datavorous/entrance-exam-dataset huggingface dataset:
What is the product obtained when nitrile is reduced with lithium aluminum hydride (LiAlH4)?

Options:
A Secondary amine
B Tertiary amine
C Primary amine
D Quaternary ammonium salt

The correct answer is:
Primary amine

Solution:
Nitriles, when reduced with LiAlH4, produce primary amines.

Topic: Chemistry, Amines, NEET
What is the product of first $2 \mathrm{n}+1$ terms of a geometric progression?

Options:
A The $(\mathrm{n}+1)$ th power of the nth term of the GP
B The $(2 \mathrm{n}+1)$ th power of the nth term of the GP
C The $(2 \mathrm{n}+1)$ th power of the $(\mathrm{n}+1)$ th term of the GP
D The nth power of the $(\mathrm{n}+1)$ th terms of the GP

The correct answer is:
The $(2 \mathrm{n}+1)$ th power of the $(\mathrm{n}+1)$ th term of the GP

Solution:
The GP is a, ar, ar $^{2}, \ldots \ldots .$ ar $^{2 \mathrm{n}}$ So, $\mathrm{P}=\mathrm{a}$. ar $\cdot \mathrm{ar}^{2} \cdot \mathrm{ar}^{3} \ldots \ldots \ldots \ldots \mathrm{ar}^{2 n}$
$=a^{2 n+1}, r^{1+2+\ldots}+2 n$
$=a^{(2 n+1)}_{r} \frac{2 n(2 n+1)}{2}=a^{2 n+1} r^{n(2 n+1)}=\left(a r^{n}\right)^{(2 n+1)}$
$=(2 n+1)$ th power of the $(\mathrm{n}+1)$ th term of $\mathrm{GP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 1)
What is the product of the binary numbers $1001.01$ and 11.1?

Options:
A $101110.011$
B $100000.011$
C $101110.101$
D 100000.101

The correct answer is:
$100000.011$

Solution:
$1001.01=1 \times 2^{3}+1 \times 2^{0}+1 \times 2^{-2}$
corresponding to number of base 10 .
$=8+1+\frac{1}{4}=\frac{37}{4}=9.25$
and $11.1=1 \times 2^{1}+1 \times 2^{0}+1 \times 2^{-1}$
$=2+1+\frac{1}{2}=\frac{7}{2}=3.5$
Corresponding to the number of base 10 $1001.01 \times 11.1=9.25 \times 3.5=32.375$
From decimal to binary (32) $_{10}=(100000)_{2}$
and $(.375)_{10}=0.25+0.125$
$=\frac{1}{4}+\frac{1}{8}=1 \times 2^{-2}+1 \times 2^{-3}=(0.011)_{2}$
$(32.375)_{10}=(100000.011)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 1)
What is the product of the perpendiculars drawn from the points $\left(\pm \sqrt{a^{2}-b^{2}}, 0\right)$ upon the line $b x \cos \alpha+$ ay sin $\alpha=a b ?$

Options:
A $a^{2}$
B $b^{2}$
C $a^{2}+b^{2}$
D $a+b$

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

Solution:
Given equation of line is $b x \cos \alpha+$ ay $\sin \alpha=a b$ Perpendicular distance from point $\left(\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ is $d_{1}=\left|\frac{b \cos \alpha \sqrt{a^{2}-b^{2}}+0-a b}{\sqrt{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}}\right|$
$\left(\because\right.$ distance from $\left(x_{1}, y_{1}\right)$ to $a x+b y+c=0$ is $=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|$
Similarly, perpendicular distance from point $\left(-\sqrt{a^{2}-b^{2}}, 0\right)$ is
$d_{2}=\left|\frac{-b \cos \alpha \sqrt{a^{2}-b^{2}}+0-a b}{\sqrt{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}}\right|$
By product of $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$, we get $=\frac{\left(\mathrm{b} \cos \alpha \sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}-\mathrm{ab}\right)\left(\mathrm{b} \cos \alpha \sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}+\mathrm{ab}\right)}{\left(\sqrt{\mathrm{b}^{2} \cos ^{2} \alpha+\mathrm{a}^{2} \sin ^{2} \alpha}\right)\left(\sqrt{ \left.\mathrm{b}^{2} \cos ^{2} \alpha+\mathrm{a}^{2} \sin ^{2} \alpha\right)}\right.}$
$=\frac{\mathrm{b}^{2} \cos ^{2} \alpha\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)-\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{~b}^{2} \cos ^{2} \alpha+\mathrm{a}^{2} \sin ^{2} \alpha}$
$=\frac{a^{2} b^{2} \cos ^{2} \alpha-b^{4} \cos ^{2} \alpha-a^{2} b^{2}}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}$
$=\frac{\mathrm{a}^{2} \mathrm{~b}^{2}\left(\cos ^{2} \alpha-1\right)-\mathrm{b}^{4} \cos ^{2} \alpha}{\mathrm{b}^{2} \cos ^{2} \alpha+\mathrm{a}^{2} \sin ^{2} \alpha}$
$=\frac{-b^{2}\left[a^{2} \sin ^{2} \alpha+b^{2} \cos ^{2} \alpha\right]}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}$
Hence, product of the perpendiculars $=-b^{2}=b^{2}$ (since, distance is positive)

