datavorous/entrance-exam-dataset huggingface dataset:
What is the area of the triangle whose vertices are $(0,0,0)$, $(1,2,3)$ and $(-3,-2,1) ?$

Options:
A $3 \sqrt{5}$ square unit
B $6 \sqrt{5}$ square unit
C 6 square unit
D 12 square unit

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

Solution:
Let $\mathrm{A}(0,0,0), \mathrm{B}(1,2,3)$ and $\mathrm{C}(-3,-2,1)$ be the vertices of a triangle.
Area of $\triangle \mathrm{ABC}=\frac{1}{2}\left\|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1\end{array}\right\|$
$=\frac{1}{2}[\hat{i}(2+6)-\hat{j}(1+9)+\hat{k}(-2+6)]$
$=\frac{1}{2}|8 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{64+16+100}$
$=\frac{1}{2}(6 \sqrt{5})=3 \sqrt{5}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
What is the area of the triangle whose vertices are $(3,0),(0,4)$ and $(3,4) ?$

Options:
A 6 sq. unit
B $7.5 \mathrm{sq}$. unit
C 9 sq. unit
D 12 sq. unit

The correct answer is:
6 sq. unit

Solution:
Required Area
$=\frac{1}{2}\left|\begin{array}{lll}3 & 0 & 1 \\ 0 & 4 & 1 \\ 3 & 4 & 1\end{array}\right|=\frac{1}{2}[3(4-4)+1(0-12)]=6$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)
What is the area of the triangle with vertices $(0,2,2),(2,0,-1)$ and $(3,4,0)$ ?

Options:
A $\frac{15}{2} \mathrm{sq}$ unit
B 15 sq unit
C $\frac{7}{2}$ sq unit
D 7 sq unit

The correct answer is:
$\frac{15}{2} \mathrm{sq}$ unit

Solution:
Let $A=(0,2,2), \mathrm{B}=(2,0,-1)$ and $C=(3,4,0)$ $\overrightarrow{A B}=(2-0,0-2,-1-2)$ and $\overrightarrow{A C}=(3-0,4-2,0-2)$
$\Rightarrow \overrightarrow{A B}=(2,-2,-3)$ and $\overrightarrow{A C}=(3,2,-2)$
$\therefore$ Area of triangle $=\frac{1}{2} \times$ magnitude of $\overrightarrow{A B} \times \overrightarrow{A C}$
$$
\begin{array}{l}
=\frac{1}{2}\|\overrightarrow{A B} \times \overrightarrow{A C}\| \\
=\frac{1}{2}\|\hat{\mathbf{i}} \quad \hat{\mathbf{j}} \quad \hat{\mathbf{k}}\| \\
=\frac{1}{2} \mid[\hat{\mathbf{i}}(4+6)-\hat{\mathbf{j}}(-4+9)+\hat{\mathbf{k}}(4+6)] \\
=\frac{1}{2}|10 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}| \\
\begin{array}{ccc}
-2 & -3 \\
3 & 2 & -2
\end{array} \| \\
=\frac{1}{2} \sqrt{(10)^{2}+(5)^{2}+(10)^{2}}=\frac{1}{2} \sqrt{225}=\frac{15}{2}
\end{array}
$$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
What is the area of the triangle with vertices $(1,2,3),(2,5,-1)$ and $(-1,1,2)$ ?

Options:
A $\frac{\sqrt{155}}{2}$ square units
B $\frac{\sqrt{175}}{2}$ square units
C $\frac{\sqrt{155}}{4}$ square units
D $\frac{\sqrt{175}}{4}$ square units

The correct answer is:
$\frac{\sqrt{155}}{2}$ square units

Solution:
Let $\overrightarrow{O A}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{O B}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and
$\overrightarrow{O C}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ be three position vectors.
Now, $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
and $\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$
$\therefore$ Area of $\Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$
Now, $\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 3 & -4 \\ -2 & -1 & -1\end{array}\right|$
$=\hat{\mathrm{i}}(-3-4)-\hat{\mathrm{j}}(-1-8)+\hat{\mathrm{k}}(-1+6)$
$=-7 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
Now, $|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{155}$
Required Area $=\frac{\sqrt{155}}{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 2)
What is the area of the triangle with vertices
$\left(x_{1}, \frac{1}{x_{1}}\right),\left(x_{2}, \frac{1}{x_{2}}\right),\left(x_{3}, \frac{1}{x_{3}}\right) ?$

