datavorous/entrance-exam-dataset huggingface dataset:
What is the $(\mathrm{m}-1)^{\text {th }}$ root of $\left[\left(\mathrm{a}^{\mathrm{m}}\right)^{\mathrm{m}}-\left(\frac{1}{\mathrm{~m}}\right)\right]^{\frac{1}{\mathrm{~m}+1}}$ ?

Options:
A $\mathrm{a}^{\mathrm{m}+(1 / \mathrm{m})}$
B $\mathrm{a}^{\mathrm{m}-(1 / \mathrm{m})}$
C a
D 1

The correct answer is:
a

Solution:
The given expression $\left[\left(a^{m}\right)^{m-\frac{1}{m}}\right]^{\frac{1}{m+1}}$
$=\left[\left(a^{m}\right)^{\frac{m^{2}-1}{m}}\right]^{\frac{1}{m+1}}=\left[\left(a^{m}\right)^{\frac{(m-1)(m+1)}{m}}\right]^{\frac{1}{m+1}}$
$=\left[\mathrm{a}^{\mathrm{m}}\right]^{\frac{\mathrm{m}-1}{\mathrm{~m}}}=\mathrm{a}^{\mathrm{m}-1}$
Its $(\mathrm{m}-1)^{t h}$ root $=\left(\mathrm{a}^{\mathrm{m}-1}\right)^{1 / \mathrm{m}-1}=\mathrm{a}$
Ience, $(\mathrm{m}-1)^{\mathrm{th}}$ root of $\left[\left(\mathrm{a}^{\mathrm{m}}\right)^{\mathrm{m}-\left(\frac{1}{\mathrm{~m}}\right)}\right]^{\frac{1}{(\mathrm{~m}+1)}}$
$=(\mathrm{m}-1)$ th root of $\mathrm{a}^{\mathrm{m}-1}=$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2006 (Phase 2)
What is the magnitude of the moment of the couple
consisting of the force $\overrightarrow{\mathbf{F}}=3 \hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}-\hat{\mathbf{k}}$ acting through the
point $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and and $-\overrightarrow{\mathbf{F}}$ acting through the
point $2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ ?

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

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

Solution:
Here, $\overrightarrow{\mathrm{F}}=\hat{3} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
and $\overrightarrow{\mathrm{r}}_{1}-\overrightarrow{\mathrm{r}_{2}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
Moment of couple $=\left(\overrightarrow{\mathrm{r}_{1}}-\overrightarrow{\mathrm{r}_{2}}\right) \times \overrightarrow{\mathrm{F}}$
$=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(\hat{3}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 3 & 2 & -1\end{array}\right|=\hat{i}(-2-4)-\hat{j}(1-6)+\hat{k}(-2-6)$
$=-6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$
Magnitude of the moment $=|-6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}|$
$=\sqrt{36+25+64}=5 \sqrt{5}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
What is the maximum area of a triangle that can be inscribed in a circle of radius a?

Options:
A $\frac{3 a^{2}}{4}$
B $\frac{\mathrm{a}^{2}}{2}$
C $\frac{3 \sqrt{3} a^{2}}{4}$
D $\frac{\sqrt{3} a^{2}}{4}$

The correct answer is:
$\frac{3 \sqrt{3} a^{2}}{4}$

Solution:
For area of triangle to be maximum, it should be equilateral triangle.
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \mathrm{ab} \sin \theta$
$=\frac{1}{2} \cdot$ a $a \cdot \sin \theta$
$=\frac{1}{2} \mathrm{a}^{2} \cdot \sin 120^{\circ}$
$=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$
Area of triangle $\mathrm{OAB}, \mathrm{OBC}, \mathrm{OAC}$
$=$ Area of triangle $\mathrm{ABC}$
$=3 \times \frac{\sqrt{3}}{4} a^{2}=\frac{3 \sqrt{3}}{4} \mathrm{a}^{2}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2017 (Phase 1)
What is the maximum number of straight lines that can be drawn with any four points in a plane such that each line contains at least two of these points?