Topic: Mathematics, Pair of Lines, NDA, NDA 2014 (Phase 2)
What is the product of the perpendiculars from the two points $\left(\pm \sqrt{b^{2}-a^{2}}, 0\right)$ to the line ax $\cos \phi+$ by $\sin \phi=a b$ ?

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

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

Solution:
Given equation of line is $a x \cos \phi+$ by $\sin \phi-a b=0$ Let $d_{1}$ be the perpendicular distance from
$\left(\sqrt{b^{2}-a^{2}}, 0\right)$ to the line $a x \cos \phi+$ by $\sin \phi-a b=0$
and $d_{2}$ from $\left(-\sqrt{b^{2}-a^{2}}, 0\right)$ to the line $a x \cos \phi+$ by sin $\phi-a b=0$
At point $\left(\sqrt{b^{2}-a^{2}}, 0\right)$
$d_{1}=\frac{a \sqrt{b^{2}-a^{2}} \cos \phi-a b}{\sqrt{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}}$
At point $\left(-\sqrt{b^{2}-a^{2}}, 0\right)$
$d_{2}=\frac{-a \sqrt{b^{2}-a^{2}} \cos \phi-a b}{\sqrt{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}}$
$\therefore d_{1} d_{2}=-\frac{\left[a^{2}\left(b^{2}-a^{2}\right) \cos ^{2} \phi-a^{2} b^{2}\right]}{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}$
$=-\frac{a^{2}\left(-b^{2} \sin ^{2} \phi-a^{2} \cos ^{2} \phi\right)}{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}=a^{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2009 (Phase 2)
What is the product of the reaction between acetone and a Grignard reagent?

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

The correct answer is:
Tertiary alcohol

Solution:
Acetone reacts to form a tertiary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
What is the product of the reaction between benzene sulphonyl chloride and a secondary amine?

Options:
A N-ethyl benzene sulfonyl amide
B N,N-diethyl benzene sulphonamide
C N-methylbenzenesulphonyl amide
D N,N-dimethyl benzenesulfonamide

The correct answer is:
N,N-diethyl benzene sulphonamide

Solution:
The reaction of benzenesulfonyl chloride with a secondary amine yields N,N-diethyl benzene sulphonamide.

Topic: Chemistry, Amines, NEET
What is the product of two parts of 20 , such that the product of one part and the cube of the other is maximum?

Options:
A 75
B 91
C 84
D 96

The correct answer is:
75

Solution:
Let 20 be divided in two parts such that first part $=\mathrm{x}$ $\therefore \quad$ Second part $=20-\mathrm{x}$
Now, assume that $\mathrm{P}=\mathrm{x}^{3}(20-\mathrm{x})$
$=20 \mathrm{x}^{3}-\mathrm{x}^{4}$
Now, $\frac{\mathrm{dP}}{\mathrm{dx}}=60 \mathrm{x}^{2}-4 \mathrm{x}^{3}$
and $\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}=120 \mathrm{x}-12 \mathrm{x}^{2}$
Put $\frac{\mathrm{dP}}{\mathrm{dx}}=0$ for maxima or minima
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dx}}=0$
$\Rightarrow 4 \mathrm{x}^{2}(15-\mathrm{x})=0 \quad \Rightarrow \mathrm{x}=0, \mathrm{x}=15$
$\therefore\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=15}=120 \times 15-12 \times(225)$
$=1800-2700=-900 < 0$
$\therefore \quad P$ is a maximum at $x=15$. $\therefore \quad$ First part $=15$
and second part $=20-15=5$ Required product $=15 \times 5=75$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 2)
What is the projection of the vector $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ on the
vector $4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$ ?