Options:
A $\left|\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(\mathrm{x}_{2}-\mathrm{x}_{3}\right)\left(\mathrm{x}_{3}-\mathrm{x}_{1}\right)\right|$
B 0
C $\left|\frac{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left(\mathrm{x}_{2}-\mathrm{x}_{3}\right)\left(\mathrm{x}_{3}-\mathrm{x}_{1}\right)}{\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3}}\right|$
D $\left|\frac{\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right)\left(x_{3}-x_{1}\right)}{2 x_{1} x_{2} x_{3}}\right|$

The correct answer is:
$\left|\frac{\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right)\left(x_{3}-x_{1}\right)}{2 x_{1} x_{2} x_{3}}\right|$

Solution:
Area $=\frac{1}{2}\left|x_{1}\left(\frac{1}{x_{2}}-\frac{1}{x_{3}}\right)+x_{2}\left(\frac{1}{x_{3}}-\frac{1}{x_{1}}\right)+x_{3}\left(\frac{1}{x_{1}}-\frac{1}{x_{2}}\right)\right|$
$=\frac{1}{2}\left|\frac{x_{1}\left(x_{3}-x_{2}\right)}{x_{2} x_{3}}+\frac{x_{2}\left(x_{1}-x_{3}\right)}{x_{1} x_{3}}+\frac{x_{3}\left(x_{2}-x_{1}\right)}{x_{1} x_{2}}\right|$
$=\frac{1}{2}\left|\frac{-x_{1}^{2}\left(x_{2}-x_{3}\right)-x_{2}^{2}\left(x_{3}-x_{1}\right)-x_{3}^{2}\left(x_{1}-x_{2}\right)}{x_{1} x_{2} x_{3}}\right|$
$=\left|\frac{\left(x_{1}-x_{2}\right)\left(x_{2}-x_{3}\right)\left(x_{3}-x_{1}\right)}{2\left(x_{1} x_{2} x_{3}\right)}\right|$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
What is the area under the curve $f(x)=x e^{x}$ above the $x$ -axis and between the lines $x=0$ and $x=1 ? \quad$

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

The correct answer is:
1 sq unit

Solution:
Given curve is $f(x)=x e^{x}, x=0$ and $x=1$.
So, Required Area $=\int_{0}^{1} f(x) d x=\int_{0}^{1} x e^{x} d x$
Let $x$ be the first function and $e^{x}$ be the second function then by parts
$=\left[x e^{x}-\int e^{x} d x\right]_{0}^{1}=\left[x e^{x}-e^{x}\right]_{0}^{1}$
$=(e-e)-(0-1)=1$ sq unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2010 (Phase 2)
What is the area under the curve $\mathrm{y}=|\mathrm{x}|+|\mathrm{x}-1|$ between $\mathrm{x}=0$ and $\mathrm{x}=1 ?$

Options:
A $\frac{1}{2}$
B 1
C $\frac{3}{2}$
D 2

The correct answer is:
1

Solution:
$|\mathrm{x}|$ for $\mathrm{x} \geq 0$
$=\mathrm{x}$ and $|\mathrm{x}-1|$ for $\mathrm{x} \leq 1$
$=-(\mathrm{x}-1)$
so, $\int_{0}^{1}(|\mathrm{x}|+|\mathrm{x}-| 1 \mid)=$ required area
$\mathrm{a}=\int_{0}^{1} \mathrm{x} \mathrm{dx}-\int_{0}^{1}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}-\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{0}^{1}=\frac{1}{2}-\left(\frac{1}{2}-1\right)=1 \mathrm{sq}$ units

Topic: Mathematics, Area Under Curves, NDA, NDA 2006 (Phase 1)
What is the argument of $(1-\sin \theta)+\mathrm{i} \cos \theta$ ?
$(i=\sqrt{-1})$

Options:
A $\frac{\pi}{2}-\frac{\theta}{2}$
B $\frac{\pi}{2}+\frac{\theta}{2}$
C $\frac{\pi}{4}-\frac{\theta}{2}$
D $\frac{\pi}{4}+\frac{\theta}{2}$

The correct answer is:
$\frac{\pi}{4}+\frac{\theta}{2}$

Solution:
Given complex number is $(1-\sin \theta)+i \cos \theta \equiv a+i b$
Argument $\equiv \tan \theta=\frac{b}{a}$
$\Rightarrow \tan \theta=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}}{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$
$=\frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}{\left(\sin \frac{\theta}{2}-\cos \frac{\theta}{2}\right)^{2}}$
$=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}=\frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}$
$\tan \theta=\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$
Hence, argument $=\frac{\pi}{4}+\frac{\theta}{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
What is the argument of the complex number $(-1-\mathrm{i})$ where $\mathrm{i}=\sqrt{-1}$ ?

