datavorous/entrance-exam-dataset huggingface dataset:
If two dice are thrown and at least one of the dice shows 5 . then the probability that the sum is 10 or more is

Options:
A $\frac{1}{6}$
B $\frac{4}{11}$
C $\frac{3}{11}$
D $\frac{2}{11}$

The correct answer is:
$\frac{3}{11}$

Solution:
A = Event of showing 5 and atleast one dice $=(1,5),(2,5),(3,5),(4,5)(5,5),(6,5),(5,1)$
$(5,2),(5,3),(5,4),(5,6)$
$n(\mathrm{~A})=11$
and $n(s)=6 \times 6=36$
$\mathrm{B}=$ Event of showing sum 10 or more when atleast one dice shows 5 $=(5,5),(5,6),(6,5)$
$n(\mathrm{~B})=3$
$\Rightarrow \quad n(A \cap B)=3$
Now, $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$
$=\frac{\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}}{\frac{n(\mathrm{~A})}{n(\mathrm{~S})}}=\frac{\frac{3}{36}}{\frac{11}{36}}=\frac{3}{11}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
If two dice are thrown, then what is the probability that the sum on the two faces is greater than or equal to $4$ ?

Options:
A $\frac{13}{18}$
B $\frac{5}{6}$
C $\frac{11}{12}$
D $\frac{35}{36}$

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

Solution:
Since two dice are thrown so number of outcomes are $36 .$
No. of ways when sum on two faces less than $4=3$. $[(1,1),(1,2),(2,1)]$
Hence prob of getting sum on two faces less than 4
$$
=\frac{3}{36}=\frac{1}{12}
$$
Thus required prob. that sum on the two faces is greater
than or equal to $4=1-\frac{1}{12}=\frac{11}{12}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
If two fair dice are rolled then what is the conditional probability that the first dice lands on 6 given that the sum of numbers on the dice is 8 ?

Options:
A $\frac{1}{3}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{5}$

Solution:
Total number of cases that sum is 8 are $(2,6),(3,5),(4,4),(5,3),(6,2)$
favourable case $=(6,2)$
$\therefore$ Probability $=\frac{1}{5}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
If two fair dice are thrown, then what is the probability that the sum is neither 8 nor $9$ ?

Options:
A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{3}{4}$
D $\frac{5}{6}$

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

Solution:
Total no of outcomes when two dice are thrown $=6 \times 6=36$
outcomes when sum is $8=(2,6)(6,2)(3,5)(5,3)(4,4)$ $=5$
Outcomes when sum is 9 $=(3,6)(6,3)(4,5)(5,4)$
$=4$
Possible number of outcomes $=36-(5+4)=36-9=27$
$\therefore \quad$ Required probability $=\frac{27}{36}=\frac{3}{4}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
If two lines of regression are perpendicular, then the correlation coefficient $\mathrm{r}$ is

Options:
A 2
B $\frac{1}{2} \quad$
C 0
D None of the above

The correct answer is:
0

Solution:
It is a property

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
If two regression lines between height $(\mathrm{x})$ and weight $(\mathrm{y})$ are $4 y-15 x+410=0$ and $30 x-2 y-825=0$, then what
will be the correlation coefficient between height ang weight?

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

The correct answer is:
$\frac{2}{3}$

Solution:
$4 y-15 x+410=0$
$\Rightarrow y-\frac{15}{4} x+\frac{410}{4}=0 \Rightarrow y=\frac{15}{4} x-\frac{410}{4}$
$\mathrm{b}_{\mathrm{yx}}=\frac{15}{4}$
$30 x-2 y-825=0 \Rightarrow x=\frac{2}{30} y+\frac{825}{30}$
$b_{x y}=\frac{2}{30}$
Correlation coefficient $=\sqrt{\left(\mathrm{b}_{\mathrm{yx}}\right)\left(\mathrm{b}_{\mathrm{xy}}\right)}$
$=\sqrt{\frac{15}{4} \times \frac{2}{30}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
If two rows of a determinant are identical, then what is the value of the determinant?

