datavorous/entrance-exam-dataset huggingface dataset:
What is the sum of the series $0.5+0.55+0.555+\ldots$ to $n$ terms?

Options:
A $\frac{5}{9}\left[n-\frac{2}{9}\left(1-\frac{1}{10^{n}}\right)\right]$
B $\frac{1}{9}\left[5-\frac{2}{9}\left(1-\frac{1}{10^{n}}\right)\right]$
C $\frac{1}{9}\left[n-\frac{5}{9}\left(1-\frac{1}{10^{n}}\right)\right]$
D $\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right]$

The correct answer is:
$\frac{5}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right]$

Solution:
Given $0.5+0.55+0.555+\ldots .$ to n
$=5[0.1+0.11+0.111+\ldots . .$ to n terms $]$
$=\frac{5}{9}[0.9+0.99+0.999+\ldots . .$ to $\mathrm{n}$ terms $]$
$=\frac{5}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots . .\right.$ to $\mathrm{n}$ terms $]$
$=\frac{5}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots . .\right]$
$=\frac{5}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)\right.$
$\left.+\ldots . .+\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right]$
$=\frac{5}{9}\left[\mathrm{n}-\left(\frac{1}{10}+\frac{1}{10^{2}}+\ldots \cdot \frac{1}{10^{\mathrm{n}}}\right)\right]$
$=\frac{5}{9}\left[\mathrm{n}-\frac{1}{10} \frac{\left\{1-\left(\frac{1}{10}\right)^{\mathrm{n}}\right\}}{\left(1-\frac{1}{10}\right)}\right]$
$=\frac{5}{9}\left[\mathrm{n}-\frac{1}{9}\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right]$

Topic: Mathematics, Sequences and Series, NDA, NDA 2015 (Phase 1)
What is the sum of the squares of direction cosines of the line joining the points $(1,2,-3)$ and $(-2,3,1) ?$

Options:
A 0
B 1
C 3
D $\frac{2}{\sqrt{26}}$

The correct answer is:
1

Solution:
Let the direction cosines be $\ell, \mathrm{m}, \mathrm{n}$. Let $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\mathrm{Q}\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ be two points which
joins a line.
$\therefore \mathrm{x}_{1}=1, \mathrm{y}_{1}=2, \mathrm{z}_{1}=-3$
$\mathrm{x}_{2}=-2, \mathrm{y}_{2}=3, \mathrm{z}_{2}=1$
Now, $\ell=\frac{\mathrm{x}_{2}-\mathrm{x}_{1}}{\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}}$
$=\frac{-3}{\sqrt{26}}$
$\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}}=\frac{1}{\sqrt{26}}$
$\mathrm{n}=\frac{4}{\sqrt{26}}$
$\therefore \ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=\frac{9}{26}+\frac{1}{26}+\frac{16}{26}=\frac{26}{26}=1$
Always equal to 1

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 1)
What is the sum of the squares of direction cosines of $\begin{array}{ll}x \text { -axis ? } & {}\end{array}$

Options:
A 0
B $\frac{1}{3}$
C 1
D 3

The correct answer is:
1

Solution:
Sum of squares of direction cosines $=(1)^{2}+(0)^{2}+(0)^{2}=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 2)
What is the sum of the squares of the roots of the equation $\mathrm{x}^{2}+2 \mathrm{x}-143=0$

Options:
A 170
B 180
C 190
D 290

The correct answer is:
290

Solution:
Let $\alpha, \beta$ be the roots of given equation
and $\alpha \beta=-$ Consider $\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$=(-2)^{2}-2(-143)=4+286=290$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
What is the sum of the two numbers $(11110)_{2}$ and $(1010)$ ?

Options:
A $(101000)$,
B $(110000)_{2}$
C $(100100)$
D $(101100)_{2}$

The correct answer is:
$(101000)$,

Solution:
Therefore relation is equivalence.
$(11110)_{2}=2^{4} \times 1+2^{3} \times 1+2^{2} \times 1+2^{1} \times 1+2^{0} \times 0$
$=16+8+4+2+0=30$
$(1010)_{2}=\left(2^{3} \times 1+2^{2} \times 0+2^{1} \times 1+2^{0} \times 0=8+0+2\right.$
$+0)=10$
Sum $=30+10=40$
$=(101000)_{3}$
\begin{array}{l|l|l}
2 & 40 & \\
\hline 2 & 20 & 0 \\
\hline 2 & 10 & 0 \\
\hline 2 & 5 & 0 \\
\hline 2 & 2 & 1 \\
\hline & 1 & 0
\end{array}

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
What is the temperature at which the clear boundary between the liquid and vapor phases disappears called?