Options:
A $\frac{5 \pi}{4}$
B $-\frac{5 \pi}{4}$
C $\frac{3 \pi}{4}$
D None of the above

The correct answer is:
$\frac{5 \pi}{4}$

Solution:
Let $z=-1-i$ Since Real part of $z < 0$ and Imaginary part of $z < 0$ therefore argument lies in IIIrd quadrant
$\operatorname{rg}(\mathrm{z})=\pi+\tan ^{-1}\left|\frac{-1}{-1}\right|=\pi+\frac{\pi}{4}=\frac{5 \pi}{4}$

Topic: Mathematics, Complex Number, NDA, NDA 2013 (Phase 1)
What is the argument of the complex number $\frac{(1+i)(2+i)}{3-i}$ where $i=\sqrt{-1} ?$

Options:
A 0
B $\frac{\pi}{4}$
C $-\frac{\pi}{4}$
D $\frac{\pi}{2}$

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

Solution:
$\frac{(1+\mathrm{i})(2+\mathrm{i})}{3-\mathrm{i}}=\frac{1+3 \mathrm{i}}{3-\mathrm{i}}$
$=\frac{1+3 \mathrm{i}}{3-\mathrm{i}} \times \frac{3+\mathrm{i}}{3+\mathrm{i}}=\frac{10 \mathrm{i}}{10}=\mathrm{i}$ or $0+\mathrm{i}$
argument, $\theta=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1}\left(\tan \frac{\pi}{2}\right)=\frac{\pi}{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 1)
What is the arithmetic mean of first 16 natural numbers with weights being the number itself?

Options:
A $\frac{17}{2}$
B $\frac{33}{2}$
C 11
D $\frac{187}{2}$

The correct answer is:
$\frac{17}{2}$

Solution:
Given natural numbers are $1,2,3,4,5, \ldots \ldots \ldots, 16$
This is an A.P. with first term $=1$ and common difference $=1, n=16$
$\therefore$
By using sum of 16 natural numbers
i.c., $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
We have
$\mathrm{S}_{16}=\frac{16}{2}[2(1)+15(1)]=8(17)=136$
$\therefore \mathrm{AM}=\frac{136}{16}=\frac{17}{2}$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 1)
What is the arithmetic mean of the series ${ }^{\mathrm{n}} \mathrm{C}_{0},{ }^{\mathrm{n}} \mathrm{C}_{1}, \ldots \ldots{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}, ?$

Options:
A $\frac{2^{\mathrm{n}}}{\mathrm{n}}$
B $\frac{2^{n}}{(n+1)}$
C $\frac{2^{(\mathrm{n}+1)}}{\mathrm{n}}$
D $\frac{2^{(\mathrm{n}+1)}}{(\mathrm{n}+1)} \quad\mathrm{}$

The correct answer is:
$\frac{2^{n}}{(n+1)}$

Solution:
Since, expansion contains $(\mathrm{n}+1)$ terms,
Required mean $=\frac{{ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{(\mathrm{n}+1)}$
$=\frac{2^{\mathrm{n}}}{\mathrm{n}+1}$

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
What is the arithmetic mean of the series ${ }^{n} C_{1},{ }^{n} C_{2},{ }^{n} C_{3} \ldots{ }^{n} C_{n}$ ?

Options:
A $\left(2^{n}-1\right) / n$
B $2^{n} /(n+1)$
C $\left(2^{n}\right) / n$
D $2^{(n+1)} /(n+1)$

The correct answer is:
$\left(2^{n}-1\right) / n$

Solution:
Arithmetic mean of the series
$=\frac{{ }^{n} C_{1}+{ }^{n} C_{2}+{ }^{n} C_{3}+\ldots+{ }^{n} C_{n}}{n}=\frac{2^{n}-1}{n}$

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
What is the atomic mass of $\mathrm{Fe}$ ?
Given abundance of ${ }^{54} \mathrm{Fe}=10 \%$
$$
\begin{aligned}
& { }^{56} \mathrm{Fe}=85 \% \\
& { }^{57} \mathrm{Fe}=5 \%
\end{aligned}
$$

Options:
A 55.65
B 55.75
C 55.85
D 55.95

The correct answer is:
55.85

Solution:
Average atomic mass of $\mathrm{Fe}=54 \times 0.1+56 \times 0.85+$ $57 \times 0.05=55.85$

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 1)
What is the binary equivalent of decimal number $(0.8125)_{10}$ ?