Options:
A 2
B 4
C 6
D 12

The correct answer is:
6

Solution:
Required number of lines $={ }^{4} C_{2}=\frac{4 !}{2 ! 2 !}=6$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 2)
What is the maximum point on the curve $x=e^{x} y$ ?

Options:
A $(1, e)$
B $\left(1, e^{-1}\right)$
C $(e, 1)$
D $\left(e^{-1}, 1\right)$

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

Solution:
Given curve is $x=e^{x} y$
Which can be rewritten as $y=x e^{-x}$ On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=-x e^{-x}+e^{-x}$
Put $\frac{d y}{d x}=0$ for maxima or minima
$\Rightarrow-x e^{-x}+e^{-x}=0$
$\Rightarrow e^{-x}(1-x)=0$
Since $e^{-x}$ can not be zero $\therefore \quad 1-x=0 \Rightarrow x=1$
Now, $\frac{d^{2} y}{d x^{2}}=-e^{-x}+x e^{-x}-e^{-x}=x e^{-x}-2 e^{-x}$
$=e^{-x}(x-2)$
$\Rightarrow\left(\frac{d^{2} y}{d x^{2}}\right) < 0$
$\therefore y$ is maximum at $x=1$. Thus, when $x=1$ then $\mathrm{y}=e^{-1}$
Hence, maximumpoint on the curve $x=e^{x} y$ is $\left(1, e^{-1}\right)$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
What is the maximum slope of the curve $y=-x^{3}+3 x^{2}+2 x-27 ?$

Options:
A 1
B 2
C 5
D $-23$

The correct answer is:
5

Solution:
The equation of curve is given as:
$y=-x^{\frac{1}{3}}+3 x^{2}+2 x-27$
On differentiating w.r.t.x, we get
$\frac{d y}{d x}=-3 x^{2}+6 x+2$
This represents slope of the curve at any point. Let $\quad A=\frac{d y}{d x}=-3 x^{2}+6 x+2$
$\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=-6 \mathrm{x}+6$
and $\frac{d^{2} A}{d x^{2}}=-6$
Put $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=0$ for maxima or minima. $-6 x+6=0$
$\Rightarrow \mathrm{x}=1$
Now, $\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1} 1=-6 < 0$
$\therefore \quad$ A is maximumat $x=1$ $\therefore \quad$ Maximum slope of curve $=-3+6+2=5$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 2)
What is the maximum value of $3 \cos x+4 \sin x+5$ ?

Options:
A 5
B 7
C 10
D 12

The correct answer is:
10

Solution:
Maximum value of
$$
\begin{aligned}
3 \cos x+4 \sin x+5 &=\sqrt{(4)^{2}+(3)^{2}}+5=\sqrt{16+9}+5 \\
&=\sqrt{25}+5=5+5=10
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
What is the maximum value of $16 \sin \theta-12 \sin ^{2} \theta$ ?

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

The correct answer is:
$\frac{16}{3}$

Solution:
$\mathrm{f}(\theta)=16 \sin \theta-12 \sin ^{2} \theta$
$f^{\prime}(\theta)=16 \cos \theta-24 \sin \theta \cos \theta$
$f^{\prime}(\theta)=0 \Rightarrow \cos \theta(16-24 \sin \theta)=0$
$\Rightarrow \cos \theta=0$ (or) $16-24 \sin \theta=0$
$\Rightarrow \theta=\frac{\pi}{2}($ or $) \sin \theta=\frac{2}{3}$
$\mathrm{f}\left(\frac{\pi}{2}\right)=16 \sin \frac{\pi}{2}-12 \sin ^{2} \frac{\pi}{2}=16-12=4$
$\mathrm{f}\left(\sin \theta=\frac{2}{3}\right)=16\left(\frac{2}{3}\right)-12\left(\frac{2}{3}\right)^{2}$
$=\frac{32}{3}-\frac{48}{9}=\frac{32}{3}-\frac{16}{3}=\frac{16}{3} .$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 1)
What is the maximum value of $\sin 3 \theta \cos 2 \theta+\cos 3 \theta \sin 2 \theta$ ?