Options:
A Boiling point
B Melting point
C Critical temperature
D Freezing point

The correct answer is:
Critical temperature

Solution:
The temperature at which the clear boundary between the liquid and vapor phases disappears is called the critical temperature.

Topic: Chemistry, States of Matter, NEET
What is the tension in the wire stretched between two rigid supports if it vibrates in its fundamental mode with a frequency of 45 Hz, given that the mass of the wire is 3.5 ×${{10}^{-2}}$ kg and its linear mass density is 4.0 × ${{10}^{-2}}$kg ${{m}^{-1}}$?

Options:
A 157 N
B 248 N
C 394 N
D 197 N

The correct answer is:
248 N

Solution:
Correct Option is : (B)
248 N

Topic: Chemistry, Waves and Sound, NEET
What is the term for a ligand that can bind through two donor atoms?

Options:
A Didendate
B Hexadentate
C Polydentate
D Unidentate

The correct answer is:
Didendate

Solution:
Ligands that can bind through two donor atoms are described as didendate ligands.

Topic: Chemistry, Coordination Compounds, NEET
What is the term for a ligand that uses two or more donor atoms simultaneously to bind a single metal ion?

Options:
A Unidentate
B Chelate
C Polydentate
D Didendate

The correct answer is:
Chelate

Solution:
Chelate ligands use two or more donor atoms simultaneously to bind a single metal ion, forming chelate complexes.

Topic: Chemistry, Coordination Compounds, NEET
What is the term for the boiling temperature at 1 atm pressure?

Options:
A Standard boiling point
B Evaporation point
C Normal boiling point
D Solidification point

The correct answer is:
Normal boiling point

Solution:
The boiling temperature at 1 atm pressure is called the normal boiling point.

Topic: Chemistry, States of Matter, NEET
What is the term for the potential difference between the two electrodes of a galvanic cell?

Options:
A Electrode potential
B Cell potential
C Standard electrode potential
D Electrolyte potential

The correct answer is:
Cell potential

Solution:
The potential difference between the two electrodes of a galvanic cell is called the cell potential.

Topic: Chemistry, Electrochemistry, NEET
What is the term for the primary structure of proteins?

Options:
A Primary conformation
B Tertiary structure
C Secondary structure
D Quaternary structure

The correct answer is:
Primary conformation

Solution:
The primary structure of proteins refers to the specific sequence of amino acids in a polypeptide chain, also known as primary conformation.

Topic: Chemistry, Biomolecules, NEET
What is the term for the shape in which a long polypeptide chain can exist in proteins?

Options:
A Tertiary structure
B Quaternary structure
C Secondary structure
D Primary structure

The correct answer is:
Secondary structure

Solution:
The secondary structure of proteins refers to the shape in which a long polypeptide chain can exist, such as alpha-helix or beta-pleated sheet.

Topic: Chemistry, Biomolecules, NEET
What is the term for the specific sequence of amino acids in a protein?

Options:
A Tertiary structure
B Quaternary structure
C Primary structure
D Secondary structure

The correct answer is:
Primary structure

Solution:
The primary structure of a protein is defined by the specific sequence of amino acids in its polypeptide chain.

Topic: Chemistry, Biomolecules, NEET
What is the term independent of $x$ in the expansion of $\left(1+x+2 x^{3}\right)\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}$ ?