Options:
A $(0.1101)_{2}$
B $(0.1001)_{2}$
C $(0.1111)_{2}$
D $(0.1011)_{2}$

The correct answer is:
$(0.1101)_{2}$

Solution:
Work with option $(0.1101)_{2}=1 \times 2^{-1}+1 \times 2^{-2}+1 \times 2^{-4}$
$=\frac{1}{2}+\frac{1}{4}+\frac{1}{16}=\frac{13}{16}=(0.8125)_{10}$
Hence option (a) gives the binary equivalent of given decimal number.

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
What is the binary equivalent of the decimal number $0.3125$ ?

Options:
A $0.0111$
B $0.1010$
C $0.0101$
D $0.1101$

The correct answer is:
$0.0101$

Solution:
$(0.3125)_{10}=(0.0101)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 1)
What is the binary number equivalent of the decimal number
$32.25$ ?

Options:
A $100010.10$
B 100000,10
C $100010.01$
D $100000.0$

The correct answer is:
$100000.0$

Solution:
$\therefore$ Required binary number equivalent to $32.25$ is $100000.01$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
What is the boiling temperature called at 1 bar pressure?

Options:
A Standard boiling point
B Evaporation temperature
C Normal boiling point
D Superheated point

The correct answer is:
Standard boiling point

Solution:
If the pressure is 1 bar, the boiling point is called the standard boiling point of the liquid.

Topic: Chemistry, States of Matter, NEET
What is the centre of the sphere $a x^{2}+b y^{2}+c z^{2}-6 x=0$ if the radius is 1 unit?

Options:
A $(0,0,0)$
B $(1,0,0)$
C $(3,0,0)$
D cannot be determined as values of $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are unknown

The correct answer is:
cannot be determined as values of $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are unknown

Solution:
In the given equation, there are three unknown parameters and no equation has been given to evaluate those, hence centre of sphere cannot be determined.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 1)
What is the characteristic of the vascular bundles in monocot stem?

Options:
A Open and radial
B Open and endarch
C Conjoin and closed
D Conjoin and endarch

The correct answer is:
Conjoin and closed

Solution:
The vascular bundles in monocot stem are conjoint and closed.

Topic: Biology, Anatomy of Flowering Plants, NEET
What is the coefficient of the middle term in the binomial expansion of $(2+3 \mathrm{x})^{4} ?$

Options:
A 6
B 12
C 108
D 216

The correct answer is:
216

Solution:
Middle term in the expansion of $(x+y)^{n}$
$=\left(\frac{n+1}{2}\right)^{\text {th }}$ term, if $n$ is odd
$=\left(\frac{n}{2}+1\right)^{\text {th }}$ term, if $n$ is even
Here $n=4$
$\therefore$ Middle term is $\left(\frac{4}{2}+1\right)^{\text {th }}=3^{\mathrm{rd}}$ term $\begin{aligned} 4 c_{2} \times 2^{2} \times 3^{2} &=6 \times 4 \times 9 \\ &=216 \\ 62 . \quad \text { (c) } \alpha={ }^{2 n} \mathrm{C}_{n} \\ \beta=^{2 n-1} \mathrm{C}_{n} \\ \gamma=^{2 n-1} \mathrm{C}_{n-1} \end{aligned}$
$\beta+\gamma={ }^{2 n-1} \mathrm{C}_{n}+{ }^{2 n-1} \mathrm{C}_{n-1}={ }^{2 n} \mathrm{C}_{n}=\alpha$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2018 (Phase 2)
What is the coefficient of $\mathrm{x}^{3}$ in $\frac{(3-2 \mathrm{x})}{(1+3 \mathrm{x})^{3}}$ ?