Options:
A 0
B 1
C $-1$
D can be any real value

The correct answer is:
0

Solution:
It is a property.

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 1)
If two unit vectors $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{q}}$ make an angle $\frac{\pi}{3}$ with each
other, what is the magnitude of $\vec{p}-\frac{1}{2} \vec{q}$ ?

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

The correct answer is:
$\frac{\sqrt{3}}{2}$

Solution:
If $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{q}}$ are unit vectors which make an angle $\frac{\pi}{3}$ with each other.
Then, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}=|\overrightarrow{\mathrm{p}}||\overrightarrow{\mathrm{q}}| \cos \frac{\pi}{3}=\frac{1}{2}$
Now, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|^{2}=|\vec{p}|^{2}+\frac{1}{4}|\vec{q}|^{2}-\frac{2}{2} \vec{p} \cdot \vec{q}$
$=1+\frac{1}{4}-\frac{1}{2} \quad[$ since $|\overrightarrow{\mathrm{p}}|=|\overrightarrow{\mathrm{q}}|=1]$
$=\frac{5}{4}-\frac{1}{2}=\frac{5-2}{4}=\frac{3}{4}$
So, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
If two variables $\mathrm{X}$ and $\mathrm{Y}$ are independent, then what is the correlation coefficient between them?

Options:
A 1
B $-1$
C 0
D None of the above

The correct answer is:
0

Solution:
The correlation coefficient of two independent events is zero.

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
If $\mathrm{u}=\mathrm{e}^{\mathrm{ax}} \sin \mathrm{bx}$ and $\mathrm{v}=\mathrm{e}^{\mathrm{ax}} \cos \mathrm{bx}$, then what is $\mathrm{u}$
$\frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}$ equal to?

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

The correct answer is:
a $\mathrm{e}^{2 \mathrm{ax}}$

Solution:
$\begin{aligned} u & \frac{d u}{d x}+v \frac{d v}{d x}=e^{a x} \sin b x\left[a e^{a x} \sin b x+b e^{a x} \cos b x\right] \\ &+e^{a x} \cos b x\left[a e^{a x} \cos b x-b e^{a x} \sin b x\right] \\=& e^{2 a x}\left[a \sin ^{2} b x+b \sin b x \cos b x+\right.\\ \left.\quad a \cos ^{2} b x-b \sin b x \cos b x\right] \\=& a e^{2 a x} \end{aligned}$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
If $\mathrm{u}=\sin ^{-1}(\mathrm{x}-\mathrm{y}), \mathrm{x}=3 \mathrm{t}, \mathrm{y}=4 \mathrm{t}^{3}$, then what is the derivative
of $\mathrm{u}$ with respect to $\mathrm{t}$?

Options:
A $3\left(1-t^{2}\right)$
B $3\left(1-t^{2}\right)^{-\frac{1}{2}}$
C $5\left(1-\mathrm{t}^{2}\right)^{\frac{1}{2}}$
D $5\left(1-t^{2}\right)$

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

Solution:
$\mathrm{u}=\sin ^{-1}(\mathrm{x}-\mathrm{y})$ and $\mathrm{x}=3 \mathrm{t}, \mathrm{y}=4 \mathrm{t}^{3}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \mathrm{t}-4 \mathrm{t}^{3}\right)$
Let $\mathrm{t}=\sin \theta \Rightarrow \theta=\sin ^{-1} \mathrm{t}$
So, $\mathrm{u}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)$
$\quad=\sin ^{-1}(\sin 3 \theta)=3 \theta$
Hence, $\mathrm{u}=3 \sin ^{-1} \mathrm{t}$
$\frac{\mathrm{du}}{\mathrm{dt}}=3 \frac{1}{\sqrt{1-\mathrm{t}^{2}}}=3\left(1-\mathrm{t}^{2}\right)^{-1 / 2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
If u, vand w (all positive) are the $\mathrm{p}^{\mathrm{th}}, \mathrm{q}^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ terms of a GP, the determinant of the matrix
$\left(\begin{array}{l}\ln \mathrm{u} \mathrm{p} 1 \\ \ln \mathrm{v} \mathrm{q} 1 \\ \ln \mathrm{w} \mathrm{r} 1\end{array}\right)$ is

