Solution:
Area of one of loop between $\mathrm{y}=\mathrm{c} \sin \mathrm{x}$ and $\mathrm{x}$ -axis
$=\mathrm{c}+\mathrm{c}=2 \mathrm{c}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2019 (Phase 1)
What is the area of the ellipse $4 x^{2}+9 y^{2}=1 \quad$

Options:
A $6 \pi$
B $\frac{\pi}{36}$
C $\frac{\pi}{6}$
D $\frac{\pi}{\sqrt{6}}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Given $4 x^{2}+9 y^{2}=1$
$\Rightarrow \frac{x^{2}}{\left(\frac{1}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$
$\therefore \quad a=\frac{1}{2}$ and $b=\frac{1}{3}$
Now, area of ellipse $=4 \int_{0}^{1 / 2} y d x$
$=4 \int_{0}^{1 / 2} \sqrt{\frac{1-4 x^{2}}{9}} d x=\frac{4}{3} \int_{0}^{1 / 2} \sqrt{1-(2 x)^{2}} d x$
$=\frac{2}{3} \int_{0}^{1} \sqrt{1-t^{2}} d t=\frac{2}{3}\left[\frac{1}{2} t \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1}(t)\right]_{0}^{1}$
$=\frac{2}{3}\left[0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{2} \sin ^{-1}(0)\right]$
$=\frac{2}{3}\left[\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{2} \times 0\right]=\frac{\pi}{6}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2008 (Phase 2)
What is the area of the largest rectangular field which can be enclosed with $200 \mathrm{~m}$ of fencing ?

Options:
A $1600 \mathrm{~m}^{2}$
B $2100 \mathrm{~m}^{2}$
C $2400 \mathrm{~m}^{2}$
D $\begin{array}{ll}\text { } 2500 \mathrm{~m}^{2} & {}\end{array}$

The correct answer is:
$\begin{array}{ll}\text { } 2500 \mathrm{~m}^{2} & {}\end{array}$

Solution:
Area of largest rectangular field for a given perimeter for is possible if length and breadth of rectangular field are equal i.e. it is a square
$\Rightarrow 4 \mathrm{x}=200 \Rightarrow \mathrm{x}=\frac{200}{4}=50 \mathrm{~m}$
$\therefore$ Area of largest rectangular field $=50 \times 50=2500 \mathrm{~m}^{2}$ Aliter:
Let length and breadth of rectangular field be $\mathrm{x}$ and $\mathrm{y}$ respectively $\therefore \quad 2(\mathrm{x}+\mathrm{y})=200$
$\Rightarrow \mathrm{y}=100-\mathrm{x}$
and area, $\mathrm{A}=\mathrm{xy}$
$=x(100-x)=100 x-x^{2}$
$\because \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=100-2 \mathrm{x}$
Put $\frac{\mathrm{dA}}{\mathrm{dx}}=0$ for maxima or minima
$100-2 x=0$
$\Rightarrow \mathrm{x}=50 \Rightarrow \mathrm{y}=50$
Now, $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=-2 < 0$, which shows maximum, indepenent of values of $\mathrm{x}$ and $\mathrm{y}$, but only when they are equal.
$\therefore \quad$ Ais maximumat $x=50$.
Hence, required area $=50(100-50)=50 \times 50=2500 \mathrm{~m}^{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
What is the area of the parabola $x^{2}=y$ bounded by the line $y=1$

Options:
A $\frac{1}{3}$ square unit
B $\frac{2}{3}$ square unıt
C $\frac{4}{3}$ square units
D 2 square units

The correct answer is:
$\frac{4}{3}$ square units

Solution:
Area of $\mathrm{ABCD}=2 \times 1=2 \mathrm{sq}$. units
Area of $\mathrm{AOD}=\int_{-1}^{1} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{\mathrm{x}^{3}}{3}\right|_{-1} ^{1}=\frac{2}{3} \mathrm{sq} .$ units
Required area $=2-\frac{2}{3}=\frac{4}{3}$ sq. units

Topic: Mathematics, Area Under Curves, NDA, NDA 2013 (Phase 2)
What is the area of the parabola $y^{2}=x$ bounded by its latus rectum?

