datavorous/entrance-exam-dataset huggingface dataset:
Consider the following:
$1. \int \ln 10 \mathrm{dx}=\mathrm{x}+\mathrm{c}$
$2. \int 10^{\mathrm{x}} \mathrm{dx}=10^{\mathrm{x}}+\mathrm{c}$
where $\mathrm{c}$ is the constant of integration. Which of the above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
(1) Let $\mathrm{I}=\int \ell \mathrm{n} 10 \mathrm{dx}=\ell \mathrm{n} 10 \int \mathrm{d} \mathrm{x}=[\ell \mathrm{n} 10] \mathrm{x}+\mathrm{c}$
(2) Let $\mathrm{I}=\int 10^{\mathrm{x}} \mathrm{dx}=\frac{10^{\mathrm{x}}}{\log _{\mathrm{e}} 10}+\mathrm{c}=\frac{10^{\mathrm{x}}}{\ln 10}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2012 (Phase 1)
Consider the following :
1) $\quad \sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\frac{\pi}{2}$
2) $\tan ^{-1} \sqrt{3}+\tan ^{-1} 1=-\tan ^{-1}(2+\sqrt{3})$
which of the above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
$\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)$
$\operatorname{If}(-1 \leq x, y \leq 1) \&\left(x^{2}+y^{2} \leq 1\right)$
$\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1} \frac{3}{5}$
$=\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right]$
$=\sin ^{-1}\left[\frac{16}{25}+\frac{9}{25}\right]=\sin ^{-1}(1)=\frac{\pi}{2}$
$\therefore$ Statement $(1)$ is correct
Again, $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1-x y}\right]$
$\begin{aligned} \text { If } ;(x>0),(y>0) \text { and }(x y>1) \\ \tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)=\pi+\tan ^{-1}\left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right] \\=\pi+\tan ^{-1}\left[\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right] \\=\pi+\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{-2}\right)=\pi+\tan ^{-1}[-(2+\sqrt{3})] \\=& \pi-\tan ^{-1}(2+\sqrt{3}) \quad \because \tan ^{-1}(-x)=-\tan ^{-1} x \end{aligned}$
$\therefore$ Statement $(2)$ is incorrect

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 2)
Consider the following :
1- $\mathrm{x}+\mathrm{x}^{2}$ is continuous at $\mathrm{x}=0$
2- $\mathrm{x}+\cos \frac{1}{\mathrm{x}}$ is discontinuous at $\mathrm{x}=0$
3- $\mathrm{x}^{2}+\cos \frac{1}{\mathrm{x}}$ is continuous at $\mathrm{x}=0$
Which of the above are correct?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 2 only

Solution:
$\mathrm{x}+\mathrm{x}^{2}$ is continuous at $\mathrm{x}=0$
$\mathrm{x}+\cos \frac{1}{\mathrm{x}}$ is discontinuous at $\mathrm{x}=0$
$\mathrm{x}^{2}+\cos \frac{1}{\mathrm{x}}$ is not continuous at $\mathrm{x}=0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
Consider the following about the Tyndall effect.
(I) It is used to distinguish between a true and colloidal solution
(II) It is possible only when the dispersed medium and dispersed phase differ much in their refractive indices
(III) It is observed only when the size of colloidal particles is much smaller than the wavelength of the light used
The correct statements are

Options:
A I & III only
B II & III only
C I, II & III
D I & II only

The correct answer is:
I & II only

Solution:
Tyndall effect is observed in large-sized colloidal particles due to scattering of light by them.
This happens when the size of the particles and the wavelength of the light are of similar magnitude.
Thus, Statement (III) is incorrect while (I) and (II) are correct.

Topic: Chemistry, Surface Chemistry, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
Consider the following compounds
(1) C6H5–NH-CH3 (2) C6H5–NH–C6H5
(3) (C6H5)3N (4) CH3–CH2–NH2

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

The correct answer is:
4 > 1 > 2 > 3

Solution:
Here the basic nature order is given here
CH3–CH2–NH2>C6H5–NH-CH3>C6H5–NH–C6H5>(C6H5)3N
(i) Aliphatic amine is more basic then aromatic amines.
(ii) Among aromatic amines, primary is more basic then secondary which is more basic than tertiary due to delocalisation of lone pair of nitrogen.