Options:
A 1
B 2
C 4
D 10

The correct answer is:
1

Solution:
Consider $\sin 3 \theta \cos 2 \theta+\cos 3 \theta \sin 2 \theta$
$=\sin (3 \theta+2 \theta) \quad\left[\begin{array}{c}\because \sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B} \\ =\sin (\mathrm{A}+\mathrm{B})\end{array}\right]$
$=\sin (5 \theta)$
We know, $-1 < \sin \theta < 1$
Hence, maximum value of given expression is $1 .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is the maximum value of $\sin \theta \cos \theta ?

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

The correct answer is:
$1 / 2$

Solution:
Let $P=\sin \theta \cos \theta$
Multiply and divide by 2 , we get $P=\frac{2 \sin \theta \cos \theta}{2}=\frac{\sin 2 \theta}{2}$
The maximum value of $\sin 2 \theta$ is 1 .
$\therefore$ Maximum value of $P=\frac{1}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
What is the maximum value of the function $f(\mathrm{x})=4 \sin ^{2} \mathrm{x}+1 ?$

Options:
A 5
B 3
C 2
D 1

The correct answer is:
5

Solution:
$\quad$ We know, $-1 \leq \sin x \leq 1$
$\Rightarrow 0 \leq \sin ^{2} x \leq 1$
$\Rightarrow 0 \leq 4 \sin ^{2} x \leq 4$
$\Rightarrow 1 \leq 4 \sin ^{2} x+1 \leq 5$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
What is the maximum value of the function $\log x-x ?$

Options:
A $-1$
B 0
C 1
D $\infty$

The correct answer is:
$-1$

Solution:
Let $y=\log x-x$ $\therefore \frac{d y}{d x}=\frac{1}{x}-1$ and $\frac{d^{2} y}{d x^{2}}=-\frac{1}{x^{2}}$
For maximum and minimum value of $y$
$\frac{d y}{d x}=\frac{1}{x}-1=0 \Rightarrow \frac{1}{x}=1 \Rightarrow x=1$
For $x=1, \frac{d^{2} y}{d x^{2}}=-v e$
Thus, the value of given function is maximum for $x=1$
So, the maximum value of the function $=\log (1)-1=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
What is the maximum value of $\mathrm{x}$. y subject to the condition $\mathrm{x}+\mathrm{y}=8 ?$

Options:
A 8
B 16
C 24
D 32

The correct answer is:
16

Solution:
The given condition is:
$x+y=8 \Rightarrow y=8-x$
and let, the product of $\mathrm{x}$ and $\mathrm{y}$ be, $\mathrm{P}=\mathrm{xy}$
$\Rightarrow P=x(8-x)=8 x-x^{2}$
Differentiating w.r.t. $\mathrm{x}$
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dx}}=8-2 \mathrm{x}$
and $\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}=-2 < 0$
Put $\frac{\mathrm{dP}}{\mathrm{dx}}=0$, for maxima or minima
$8-2 x=0$
$\Rightarrow x=\frac{8}{2}=4$ and $y=4$
and $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right) < 0$ when $\mathrm{x}=\mathrm{y}$
$\therefore$ P is maximumat $\mathrm{x}=4$ Maximum value of $\mathrm{P}=4 \cdot(8-4)=4.4=16$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
What is the mean deviation about the mean for the data 4, $8,9,10,12,13,17 ?$