Options:
A 0
B 1
C $(\mathrm{p}-\mathrm{q})(\mathrm{q}-\mathrm{r})(\mathrm{r}-\mathrm{p})$
D $\ln \mathrm{u} \times 1 \mathrm{n} \mathrm{v} \times \ln \mathrm{w}$

The correct answer is:
0

Solution:
Let $u, v$ and $w$ are $\mathrm{p}^{\text {th }}, q^{\text {th }}$ and $\mathrm{r}^{\text {th }}$ term of the G.P. with first term ' $a$ ' and common ratio ' $d$ '.
then, $u=a .(d)^{p-1} \Rightarrow \operatorname{In} u=\operatorname{In}(a)+(p-1) \operatorname{In}(d)$
$v=a \cdot(d)^{q-1} \Rightarrow \operatorname{In} v=\operatorname{In}(a)+(q-1) \operatorname{In}(d)$
$w=a .(d)^{r-1} \Rightarrow \operatorname{In} w=\operatorname{In}(a)+(r-1) \operatorname{In}(d)$
Now, $\operatorname{In} u-\operatorname{In} v=(p-q) \operatorname{In}(d)$
In $u-\operatorname{In} w=(p-r) \operatorname{In}(d)$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} v & q & 1 \\ \operatorname{In} w & r & 1\end{array}\right|$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ \operatorname{In} u-\operatorname{In} v & (p-q) & 0 \\ \operatorname{In} u-\operatorname{In} w & (p-r) & 0\end{array}\right|$
\{Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right\}$
$\left|\begin{array}{ccc}\operatorname{In} u & p & 1 \\ (p-q) \operatorname{In}(d) & (p-q) & 0 \\ (p-r) \operatorname{In}(d) & (p-r) & 0\end{array}\right|$
$=(p-q)(p-r)[\operatorname{In}(d)-\operatorname{In}(d)]=0$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 2)
If $V_0$ is the volume of a standard unit cell of germanium crystal containing $N_0$ atoms, then the expression for the mass $m$ of a volume $V$ in terms of $V_0, N_0, M_{\text {mol }}$ and $N_A$ is [here, $M$ is the molar mass of germanium and $N_A$ is the Avogadro's constant]

Options:
A $M \frac{V}{V_0} \frac{N_A}{N_0}$
B $\frac{N_{\mathrm{A}}}{N_0} \frac{V_0}{V} M$
C $M \frac{V}{V_0} \frac{N_0}{N_A}$
D $M \frac{V_0}{V} \frac{N_0}{N_A}$

The correct answer is:
$M \frac{V}{V_0} \frac{N_0}{N_A}$

Solution:
Number of unit cells in volume $V$
$=\frac{\text { Total volume }}{\text { Volume of a unit cell }}=\frac{V}{V_0}$
Number of atoms in volume $V$
$=$ Number of unit cells $\times$ Number of atoms in 1 unit cell $=\frac{V}{V_0} \times N_0$
Number of moles in volume $V$
$=\frac{\text { Number of a toms }}{\text { Avagadro number }}=\frac{V}{V_0} \times \frac{N_0}{N_A}$
Mass $n$ of given sample volume
$=$ Number of moles $\times$ Molar mass
$\Rightarrow m=\frac{V}{V_0} \times \frac{N_0}{N_A} \times M$

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2018 (05 May Shift 1)
If we add $3 \mathrm{~kg}$ load to the hanger of sonometer, the fundamental frequency
becomes two times its initial value. The initial load must be

Options:
A $2 \mathrm{~kg}$
B $1 \cdot 5 \mathrm{~kg}$
C $2 \cdot 5 \mathrm{~kg}$
D $1 \mathrm{~kg}$