Topic: Chemistry, General Organic Chemistry, NEET
Consider the following data: $\quad[2010-\mathrm{II}]$ \begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{x} & 5 & 7 & 8 & 4 & 6 \\
\hline \mathrm{y} & 2 & 4 & 3 & 2 & 4 \\
\hline
\end{array} What is the regression equation of $y$ on $x ?$

Options:
A $y=0.6+0.4 x$
B $y=0.7+0.3 x$
C $y=6+5 x$
D $y=4+9 x$.

The correct answer is:
$y=0.6+0.4 x$

Solution:
Given table can be rewritten as
\begin{array}{|c|c|c|c|c|}
\hline \mathbf{x} & \mathbf{y} & \mathbf{x}^{2} & \mathbf{y}^{2} & \mathbf{x y} \\
\hline 5 & 2 & 25 & 4 & 10 \\
\hline 7 & 4 & 49 & 16 & 28 \\
\hline 8 & 3 & 64 & 9 & 24 \\
\hline 4 & 2 & 16 & 4 & 8 \\
\hline 6 & 4 & 36 & 16 & 24 \\
\hline\Sigma x=30 & \sum y=15 & \sum x^{2}=190 & \sum y^{2}=49 & \sum x y=94 \\
\hline
\end{array}
$\overline{\mathrm{x}}=\frac{30}{5}=6$ and $\overline{\mathrm{y}}=\frac{15}{5}=3$
$b_{y x}=\frac{n \sum x y-\left(\sum x\right)\left(\sum y\right)}{n \sum x^{2}-\left(\sum x\right)^{2}}$
$=\frac{5 \times 94-30 \times 15}{5 \times 190-(30)^{2}}=\frac{2}{5}=0.4$
Hence, line of regression is $y-3=0.4(x-6) \Rightarrow y=0.4 x+0.6$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
Consider the following data: $\quad\mathrm{}$ \begin{array}{|l|c|c|}
\hline & Factory - A & Factory - B \\
\hline Mean wage of workers & â‚¹ 540 & â‚¹ 620 \\
\hline Standard deviation of wages & â‚¹ 40.50 & â‚¹ 31 \\
\hline
\end{array} What is the variability in the wages of the workers in Factory $-A ?$

Options:
A $100 \%$ more than the variability in the wages of the workers in Factory $-B$
B $50 \%$ more than the variability in the wages of the workers in Factory $-B$
C $50 \%$ less than the variability in the wages of the workers in Factory-B
D $150 \%$ more than the variability in the wages of the workers in Factory- $B$

The correct answer is:
$50 \%$ less than the variability in the wages of the workers in Factory-B

Solution:
Coefficient of variation $=\frac{\text { S.D }}{\text { Mean }} \times 100$
For factory $\mathrm{A}=\frac{40 \cdot 50}{540} \times 100=7.5$
For factory $\mathrm{B}=\frac{31}{620} \times 100=5$
Variability in wages of $\mathrm{A}$ is $50 \%$ more than the variability in wages of B.

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
Consider the following data:
The coefficient of variation of the above distribution of wages, if its standard deviation is 14.72 , is

Options:
A 14.73
B 23.73
C 26.91
D 20.82

The correct answer is:
26.91

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 2)
Consider the following Eo values
EFe3+/Fe2+o=0.77V
ESn2+/Sno=-0.14V
Under standard conditions the potential for the reaction
Sn(s)+2Fe3+(aq)â2Fe2+(aq)+Sn2+(aq)

Options:
A 1.68 V
B 0.91 V
C 0.63 V
D 1.46 V

The correct answer is:
0.91 V

Solution:
Sn(s)+2Fe3+(aq)â2Fe2+(aq)+Sn2+(aq)
E cell o = E reduction o cathode − E reduction o anode
E cell o =0.77−(−0.14)
∴  E cell o =−0.91 V