Options:
A 25
B 3
C $3.5$
D 4

The correct answer is:
3

Solution:
Mean $=(\bar{x})-\frac{\sum x_{1}}{N}$
Here $\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=16$
Mean deviation about mean $=\frac{\sum\left|\mathrm{x}_{i}-\overline{\mathrm{x}}\right|}{\mathrm{N}}$
$|4-10|+|7-10|+|8-10|+|9-10|+|10-10|$
$=\frac{+|12-10|+|13-10|+|17-10|}{8}$
$=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}=3$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 2)
What is the mean deviation from the mean of the numbers $10,9,21,16,24 ?$

Options:
A $5-2$
B $5.0$
C $4.5$
D $4.0$

The correct answer is:
$5-2$

Solution:
Given numbers $-10,9,21,16,24$ Mean $=\frac{10+9+21+16+24}{5}=\frac{80}{5}=16$
\begin{array}{|c|c|}
\hline Numbers & Distance (d) from mean (16) \\
\hline 10 & 6 \\
\hline 9 & 7 \\
\hline 21 & 5 \\
\hline 16 & 0 \\
\hline 24 & 8 \\
\hline\Sigma d=26 & \\
\hline
\end{array}
Mean deviation $=\frac{\Sigma d}{5}=\frac{26}{5}=5.2$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 1)
What is the mean deviation of the data $2,9,9,3,6,9,4 ?$

Options:
A $2.23$
B $257 \quad$
C $3.23$
D $3.57$

The correct answer is:
$257 \quad$

Solution:
Given data is $2,9,9,3,6,9,4$. We know,
bservatuon Mean $=\frac{\text { Sum or an }}{\text { Total number of }}$
Mean $=\frac{2+9+9+3+6+9+4}{7}=\frac{42}{7}=6$
Mean deviation
$=\frac{|2-6|+3|9-6|+|3-6|+|6-6|+|4-6|}{7}$
$=\frac{4+9+3+0+2}{7}=\frac{18}{7}=2.57$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
What is the mean of first $\mathrm{n}$ odd natural numbers

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

The correct answer is:
$\mathrm{n}$

Solution:
Sum of first n odd natural numbers $=\mathrm{n}^{2}$.
Now, mean $=\frac{\mathrm{n}^{2}}{\mathrm{n}}=\mathrm{n}$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)
What is the mean of the squares of the first 20 natural numbers ? $\quad$

Options:
A $151.5$
B $143.5$
C 65
D 72

The correct answer is:
$143.5$

Solution:
Mean of the squares of the first 20 natural number
$=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}=\frac{21 \times 41}{6}=143.5$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
What is the meaning of vexillary or papilionaceous aestivation?

Options:
A When the margins of sepals or petals overlap one another but not in any particular direction
B When sepals or petals overlap one another in a particular direction
C When the largest petal overlaps the two lateral petals which in turn overlap the two smallest anterior petals
D When sepals or petals just touch one another at the margin, without overlapping

The correct answer is:
When the largest petal overlaps the two lateral petals which in turn overlap the two smallest anterior petals

Solution:
Vexillary or papilionaceous aestivation refers to a situation where there are five petals in a flower, and the largest petal (standard) overlaps the two smallest anterior petals.

Topic: Biology, Morphology of Flowering Plants, NEET
What is the measure of the angle $114^{\circ} 35^{\prime} 30^{\prime}$ in radian?

Options:
A $1 \mathrm{rad}$
B $2 \mathrm{rad}$
C $3 \mathrm{rad}$
D $4 \mathrm{rad}$