The correct answer is:
$1 \mathrm{~kg}$

Solution:
$\mathrm{n} \propto \sqrt{\mathrm{T}} \quad \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}$
$\frac{1}{2}=\left(\frac{T}{T+3}\right)^{1 / 2} \Rightarrow \frac{1}{4}=\frac{T}{T+3}$
$1+\frac{3}{T}=4$
$\frac{3}{T}=3 \Rightarrow T=1 \mathrm{~kg}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (20 Oct Shift 1)
If we define a relation $\mathrm{R}$ on the set $\mathrm{N} \times \mathrm{N}$ as $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ for $a l l(a, b),(c, d) \in N \times N$, then
the relation is:

Options:
A symmetric only
B symmetric and transitive only
C equivalence relation
D reflexive only

The correct answer is:
equivalence relation

Solution:
$(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) \Leftrightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$
(i) $\mathrm{a}+\mathrm{a}=\mathrm{a}+\mathrm{a}$.
$(\mathrm{a}, \mathrm{a}) \mathrm{R}(\mathrm{a}, \mathrm{a}) \Rightarrow \mathrm{R}$ is reflexive
(ii) $(a, b) R(c, d) \Rightarrow a+d=b+c$
$(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{c}+\mathrm{b}=\mathrm{d}+\mathrm{a}$
$\mathrm{R}$ is symmetric.
(iii) Let $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ and $(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})$
$\Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ and $\mathrm{c}+\mathrm{f}=\mathrm{d}+\mathrm{e}$
$\Rightarrow a+d+c+f=b+c+d+e$
$\Rightarrow \mathrm{a}+\mathrm{f}=\mathrm{b}+\mathrm{e}$
$\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}, \mathrm{f})$
R is transitive.
from (i), (ii), (iii) R is anequivalence relation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 2)
If we join the mid points of the upper horizontal sides of each rectangle of a histogram by straight lines, what is the figure so obtained known as?

Options:
A Frequency curve
B Frequency polygon
C Ogive $(>)$
D Ogive $( < ) \quad$

The correct answer is:
Frequency polygon

Solution:
Joining the mid points of the upper horizontal sides of each rectangle of a histogram by straight lines, the figure so obtained is known as frequency polygon.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 2)
If we study the vibration of a pipe open at both ends, then the following statements is not true

Options:
A Open end will be anti-node
B Odd harmonics of the fundamental frequency will be generated
C All harmonics of the fundamental frequency will be generated
D Pressure change will be maximum at both ends

The correct answer is:
Pressure change will be maximum at both ends

Solution:
Statement (d) is not true, because at the open ends pressure change will be zero.

Topic: Physics, Waves and Sound, NEET, NEET 2013 (All India)
If $\mathrm{x} \in[0,5]$, then what is the probability that $\mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0$ ?

Options:
A $\frac{4}{5}$
B $\frac{1}{5}$
C $\frac{2}{5}$
D $\frac{3}{5}$

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

Solution:
Let $x^{2}-3 x+2=0$
$\Rightarrow \mathrm{x}=1.2$
$\therefore \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0$ for $\mathrm{x} \in[0,1] \cup[2,3] \cup[3,4] \cup[4,5] .$
It is given that :
$\mathrm{x} \in[0,1] \cup[1,2] \cup[2,3] \cup[3,4] \cup[4,5]$
$\therefore$ Required probability $=\frac{4}{5}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
If $\left|\begin{array}{lll}\mathrm{x} & \mathrm{y} & 0 \\ 0 & \mathrm{x} & \mathrm{y} \\ \mathrm{y} & 0 & \mathrm{x}\end{array}\right|=0$, then which one of the following is correct?