Options:
A $\frac{\sqrt{5}}{2}$
B $\frac{19}{9}$
C $\frac{\sqrt{5}}{4}$
D $\frac{11}{3}$

The correct answer is:
$\frac{19}{9}$

Solution:
Let $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}=4 \hat{i}-4 \hat{j}+7 \hat{k}$
$\therefore \quad$ Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Now, $\vec{a} \cdot \vec{b}=4(1)+4(2)+1(7)=19$
and $|\vec{b}|=\sqrt{(4)^{2}+(4)^{2}+(7)^{2}}=\sqrt{81}=9$
$\therefore \quad$ Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{19}{9}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
What is the propability that there are 5 Mondays in the month of February 2016 ?

Options:
A 0
B $\frac{1}{7}$
C $\frac{2}{7}$
D None of these

The correct answer is:
$\frac{1}{7}$

Solution:
February 2016 has 29 days. $=4$ weeks $+1$ odd day.
Now, probability that 1 odd day is monday $=\frac{1}{7}$.

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
What is the purpose of computing the average atomic mass of an element?

Options:
A To determine the total number of atoms in the element
B To establish the most common isotope of the element
C To account for the presence of isotopes and their relative abundances
D To simplify the naming of elements

The correct answer is:
To account for the presence of isotopes and their relative abundances

Solution:
Computing the average atomic mass of an element is done to account for the presence of isotopes and their relative abundances.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
What is the purpose of ethical standards in genetic modification of organisms?

Options:
A To make decisions about the validity of GM research
B To evaluate the morality of human activities that might help or harm living organisms
C To regulate the manipulation of living organisms by the human race
D To control the problems with patents granted for the usage of living organisms for public services

The correct answer is:
To evaluate the morality of human activities that might help or harm living organisms

Solution:
Ethical standards are required to evaluate the morality of human activities that might help or harm living organisms, beyond the biological significance of such things.

Topic: Biology, Biotechnology and Its Applications, NEET
What is the purpose of ex situ conservation?

Options:
A To protect species at risk of extinction in the wild.
B To protect a specific region's biodiversity.
C To save the entire ecosystem.
D To address global conservation problems.

The correct answer is:
To protect species at risk of extinction in the wild.

Solution:
Ex situ conservation is the desirable approach when there are situations where an animal or plant is endangered or threatened and needs urgent measures to save it from extinction.

Topic: Biology, Biodiversity and its Conservation, NEET
What is the purpose of nomenclature?

Options:
A To standardize the naming of living organisms
B To facilitate the study of organisms
C To provide scientific names for plants
D To assign a name to each known organism

The correct answer is:
To standardize the naming of living organisms

Solution:
The purpose of nomenclature is to standardize the naming of living organisms such that a particular organism is known by the same name all over the world.

Topic: Biology, The Living World, NEET
What is the purpose of pumping air into the effluent in secondary treatment?

Options:
A To reduce the BOD
B To increase the pollution potential
C To promote the growth of useful aerobic microbes
D To increase the physical removal of particles

The correct answer is:
To promote the growth of useful aerobic microbes

Solution:
Air is pumped into the effluent in secondary treatment, allowing vigorous growth of useful aerobic microbes into flocs.

Topic: Biology, Microbes in Human Welfare, NEET
What is the purpose of secondary treatment of sewage?

Options:
A Physical removal of particles
B Removal of organic matter
C Reduction of BOD
D Increase in pollution potential

The correct answer is:
Removal of organic matter

Solution:
The primary effluent is passed into large aeration tanks where it is agitated mechanically and air is pumped into it, allowing vigorous growth of useful aerobic microbes.