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

Solution:
From relation between minute 6 seconds measure $60^{13}=\mathrm{I}^{\circ}$
$\Rightarrow 30^{\prime \prime}=\frac{1^{\prime}}{2}$
$\Rightarrow 35^{\prime} 30^{\prime \prime}=\left(35+\frac{1}{2}\right)^{\prime}=\left(\frac{71}{2}\right)^{\prime}$
Also, $60^{\circ}=1^{\circ}$
$\therefore 1^{\prime}=\left(\frac{1}{60}\right)^{\circ}$
$\Rightarrow\left(\frac{71}{2}\right)^{\prime}=\left(\frac{71}{2} \times \frac{1}{60}\right)^{\circ}=\left(\frac{71}{120}\right)^{\circ}$
$\therefore \quad 114^{\circ} 35^{\circ} 30^{\prime}=\left(114+\frac{71}{120}\right)^{\circ}=\left(\frac{13751}{120}\right)^{\circ}$
We know that, $2 \pi \mathrm{rad}=360^{\circ}$
$\Rightarrow 1^{\circ}=\frac{2 \pi}{360} \mathrm{rad}$
$\Rightarrow\left(\frac{13751}{120}\right)^{\circ}=\frac{2 \pi}{360^{\circ}} \times \frac{13751}{120} \mathrm{rad}$
$=\frac{2 \times 22 \times 13751}{7 \times 360 \times 120} \mathrm{rad}=2.0008069 \mathrm{rad}$
$\Rightarrow 114^{\circ} 35^{\prime} 30^{\prime \prime}=2 \mathrm{rad}($ approx. $)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
What is the measured emf of the cell with the representation Pt(s) |H2(g, 1 bar) |H+(aq, 1 M) | |Cu2+ (aq,1 M) | Cu?

Options:
A 0.34 V
B 1.10 V
C -0.76 V
D 1.44 V

The correct answer is:
0.34 V

Solution:
The measured emf of the cell:Pt(s) |H2(g, 1 bar) |H+(aq, 1 M) | |Cu2+ (aq,1 M) | Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction: Cu2+ (aq, 1M) + 2 e- → Cu(s).

Topic: Chemistry, Electrochemistry, NEET
What is the measured emf of the cell with the representation Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) | | Zn2+ (aq, 1M) | Zn?

Options:
A 0.34 V
B 1.10 V
C -0.76 V
D -1.44 V

The correct answer is:
-0.76 V

Solution:
The measured emf of the cell: Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) | | Zn2+ (aq, 1M) | Zn is -0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e- → Zn(s).

Topic: Chemistry, Electrochemistry, NEET
What is the median of the distribution $3,7,6,9,5,4,2$ ?

Options:
A 5
B 6
C 7
D 8

The correct answer is:
5

Solution:
$\therefore$ Required Median $=\left(\frac{n+1}{2}\right)^{\mathrm{th}}$ obs $=4^{\mathrm{th}} \mathrm{obs}=5$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
What is the median of the numbers $4.6,0,9.3,-4.8,7.6,2$. $12.7,3.5,8.2,6.1,3.9,5.2 ?$

Options:
A $3.8$
B $4.9$
C $5.7$
D $6.0$

The correct answer is:
$4.9$

Solution:
By arranging the given numbers in ascending order, $-4.8,0,2.3,3.5,3.9, \frac{4.6,5.2}{\downarrow}, 6.1,7.6,8.2,9.3,12.7$
Median $=\frac{4.6+5.2}{2}$
$=\frac{9.8}{2}=4.9$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
What is the middle term in the expansion of $\left(1-\frac{x}{2}\right)^{8}$ ?

Options:
A $\frac{35 x^{4}}{8}$
B $\frac{17 x^{5}}{8}$
C $\frac{35 x^{5}}{8}$
D None of these

The correct answer is:
$\frac{35 x^{4}}{8}$

Solution:
Since $n=8$ is even number therefore middle term
$=\left(\frac{n}{2}+1\right)^{\mathrm{th}}$ term $=(4+1)=5^{\mathrm{th}}$ term
Hence, $T_{5}={ }^{8} C_{4}(1)^{4}\left(-\frac{x}{2}\right)^{4}$
$=\frac{8 !}{4 ! 4 !} \times \frac{x^{4}}{16}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot \frac{x^{4}}{16}=\frac{70 x^{4}}{16}=\frac{35 x^{4}}{8}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2011 (Phase 1)
What is the middle term in the expansion of $\left(\frac{\mathrm{x} \sqrt{\mathrm{y}}}{3}-\frac{3}{\mathrm{y} \sqrt{\mathrm{x}}}\right)^{12}$ ?