Options:
A $\frac{x}{y}$ is one of the cube roots of unity
B $\mathrm{x}$ is one of the cube roots of unity
C $\mathrm{y}$ is one of the cube roots of unity
D $\frac{x}{y}$ is one of the cube roots of $-1$

The correct answer is:
$\frac{x}{y}$ is one of the cube roots of $-1$

Solution:
$\left|\begin{array}{lll}x & y & 0 \\ 0 & x & y \\ y & 0 & x\end{array}\right|=0 \Rightarrow x\left(x^{2}-0\right)-y\left(0-y^{2}\right)+0=0$
$\Rightarrow x^{3}+y^{3}=0$
$\Rightarrow x^{3}=-y^{3}$
$\Rightarrow \frac{x^{3}}{y^{3}}=-1 \Rightarrow\left(\frac{x}{y}\right)^{3}=-1 \Rightarrow \frac{x}{y}=\sqrt{-1}$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
If $\mathrm{X}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]$, and $\mathrm{I}$ is a $2 \times 2$ identity matrix, then $\mathrm{X}^{2}-2 \mathrm{X}$ $+3$ I equals to which one of the following?

Options:
A $-\mathrm{I}$
B $-2 \mathrm{X}$
C $2 X$
D $4 \mathrm{X}$

The correct answer is:
$2 X$

Solution:
Given matrix is :
$\mathrm{X}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]$
$\therefore \mathrm{X}^{2}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]$
$=\left[\begin{array}{cc}1 & -2-6 \\ 0 & 9\end{array}\right]=\left[\begin{array}{cc}1 & -8 \\ 0 & 9\end{array}\right]$
So, the given expression is:
$\mathrm{X}^{2}-2 \mathrm{X}+3 \mathrm{I}=\left[\begin{array}{cc}1 & -8 \\ 0 & 9\end{array}\right]-2\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]+3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}1 & -8 \\ 0 & 9\end{array}\right]+\left[\begin{array}{cc}-2 & +4 \\ 0 & -6\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$
$=\left[\begin{array}{cc}1-2+3 & -8+4 \\ 0 & 9-6+3\end{array}\right]=\left[\begin{array}{cc}2 & -4 \\ 0 & 6\end{array}\right]=2\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]=2 \mathrm{X}$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 1)
If $x=1+\frac{y}{2}+\left(\frac{y}{2}\right)^{2}+\left(\frac{y}{2}\right)^{3}+\ldots .$ where $|y| < 2$, what is $1 y$
equal to ?

Options:
A $\frac{x-1}{x}$
B $\frac{x-1}{2 x}$
C $\frac{2 x-2}{x} \quad$
D $\frac{2 x+1}{2 x}$

The correct answer is:
$\frac{2 x-2}{x} \quad$

Solution:
$\mathrm{x}=1+\frac{\mathrm{y}}{2}+\left(\frac{\mathrm{y}}{2}\right)^{2}+\left(\frac{\mathrm{y}}{2}\right)^{3}+\ldots . .$
Here, $\frac{\mathrm{y}}{2} < 1$ and this a G. with first term $=1$ and common
ratio $=\frac{\mathrm{y}}{2}$ so,
$\Rightarrow \quad x=\frac{1}{1-\frac{y}{2}} \Rightarrow x=\frac{2}{2-y}$
$\Rightarrow 2 x-x y=2 \Rightarrow y=\frac{2 x-2}{x}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 1)
If $x+1,4 x+1$, and $8 x+1$ are in geometric progression, then what is the non-trivial value of $\mathrm{x}$ ?

Options:
A $-1$
B 1
C $\frac{1}{8}$
D $\frac{1}{4}$

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

Solution:
$\operatorname{If}(\mathrm{x}+1),(4 \mathrm{x}+1)$ and $(8 \mathrm{x}+1)$ are in GP.
then $(4 x+1)^{2}=(x+1)(8 x+1)$
$\Rightarrow 16 x^{2}+8 x+1=8 x^{2}+x+8 x+1$
$\Rightarrow 8 x^{2}-x=0 \Rightarrow x(8 x-1)=0$
$\Rightarrow \mathrm{x}=0, \frac{1}{8},\left[\frac{1}{8}\right.$ is non-trivial value]

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
If $x>1$ and $\log _{2} x, \log _{3} x, \log _{x} 16$ are in GP, then what is $x$
equal to?