Topic: Biology, Microbes in Human Welfare, NEET
What is the purpose of the formation of the ‘ES’ complex?

Options:
A To stabilize the substrate
B To form the transition state structure
C To catalyze the reaction
D To release the product

The correct answer is:
To catalyze the reaction

Solution:
The formation of the ‘ES’ complex, where E stands for enzyme, is mandatory for the catalyzation of the reaction. The complex formation helps in catalyzing the reaction.

Topic: Biology, Biomolecules (B), NEET
What is the purpose of the zona pellucida formed around the secondary oocyte?

Options:
A To protect the secondary oocyte
B To facilitate fertilization
C To facilitate meiosis
D None of the above

The correct answer is:
To facilitate fertilization

Solution:
The secondary oocyte forms a new membrane called zona pellucida surrounding it, which helps in the process of fertilization.

Topic: Biology, Human Reproduction, NEET
What is the purpose of ZnCl2 in the reaction of primary and secondary alcohols with HCl?

Options:
A ZnCl2 is a reactant in the reaction.
B ZnCl2 catalyses the reaction.
C ZnCl2 neutralises HCl.
D ZnCl2 produces a colour change.

The correct answer is:
ZnCl2 catalyses the reaction.

Solution:
ZnCl2 acts as a catalyst in the reaction of primary and secondary alcohols with HCl.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
What is the radius of the circle passing through the point $(2,4)$ and having centre at the intersection of the lines $\mathrm{x}-\mathrm{y}$ $=4$ and $2 x+3 y+7=0 ?$

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

The correct answer is:
$5 \sqrt{2}$ units

Solution:
We have $x-y=4 \& 2 x+3 y+7=0$
On solving, we get, $x=1 \& y=-3$
(these are coordinates of centre of the circle)
$\Rightarrow$ radius $=\sqrt{(2-1)^{2}+(4+3)^{2}}=5 \sqrt{2}$

Topic: Mathematics, Circle, NDA, NDA 2016 (Phase 2)
What is the radius of the circle passing through the points $(0,0),(a, 0)$ and $(0, b) ?$

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

The correct answer is:
$\frac{1}{2} \sqrt{a^{2}+b^{2}}$

Solution:
Let $(\mathrm{h}, \mathrm{k})$ be the centre of the circle. Since, circle is passing through $(0,0),(\mathrm{a}, 0)$ and $(0, \mathrm{~b})$, distance between centre and these points would be same and equal to radius. Hence, $h^{2}+k^{2}=(h-a)^{2}+k^{2}=h^{2}+(k-b)^{2}$
$$
\begin{array}{l}
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{b}^{2}-2 \mathrm{bk} \\
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah} \\
\Rightarrow \mathrm{h}=\frac{\mathrm{a}}{2}
\end{array}
$$
Similarly, $\mathrm{k}=\frac{\mathrm{b}}{2}$
$\therefore \quad$ Radius of circle $=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$

Topic: Mathematics, Circle, NDA, NDA 2006 (Phase 1)
What is the radius of the circle touching $x$ -axis at $(3,0)$ and $y$ -axis at $(0,3)$ ?

Options:
A 3 units
B 4 units
C 5 units
D 6 units

The correct answer is:
3 units

Solution:
Since, circle is touching $x$ -axis at $(3,0)$ and y-axis at $(0$,
3) therefore radius $=3$ unit

Topic: Mathematics, Circle, NDA, NDA 2011 (Phase 2)
What is the radius of the sphere $x^{2}+y^{2}+z^{2}-6 x+8 y-10 z$ $+1=0 ?$

Options:
A 5
B 2
C 7
D $\underline{3}$

The correct answer is:
7

Solution:
The equation of sphere is $x^{2}+y^{2}+z^{2}-6 x+8 y-10 z+1=0$
Comparing the equation with general form of sphere, $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z-d-0$
we get, $u=\frac{-6}{2}=-3, v=\frac{8}{2}=4, w=\frac{-10}{2}=5, d=1$
Radius $=\sqrt{u^{2}+v^{2}+w^{2}-d}$
$=\sqrt{(-3)^{2}+(4)^{2}+(5)^{2}-1}$
$=\sqrt{9+16+25-1}=\sqrt{49}=7$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2019 (Phase 1)
What is the radius of the sphere $x^{2}+y^{2}+z^{2}-x-y-z=0 ?$