Topic: Chemistry, Electrochemistry, NEET
Consider the following electrodes $P=\mathrm{Zn}^{2+}(0.0001 \mathrm{M}) / \mathrm{Zn}, Q=\mathrm{Zn}^{2+}(0.1$ M) $/ \mathrm{Zn}$ $R=\mathrm{Zn}^{2+}(0.01 \mathrm{M}) / \mathrm{Zn}, S=\mathrm{Zn}^{2+}(0.001 \mathrm{M}) / \mathrm{Zn}$ $E^{\circ}\left(\mathrm{Zn} / \mathrm{Zn}^{2+}\right)=-0.76 \mathrm{~V}$ electrode potentials of the above electrodes in volts are in the order

Options:
A $P>S>R>Q$
B $S>R>Q < P$
C $Q>R>S>P$
D $P>Q>R>S$

The correct answer is:
$Q>R>S>P$

Solution:
The standard potential of $\mathrm{Zn} \mid \mathrm{Zn}^{2+}$ half-cell $=-0.76 \mathrm{~V}$
Half-cell reaction,
$\overline{\mathrm{Z}} \mathrm{n}^{2 *}(a q)+\overline{\mathcal{2}} e^{=} \longrightarrow \overline{\mathrm{Z}}(s)$
$E_{\text {red }}=E_{\text {red }}^{\circ}-\frac{0.059}{n} \log \left[\frac{1}{\mathrm{Zn}^{2+}}\right]$
$=-0.76+\frac{0.059}{n} \log \left[\mathrm{Zn}^{2+}\right]$
Lower the concentration of $\mathrm{Zn}^{2+}$, lower is the $E_{\text {red }}$ or vice-versa.
Hence, the correct order is $Q>R>S>P$.

Topic: Chemistry, Electrochemistry, KCET, KCET 2021
Consider the following expressions:
1.$x+x^{2}-\frac{1}{x}$
2.$\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{x}-\mathrm{c}+\frac{\mathrm{d}}{\mathrm{c}}-\frac{\mathrm{e}}{\mathrm{x}^{2}}}$
3.$3 x^{2}-5 x+a b$
4.$\frac{2}{x^{2}-a x+b^{3}}$
5.$\frac{1}{\mathrm{x}}-\frac{2}{\mathrm{x}+5}$
Which of the above are rational expressions?

Options:
A 1,4 and 5 only
B $1,3,4$ and 5 only
C 2,4 and 5 only
D 1 and 2 only

The correct answer is:
$1,3,4$ and 5 only

Solution:
A rational expression isnothing more than a fraction in which the numerator and denominator are polynomials. Here are some example of rational expressions are $\left(x+x^{2}+\frac{1}{x}\right),\left(3 x^{2}-5 x+a b\right), \frac{2}{x^{2}-a x+a b}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 2)
Consider the following for any three non-empty sets $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.
1.$A-(B \cup C)=(A-B) \cup(A-C)$
2.$2 \quad A-B=A-(A \cap B)$
3.$\mathrm{A}=(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A}-\mathrm{B})$
Which of the above is/are correct?

Options:
A Only 1
B 2 and 3
C 1 and 2
D 1 and 3

The correct answer is:
2 and 3

Solution:
A Venndiagram ofthe three non-empty andintersecting sets is drawn, dividing into 7 regions, a to gas showr below and consider statements one by one Statement $(1): \mathrm{A}-(\mathrm{B} \cup \mathrm{C})=(\mathrm{A}-\mathrm{B}) \cup(\mathrm{A}-\mathrm{C})$
LHS = regions $a, b, c, d-b, c, d, e, f, g=a$.
$\mathrm{RHS}=\mathrm{a}, \mathrm{d}, \cup \mathrm{a}, \mathrm{b}=\mathrm{a}, \mathrm{b}, \mathrm{d}$
So, statement $(1)$ is not correct. Statement $(2): A-B=A-(A \cap B)$
LHS=region a, $\mathrm{d}$. $\mathrm{RHS}=$ region $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}-($ region $\mathrm{b}, \mathrm{c},=\mathrm{a}, \mathrm{d}$
So, statement $(2)$ iscorrect. Statement $(3): A=(A \cap B) \cup(A-B)$
LHS = regions $a, b, c, d$. $\mathrm{RHS}=$ regions $\mathrm{b}, \mathrm{c}, \cup$ region $\mathrm{a}, \mathrm{d}=\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ so,
statement (3) is correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
Consider the following for triangle ABC: $\quad[2017-I]$
1) $\quad \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\cos \left(\frac{\mathrm{A}}{2}\right)$
2) $\tan \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\cot \left(\frac{\mathrm{A}}{2}\right)$
3) $\sin (B+C)=\cos A$
4) $\tan (\mathrm{B}+\mathrm{C})=-\cot \mathrm{A}$
Which of the above are correct?