Options:
A 9
B 8
C 4
D 2

The correct answer is:
9

Solution:
$\log _{2} x, \log _{3} x, \log _{x} 16$ are in G.P.
$\therefore \quad \frac{\log _{3} x}{\log _{2} x}=\frac{\log _{x} 16}{\log _{3} x}$
$\Rightarrow\left(\log _{3} x\right)^{2}=\log _{2} x \cdot \log x 16$
$\Rightarrow 2 \times \log _{3} x=\log 2 x \cdot \log x 2^{4}$
$\Rightarrow 2 \times \log _{3} x=4 \times \log _{2} x \cdot \log _{x} 2$
$\Rightarrow \log _{3} x=2\left(\log _{2} x \times \log _{x} 2\right)$
$\Rightarrow \log _{3} x=2\left[\frac{\log _{2} x}{\log _{2} x}\right] \quad\left(\because \log _{b} a=\frac{1}{\log _{a} b}\right)$
$\Rightarrow \log _{3} x=2 \Rightarrow x=3^{2}=9$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 1)
If $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is

Options:
A $x_{1}+x_{2}>2 \sqrt{x_{1} x_{2}}$
B $\sqrt{\mathrm{x}_{1}}+\sqrt{\mathrm{x}_{2}}>\sqrt{2}$
C $\left|\sqrt{\mathrm{x}_{1}}+\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$
D $x_{1}+x_{2} < 2\left(\sqrt{x_{1} x_{2}}+1\right)$

The correct answer is:
$\left|\sqrt{\mathrm{x}_{1}}+\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$

Solution:
Arithmetic mean of $\mathrm{x}_{1}, \mathrm{x}_{2}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}$
Geometric mean of $\mathrm{x}_{1}, \mathrm{x}_{2}=\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}$
Given, $\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}-\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>1$
$\Rightarrow \frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}>\sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+1$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}>2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}+2$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}-2 \sqrt{\mathrm{x}_{1} \mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}\right)^{2}+\left(\sqrt{\mathrm{x}_{2}}\right)^{2}-2 \sqrt{\mathrm{x}_{1}} \sqrt{\mathrm{x}_{2}}>2$
$\Rightarrow\left(\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right)^{2}>2$
$\Rightarrow\left|\sqrt{\mathrm{x}_{1}}-\sqrt{\mathrm{x}_{2}}\right|>\sqrt{2}$
Correct option (c).

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
If x1,x2 and x3 are enthalpies of H-H,O=O and O-H bonds respectively, and x4 is the enthalpy of vaporization of water, estimate the standard enthalpy of combustion of hydrogen.

Options:
A x1+x22-2x3+x4
B x1+x22-2x3-x4
C x1+x22-x3+x4
D 2x3-x1-x22-x4

The correct answer is:
x1+x22-2x3-x4

Solution:
H2(g)+12O2(g)→H2O(l) …(i)
H2(g)→2H(g); ΔH=x1
O2(g)→2O(g); ΔH=x2
H2O(l)→H2O(g); ΔH=x4
H2O(g)→2H(g)+O(g); ΔH=2x3
Using all four, we (find) (i)
x1+x22-x4-2x3

Topic: Chemistry, Thermodynamics (C), NEET
If $\mathrm{x} \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+\mathrm{x}}=0$, what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?

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

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

Solution:
Given equation
$\mathrm{x} \sqrt{1+\mathrm{y}}+\mathrm{y} \sqrt{1+\mathrm{x}}=0$
Can be written as:
$\mathrm{x} \sqrt{1+\mathrm{y}}=-\mathrm{y} \sqrt{1+\mathrm{x}}$
Squaring both sides, we get $\begin{aligned} & x^{2}(1+y)=y^{2}(1+x) \\ \Rightarrow & x^{2}+x^{2} y=y^{2}+y^{2} x \quad \Rightarrow \quad x^{2}-y^{2}=y^{2} x-x^{2} y \\ \Rightarrow &(x-y)(x+y)=-x y(x-y) \\ \Rightarrow & x+y=-x y \quad \Rightarrow y(1+x)=-x \end{aligned}$
$\mathrm{y}=\frac{-\mathrm{x}}{1+\mathrm{x}}$ which is in explicit form.
Differentiating w.r.t.x, we get
$\frac{d y}{d x}=\frac{(1+x)(-1)+x(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
If $\mathrm{x}=1-\mathrm{y}+\mathrm{y}^{2}-\mathrm{y}^{3}+\ldots$ up to infinite terms, where $|\mathrm{y}| < 1$, then which one of the following is correct?