Options:
A $\sqrt{\frac{3}{4}}$
B $\sqrt{\frac{1}{2}}$
C $\sqrt{\frac{3}{2}}$
D $\frac{1}{3}$

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

Solution:
The given equation of sphere is
$x^{2}+y^{2}+z^{2}-x-y-z=0$
On comparing with
$x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z+d=0$
we get, $u=-\frac{1}{2}, v=-\frac{1}{2}, w=-\frac{1}{2}, d=0$
Radius of sphere $=\sqrt{u^{2}+v^{2}+w^{2}-d}$
$=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{3}{4}}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 1)
What is the range of $f(x)=\cos 2 x-\sin 2 x$ ?

Options:
A $[2,4]$
B $\quad[-1,1]$
C $[-\sqrt{2}, \sqrt{2}]$
D $(-\sqrt{2}, 2)$

The correct answer is:
$[-\sqrt{2}, \sqrt{2}]$

Solution:
Let $f(x)=\cos 2 x-\sin 2 x$
$f(x)=\frac{1}{\sqrt{2}}[\sqrt{2} \cos 2 x-\sin 2 x]$
$f(x)=\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos 2 x-\frac{1}{\sqrt{2}} \sin 2 x\right]$
$f(x)=\sqrt{2}\left[\cos \frac{\pi}{4} \cos 2 x-\sin \frac{\pi}{4} \sin 2 x\right]$
$f(x)=\sqrt{2}\left[\cos \left(\frac{\pi}{4}+2 x\right)\right]$
We know,
$-1 \leq \cos \left(\frac{\pi}{4}+2 x\right) \leq 1$
$\Rightarrow-\sqrt{2} \leq \sqrt{2} \cos \left(\frac{\pi}{4}+2 x\right) \leq \sqrt{2}$
$\Rightarrow-\sqrt{2} \leq f(x) \leq \sqrt{2}$
$\therefore \quad$ Range of $f(x)=[-\sqrt{2}, \sqrt{2}]$

Topic: Mathematics, Functions, NDA, NDA 2011 (Phase 1)
What is the range of the function $\mathrm{f}(\mathrm{x})=\frac{|\mathrm{x}|}{\mathrm{x}}, \mathrm{x} \neq 0$?

Options:
A Set of all real numbers
B Set of all integers
C $\{-1,1\}$
D $\{-1,0,1\}$

The correct answer is:
$\{-1,1\}$

Solution:
As we know
$|\mathrm{x}|=\left\{\begin{array}{ccc}\mathrm{x} & \text { if } & \mathrm{x}>0 \\ -\mathrm{x} & \text { if } & \mathrm{x} < 0\end{array}\right.$
$\therefore f(x)=\frac{|x|}{x}=\left\{\begin{array}{lll}\frac{x}{x} & \text { if } & x>0 \\ \frac{-x}{x} & \text { if } & x < 0\end{array}\right.$
$=\left\{\begin{array}{ccc}1 & \text { if } & x>0 \\ -1 & \text { if } & x < 0\end{array}\right.$
Hence, Range $=\{-1,1\}$.

Topic: Mathematics, Functions, NDA, NDA 2013 (Phase 1)
What is the rangeofthe function $\mathrm{y}=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$, where $\mathrm{X} \in \mathbf{R}$ ?

Options:
A $[0,1)$
B $[0,1]$
C $(0,1$)
D $(0,1]$

The correct answer is:
$[0,1)$

Solution:
Clearly $0 \leq y < 1$
$\Rightarrow y \in[0,1)$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 1)
What is the rate of change of $\sqrt{\mathrm{x}^{2}+16}$ with respect to $\mathrm{x}^{2}$ at $\mathrm{x}=3 ?$