Options:
A $1 / 3$
B $17 / 54$
C $1 / 4$
D No such term exists in the expansion

The correct answer is:
$17 / 54$

Solution:
Given expansion is
$\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$
$=\left(1+x+2 x^{3}\right)\left[\left(\frac{3}{2} x^{2}\right)^{9}-{ }^{9} C_{1}\left(\frac{3}{2} x^{2}\right)^{8} \cdot \frac{1}{3 x}\right.$
$\left.+{ }^{9} C_{6}\left(\frac{3}{2} x^{2}\right)^{3}\left(\frac{1}{3 x}\right)^{6}-{ }^{9} C_{7}\left(\frac{3}{2} x^{2}\right)^{2}\left(\frac{1}{3 x}\right)^{7} \ldots \ldots \ldots .\right]$
In the second bracket we have to search out terms of $x^{\circ}$
and $\frac{1}{x^{3}}$ which when multiplied with the terms 1 and
$2 x^{3}$ in the first bracket will give a term independent of
$x$. The term containing $\frac{1}{x}$ will not occur in the 2 nd
bracket.
- Term independent of $\mathrm{x}$
$=1-\left[{ }^{9} C_{6} \frac{3^{3}}{2^{3}} \cdot \frac{1}{3^{6}}\right]-2 x^{3}\left[{ }^{9} C_{7} \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}} \cdot \frac{1}{x^{3}}\right]$
$=\left[\frac{9.8 .7}{1.2 .3} \cdot \frac{1}{8.27}\right]-2\left[\frac{9.8}{1.2} \cdot \frac{1}{4.243}\right]$
$=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2009 (Phase 1)
What is the term used for the atoms within the square brackets in the compound formulation shown in Table 5.1?

Options:
A Primary valence
B Secondary valence
C Tertiary valence
D Quaternary valence

The correct answer is:
Primary valence

Solution:
The atoms within the square brackets in the compound formulation are referred to as "primary valence."

Topic: Chemistry, Coordination Compounds, NEET
What is the term used for the compounds' behavior under the reaction conditions as described in the paragraph?

Options:
A Primary valence
B Secondary valence
C Tertiary valence
D Quaternary valence

The correct answer is:
Quaternary valence

Solution:
The term "quaternary valence" is not mentioned in the paragraph. It is not used to describe the compounds' behavior under the reaction conditions.

Topic: Chemistry, Coordination Compounds, NEET
What is the term used for the electrode potential when the concentrations of all species in a half-cell are unity?

Options:
A Standard electrode potential
B Cell potential
C Electrolyte potential
D Reduction potential

The correct answer is:
Standard electrode potential

Solution:
When the concentrations of all species in a half-cell are unity, the electrode potential is known as standard electrode potential.

Topic: Chemistry, Electrochemistry, NEET
What is the term used for the lateral roots produced by the primary root in dicotyledonous plants?

Options:
A Secondary roots
B Tertiary roots
C Quaternary roots
D Quintary roots

The correct answer is:
Secondary roots

Solution:
The lateral roots produced by the primary root in dicotyledonous plants are referred to as secondary roots. They continue to grow and develop, forming more branches and helping the plant to absorb water and nutrients from a wider area of the soil.

Topic: Biology, Morphology of Flowering Plants, NEET
What is the term used for the potential difference between the electrode and the electrolyte in a half-cell?

Options:
A Cell potential
B Standard electrode potential
C Electrode potential
D Electrolyte potential

The correct answer is:
Electrode potential

Solution:
A potential difference develops between the electrode and the electrolyte which is called electrode potential.

Topic: Chemistry, Electrochemistry, NEET
What is the term used to describe the additional potential required for slow kinetically feasible electrochemical processes?

Options:
A Electrolyte potential
B Overpotential
C Standard potential
D Kinetic potential

The correct answer is:
Overpotential

Solution:
Slow kinetically feasible processes, an extra potential called "overpotential" has to be applied.

Topic: Chemistry, Electrochemistry, NEET
What is the total number of combination of $n$ different things taken $1,2,3, \ldots,, \mathrm{n}$ at a time?

Options:
A $2^{n+1}$
B $2^{2 n+1}$
C $2^{n-1}$
D $2^{n}-1$

The correct answer is:
$2^{n}-1$

Solution:
Since, combinations of taking $1,2,3, \ldots \ldots \ldots, n$ things at a times are ${ }^{n} C_{1},{ }^{n} C_{2}, \ldots \ldots . .,{ }^{n} C_{n}$
$\therefore \quad$ Total number of combinations $={ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots . .+{ }^{n} C_{n}$
$=1+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots . .+{ }^{n} C_{n}-1$
$=2^{n}-1$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
What is the unique behaviour of Cu with regard to its standard electrode potential (Eo)?

Options:
A Cu has a negative Eo.
B Cu has a positive Eo.
C Cu's E_ is zero.
D Cu's E_ depends on the temperature.