Options:
A $-272$
B $-540$
C $-870$
D $-918$

The correct answer is:
$-918$

Solution:
$\frac{(3-2 \mathrm{x})}{(1+3 \mathrm{x})^{3}}=(3-2 \mathrm{x})(1+3 \mathrm{x})^{-3}$
$=(3-2 \mathrm{x})\left(1-9 \mathrm{x}+\frac{(-3)(-4)}{2 !} \cdot 9 \mathrm{x}^{2}\right.$
$\left.\quad+\frac{(-3)(-4)(-5)}{3 !} \cdot 27 \mathrm{x}^{3}+\ldots .\right)$
$\left[\operatorname{Expanding}(1+3 \mathrm{x})^{-3}\right]$
$=(3-2 \mathrm{x})\left(1-9 \mathrm{x}+54 \mathrm{x}^{2}-270 \mathrm{x}^{3}+\ldots \ldots \ldots\right)$
$\therefore$ Coefficient of $\mathrm{x}^{3}=-270 \times 3-2 \times 54$
$=-810-108=-918$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2006 (Phase 1)
What is the coefficient of $x^{3} y^{4}$ in $\left(2 x+3 y^{2}\right)^{5}$ ?

Options:
A 240
B 360
C 720
D 1080

The correct answer is:
720

Solution:
$\mathrm{T}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2 \mathrm{x})^{\mathrm{r}-1}\left(3 \mathrm{y}^{2}\right)^{\mathrm{n}-\mathrm{r}+1}$
$\mathrm{~T}_{4}=={ }^{5} \mathrm{C}_{3}(2 \mathrm{x})^{3}\left(3 \mathrm{y}^{2}\right)^{2}$
$=\frac{5 !}{3 ! 2 !} 2^{3} \cdot \mathrm{x}^{3} \cdot 9 \mathrm{y}^{4}=\frac{5.4}{2.1} \times 8 \times 9 \times \mathrm{x}^{3} \mathrm{y}^{4}=720 \mathrm{x}^{3} \mathrm{y}^{4}$
$\therefore$ Coefficient of $x^{3} y^{4}=720$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2008 (Phase 1)
What is the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+3 x^{2}+4 x^{3}+\ldots .\right)^{1 / 2} ?$

Options:
A 1/4
B 1/16
C 1
D 1/128

The correct answer is:
1

Solution:
Consider $\left(1+2 \mathrm{x}+3 \mathrm{x}^{2}+4 \mathrm{x}^{3}+\ldots\right)^{1 / 2}=\left(1-\mathrm{x}^{-2}\right)^{1 / 2}$
As we know that $(1-\mathrm{x})^{-2}=1+2 \mathrm{x}+3 \mathrm{x}^{2}+4 \mathrm{x}^{3}+\ldots$
$\Rightarrow(1-x)^{-1}=1+x+x^{2}+x^{3}+x^{4}+\ldots$
$\therefore$ Required coefficient of $x^{4}$ is 1 .

Topic: Mathematics, Binomial Theorem, NDA, NDA 2009 (Phase 2)
What is the coefficient of $x^{4}$ in the expansion of $\left(\frac{1-x}{1+x}\right)^{2}$ ?

Options:
A -16
B 16
C 8
D -8

The correct answer is:
16

Solution:
Consider $\left(\frac{1-x}{1+x}\right)^{2}=(1-x)^{2}(1+x)^{-2}$
$=\left(1-2 x+x^{2}\right)(1+x)^{-2}$
$=\left(1-2 x+x^{2}\right)\left(1-2 x+3 x^{2}-4 x^{3}+5 x^{4}-\ldots . .\right)$
$\therefore \quad$ Coefficient of $x^{4}$ in $\left(\frac{1-x}{1+x}\right)^{2}=5+8+3=16$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2010 (Phase 2)
What is the coefficient of $\mathrm{x}^{5}$ in the expansion $\left(1-2 \mathrm{x}+3 \mathrm{x}^{2}-4 \mathrm{x}^{3}+\ldots \ldots . \infty\right)^{-5} ?$

Options:
A $(10 !) /(5 !)^{2}$
B $5^{-5}$
C $5^{5}$
D $10 ! /\{6 !)(4 !)\}$

The correct answer is:
$(10 !) /(5 !)^{2}$

Solution:
$1-2 \mathrm{x}+3 \mathrm{x}^{2}-4 \mathrm{x}^{3}+\ldots$
$=(1+\mathrm{x})^{-2}, \mathrm{so},\left(1-2 \mathrm{x}+3 \mathrm{x}^{2}-4 \mathrm{x}^{3}+\ldots \infty\right)^{-5}$
$=\left((1+\mathrm{x})^{-2}\right)^{-5}=(1+\mathrm{x})^{10} \Rightarrow \mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}$
Putting $\mathrm{r}=5$, coefficient of $\mathrm{x}^{5}={ }^{10} \mathrm{C}_{5}=\frac{10 !}{5 ! 5 !}=\frac{10 !}{(5 !)^{2}}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2007 (Phase 1)
What is the coefficient of $x^{17}$ in the expansion of $\left(3 x-\frac{x^{3}}{6}\right)^{9} ?$

