datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following is correct? If $\frac{1}{b-c}+\frac{1}{b-a}=\frac{1}{a}+\frac{1}{c}$, then $a, b, c$ are in

Options:
A AP
B $\mathrm{HP}$
C GP
D None of these

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

Solution:
As given : $\frac{1}{\mathrm{~b}-\mathrm{c}}+\frac{1}{\mathrm{~b}-\mathrm{a}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}$
$\Rightarrow \frac{1}{b-c}-\frac{1}{a}+\frac{1}{b-a}-\frac{1}{c}=0$
$\Rightarrow \frac{a-b+c}{a(b-c)}+\frac{c-b+a}{c(b-a)}=0$
$\Rightarrow(a-b+c)\left\{\frac{1}{a(b-c)}+\frac{1}{c(b-a)}\right\}=0$
$\Rightarrow \frac{\mathrm{cb}-\mathrm{ac}+\mathrm{ab}-\mathrm{ac}}{\mathrm{ac}(\mathrm{b}-\mathrm{c})(\mathrm{b}-\mathrm{a})}=0$
$\Rightarrow \mathrm{cb}+\mathrm{ab}=2 \mathrm{ac}$
Dividing both sides by abc
$\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}=\frac{2}{\mathrm{~b}} \Rightarrow \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}$ are in $\mathrm{AP.} \Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{HP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 1)
Which one of the following is correct? If the positive numbers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are in AP, then bed, cda, dab, abc

Options:
A are in AP
B are in GP
C are in HP
D are in none of the above progressions

The correct answer is:
are in HP

Solution:
As given : $a, b, c, d$ are in AP
$\Rightarrow \frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}, \frac{1}{\mathrm{~d}}$ are in $\mathrm{HP}$
Multiplying by abcd throughout
$\Rightarrow \frac{\text { abcd }}{\mathrm{a}}, \frac{\mathrm{abcd}}{\mathrm{b}}, \frac{\mathrm{abcd}}{\mathrm{c}}, \frac{\mathrm{abcd}}{\mathrm{d}}$, are in HP
$\Rightarrow$ bed, acd, abd, abc are in HP.

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
Which one of the following is correct? If the vector $\vec{c}$ is
normal to the vectors $\vec{a}$ and $\vec{b}$, then $\vec{c}$, is :

Options:
A parallel to both $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
B normal to $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$ and parallel to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
C normal to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and parallel to $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
D normal to both $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$

The correct answer is:
normal to both $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$

Solution:
As given
$\vec{c}$ is normal to the vectors $\vec{a}$ and $\vec{b}$
$\begin{array}{ll}\Rightarrow & \vec{c} \cdot \vec{a}=0 \text { and } \vec{c} \cdot \vec{b}=0 \\ \Rightarrow & \vec{c} \cdot(\vec{a}+\vec{b})=\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}=0 \\ \Rightarrow & \text { Also } \vec{c} \cdot(\vec{a}-\vec{b})=\vec{c} \cdot \vec{a}-\vec{c} \cdot \vec{b}=0\end{array}$
$\overrightarrow{\mathrm{c}}$ is normal to $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 2)
Which one of the following is correct? If $z$ and w are complex numbers and $\overline{\mathrm{w}}$ denotes the conjugate of $\mathrm{w}$, then $\mathrm{z}+\mathrm{w}$ $=|\mathrm{z}-\mathrm{w}|$ holds only, if

Options:
A $\quad z=0$ or $\mathrm{w}=0$
B $\quad z=0$ and $w=0$
C $\mathrm{z} \cdot \overline{\mathrm{w}}$ is purelyreal
D $\mathrm{z} \cdot \overline{\mathrm{w}}$ is purelyimaginary

The correct answer is:
$\quad z=0$ or $\mathrm{w}=0$

Solution:
Given that, $|\mathrm{z}+\mathrm{w}|=|\mathrm{z}-\mathrm{w}| \Rightarrow$ either $\mathrm{z}$ or, $\mathrm{w}$ is zero.