Options:
A $1 / 5$
B $1 / 10$
C $1 / 20$
D None of the above

The correct answer is:
$1 / 10$

Solution:
Let $\mathrm{u}=\sqrt{\mathrm{x}^{2}+16}, \mathrm{v}=\mathrm{x}^{2}$
$\Rightarrow \mathrm{u}^{2}=\mathrm{x}^{2}+16, \quad \frac{\mathrm{dv}}{\mathrm{dx}}=2 \mathrm{x}$
$\Rightarrow 2 \mathrm{u} \frac{\mathrm{du}}{\mathrm{dx}}=2 \mathrm{x}$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+16}}$
Now, required rate of change $=\frac{\mathrm{du}}{\mathrm{dv}}$
$=\frac{d u}{d x} \times \frac{d x}{d v}=\frac{x}{\sqrt{x^{2}+16}} \times \frac{1}{2 x}=\frac{1}{2 \sqrt{x^{2}+16}}$
Now, $\left.\frac{d u}{d v}\right|_{x=3}=\frac{1}{2 \times 5}=\frac{1}{10}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 1)
What is the ratio in which the line joining the points $(2,4,5)$ and $(3,5,-4)$ is internally divided by the $x y$ -plane?

Options:
A $5: 4$
B $3: 4$
C $1: 2$
D $7: 5$

The correct answer is:
$5: 4$

Solution:
Let the line joining the points $(2,4,5)$ and $(3,5,-4)$ is internally divided by the $x y$ - plane in the ratio $k: 1$.
$\therefore \quad$ For xy plane, $z=0$
$\Rightarrow 0=\frac{-\mathrm{k} \times 4+5}{\mathrm{k}+1} \Rightarrow 4 \mathrm{k}=5 \Rightarrow \mathrm{k}=\frac{5}{4}$.
$\mathrm{k}=\frac{5}{4}$ so, ratio is $5: 4$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
What is the ratio in which the point $C\left(-\frac{2}{7},-\frac{20}{7}\right)$ divides the line joining the points $\mathrm{A}(-2,-2)$ and $\mathrm{B}(2,-4)$ ?

Options:
A $1: 3$
B $3: 4$
C $1: 2$
D $2: 3$

The correct answer is:
$3: 4$

Solution:
$\quad \mathrm{C}=\left(\frac{2 \mathrm{~K}-2}{\mathrm{~K}+1}, \frac{-4 \mathrm{~K}+2}{\mathrm{~K}+1}\right)$
So, $\frac{2 \mathrm{~K}-2}{\mathrm{~K}+1}=\frac{-2}{7}$
$\mathrm{C}\left(\frac{-2}{7}, \frac{-20}{7}\right)$
$\Rightarrow 7 \mathrm{~K}-7=-\mathrm{K}-1$
$\Rightarrow 8 \mathrm{~K}=6 \Rightarrow \mathrm{K}=\frac{3}{4}$
$\therefore \mathrm{K}: 1=\frac{3}{4}: 1=3: 4$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)
What is the ratio of coefficient of $x^{15}$ to the term independent
of $x$ in $\left(x^{2}+\frac{2}{x}\right)^{15} ?$

Options:
A $1 / 64$
B $1 / 32$
C $1 / 16$
D $1 / 4$

The correct answer is:
$1 / 32$

Solution:
Given expansion is $\left(x^{2}+\frac{2}{x}\right)^{15}$
$\mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{r}\left(x^{2}\right)^{15-r}\left(\frac{2}{x}\right)^{r}$
$={ }^{15} \mathrm{C}_{r} x^{30-2 r} 2^{r} \cdot x^{-r}={ }^{15} \mathrm{C}_{\mathrm{r}} . \mathrm{x}^{30-3 r} \cdot 2^{r}$
Now, Above term will be independent of $x$ when $30-3 r=0 \Rightarrow r=10$
$\therefore$ Term independent of $\mathrm{x}={ }^{15} \mathrm{C}_{10} 2^{10}$
Now, coeff of $\mathrm{x}^{15}$ When $30-3 r=15 \Rightarrow r=5$
$\therefore$ Required coeff $={ }^{15} \mathrm{C}_{5} 2^{5}$
Thus, Required Ratio $=\frac{15 \mathrm{C}_{5} \cdot 2^{5}}{15 \mathrm{C}_{10} \cdot 2^{10}}$
$=\frac{\frac{15 !}{5 !(10 !)}}{\frac{15 !}{10 ! 5 !} \times 2^{5}}=\frac{1}{2^{5}}=\frac{1}{32}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2011 (Phase 2)
What is the real part of $(\sin x+\mathrm{i} \cos x)^{3}$ where $i=\sqrt{-1}$ ?