Options:
A $\frac{189}{8}$
B $\frac{567}{2}$
C $\frac{21}{16}$
D None of these

The correct answer is:
$\frac{189}{8}$

Solution:
Given expansion is
$\left(3 x-\frac{x^{3}}{6}\right)^{9}$ where $a=3 x, \mathrm{~b}=\frac{-x^{3}}{6}, \mathrm{n}=9$
Now, General Term $=\mathrm{T}_{r+1}={ }^{n} C_{r}(\mathrm{a})^{\mathrm{n}-\mathrm{r}} \cdot b^{2}$
$={ }^{9} \mathrm{C}_{r}(3 x)^{9-\mathrm{r}}\left(\frac{-x^{3}}{6}\right)^{r}={ }^{9} \mathrm{C}_{r} \cdot 3^{9-r} \mathrm{x}^{9-r} \cdot \frac{(-1)^{r} x^{3 r}}{6^{r}}$
$={ }^{9} C_{r} 3^{9-r}(-1)^{r} \frac{x^{9+2 r}}{6^{r}}$
We can get coeff of $x^{17}$ when $9+2 r=17$
$\Rightarrow \quad 2 r=17-9$
$\Rightarrow \quad r=\frac{8}{2}=4$
Hence, required coefficient
$={ }^{9} \mathrm{C}_{4} \frac{3^{5}}{6^{4}}=\frac{126 \times 3}{16}=\frac{189}{8}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2010 (Phase 2)
What is the condition that one root of the equation $a x^{2}+b$
$+c=0, a \neq 0$ should be double the other?

Options:
A $2 a^{2}=9 b c$
B $2 b^{2}=9 a c$
C $2 c^{2}=9 a b$
D None of these

The correct answer is:
$2 b^{2}=9 a c$

Solution:
Let the roots of the equation $a x^{2}+b x+c=0$ be $\alpha$ and $2 \alpha$
Sum $=\alpha+2 \alpha=\frac{-b}{a}$ and product $=\alpha \cdot 2 \alpha=\frac{c}{a}$
$\Rightarrow \alpha=\frac{-b}{3 a}$ and $\alpha^{2}=\frac{c}{2 a}$
$\Rightarrow\left(\frac{-b}{3 a}\right)^{2}=\frac{c}{2 a} \Rightarrow \frac{b^{2}}{9 a^{2}}=\frac{c}{2 a}$
$\Rightarrow 2 b^{2}=9 a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
What is the conjugate of $\left(\frac{1+2 i}{2+i}\right)^{2}$ ?

Options:
A $\frac{7}{25}+i \frac{24}{25}$
B $-\frac{7}{25}-i \frac{24}{25}$
C $-\frac{7}{25}+i \frac{24}{25}$
D $\frac{7}{25}-i \frac{24}{25}$

The correct answer is:
$\frac{7}{25}-i \frac{24}{25}$

Solution:
Consider $\left(\frac{1+2 i}{2+i}\right)^{2}=\left(\frac{(1+2 i)(2-i)}{(2+i)(2-i)}\right)^{2}$
$=\left(\frac{2+4 i-i-2 i^{2}}{4-i^{2}}\right)^{2}=\left(\frac{4+3 i}{5}\right)^{2}$
$=\frac{1}{25}\left(16+9 i^{2}+24 i\right)=\frac{7}{25}+\frac{24 i}{25}$
$\therefore \quad$ Conjugate of $\left(\frac{1+2 i}{2+i}\right)^{2}=\frac{7}{25}-\frac{24}{25} i$
$(\because$ conjugate of $a+i b=a-i b)$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
What is the conventionally chosen standard hydrogen electrode represented as?

Options:
A Pt(s) |H2(g) |H+(aq)
B Pt(s) | H+(aq) |H2(g)
C H2(g) |Pt(s) |H+(aq)
D H+(aq) |H2(g) |Pt(s)

The correct answer is:
Pt(s) |H2(g) |H+(aq)

Solution:
The representation of the standard hydrogen electrode is Pt(s)| H2(g)| H+(aq).