The correct answer is:
Cu has a positive Eo.

Solution:
The unique behaviour of Cu, having a positive E_, accounts for its inability to liberate H2 from acids.

Topic: Chemistry, d and f Block Elements, NEET
What is the value of $\int_{0}^{1}(x-1) e^{-x} d x$ ?

Options:
A 0
B e
C $\frac{1}{\mathrm{e}}$
D $\frac{-1}{\mathrm{e}}$

The correct answer is:
$\frac{-1}{\mathrm{e}}$

Solution:
Given integral is $\mathrm{I}=\int_{0}^{1}(\mathrm{x}-1) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}$
Integrating by parts taking $(\mathrm{x}-1)$ as first function
We get, $\mathrm{I}=\left[(\mathrm{x}-1)\left\{-\mathrm{e}^{-\mathrm{x}}\right\}\right]_{0}^{1}-\int_{0}^{1} 1 .\left(-\mathrm{e}^{-\mathrm{x}}\right) \mathrm{dx}$
$=-(1-1) \frac{1}{\mathrm{e}}+(-1) \mathrm{e}^{0}+\left[-\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=-1-\frac{1}{\mathrm{e}}+1=-\frac{1}{\mathrm{e}}$

Topic: Mathematics, Definite Integration, NDA, NDA 2006 (Phase 2)
What is the value of $\int_{0}^{1} x e^{x^{2}} d x$

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

The correct answer is:
$\frac{(\mathrm{e}-1)}{2}$

Solution:
Let $\mathrm{I}=\int_{0}^{1} \mathrm{xe}^{\mathrm{x}^{2}} \cdot \mathrm{dx}$
Let $x^{2}=t$
$\Rightarrow 2 x d x=d t$
$\Rightarrow x d x=\frac{d t}{2}$
when $\mathrm{x}=0, \mathrm{t}=0$ then $\mathrm{x}=1, \mathrm{t}=1$
$\Rightarrow \mathrm{I}=\frac{1}{2} \int_{0}^{1} \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\frac{1}{2}\left[\mathrm{e}^{\mathrm{t}}\right]_{0}^{1}$
$=\frac{1}{2}\left[\mathrm{e}^{\mathrm{x}^{2}}\right]_{0}^{1}=\frac{1}{2}\left[\mathrm{e}-\mathrm{e}^{0}\right]=\frac{\mathrm{e}-1}{2}$

Topic: Mathematics, Definite Integration, NDA, NDA 2008 (Phase 1)
What is the value of $0 . \overline{2}+0 . \overline{23}$ ?

Options:
A $0 . \overline{43}$
B $0 . \overline{45}$
C $0 . \overline{223}$
D $0.2 \overline{23}$

The correct answer is:
$0 . \overline{45}$

Solution:
$0 . \overline{2}+0 . \overline{23}$
$=\frac{2}{9}+\frac{23}{99}=\frac{22+23}{99}=\frac{45}{99}=0 . \overline{45}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
What is the value of $\int_{0}^{\pi / 2} \cos ^{8} x d x$ ?

Options:
A $\frac{35 \pi}{256}$
B $\frac{70}{256}$
C $\frac{16}{35}$
D $\frac{8 \pi}{35}$

The correct answer is:
$\frac{35 \pi}{256}$

Solution:
Let, $I=\int_{0}^{\pi / 2} \cos ^{8} x d x$
Given integral can be also be written as:
$\mathrm{I}=\int_{0}^{\pi / 2} \sin ^{\circ} \mathrm{x} \cos ^{8} \mathrm{x} \mathrm{dx}$, which is known as Gamma
function.
$$
\begin{array}{l}
\text { Solution is : } \int_{0}^{\pi / 2} \sin ^{\mathrm{m}} x \cos ^{\mathrm{n}} x \mathrm{dx} \\
=\frac{[(\mathrm{m}-1)(\mathrm{m}-3) \ldots 2 \operatorname{or} 1][(\mathrm{n}-1)(\mathrm{n}-3) \ldots 2 \mathrm{or} 1]}{[(\mathrm{m}+\mathrm{n})(\mathrm{m}+\mathrm{n}-2) \ldots .2 \mathrm{or} 1]} \cdot \theta
\end{array}
$$
If $m$ and $n$ both are even then RHS should be multiplied
by $\frac{\pi}{2}$ here, $\mathrm{m}=0, \mathrm{n}=8$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{I} &=\frac{(8-1)(8-3)(8-5) \ldots(8-7)}{8 .(8-2)(8-4)(8-6)} \frac{\pi}{2} \\
&=\frac{7.5 .3 .1}{8.6 .4 .2} \frac{\pi}{2}=\frac{35 \pi}{256}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)
What is the value of $\int_{0}^{\pi / 2} \log (\tan x) d x ? \quad$