Options:
A $\mathrm{x}=\frac{1}{1+\mathrm{y}}$
B $x=\frac{y}{1-y}$
C $\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}$
D $x=\frac{y}{1-y}$

The correct answer is:
$\mathrm{x}=\frac{1}{1+\mathrm{y}}$

Solution:
$\mathrm{x}=1-\mathrm{y}+\mathrm{y}^{2}-\mathrm{y}^{3}+\ldots$ up to infinite terms.
We can see that the given series is geometric progression, with $\mathrm{a}=1$ and $\mathrm{r}=-\mathrm{y}$
$\mathrm{S}_{\infty} \frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-(-\mathrm{y})}=\frac{1}{1+\mathrm{y}}$
$\therefore \mathrm{x}=\frac{1}{1+\mathrm{y}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 1)
If $|\mathrm{x}| < \frac{1}{2}$, what is the value of
$1+n\left[\frac{x}{1-x}\right]+\left[\frac{n(n+1)}{2 !}\right]\left[\frac{x}{1-x}\right]^{2}+\ldots \ldots \ldots \ldots \ldots \infty ?$

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

The correct answer is:
$\left[\frac{1-x}{1-2 x}\right]^{n}$

Solution:
Given that $1+\mathrm{n}\left[\frac{\mathrm{x}}{1-\mathrm{x}}\right]+\frac{\mathrm{n}(\mathrm{n}+1)}{2 !}\left[\frac{\mathrm{x}}{1-\mathrm{x}}\right]^{2}+\ldots \infty$ is
expansion of $\left[1-\frac{x}{1-x}\right]^{-n}$.
So, it is $=\left[1-\frac{x}{1-x}\right]^{-n}=\left[\frac{1-x-x}{1-x}\right]^{-n}=\left[\frac{1-x}{1-2 x}\right]^{n}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
If $\mathrm{x}=2+2^{1 / 3}+2^{2 / 3}$, then what is the value of $x^{3}-6 x^{2}+6 x ?$

Options:
A 1
B 2
C -2
D 3

The correct answer is:
2

Solution:
Let $\begin{aligned} x=2+2^{1 / 3}+2^{2 / 3} \\ &=2+2^{1 / 3}+\left(2^{1 / 3}\right)^{2}=2+2^{1 / 3}\left(1+2^{1 / 3}\right) \\ & \Rightarrow x-2=2^{1 / 3}\left(1+2^{1 / 3}\right) \end{aligned}$
On cubing both sides, we get
$x^{3}-8-6 x^{2}+12 x=2\left(1+2+3.2^{1 / 3}+3.2^{2 / 3}\right)$
$\Rightarrow x^{3}-6 x^{2}+6 x=14+6.2^{1 / 3}+6.2^{2 / 3}-6 x$
$\quad=14+6.2^{1 / 3}+6.2^{2 / 3}-6\left(2+2^{1 / 3}+2^{2 / 3}\right)=2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)
If $-\mathrm{x}^{2}+3 \mathrm{x}+4>0$, then which one of the following is correct?

Options:
A $x \in(-1,4)$
B $x \in[-1,4]$
C $\mathrm{x} \in(\infty,-1) \cup(4, \infty)$
D $\mathrm{x} \in(-\infty,-1] \cup[4, \infty)$

The correct answer is:
$x \in(-1,4)$

Solution:
$-x^{2}+3 x+4>0$
$\Rightarrow x^{2}-3 x-4 < 0 \Rightarrow(x-4)(x+1) < 0$
Noting the sign of expression around $-1$ and 4 , we use way curve method.
$\Rightarrow \mathrm{x} \in(-1,4)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)