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

The correct answer is:
1 and 2

Solution:
In triangle $\mathrm{ABC}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$
$\Rightarrow \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}=\frac{\pi}{2} \Rightarrow \frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}+\frac{\mathrm{C}}{2}=\frac{\pi}{2}$
$\Rightarrow \frac{\mathrm{B}}{2}+\frac{\mathrm{C}}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2}$
$\Rightarrow \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right)=\cos \frac{\mathrm{A}}{2}$
Also, from (i), $\tan \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right)=\cot \frac{\mathrm{A}}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2017 (Phase 1)
Consider the following four statements (i) ,(ii) ,(iii) and (iv) and select the right option.

(i) In vexillary aestivation, the large posterior petal is called - standard, two lateral ones are wings and two small anterior petals are termed keel.
(ii) The floral formula for liliaceae is â o + â P 3 + 3 A 3 + 3 + G 3 Â¯
(III) In pea flower the stamens are monadelphous
(iv) The floral formula for solanaceae is â o + â P ( 3 ) C ( 3 ) A ( 4 ) + G Â¯ ( 2 )

Options:
A (i) and (iii)
B (i) and (ii)
C (ii) and (iii)
D (iii) and (iv)

The correct answer is:
(i) and (ii)

Solution:
Vexillary or Descending imbricate: The posterior petal is outermost and largest. In it standard or vexillum covers two lateral petals. These two laterals covers two anterior ones. Vexillary arrangement is present in pea family. E.g. Pea, Bean.
The general characters of Liliaceae family is the flowers are bracteate, bisexual, complete, actinomorphic symmetry, hypogynous and trimerous, epiphyllous and in some plants may be epipetalous. They have polyandrous andoecium and superior ovary.

Topic: Biology, Morphology of Flowering Plants, NEET
Consider the following frequency distribution:
\begin{array}{|c|c|c|c|c|c|}
\hline Class interval & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline Frequency & 14 & \mathrm{x} & 27 & \mathrm{y} & 15 \\
\hline
\end{array} If the total of the frequencies is 100 and mode is 25, then which one of the following is correct?

Options:
A $x=2 y$
B $2 x=y$
C $x=y$
D $x=3 y$

The correct answer is:
$x=y$

Solution:
Since the total of the frequencies $=100$ $14+x+27+y+15=100$
$\Rightarrow x+y=44$
Now, to exist the mode $x$ and $y$ should be equal $\mathrm{x}=\mathrm{y}$ is only possibility

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
Consider the following function $f: R \rightarrow R$ such that
following is correct?

Options:
A $f(x)$ is continuous at every $x \in R$
B $f(x)$ is continuous at $x=0$ only
C $f(x)$ is discontinous at $x=0$ only
D $f(x)$ is discontinuous at every $x \in R$

The correct answer is:
$f(x)$ is continuous at every $x \in R$

Solution:
Given, $f(x)=\left\{\begin{aligned} x, & x \geq 0 \\-x^{2}, & x < 0 \end{aligned}\right.$
$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=-\lim _{x \rightarrow 0} x^{2}=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} x=0$
and $f(0)=0$ Since, $\mathrm{LHL}=\mathrm{RHL}=f(0)$
$\therefore f(x)$ is continuous at $x=0$. Also, $f(x)$ is continuous in the given interval ie. $\mathrm{R}$ Hence, $f(x)$ is continuous in every $x \in R$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2009 (Phase 2)
Consider the following functions:
1- $f(x)=e^{x}$, where $x>0$
2- $g(x)=|x-3|$
Which of the above functions is/are continuous?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$, where $\mathrm{x}>0$
According to graph, Graph of $\mathrm{e}^{\mathrm{x}}$ is not breaking when $\mathrm{x}>0$ Therefore, graph is continuous at $\mathrm{x}>0$ Statement II $g(x)=|x-3|$
Graph of $|x-3|$ is not breaking but have sharp turn at $x$ $=3 .$
So, it is continuous