Options:
A 0
B 1
C -1
D $\pi / 4$

The correct answer is:
0

Solution:
Let $I=\int_{0}^{\pi / 2} \log (\tan x) d x$
and $I=\int_{0}^{\pi / 2} \log \left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x$
[By the property of definite integral which says
$\left.\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$\Rightarrow I=\int_{0}^{\pi / 2} \log (\cot x) d x$
By adding equation (i) and (ii), we get $2 \mathrm{I}=\int_{0}^{\pi / 2} \log (\tan x) d x+\int_{0}^{\pi / 2} \log (\cot x) d x$
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \log (\tan x \cot x) d x$
$[\because \log m+\log n=\log (m n)]$
$=\int_{0}^{\frac{\pi}{2}} \log \left(\tan x \cdot \frac{1}{\tan x}\right) d x=\int_{0}^{\frac{\pi}{2}} \log 1 d x=0$
$\Rightarrow \mathrm{I}=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2009 (Phase 1)
What is the value of $\int_{-1}^{1} \mathrm{x}|\mathrm{x}| \mathrm{dx}$ ?

Options:
A 2
B 1
C $\frac{1}{4} \quad$
D $0 \quad\}$

The correct answer is:
$0 \quad\}$

Solution:
Let $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|$
$\mathrm{f}(-\mathrm{x})=-\mathrm{x}|-\mathrm{x}|=-\mathrm{x}|\mathrm{x}|=-\mathrm{f}(\mathrm{x})$
$\Rightarrow \mathrm{x}|\mathrm{x}|$ is an odd function
Hence, $\int_{-1}^{1} x|x| d x=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)
What is the value of
$\left|\begin{array}{ccc}1 & \omega & 2 \omega^{2} \\ 2 & 2 \omega^{2} & 4 \omega^{3} \\ 3 & 3 \omega^{3} & 6 \omega^{4}\end{array}\right|$
where $\omega$ is the cube root of unity?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Consider $\left|\begin{array}{ccc}1 & \omega & 2 \omega^{2} \\ 2 & 2 \omega^{2} & 4 \omega^{3} \\ 3 & 3 \omega^{3} & 6 \omega^{4}\end{array}\right|=\left|\begin{array}{ccc}1 & \omega & 2 \omega^{2} \\ 2 & 2 \omega^{2} & 4 \\ 3 & 3 & 6 \omega\end{array}\right|$
$\left(\because \omega^{3}=1\right.$ and $\left.\omega^{4}=\omega\right)$
$=2 \times 3\left|\begin{array}{ccc}1 & \omega & 2 \omega^{2} \\ 1 & \omega^{2} & 2 \\ 1 & 1 & 2 \omega\end{array}\right|$
$=6\left[1\left(2 \omega^{3}-2\right)-\omega(2 \omega-2)+2 \omega^{2}\left(1-\omega^{2}\right)\right]$
$=6\left[0-2 \omega^{2}-2 \omega+2 \omega^{2}-2 \omega\right]=0$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 2)
What is the value of $\int_{1}^{2} e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x^{\prime} ? \quad$

Options:
A $e\left(\frac{e}{2}-1\right)$
B $\mathrm{e}(\mathrm{e}-1)$
C $e-\frac{1}{e}$
D 0

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

Solution:
Let $\mathrm{I}=\int_{1}^{2} e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x$
$=\int_{1}^{2} e^{x}\left(f(x)+f^{\prime}(x)\right) d x$ where $f(x)=\frac{1}{x}$
$=\left.e^{x} f(x)\right|_{1} ^{2}$
$\therefore \quad \mathrm{I}=\left.\frac{e^{x}}{x}\right|_{1} ^{2}=\frac{e^{2}}{2}-e=e\left(\frac{e}{2}-1\right)$

Topic: Mathematics, Definite Integration, NDA, NDA 2010 (Phase 2)
What is the value of $(-\sqrt{-1})^{4 \mathrm{n}+3}+\left(\mathrm{i}^{41}+i^{-257}\right)^{9}$,
where $n \in N ?$

Options:
A 0
B 1
C i.
D -i

The correct answer is:
i.