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 1)
Which one of the following is correct in respect of the cube roots of unity?

Options:
A They are collinear
B They lie on a circle of radius $\sqrt{3}$
C They form an equilateral triangle
D None of the above

The correct answer is:
They form an equilateral triangle

Solution:
$\Delta$ is equilateral.

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 2)
Which one of the following is correct in respect of the
function $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}^{+}$ defined as $\mathrm{f}(\mathrm{x})=|\mathrm{x}+1|$ ?

Options:
A $\quad f\left(x^{2}\right)=[f(x)]^{2}$
B $\mathrm{f}(\mid \mathrm{x})=|\mathrm{f}(\mathrm{x})|$
C $\quad f(x+y)=f(x)+f(y)$
D None of the above

The correct answer is:
None of the above

Solution:
Given, $\mathrm{f}(\mathrm{x})=|\mathrm{x}+1|$.
Let us check all options.
$\begin{array}{ll}\text { (a) } & \mathrm{f}(\mathrm{x})^{2}=\left|\mathrm{x}^{2}+1\right| \\ & (\mathrm{f}(\mathrm{x}))^{2}=(\mathrm{x}+1)^{2} \\ & \text { So, } \mathrm{f}(\mathrm{x})^{2} \neq(\mathrm{f}(\mathrm{x}))^{2} \\ \text { (b) } & \mathrm{f}(|\mathrm{x}|)=\| \mathrm{x}|+1| \\ & |\mathrm{f}(\mathrm{x})|=\|\mathrm{x}+1\|=|\mathrm{x}+1| \\ & \mathrm{f}(\mathrm{x}) \neq|\mathrm{f}(\mathrm{x})| \\ \text { (c) } & \mathrm{f}(\mathrm{x}+\mathrm{y})=|\mathrm{x}+\mathrm{y}+1| \\ & \mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})=|\mathrm{x}+1|+|\mathrm{y}+1| \\ & \mathrm{f}(\mathrm{x}+\mathrm{y}) \neq \mathrm{f}(\mathrm{x}+\mathrm{f}(\mathrm{y}))\end{array}$

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 1)
Which one of the following is correct in respect of the function $f(x)=x(x-1)(x+1) ?$

Options:
A Thelocalmaximumvalueislarger than local minimumvahe
B Thelocal maximum value is smaller than local minimum value
C The function has no local maximum
D The function has no local minimum

The correct answer is:
Thelocalmaximumvalueislarger than local minimumvahe

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)=\mathrm{x}\left(\mathrm{x}^{2}-1\right)$
Differentiating, we get $f^{\prime}(x)=x(2 x)+\left(x^{2}-1\right)=2 x^{2}+x^{2}-1=3 x^{2}-1$
Again differentiating $\mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}$
At $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 3 \mathrm{x}^{2}-1=0 \Rightarrow \mathrm{x}^{2}=\frac{1}{3} \Rightarrow \mathrm{x}=\pm \frac{1}{\sqrt{3}}$
Maximum value of $\mathrm{f}(\mathrm{x})=\mathrm{f}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-1}{\sqrt{3}}\left(\left(\frac{-1}{\sqrt{3}}\right)^{2}-1\right)$
$=\frac{-1}{\sqrt{3}}\left(\frac{1}{3}-1\right)=\frac{-1}{\sqrt{3}}\left(\frac{-2}{3}\right)=\frac{2}{3 \sqrt{3}}$
Minimumvalue of $f(x)=f\left(\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}\left(\frac{1}{3}-1\right)=\frac{-2}{3 \sqrt{3}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
Which one of the following is correct in respect of the
function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{|\mathrm{x}|}$ for $\mathrm{x} \neq 0$ and $\mathrm{f}(0)=0 ?$

Options:
A $\mathrm{f}(\mathrm{x})$ is discontinuous every where
B $\mathrm{f}(\mathrm{x})$ is continuous every where
C $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ only
D $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$ only