Topic: Mathematics, Functions, NDA, NDA 2013 (Phase 2)
Consider the following functions:
1- $\quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{3}, x \in \mathbb{R}$
2- $\quad f(x)=\sin x, 0 < x < 2 \pi$
3- $\quad f(x)=e^{x}, x \in \mathbb{R}$
Which of the above functions have inverse defined on thein ranges?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
No Solution

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 1)
Consider the following functions: $\quad[2015-I I]$
1- $\quad f(x)=\left\{\begin{array}{lll}\frac{1}{x} & \text { if } & x \neq 0 \\ 0 & \text { if } & x=0\end{array}\right.$
2- $\quad f(x)=\left\{\begin{array}{cll}2 x+5 & \text { if } & x>0 \\ x^{2}+2 x+5 & \text { if } & x \leq 0\end{array}\right.$
Which of the above functions is/are derivable at $\mathrm{x}=0$ ?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{1}{\mathrm{x}} & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.$
as there is a discontinuity at $\mathrm{x}=0$, so function is not differentiable at $\mathrm{x}=0$
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}2 \mathrm{x}+5 & \mathrm{x}>0 \\ \mathrm{x}^{2}+2 \mathrm{x}+5 \quad \mathrm{x} \leq 0\end{array}\right.$
$\mathrm{f}(0)=5$
$\mathrm{LHD}=\lim _{\mathrm{x} \rightarrow 0^{-}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{x}^{2}+2 \mathrm{x}+5-5}{\mathrm{x}}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x}+2=2$
$\mathrm{RHD}=\lim _{\mathrm{x} \rightarrow 0^{+}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{x}+5-5}{\mathrm{x}}=2$
$\therefore$ It is differentiable at $\mathrm{x}=0$
$\therefore$ Only $(2)$ is differentiable at $\mathrm{x}=0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 2)
$\quad$ Consider the following
I. $\quad \operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right)=-\frac{\pi}{3}$
II. $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}$
Which of the above is/are correct?

Options:
A Only I
B Only II
C Both I and II
D Neither Inor II

The correct answer is:
Both I and II

Solution:
Consider (I):
$$
\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3}
$$
Let $\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\theta$
$$
\begin{array}{l}
\Rightarrow \frac{-2}{\sqrt{3}}=\operatorname{cosec} \theta \Rightarrow \frac{-\sqrt{3}}{2}=\frac{1}{\operatorname{cosec} \theta} \\
\Rightarrow \frac{-\sqrt{3}}{2}=\sin \theta \Rightarrow \theta=\frac{-\pi}{3}
\end{array}
$$
Now, consider (II):
$$
\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}
$$
Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta$
$\Rightarrow \frac{2}{\sqrt{3}}=\sec \theta \Rightarrow \frac{\sqrt{3}}{2}=\frac{1}{\sec \theta}$
$\Rightarrow \frac{\sqrt{3}}{2}=\cos \theta \quad \Rightarrow \theta=\frac{\pi}{6}$
Hence, both statements I and II are correct.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2011 (Phase 1)
Consider the following:

i. For a solution at atmospheric pressure Î¨w = Î¨s
ii. Î¨w (water potential) of a cell is affected by solute potential only
iii. Î¨p (pressure potential) is usually positive in xylem during the day

Which of the statements given above is/are correct?

Options:
A ii & iii
B i only
C ii only
D i, ii & iii

The correct answer is:
i only

Solution:
Water potential of a solution is the sum total of four components: gravitational potential, osmotic(solute) potential, matric potential and pressure potential. But, since matric potential and gravitational potential are negligible, water potential is generally more affected by solute and pressure potential. Pressure potential of water during the standard stage is 0 MPa. The positive value of water potential within the cell is referred to as turgor pressure. During the day when transpiration is high, pressure potential of the xylem mostly is negative since water is evaporated and its turgor pressure is less.