Solution:
Consider $(-\sqrt{-1})^{4 n+3}+\left(i^{41}+i^{-257}\right)^{9}$
$=(-i)^{4 n+3}+\left[\left(i^{4}\right)^{10} . i^{1}+\left(i^{3}\right)^{-85} \cdot i^{-2}\right]^{9}$
$=(-i)^{4 n+3}+\left[i+\frac{1}{\left(i^{3}\right)^{85}} \cdot \frac{1}{i^{2}}\right]^{9}=(-i)^{4 n+3}+\left(i+\frac{1}{i}\right)^{9}$
$=-(-1)^{4 n+3}(i)^{4 n}(i)^{3}+(i-i)^{9}=-(1)(-i)+0=i$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 2)
What is the value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) ?$

Options:
A $\frac{1}{2}$
B $\frac{1}{2}+\frac{1}{2 \sqrt{2}}$
C $\frac{1}{2}-\frac{1}{2 \sqrt{2}}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{8}$

Solution:
$\left[1+\cos \frac{\pi}{8}\right]\left[1+\cos \frac{3 \pi}{8}\right]\left[1+\cos \frac{5 \pi}{8}\right]\left[1+\frac{\cos 7 \pi}{8}\right]$
We have, $\cos \frac{7 \pi}{8}=\cos \left[\pi-\frac{\pi}{8}\right]=-\cos \frac{\pi}{8}$
and $\cos \frac{5 \pi}{8}=\cos \left[\pi-\frac{3 \pi}{8}\right]=-\cos \frac{3 \pi}{8}$
$\therefore=\left[1+\cos \frac{\pi}{8}\right]\left[1+\cos \frac{3 \pi}{8}\right]\left[1-\cos \frac{\pi}{8}\right]\left[1-\cos \frac{3 \pi}{8}\right]$
$=\left[1-\cos ^{2} \frac{\pi}{8}\right]\left[1-\cos ^{2} \frac{3 \pi}{8}\right]=\sin ^{2} \frac{\pi}{8} \cdot \sin ^{2} \frac{3 \pi}{8}$
$=\frac{1}{4}\left[2 \sin ^{2} \frac{\pi}{8} \cdot 2 \sin ^{2} \frac{3 \pi}{8}\right]$
$=\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right)\right]$
$\left(\because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}\right)$
$=\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right)\right]=\frac{1}{8}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
What is the value of $\left|\begin{array}{ccc}1-i & \omega^{2} & -\omega \\ \omega^{2}+i & \omega & -i \\ 1-2 i-\omega^{2} & \omega^{2}-\omega & i-\omega\end{array}\right|$, where $\omega$ is the cube root of unity?

Options:
A $-1$
B 1
C 2
D 0

The correct answer is:
0

Solution:
Let $\Delta=\left|\begin{array}{ccc}1-i & \omega^{2} & -\omega \\ \omega^{2}+i & \omega & -i \\ 1-2 i-\omega^{2} & \omega^{2}-\omega & i-\omega\end{array}\right|$
Applying $R_{3} \rightarrow R_{1}-R_{2}-R_{3}$
$=\left|\begin{array}{ccc}1-i & \omega^{2} & -\omega \\ \omega^{2}+i & \omega & -i \\ 0 & 0 & 0\end{array}\right|=0$
$(\because$ one row of determinant is zero)

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 2)
What is the value of
$$
1+i^{2}+i^{4}+i^{6}+\ldots . .+i^{100}
$$
where $i=\sqrt{-1}$ ? $[2010-I I]$

Options:
A 0
B 1
C -1
D None of these

The correct answer is:
1

Solution:
Consider $1+\mathrm{i}^{2}+\mathrm{i}^{4}+\mathrm{i}^{6}+\ldots+\mathrm{i}^{100}$
$=1+\left[i^{2}+i^{4+} i^{6}+\ldots+i^{100}\right]$
$=1+[(-1)+(1)-1+\ldots+1]$
$=1-1+1-1+\ldots+1=1$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
What is the value of
$\left.[-1+\mathrm{i} \sqrt{3}) / 2]^{10}+[-1-\mathrm{i} \sqrt{3}) / 2\right]^{10}$