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is continuous every where

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\mathrm{x}^{2}}{\mathrm{x}}, & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.$
$=\left\{\begin{array}{ll}\frac{x^{2}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{x^{2}}{-x}=-x, & x < 0\end{array}\right.$
Now, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(-x)=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x)=0$
and $f(0)=0$ So, $f(x)$ is continuous at $x=0$ Also, $\mathrm{f}(\mathrm{x})$ is continuous for all other values of $\mathrm{x}$. Hence, $\mathrm{f}(\mathrm{x})$ is continuous everywhere.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2012 (Phase 2)
Which one of the following is correct in respect of the
function $f(x)=|x|+x^{2}$

Options:
A $f(x)$ is not continuous at $x=0$
B $f(x)$ is differentiable at $x=0$
C $f(x)$ is continuous but not differentiable at $x=0$
D None of the above

The correct answer is:
$f(x)$ is continuous but not differentiable at $x=0$

Solution:
$\because \quad f(x)=|x|+x^{2}$
$\Rightarrow f(x)=\left\{\begin{array}{ll}x^{2}+x, & x \geq 0 \\ x^{2}-x, & x < 0\end{array}\right.$
LHL $=\lim _{x \rightarrow 0^{-}} f(x)$
$=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}(0-h)^{2}-(0-h)$
$=\lim _{h \rightarrow 0} h^{2}+h=0$
and $\operatorname{RHL}=\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0} f(0+h)$
$=\lim _{h \rightarrow 0}(0+h)^{2}+(0+h)=\lim _{h \rightarrow 0} h^{2}+h=0$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$
$\Rightarrow f(x)$ is continuous at $x=0$
Now, $\mathrm{LHD}=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$
$=\lim _{h \rightarrow 0} \frac{h^{2}+h}{-h}=-\lim _{h \rightarrow 0} h+1=-1$
and, $\mathrm{RHD}=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{h^{2}+h}{h}=\lim _{h \rightarrow 0} h+1=1$
Thus, $\mathrm{LHD} \neq \mathrm{RHD}$
$\Rightarrow f(x)$ is not differentiable at $x=0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2011 (Phase 1)
Which one of the following is correct in respect of the matrix
$A=\left[\begin{array}{ccc}
0 & 0 & -1 \\
0 & -1 & 0 \\
-1 & 0 & 0
\end{array}\right] ?$

Options:
A $A^{-1}$ does not exist
B $A=(-1) I$
C $A$ is a unit matrix
D $A^{2}=I$

The correct answer is:
$A^{2}=I$

Solution:
Given, $A=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]$
$\therefore \quad|A|=\left|\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right|=-1(-1)=1 \neq 0$
$\therefore \quad \mathrm{A}^{-1}$ exists
Now, $A^{2}=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A^{2}=I$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 1)
Which one of the following is correct? The eccentricity of the conic
$$
\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}=1,(\lambda \geq 0)
$$

Options:
A increases with increase in $\lambda$
B decreases with increase in $\lambda$
C does not change with $\lambda$
D None of the above

The correct answer is:
decreases with increase in $\lambda$

Solution:
Equation of the given conic is an equation of ellipse
$\frac{x^{2}}{a^{2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}(x \geq 0)$
$\Rightarrow \quad A^{2}=a^{2}+\lambda$ and $B^{2}=b^{2}+\lambda$
Eccentricity, $\mathrm{e}=\sqrt{1-\frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}+\lambda}{\mathrm{a}^{2}+\lambda}}$
$=\sqrt{\frac{a^{2}+\lambda-b^{2}-\lambda}{a^{2}+\lambda}}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}+\lambda}}$
$\lambda$ is in the denominator so, when $\lambda$ increases, the eccentricity decreases.