Topic: Biology, Transport in Plants, NEET
Consider the following
If $\vec{a}$ and $\vec{b}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then
$1.$ $\overrightarrow{\mathrm{CE}}=\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}$
$2.$ $\overrightarrow{\mathrm{AE}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$3.$ $\quad \overrightarrow{\mathrm{FA}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
Which of the above are correct?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1,2 and 3

Solution:
Let ABCDEF be the regular hexagen as shown in the figure.
Let $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
Join $\mathrm{AD}, \mathrm{FC}$ and $\mathrm{EB}$. They meet at a common point $\mathrm{O}$, which is the centre of hexagon.
$\mathrm{AO} \| \mathrm{BC}$ so, $\overrightarrow{\mathrm{AO}}=\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
$\mathrm{OC} \| \mathrm{AB} \mathrm{so}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$
OAB forms a triangle, $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}$
$\Rightarrow \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\mathrm{BO}=\mathrm{OE}$ and they are on the same line,
So, $\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\operatorname{In} \Delta \mathrm{OCE}, \overrightarrow{\mathrm{CO}}+\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{CE}}$
$\Rightarrow \overrightarrow{\mathrm{CE}}=-\overrightarrow{\mathrm{OC}}+\mathrm{OE}=-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}-\overrightarrow{2 \mathrm{a}}$
So, (1) is correct.
$\overrightarrow{\mathrm{BE}}=2 \overrightarrow{\mathrm{OB}}$ In $\Delta \mathrm{AEB}, \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{AB}}+2 \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{A}}+2(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
So, $(2)$ is also correct.
$\mathrm{FA} \| \mathrm{OB} \Rightarrow \overrightarrow{\mathrm{FA}}=-\overrightarrow{\mathrm{BO}}=-(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, (3) is also correct.
So, $(1),(2) \&(3)$ are correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 1)
Consider the following in respect of matrices $\mathrm{A}$ and $\mathrm{B}$ of same order: $\quad[2018-I I]$
$1.$ $\mathrm{A}^{2}-\mathrm{B}^{2}=(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})$
$2.$ $(A-I)(I+A)=0 \Leftrightarrow A^{2}=I$
Where I is the identity matrix and $\mathrm{O}$ is the null matrix.
Which of the above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
Matrix product is commutative if both are diagonal matrices of same order.
$\Rightarrow \quad A^{2}-B^{2}=(A+B)(A-B)$ is not true.
Next, $(\mathrm{A}-\mathrm{I})(\mathrm{A}+\mathrm{I})=0$
$\Rightarrow \quad \mathrm{A}^{2}+\mathrm{AI}-\mathrm{IA}-\mathrm{I}^{2}=0(\because \mathrm{AI}=\mathrm{IA})$
$\Rightarrow \mathrm{A}^{2}=\mathrm{I}$ is correct.

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
Consider the following in respect of matrices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of same order:
$1.$ $(\mathrm{A}+\mathrm{B}+\mathrm{C})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}+\mathrm{C}^{\prime}$
$2.$ $(\mathrm{AB})^{\prime}=\mathrm{A}^{\prime} \mathrm{B}^{\prime}$
$3.$ $(\mathrm{ABC})^{\prime}=\mathrm{C}^{\prime} \mathrm{B}^{\prime} \mathrm{A}^{\prime}$
Where $\mathrm{A}^{\prime}$ is the transpose of the matrix $\mathrm{A}$. Which of the above are correct?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
By property, statement 1 and 3 are correct.

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
Consider the following in respect of sets $\mathrm{A}$ and $\mathrm{B}$
1.$(\mathrm{A}-\mathrm{B}) \cup \mathrm{B}=\mathrm{A}$
2.$\quad(\mathrm{A}-\mathrm{B}) \cup \mathrm{A}=\mathrm{A}$
3.$(\mathrm{A}-\mathrm{B}) \cap \mathrm{B}=\phi$
4.$\quad \mathrm{A} \subset \mathrm{B} \Rightarrow \mathrm{A} \cup \mathrm{B}=\mathrm{B}$
Which of the above are correct?

Options:
A 1,2 and 3
B 2,3 and 4
C 1,3 and 4
D 1,2 and 4

The correct answer is:
2,3 and 4

Solution:
Venn diagram
from, Venn diagram we can observe that $\mathrm{A}-\mathrm{B}$ is the shaded part.
$(\mathrm{A}-\mathrm{B}) \cup \mathrm{A}=\mathrm{A}$
$(\mathrm{A}-\mathrm{B}) \cap \mathrm{B}=\phi$
$A \subseteq B \Rightarrow A \cup B=B$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 1)