Options:
A $\mathrm{C}(12,7) \mathrm{x}^{3} \mathrm{y}^{-3}$
B $\mathrm{C}(12,6) \mathrm{x}^{-3} \mathrm{y}^{3}$
C $\mathrm{C}(12,7) \mathrm{x}^{-3} \mathrm{y}^{3}$
D $C(12,6) \times^{3} y^{-3}$

The correct answer is:
$C(12,6) \times^{3} y^{-3}$

Solution:
In the expansion of $\left(\frac{\mathrm{x} \sqrt{\mathrm{y}}}{3}-\frac{3}{\mathrm{y} \sqrt{\mathrm{x}}}\right)^{12}$, then middle term
is $\frac{12}{2}+1=7^{\text {th }}$ term. $(\mathrm{r}+1)_{\mathrm{th}}$ term
$\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left[\frac{\mathrm{x} \sqrt{\mathrm{y}}}{3}\right]^{12-\mathrm{r}} \cdot\left(-\frac{3}{\mathrm{y} \sqrt{\mathrm{x}}}\right)^{\mathrm{r}}$
$\therefore \quad \mathrm{T}_{7}=\mathrm{T}_{6+1}={ }^{12} \mathrm{C}_{6}\left(\frac{\mathrm{x} \sqrt{\mathrm{y}}}{3}\right)^{6}\left(-\frac{3}{\mathrm{y} \sqrt{\mathrm{x}}}\right)^{6}$
$={ }^{12} \mathrm{C}_{6} \frac{\mathrm{x}^{6} \mathrm{y}^{3}}{\mathrm{y}^{6} \mathrm{x}^{3}}={ }^{12} \mathrm{C}_{6} \mathrm{x}^{3} \mathrm{y}^{-3}=\mathrm{C}(12,6) \mathrm{x}^{3} \mathrm{y}^{-3}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2007 (Phase 1)
What is the minimum pH required to prevent the precipitation of ZnS in a solution that is 0.01 M ZnCl2 and saturated with 0.10 M H2S ?

Given Ksp=10-21, Ka1ÃKa2=10-20

Options:
A 0
B 1
C 2
D 4

The correct answer is:
1

Solution:
ZnSâZn2++S2-
Ksp=Zn2+S2-
ZnCl2âZn2++2Cl-
S2-=KspZn2+=10-210.01
âS2-=10-19
Ka1.Ka2=H+2S2-H2S
Where, Ka1=[HS-][H+]H2S
Ka2=H+S2-HS-
So, 10-20=H+2S2-H2S=H+210-190.1
â H+=0.1
âpH=-logâ¡H+=1

Topic: Chemistry, Ionic Equilibrium, NEET
What is the minimum value of $2 x^{2}-3 x+5 ? \quad$

Options:
A 0
B $3 / 4$
C $31 / 4$
D $31 / 8$

The correct answer is:
$31 / 8$

Solution:
Let $y=2 x^{2}-3 x+5$
$\Rightarrow \frac{d y}{d x}=4 x-3$
and $\frac{d^{2} y}{d x^{2}}=4$
For maximum and minimum value of $y$
$\frac{d y}{d x}=0$
$\Rightarrow 4 x-3=0$
$\Rightarrow \quad x=\frac{3}{4}$
Since, for every value of $x, \frac{d^{2} y}{d x^{2}}=+v e$ So, minimum value of $2 x^{2}-3 x+5$ is minimum
$x=\frac{3}{4}$
$\therefore \quad$ The minimum value $=2\left(\frac{3}{4}\right)^{2}-3\left(\frac{3}{4}\right)+5$
$=\frac{9}{8}-\frac{9}{4}+5=\frac{9-18+40}{8}=\frac{31}{8}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
What is the minimum value of $a^{2} x+b^{2} y$ where $x y=c^{2} ?$