Options:
A $\frac{1}{12}$ square unit
B $\frac{1}{6}$ square unit
C $\frac{1}{3}$ square unit
D None of the above

The correct answer is:
$\frac{1}{6}$ square unit

Solution:
Required Area $=2 \int_{0}^{\frac{1}{4}} \sqrt{\mathrm{x}} \mathrm{dx}$
$=2 \cdot \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1 / 4}=\frac{4}{3}\left[\frac{1}{8}-0\right]=\frac{1}{6} \mathrm{sq} .$ unit.

Topic: Mathematics, Area Under Curves, NDA, NDA 2013 (Phase 1)
What is the area of the parabola $y^{\prime}-4 b x$ bounded by its latus rectum?

Options:
A $2 b^{2} / 3$ square unit
B $1 b^{2} / 3$ square unit
C $b^{2}$ square unit
D $8 b^{2} / 3$ square uni

The correct answer is:
$8 b^{2} / 3$ square uni

Solution:
Given equation $y^{2}=4 b x=2 \int_{0}^{b} \sqrt{4 b x} d x$
$=4 \sqrt{\mathrm{b}} \times \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{\mathrm{b}}=\frac{8 \sqrt{\mathrm{b}}}{3}\left[\mathrm{~b}^{\frac{3}{2}}-0\right]$
$\therefore$ area of parabola bounded by its latus rectum
$=\frac{8 \mathrm{~b}^{2}}{3}$ sq. units

Topic: Mathematics, Area Under Curves, NDA, NDA 2014 (Phase 2)
What is the area of the parallelogram having diagonals $3 \hat{i}+\hat{j}-2 \hat{k}$ and $\hat{i}-3 \hat{j}+4 \hat{k} ?$

Options:
A $5 \sqrt{5}$ square units
B $4 \sqrt{5}$ square units
C $5 \sqrt{3}$ square units
D $15 \sqrt{2}$ square units

The correct answer is:
$5 \sqrt{3}$ square units

Solution:
Diagonal $d_{1}, \overline{A C}=3 i+j-2 k$
Diagonal $d_{2}, \overrightarrow{B D}=i-3 j+4 k$
Area of parallelogram is $\frac{1}{2}\left|\overrightarrow{d_{1}} \times \overrightarrow{d_{2}}\right|$
$\begin{aligned} \text { Hence area } &=\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| \\ &=\frac{1}{2}|[\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)]| \\ &=\frac{1}{2}|-2 \hat{i}-14 \hat{j}-10 \hat{k}| \\ &=\frac{1}{2} \sqrt{4+196+100} \\ &=\frac{10 \sqrt{3}}{2}=5 \sqrt{3} \text { square units } \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 1)
What is the area of the portion of the curve $y=\sin x$, lying between $\mathrm{x}=0, \mathrm{y}=0$ and $\mathrm{x}=2 \pi$

Options:
A 1 square unit
B 2 square units
C 4 square units
D 8 square units

The correct answer is:
2 square units

Solution:
$\begin{aligned} \text { Required area }=\int_{0}^{2 \pi} \sin x \mathrm{dx} \\=&-\left.\cos \mathrm{x}\right|_{0} ^{2 \pi}=-\cos 2 \pi-(-\cos 0) \\=-\cos (\pi+\pi)+1=-[-\cos \pi]+1 \\=+\cos \left(\frac{\pi}{2}+\frac{\pi}{2}\right)+1=\sin \frac{\pi}{2}+1=1+1=2 \mathrm{sq} . \text { units. } \end{aligned}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2012 (Phase 2)
What is the area of the rectangle having vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and
$\mathrm{D}$ with positive vectors $-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $-\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}} ?$

Options:
A $1 / 2$ square unit
B 1 square unit
C 2 square unit
D 4 square unit

The correct answer is:
2 square unit

Solution:
From the given vectors we can conclude that
$\mathrm{A}\left(-1, \frac{1}{2}, 4\right), \mathrm{B}\left(1, \frac{1}{2}, 4\right), \mathrm{C}\left(1, \frac{-1}{2}, 4\right), \mathrm{D}\left(-1,-\frac{1}{2}, 4\right)$
Length $=\mathrm{AB}=2, \mathrm{BC}=1$
Area $=\mathrm{AB} \times \mathrm{BC}=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 1)
What is the area of the rectangle of which $\overrightarrow{\mathrm{r}}=\mathrm{a} \overrightarrow{\mathrm{i}}+\mathrm{b} \overrightarrow{\mathrm{j}}$ is a semidiagonal?