Options:
A 1
B -1
C 2
D 0

The correct answer is:
-1

Solution:
We know that $\frac{-1+\mathrm{i} \sqrt{3}}{2}=\omega$,
so, $\frac{-1-i \sqrt{3}}{2}=\omega^{2}$
$\therefore\left[\frac{-1+\mathrm{i} \sqrt{3}}{2}\right]^{10}+\left[\frac{-1-\mathrm{i} \sqrt{3}}{2}\right]^{10}$
$=\omega^{10}+\omega^{20}=\omega^{3 \times 3+1}+\omega^{3 \times 6+2}$
$=\omega^{3} \omega^{3} \omega+\omega^{3} \omega^{6} \omega^{2}=\omega+\omega^{2} \quad$ (since $\left.\omega^{3}=1\right)$
and $\omega+\omega^{2}=-1 \quad\left(1+\omega+\omega^{2}=0\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 1)
What is the value of $(-1+i \sqrt{3})^{48}$ ?

Options:
A 1
B 2
C $2^{24}$
D $2^{48}$

The correct answer is:
$2^{48}$

Solution:
We know, $\omega=\frac{1+i \sqrt{3}}{2}$ and
$\omega^{2}=\frac{1-i \sqrt{3}}{2} \Rightarrow 1-i \sqrt{3}=2 \omega^{2}$
Consider $(-1+i \sqrt{3})^{48}=[-(1-i \sqrt{3})]^{48}$
$=(1-i \sqrt{3})^{48}$
$\Rightarrow \quad\left(2 \omega^{2}\right)^{48}=2^{48} \cdot \omega^{96}=2^{48}\left(\omega^{3}\right)^{32}=2^{48}$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
What is the value of $(1+i)^{5}+(1-i)^{5}$ where $i=\sqrt{-1}$ ?

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

The correct answer is:
-8

Solution:
$\begin{aligned} &(1+i)^{5}={ }^{5} \mathrm{C}_{0}(1)^{5-0} i^{0}+{ }^{5} \mathrm{C}_{1}(1)^{5-1}(i)^{1}+{ }^{5} \mathrm{C}_{2}(1)^{5-2}(i)^{2} \\ &+{ }^{5} \mathrm{C}_{3}(1)^{5-3}(i)^{3}+{ }^{5} \mathrm{C}_{4}(1)^{5-4}(i)^{4}+{ }^{5} \mathrm{C}_{5}(1)^{5-5}(i)^{5} \\ &=1+5 i+10 i^{2}+10 i^{3}+5 i^{4}+i^{5} \\ &(1-i)^{5}={ }^{5} \mathrm{C}_{0} i^{0}-{ }^{5} \mathrm{C}_{1} i+{ }^{5} \mathrm{C}_{2} i^{2}-{ }^{5} \mathrm{C}_{3} i^{3} \\ & \quad+{ }^{5} \mathrm{C}_{4} i^{4}-{ }^{5} \mathrm{C}_{5} i^{5} \\ &=1-5 i+10 i^{2}-10 i^{3}+5 i^{4}-i^{5} \ldots \ldots(2) \end{aligned}$
By adding $(1)$ and $(2)$, we get $(1+i)^{5}+(1-i)^{5}=1+5 i+10 i^{2}+10 i^{3}+5 i^{4}+i^{5}$
$\quad+1-5 i+10 i^{2}-10 i^{3}+5 i^{4}-\mathrm{i}^{5}$
$=2+20 i^{2}+10 i^{4}=2-20+10=-8$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
What is the value of $\frac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}$ ?

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

The correct answer is:
$\sqrt{3}$

Solution:
Consider $\frac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}=\frac{\tan 45^{\circ}+\tan 15^{\circ}}{1-\tan 45^{\circ} \tan 15^{\circ}}\left(\because \tan 45^{\circ}=1\right)$
$\begin{aligned}=& \tan \left(45^{\circ}+15^{\circ}\right) \quad\left(\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\right) \\ &=\tan 60^{\circ}=\sqrt{3} \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)