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

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

Solution:
Let $p=a^{2} x+b^{2} y$ and $x y=c^{2}$
$\Rightarrow y=\frac{c^{2}}{x}$ ...(1)
$\Rightarrow P=a^{2} x+b^{2}\left(\frac{c^{2}}{x}\right)$
Now, $\frac{d P}{d x}=0 \Rightarrow a^{2}-\frac{b^{2} c^{2}}{x^{2}}=0$
$\therefore y=\frac{c^{2}}{\frac{b c}{a}}=\frac{a c^{2}}{b c}=\frac{a c}{b}$
$\Rightarrow a^{2}=\frac{b^{2} c^{2}}{x^{2}}$
$\Rightarrow x=\frac{b c}{a} \therefore P_{\min }=a^{2}\left(\frac{b c}{a}\right)+b^{2}\left(\frac{a c}{b}\right)$
$=a b c+a b c=2 a b c .$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
What is the minimum value of $\cos \theta+\cos 2 \theta$ ?

Options:
A $-2$
B $-\frac{9}{8}$
C 0
D $-\frac{9}{16}$

The correct answer is:
$-\frac{9}{8}$

Solution:
Let $\mathrm{A}=\cos \theta+\cos 2 \theta$
$\therefore$ On differentiating w.r.t. to $\theta$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \theta}=-\sin \theta-2 \sin 2 \theta$
Put $\frac{\mathrm{dA}}{\mathrm{d} \theta}=0$ for maximaor minima. $\sin \theta+2 \sin 2 \theta=0 \Rightarrow \sin \theta+4 \sin \theta \cos \theta$
$\Rightarrow \sin \theta(1+4 \cos \theta)=0$
$\Rightarrow \sin \theta=0$, or $4 \cos \theta+1=0$
$\Rightarrow \cos \theta=1$ or $\cos \theta=-\frac{1}{4}$
Now, $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=-\cos \theta-4 \cos 2 \theta$
$=-\cos \theta-4\left(2 \cos ^{2} \theta-1\right)$
For $\cos \bar{\theta}=1$
$\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=-\cos \theta-4\left(2 \cos ^{2} \theta-1\right)$
$=-1-4(2(1)-1)=-1-4=-5 < 0$
So, A is maximum at $\cos \theta=1$
$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}\right)_{\cos \theta=\frac{-1}{4}}=\frac{1}{4}-4\left(2 \cdot \frac{1}{16}-1\right)>0$
$\quad\left[\right.$ Since $\left.\cos 2 \theta=2 \cos ^{2} \theta-1\right]$
$\therefore \mathrm{A}$ is minimum at $\theta=\cos ^{-1}\left(\frac{-1}{4}\right)$
Now minimum value of $\cos \theta+\cos 2 \theta$
or of $\cos \theta+2 \cos ^{2} \theta-1$
$=\left(\frac{-1}{4}\right)+2\left(\frac{1}{16}\right)-1$
$=\frac{-1}{4}+\frac{1}{8}-1=\frac{-2+1-8}{8}=\frac{-9}{8}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
What is the minimum value of $\mathrm{px}+\mathrm{qy}(\mathrm{p}>0, \mathrm{q}>0)$ when $\mathrm{xy}=\mathrm{r}^{2} ?$

Options:
A $2 r \sqrt{p q}$
B $2 \mathrm{pq} \sqrt{\mathrm{r}}$
C $-2 \mathrm{r} \sqrt{\mathrm{pq}}$
D $2 \mathrm{rpq}$