Topic: Mathematics, Ellipse, NDA, NDA 2008 (Phase 1)
Which one of the following is correct?
The function $\mathrm{f}: \mathrm{A} \rightarrow$ R where $A=\left\{x \in R,-\frac{\pi}{2} < x < \frac{\pi}{2}\right\}$ defined by $f(x)=\tan x$.

Options:
A Injective
B Not injective
C Bijective
D Not bijective

The correct answer is:
Injective

Solution:
$f(x)=tan x$
$\mathrm{f}(\mathrm{x})$ is increasing in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Hence, $\mathrm{f}(\mathrm{x})$ is injective in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
Which one of the following is correct? The function $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+1$ is $\quad$

Options:
A negative for all $\mathrm{x}>0$
B positive for all $x>0$
C increasing for all $\mathrm{x}$
D decreasing for all $x$

The correct answer is:
increasing for all $\mathrm{x}$

Solution:
The given function is $f(x)=(x-1) e^{x}+1$
$\begin{aligned} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) &=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}} \\ &=(\mathrm{x}-1+1) \mathrm{e}^{\mathrm{x}}=\mathrm{xe}^{\mathrm{x}} \end{aligned}$
Thus, it is clear that $\mathrm{f}(\mathrm{x})$ is increasing for all $\mathrm{x}$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 2)
Which one of the following is correct? The real number $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ is:

Options:
A an integer
B a rational number but not an integer
C an irrational number
D none of the above

The correct answer is:
a rational number but not an integer

Solution:
The given number
$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
can be written as
$=(2+\sqrt{5})^{1 / 3}+(2-\sqrt{5})^{1 / 3}$
$=2^{1 / 3}\left[1+\frac{1}{2} \sqrt{5}\right]^{1 / 3}+2^{1 / 3}\left[1-\frac{1}{2} \sqrt{5}\right]^{1 / 3}$
$=2^{1 / 3}\left[1+\frac{1}{6} \sqrt{5}+\ldots+1-\frac{1}{6} \sqrt{5}+\ldots\right]$
Thus the given number is a rational number but not an integer

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
Which one of the following is correctly matched regarding an Institute and its location?

Options:
A National Institute of Virology - Pune
B National Institute of Communicable Diseases - Lucknow
C Central Drug Research Institute - Kasauli
D National Institute of Nutrition - Mumbai

The correct answer is:
National Institute of Virology - Pune

Solution:
The Virus Reselarch Centre (VRC), Pune came to existance in 1952 under the joint auspices of the ICMR and the Rockefeller Foundation, as a part of the global programme of investigations on the arthropod - borne group of viruses. In view of its expanded scope and activities, the VRC was undesignated as the National Institute of Virology (NIV) in 1978. The NIV is identified today as the WHO collaborating centre for arbovirus reference and research. NIV also acts as the National Monitoring Centre for Influenza and Hepatitis.

Topic: Biology, Biological Classification, AIIMS, AIIMS 2004
Which one of the following is not a characteristic feature of biodiversity hot spots?

Options:
A Large number of species
B Abundance of endemic species
C Mostly located in the polar regions
D Mostly located in the tropics

The correct answer is:
Mostly located in the polar regions

Solution:
Characteristic feature of biodiversity hot spots are large number of species, abundance of endemic species and mostly located in the tropics.

Topic: Biology, Biodiversity and its Conservation, NEET
Which one of the following is not a feature of biodiversity hotspots?

Options:
A
Large number of species

B
Abundance of endemic species

C
Mostly located in the tropics

D
Mostly located in the polar regions

The correct answer is:
Mostly located in the polar regions

Solution:
Biodiversity hotspots are characterised by large number of flora and fauna which is abundant in endemic species and also have large number of alien or exotic species. They are mostly found in tropical and temperate regions. There are no biodiversity hotspots in polar regions.

Topic: Biology, Biodiversity and its Conservation
Which one of the following is not correct?