Options:
A $a^{2}+b^{2}$
B $2\left(a^{2}+b^{2}\right)$
C $4\left(a^{2}+b^{2}\right)$
D $4 \mathrm{ab}$

The correct answer is:
$4 \mathrm{ab}$

Solution:
As given Semidiagonal is $\overrightarrow{\mathrm{r}}=\mathrm{a} \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{b}}$
So diagonal is $2 \overrightarrow{\mathrm{r}}=2 \mathrm{a} \overrightarrow{\mathrm{i}}+2 \mathrm{~b} \overrightarrow{\mathrm{j}}$
$\Rightarrow$ Sides of rectangle are $2 \mathrm{a}$ and $2 \mathrm{~b}$ Hence, area of rectangle $=2 \mathrm{a} \times 2 \mathrm{~b}=4 \mathrm{ab}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
What is the area of the region bounded by the curve
$f(x)=1-\frac{x^{2}}{4}, x \in[-2,2]$, and the $x$ -axis? $\quad$

Options:
A $\frac{8}{3} \mathrm{sq}$ unit
B $\frac{4}{3} \mathrm{sq}$ unit
C $\frac{2}{3} \mathrm{sq}$ uni
D $\frac{1}{3}$ sq unit

The correct answer is:
$\frac{8}{3} \mathrm{sq}$ unit

Solution:
Required area $=\int_{-2}^{2}\left(1-\frac{x^{2}}{4}\right) \mathrm{dx}$
Since, $\left(1-\frac{\mathrm{x}^{2}}{4}\right)$ is an even function therefore
$\int_{-2}^{2}\left(1-\frac{x^{2}}{4}\right) d x=2 \int_{0}^{2}\left(1-\frac{x^{2}}{4}\right) d x$
$=2\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=2\left[2-\frac{2^{3}}{12}\right]$
$=2\left(2-\frac{2}{3}\right)=\frac{8}{3} \mathrm{sq}$ unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2010 (Phase 2)
What is the area of the region bounded by the line $3 \mathrm{x}-5 \mathrm{y}=15$, $\mathrm{x}=1, \mathrm{x}=3$ and $\mathrm{x}$ -axis in sq unit?

Options:
A $\frac{36}{5}$
B $\frac{18}{5}$
C $\frac{9}{5}$
D $\frac{3}{5}$

The correct answer is:
$\frac{18}{5}$

Solution:
The given equation of line can be rewritten as
$\frac{x}{5}-\frac{y}{3}=1$ and $y=\frac{3 x-15}{5}$
$\begin{array}{ll}\therefore \quad & \text { Required area }=\int_{1}^{3} \mathrm{ydx} \\ & =\int_{1}^{3}\left(\frac{3 \mathrm{x}-15}{5}\right) \mathrm{d} \mathrm{x}=\frac{1}{5} \int_{1}^{3}(3 \mathrm{x}-15) \mathrm{d} \mathrm{x} \\ & =\frac{1}{5}\left[\frac{3 \mathrm{x}^{2}}{2}-15 \mathrm{x}\right]_{1}^{3}=\frac{1}{5}\left[\frac{27}{2}-45-\frac{3}{2}+15\right]\end{array}$
$=\frac{1}{5}\left[\frac{24}{3}-30\right]=\frac{1}{5}[12-30]$
$=\frac{-18}{5}=\frac{18}{5}$ sq. unit (neglecting -ve sign)

Topic: Mathematics, Area Under Curves, NDA, NDA 2008 (Phase 1)
What is the area of the region bounded by the lines $y=$ $\mathrm{y}=0$ and $\mathrm{x}=4 ?$

Options:
A 4 square units
B 8 square units
C 12 squate units
D 16 syuare urtib

The correct answer is:
8 square units

Solution:
Area of shaded region $=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$
$=\frac{1}{2} \times 4 \times 4=8$ sq. units.