The correct answer is:
$2 r \sqrt{p q}$

Solution:
Given that $\mathrm{xy}=\mathrm{r}^{2}$ $\Rightarrow \quad \mathrm{y}=\frac{\mathrm{r}^{2}}{\mathrm{x}}$
Let $S=p x+q y=p x+\frac{q r^{2}}{x}$
$\Rightarrow \quad \frac{\mathrm{dS}}{\mathrm{dx}}=\mathrm{p}-\frac{\mathrm{qr}^{2}}{\mathrm{x}^{2}}$
$\frac{\mathrm{dS}}{\mathrm{dx}}=0$ for maximum or minimum.
So, $0=p-\frac{q r^{2}}{x^{2}}$
$\Rightarrow \quad x^{2}=\frac{q r^{2}}{p} \Rightarrow x=\pm \sqrt{\frac{q}{p}} \cdot$
Now, $\frac{d^{2} S}{d x^{2}}=\frac{2 q r^{2}}{x^{3}}$
At $x=+\sqrt{\frac{q}{p}} \cdot r \quad \frac{d^{2} S}{d x^{2}}>0$
Hence, $\mathrm{S}$ is minimum at $\mathrm{x}=\sqrt{\frac{\mathrm{q}}{\mathrm{p}}} \mathrm{r}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{r}^{2}}{\sqrt{\frac{\mathrm{q}}{\mathrm{p}} \cdot \mathrm{r}}}=\sqrt{\frac{\mathrm{p}}{\mathrm{q}} \cdot \mathrm{r}}$
Minimum value of $p x+q y=p \cdot \sqrt{\frac{q}{p}} \cdot r+q \sqrt{\frac{p}{q}} \cdot r$
$=\sqrt{\mathrm{pq}} \mathrm{r}+\sqrt{\mathrm{pq}} \mathrm{r}=2 \mathrm{r} \sqrt{\mathrm{pq}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the minimum value of $[\mathrm{x}(\mathrm{x}-1)+1]^{3}$, where $0 < \mathrm{x} < 1$ ?

Options:
A $\left(\frac{3}{4}\right)^{\frac{1}{3}}$
B 1
C $\frac{1}{3}$
D $\left(\frac{3}{8}\right)^{\frac{1}{3}}$

The correct answer is:
$\left(\frac{3}{4}\right)^{\frac{1}{3}}$

Solution:
$[x(x-1)+1]^{1 / 3}$ is minimum when $x^{2}-x+1$ is minimum, $0 \leq x \leq 1$
$x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\left(1-\frac{1}{4}\right)$
$=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}$
$\therefore \quad$ minimum value of $x^{2}-x+1=\frac{3}{4}$
$\Rightarrow$ Required value $=\left(\frac{3}{4}\right)^{1 / 3}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 2)
$\begin{array}{ll}\text {What is the minimum value of }|\mathrm{x}| ? & {}\end{array}$

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

The correct answer is:
0

Solution:
Minimum value of $|\mathrm{x}|$ is 0 .

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
What is the mode for the data $20,20,20,21,21,21,21,21,22$, $22,22,22,22,22,22,23,23,23,23,23,24,24,25 ?$

Options:
A 7
B 21
C 22
D 25

The correct answer is:
22

Solution:
Since observation 22 occurs maximum time. mode $=22$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 1)
What is the modulus of $\frac{1+2 i}{1-(1-i)^{2}}$ equal to?

Options:
A 5
B 4
C 3
D 1

The correct answer is:
1

Solution:
Let $z=\frac{1+2 i}{1-(1-i)^{2}}$
$\Rightarrow|z|=\frac{|1+2 i|}{\left|1-(1-i)^{2}\right|}=\frac{|1+2 i|}{\left|1-1-i^{2}+2 i\right|}=\frac{|1+2 i|}{|1+2 i|}=1$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 2)
What is the modulus of $\left|\frac{1+2 i}{1-(1-i)^{2}}\right|$ ?

Options:
A 1
B $\sqrt{5}$
C $\sqrt{3}$
D 5

The correct answer is:
1

Solution:
Consider $\frac{1+2 i}{1-(1-i)^{2}}=\frac{1+2 i}{1-(1-1-2 i)}=\frac{1+2 i}{1+2 i}=1$
$\therefore\left|\frac{1+2 i}{1-(1-i)^{2}}\right|=1$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 1)