Options:
A $\sin ^{-1}\{\sin (5 \pi / 4)\}=-\pi / 4$
B $\sec ^{-1}\{\sec (5 \pi / 4)\}=3 \pi / 4$
C $\tan ^{-1}\{\tan (5 \pi / 4)\}=\pi / 4$
D $\cos e c^{-1}\{\operatorname{cosec}(7 \pi / 4)\}=\pi / 4$

The correct answer is:
$\cos e c^{-1}\{\operatorname{cosec}(7 \pi / 4)\}=\pi / 4$

Solution:
Option (a) $\sin ^{-1}\left(\sin \frac{5 \pi}{4}\right)=\frac{-\pi}{4}$
$\Rightarrow \sin \frac{5 \pi}{4}=\sin \left(\frac{-\pi}{4}\right)$
$\Rightarrow \sin \left(\pi+\frac{\pi}{4}\right)=-\sin \frac{\pi}{4}$
$\Rightarrow-\sin \left(\frac{\pi}{4}\right)=-\sin \left(\frac{\pi}{4}\right)$
Hence it is correct Option (b)
$\sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)=\frac{3 \pi}{4}$
$\Rightarrow \sec \frac{5 \pi}{4}=\sec \frac{3 \pi}{4}$
$\Rightarrow \sec \left(2 \pi-\frac{3 \pi}{4}\right)=\sec \frac{3 \pi}{4}$
$\Rightarrow \sec \frac{3 \pi}{4}=\sec \frac{3 \pi}{4}$
Hence, it is correct
Option (c)
$\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)=\frac{\pi}{4}$
$\Rightarrow \tan \frac{5 \pi}{4}=\tan \frac{\pi}{4}$
$\Rightarrow \tan \left(\pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4}$
$\Rightarrow \tan \frac{\pi}{4}=\tan \frac{\pi}{4}$
Hence it is correct Option (d). $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{7 \pi}{4}\right)=\frac{\pi}{4}$
$\Rightarrow \operatorname{cosec} \frac{7 \pi}{4}=\operatorname{cosec} \frac{\pi}{4}$
$\Rightarrow \operatorname{cosec}\left(2 \pi-\frac{\pi}{4}\right)=\operatorname{cosec} \frac{\pi}{4}$
$\Rightarrow-\operatorname{cosec} \frac{\pi}{4}=\operatorname{cosec} \frac{\pi}{4}$
Hence it is not correct.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 2)
Which one of the following is not included under in-situ conservation?

Options:
A national park
B sanctuary
C botanical garden
D biosphere reserve

The correct answer is:
botanical garden

Solution:
In situ is the methods of conserving all the living species, especially the wild and endangered species in their natural habitats and environment. Ex-situ is the methods of conserving all the living species in the artful habitats that reflect their natural living habitats.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2006
Which one of the following is not included under In-situ conservation?

Options:
A Botanical garden
B Biosphere reserve
C National Park
D Sanctuary

The correct answer is:
Botanical garden

Solution:
In-situ conservation means preserving endangered species of animals or plants in their natural environment. In-situ conservation methods include hotspots, national parks, sanctuaries, biosphere reserve, etc.. Ex-situ conservation includes a botanical garden, cryopreservation, zoological parks, etc.

Topic: Biology, Biodiversity and its Conservation, NEET
Which one of the following is one of the roots of the equation $(b-c) x^{2}+(c-a) x+(a-b)=0 ?$

Options:
A $(c-a) /(b-c)$
B $(a-b) /(b-c)$
C $(b-c) /(a-b)$
D $(c-a) /(a-b)$

The correct answer is:
$(a-b) /(b-c)$

Solution:
Given eqn is $(b-c) x^{2}+(c-a) x+(a-b)=0$ $\Rightarrow(b-c) x^{2}-(b-c-b+a) x+(a-b)=0$
$\Rightarrow(b-c) x(x-1)-(a-b)(x-1)=0$
$\Rightarrow\{(b-c) x-(a-b)\}\{x-1\}=0$
$\Rightarrow x=\frac{a-b}{b-c}$ and $x=1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
Which one of the following is one of the solutions of the equation of the equation $\tan 2 \theta \cdot \tan \theta=1 ?