Topic: Mathematics, Area Under Curves, NDA, NDA 2012 (Phase 2)
What is the area of the region bounded by the parabolas $y^{2}=6(x-1)$ and $y^{2}=3 x ?$

Options:
A $\frac{\sqrt{6}}{3}$
B $\frac{2 \sqrt{6}}{3}$
C $\frac{4 \sqrt{6}}{3}$
D $\frac{5 \sqrt{6}}{3}$

The correct answer is:
$\frac{4 \sqrt{6}}{3}$

Solution:
Given parabolas are $y^{2}=6(x-1)$....(1) and $y^{2}=3 x$
$y^{2}=6(x-1)$
$\Rightarrow \quad 6 x-6=y^{2}$
$\Rightarrow x=\frac{y^{2}+6}{6}=\frac{y^{2}}{6}+1$
Also, $\mathrm{y}^{2}=3 \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{y}^{2}}{3}$.
Solving $(1),(2), 3 x=6(x-1)$
$\Rightarrow 3 x=6 x-6$
$\Rightarrow 3 x=6 \Rightarrow x=2 .$
$y^{2}=6 \Rightarrow y=\pm \sqrt{6}$
$\therefore$ Area $=$
$\int_{-\sqrt{6}}^{\sqrt{6}}\left(\frac{y^{2}}{6}+1-\frac{y^{2}}{3}\right) \cdot d y=2 \int_{0}^{\sqrt{6}}\left(1-\frac{y^{2}}{6}\right) d y=2\left[y-\frac{y^{3}}{18}\right]_{0}^{\sqrt{6}}$
$=2\left[\sqrt{6}-\frac{(\sqrt{6})^{3}}{18}\right]=2 \times \frac{2 \sqrt{6}}{3}=\frac{4 \sqrt{6}}{3}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2018 (Phase 1)
What is the area of the region bounded by $x$ -axis, the curve
$\mathrm{y}=\mathrm{f}(\mathrm{x})$ and the two ordinates $\mathrm{x}=1$ and $\mathrm{x}=\frac{3}{2}$ ?
Consider the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^{2}$, where $\mathrm{x} \in \mathbf{R}$.

Options:
A $\frac{5}{12}$ square unit
B $\frac{7}{12}$ square unit
C $\frac{2}{3}$ square unit
D $\frac{11}{12}$ square unit

The correct answer is:
$\frac{11}{12}$ square unit

Solution:
Area required for given region is
$\begin{aligned} A_{2} &=\int_{1}^{3 / 2} f(x) d x \\ &=\int_{1}^{3 / 2}\left[x^{2}+x-1\right] d x \\ &=\left[\frac{1}{3} x^{3}+\frac{1}{2} x^{2}-x\right]_{1}^{3 / 2} \\ A_{2} &=\left(\frac{1}{3}\left(\frac{27}{8}\right)+\frac{1}{2}\left(\frac{9}{4}\right)-\frac{3}{2}\right)-\left(\frac{1}{3}(1)+\frac{1}{2}(1)-1\right) \\ A_{2} &=\frac{11}{12} \quad \text { square unit } \end{aligned}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2016 (Phase 1)
What is the area of the region bounded by x-axis, the curve
$\mathrm{y}=\mathrm{f}(\mathrm{x})$ and the two ordinates $\mathrm{x}=\frac{1}{2}$ and $\mathrm{x}=1 ?$
Consider the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^{2}$, where $\mathrm{x} \in \mathbf{R}$.