Topic: Chemistry, Electrochemistry, NEET
What is the correct sequence of sperm formation?

Options:
A Spermatid, Spermatocyte, Spermatogonia, Spermatozoa
B Spermatogonia,
Spermatocyte, Spermatozoa, Spermatid
C Spermatogonia, Spermatozoa, Spermatocyte, Spermatid
D Spermatogonia, Spermatocyte, Spermatid, Spermatozoa

The correct answer is:
Spermatogonia, Spermatocyte, Spermatid, Spermatozoa

Solution:
Spermatogonia $\rightarrow$ Spermatocyte $\rightarrow$ Spermatid $\rightarrow$ Spermatozoa
Spermatogonia are present on the inside wall of seminiferous tubule multiplied by mitotic division and increase in number. Spermatocytes are some of the spermatogonia, which periodically undergo meiosis. The secondary spermatocytes undergo the second meiatic division to produce four, equal haploid spermatids. The spermatids are transformed into spermatozoa (sperm).

Topic: Biology, Human Reproduction, NEET, NEET 2013 (All India)
What is the correct sequence of the following values?

Options:
A $3>2>1$
B $1>2>3$
C $1>3>2$
D $3>1>2

The correct answer is:
$3>2>1$

Solution:
We work out the given statements.
1. $\sin \frac{\pi}{12}=\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
2. $\cos \frac{\pi}{12}=\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
3. $\cot \frac{\pi}{12}=\cot 15^{\circ}=2+\sqrt{3}$
So, correct sequence is $3>2>1$.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
What is the cosine of angle between the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{l}=0$ and $2 \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+1=0 ?$

Options:
A $1 / 2$
B $1 / 3$
C $2 / 3$
D None of the above

The correct answer is:
$1 / 3$

Solution:
If $\theta$ be the angle between the planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0$ and $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
then $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\begin{aligned} \cos \theta &=\frac{1(2)+1(-2)+1(2)}{\sqrt{1^{2}+1^{2}+1^{2}}} \cdot \sqrt{2^{2}+(-2)^{2}+2^{2}} \\ &=\frac{2}{\sqrt{3} 2 \sqrt{3}}=\frac{1}{3} \end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 1)
What is the cumulative frequency curve of statistical data commonly called?

Options:
A Cartogram
B Histogram
C Ogive
D Pictogram

The correct answer is:
Ogive

Solution:
The cumulative frequency curve of statistical data is called Ogive.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
What is the curve which passes through the point $(1,1)$ and
whose slope is $\frac{2 y}{x} ?$

Options:
A Circle
B Parabola
C Ellipse
D Hyperbola

The correct answer is:
Parabola

Solution:
$\frac{d y}{d x}=\frac{2 y}{x}$
$\Rightarrow \frac{d y}{y}=2 \frac{d x}{x}$
On integration
$\int \frac{d y}{y}=2 \int \frac{d x}{x}$
$\Rightarrow \log y=2 \log x+\log a$
$\Rightarrow \log y=\log x^{2}+\log a$
$\Rightarrow \log y=\log \left(x^{2} \cdot a\right)$
$\Rightarrow y=x^{2} a$
at $(1,1) ; a=1$
$\Rightarrow x^{2}=y=4\left(\frac{1}{4}\right) y$
$\Rightarrow$ the curve is parabola.

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
What is the decimal equivalent of $(101.101)$, ?

Options:
A $(5.225)_{10}$
B $(5.525)_{10}$
C $(5.625)_{10}$
D $(5.65)_{10}$

The correct answer is:
$(5.625)_{10}$

Solution:
Consider $(101.101)_{2}=1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}+$
$=4+1+\frac{1}{2}+\frac{1}{8}=\frac{40+4+1}{8}=\frac{45}{8}=(5.625)_{10}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
What is the decimal number representation of the binary

Options:
A $30.125$
B $29.025$
C $29.125$
D $28.02^{4}$

The correct answer is:
$29.125$

Solution:
$\begin{aligned} 11101.001 &=1 \times 2^{4}+1 \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0} \cdot 0 \times 2^{-1} \\+0 \times 2^{-2}+1 \times 2^{-3} \end{aligned}$
$=(16+8+4+0+1) \cdot\left(0+0+\frac{1}{8}\right)=29.125$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)