Options:
A $-\cos 3 x$
B $-\sin 3 x$
C $\sin 3 x$
D $\cos 3 x$

The correct answer is:
$-\sin 3 x$

Solution:
$\begin{aligned} &(\sin x+i \cos x)^{3} \\ &=\sin ^{3} x+(i)^{3} \cos ^{3} x+3 i(\sin x)(\cos x)(\sin x+i \cos x) \\ &=\sin ^{3} x-i \cos ^{3} x+3 i \sin ^{2} x \cos x-3 \sin x \cos ^{2} x \\ &=\sin ^{3} x-3 \sin x \cos ^{2} x-i \cos x\left(\cos ^{2} x+\sin ^{2} x\right) \\ &=\sin x\left(\sin ^{2} x-3 \cos ^{2} x\right)-i \cos x\left(\cos ^{2} x+3 \sin ^{2} x\right) \\ & \text { Real part of }(\sin x+i \cos x)^{3} \\ &=\sin x\left(\sin ^{2} x-3 \cos ^{2} x\right) \\ &=\sin x\left[\sin ^{2} x-3\left(1-\sin ^{2} x\right)\right] \\ &=\sin x\left[4 \sin ^{2} x-3\right] \\ &=4 \sin ^{3} x-3 \sin x \\ &=-\left(3 \sin x-4 \sin ^{3} x\right)=-\sin 3 x \end{aligned}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 1)
What is the region that represents $\mathrm{A} \cap \mathrm{B}$ if $\mathrm{A}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}+\mathrm{y} \leq 4\}$ and $\mathrm{B}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}+\mathrm{y} \leq 0\} ?$

Options:
A $\{(x, y) \mid x+y \leq 2\}$
B $\{(x, y) \mid 2 x+y \leq 4\}$
C $\{(x, y) \mid x+y \leq 0\}$
D $\{(x, y) \mid x+y \leq 4\}$

The correct answer is:
$\{(x, y) \mid x+y \leq 0\}$

Solution:
Asgiven $\mathrm{A}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}+\mathrm{y} \leq 4\}$
and $B=\{(x, y) \mid x+y \leq 0\}$
Set Acontains all the pairs in the interval $(-\propto, 2)$ andset B contains all the pairs in the interval $(-\infty, 0)$ so, $\mathrm{A} \cap \mathrm{B}$ shows a set containing all the pairs in the interval $[-\infty, 0]$ So, $A \cap B=\{(x, y) \mid x+y \leq 0\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
What is the relationship between RNA and DNA?

Options:
A RNA is more abundant than DNA
B DNA is more abundant than RNA
C RNA and DNA have no relationship
D They are both genetic materials

The correct answer is:
They are both genetic materials

Solution:
Both RNA and DNA are genetic materials found in living systems. However, DNA acts as the genetic material in most organisms, while RNA mostly functions as a messenger.

Topic: Biology, Molecular Basis of Inheritance, NEET
What is the relationship between the emf of the cell (E(cell)) and the standard electromotive force (Eo(cell)) when the concentration of all reacting species is unity?

Options:
A E(cell) = Eo(cell)
B E(cell) = 2 Eo(cell)
C E(cell) = 0.5Eo(cell)
D E(cell) = ΔrGo / RT

The correct answer is:
E(cell) = Eo(cell)

Solution:
When the concentration of all reacting species is unity, E(cell) = Eo(cell).

Topic: Chemistry, Electrochemistry, NEET
What is the relationship between the Gibbs energy change (∆rG°) and the equilibrium constant (K) for a reaction at standard state conditions?

Options:
A ∆rG° = RT ln K
B ∆rG° = – RT ln K
C ∆rG° = 2.303 RT log K
D ∆rG° = – 2.303 RT log K

The correct answer is:
∆rG° = – RT ln K

Solution:
The relationship between the Gibbs energy change (∆rG°) and the equilibrium constant (K) for a reaction at standard state conditions is given by ∆rG° = – RT ln K.

Topic: Chemistry, Thermodynamics (C), NEET