Options:
A $\frac{5}{12}$ square unit
B $\frac{5}{6}$ square unit
C $\frac{7}{-}$ square units
D 2 square units

The correct answer is:
$\frac{5}{12}$ square unit

Solution:
$\quad f(x)=\left\{\begin{array}{ll}x^{2}+x-1 & x \geq 1 \quad \forall x \in R \\ x^{2}-x+1 & x < 1\end{array}\right.$
Hence area required for given region is
$\begin{aligned} A_{1} &=\int_{1 / 2}^{1} f(x) d x \\ &=\int_{1 / 2}^{1}\left(x^{2}-x+1\right) d x \\ &=\left[\frac{1}{3} x^{3}-\frac{x^{2}}{2}+x\right]_{1 / 2}^{1} \\ &=\left(\frac{1}{3} \times 1-\frac{1}{2}+1\right)-\left(\frac{1}{3} \times \frac{1}{8}-\frac{1}{2} \times \frac{1}{4}+\frac{1}{2}\right) \\ A_{1}=\frac{5}{12} & \text { square unit. } \end{aligned}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2016 (Phase 1)
What is the area of the region cnclosed by $\mathrm{y}-2|\mathrm{x}|$ and $\mathrm{y}=4 ?$

Options:
A 2 square unit
B 4 square unit
C 8 square unit
D 16 squareunit

The correct answer is:
8 square unit

Solution:
Required Area $=2 \int_{y=0}^{4} \frac{y}{2} d y=\left.\frac{y^{2}}{2}\right|_{0} ^{4}=\frac{16}{2}=8$ sq. unit.
or Area $=2 \times 4=8$ sq. unit.

Topic: Mathematics, Area Under Curves, NDA, NDA 2013 (Phase 1)
What is the area of the triangle bounded by the side $\mathrm{x}=0$ $\mathrm{y}=0$, and $\mathrm{x}+\mathrm{y}=2 ?$

Options:
A 1 square unit
B 2 square unit
C 4 square unit
D 8 square unit

The correct answer is:
2 square unit

Solution:
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \times \mathrm{OB} \times \mathrm{OA}$
$=\frac{1}{2} \times 2 \times 2=2$ square unit.

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)
What is the area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12 y$ to the ends of the latus rectum?

Options:
A 9 square units
B 12 square units
C 14 square units
D 18 square units

The correct answer is:
18 square units

Solution:
Given parabola is $\mathrm{x}^{2}=12 y$ which is of the form $\mathrm{x}^{2}=4 a y$.
$\Rightarrow 4 a=12 \Rightarrow a=3$
Now, $L M$ is the latus rectum whose length $=4 a=4 \times 3$
=12
So, area of $\Delta L M V=\frac{1}{2} \times L M \times V F$.
$=\left(\frac{1}{2} \times 12 \times 3\right)$ sq. unit
$=18$ square unit

Topic: Mathematics, Parabola, NDA, NDA 2011 (Phase 2)
What is the area of the triangle formed by the lines $y-x=0, y+x=0, x=c ?$

Options:
A $c / 2$
B $c^{2}$
C $2 c^{2}$
D $c^{2} / 2$

The correct answer is:
$c^{2}$

Solution:
The shaded portion shows the required area. Since the area is symmetric about the $x$ -axis therefore Required area $=2$ area $(\Delta A O D)$
$$
\begin{array}{l}
=2 \times \frac{1}{2} \times \text { base } \times \text { height } \\
=2 \times \frac{1}{2} \times O D \times A D \\
=c \times c \\
=c^{2}
\end{array}
$$

Topic: Mathematics, Straight Lines, NDA, NDA 2009 (Phase 1)
What is the area of the triangle OAB where $\mathrm{O}$ is the origin, $\overrightarrow{\mathrm{OA}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ ?

Options:
A $5 \sqrt{6}$ square unit
B $\frac{5 \sqrt{6}}{2}$ square unit
C $\sqrt{6}$ square unit
D $\sqrt{30}$ square unit

The correct answer is:
$\frac{5 \sqrt{6}}{2}$ square unit

Solution:
Area of $\Delta \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|$
$\therefore \overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}=\left|\begin{array}{ccc}\hat{i} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ 2 & 1 & -3\end{array}\right|$
$=\hat{\mathrm{i}}[3-1]-\hat{\mathrm{j}}[-9-2]+\hat{\mathrm{k}}[3+2]$
$=2 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
$\therefore|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|=\sqrt{2^{2}+11^{2}+5^{2}}=\sqrt{150}=5 \sqrt{6}$
$\therefore$ Required area $=\frac{1}{2} \times 5 \sqrt{6}=\frac{5 \sqrt{6}}{2} \mathrm{sq} .$ units

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)