Options:
A $\pi / 12$
B $\pi / 6$
C $\pi / 4$
D $\pi / 3$

The correct answer is:
$\pi / 6$

Solution:
$\tan 2 \theta \cdot \tan \theta=1$
$\Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta} \cdot \tan \theta=1$
$\Rightarrow 2 \tan ^{2} \theta=1-\tan ^{2} \theta \Rightarrow 3 \tan ^{2} \theta=1$
$\Rightarrow \tan ^{2} \theta=\frac{1}{3}=\left(\frac{1}{\sqrt{3}}\right)^{2}$
$\Rightarrow \tan ^{2} \theta=\tan ^{2}\left(30^{\circ}\right)=\tan ^{2}\left(\frac{\pi}{6}\right)\left[\because \theta=\mathrm{n} \pi \pm \frac{\pi}{6}\right]$
$\therefore \theta=\frac{\pi}{6}$
$$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2014 (Phase 2)
Which one of the following is positive in the third quadrant?

Options:
A $\sin \theta$
B $\cos \theta$
C $\tan \theta$
D $\sec \theta$

The correct answer is:
$\tan \theta$

Solution:
$\tan \theta$ is positive in third quadrant

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
Which one of the following is resistant enzyme action?

Options:
A Cork
B Wood fibre
C Pollen exine
D Leaf cuticle

The correct answer is:
Pollen exine

Solution:
The pollen wall is consisted of two layers the outer exine and inner intine. The exine is chiefly made up of sporopollenin, which is derived by the oxidative polymerization of carotenoids. Sporopollenin is one of the most resistant biological materials known. Exine is thin in begining but become very thick with maturity.
During secondary growth in extrastelar region cork cambium or phellogen cells divide periclinally, cutting off cells towards the outside and inside. The cells cutoff towards outside become suberised and dead and form cork. Cork is impervious to water due to suberin and provides protection to underlying tissues.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2008 (Screening)
Which one of the following is the correctly matched pair of an endangered animal and a national park?

Options:
A Great Indian : Keoladeo national Bustard park
B Lion : Corbett national park
C Rhinoceros : Kaziranga national park
D Wild Ass : Dudhwa national park

The correct answer is:
Rhinoceros : Kaziranga national park

Solution:
Rhinoceros (Rhinoceros unicornis)
- Kaziranga National Park, Assam
Siberian Crane - Keoladeo National Park, Rajasthan
Tiger - Jim Corbett National Park, Shimla
Tiger - Dudhwa Tiger Reserve, UP
Related Theory
National Parks are the areas that are set by the government to conserve the natural environment. Sanctuary is a place of refuge where injured, abandoned and abused wildlife is allowed to live in peace in their natural environment without any human intervention.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2006
Which one of the following is the differential equation to family of circles having centre at the origin? $\quad$

Options:
A $\left(x^{2}-y^{2}\right) \frac{d y}{d x}=2 x y$
B $\left(x^{2}+y^{2}\right) \frac{d y}{d x}=2 x y$
C $\frac{d y}{d x}=\left(x^{2}+y^{2}\right)$
D $x d x+y d y=0$

The correct answer is:
$x d x+y d y=0$

Solution:
The equation of family of circles having centres at the origin is $x^{2}+y^{2}=r^{2}$
where $r^{\prime}$ is the radius. Differentiate both side w.r.t. $x$, we get
$$
2 x+2 y \frac{d y}{d x}=0
$$
$$
\begin{aligned}
& 2 x d x+2 y d y=0 \\
\Rightarrow \quad & x d x+y d y=0
\end{aligned}
$$
which is required differential equation.

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 2)
Which one of the following is the mean of the data given below?
\begin{array}{|c|c|c|c|c|c|c|c|}
\hlinex_{i} & 6 & 10 & 14 & 18 & 24 & 28 & 30 \\
\hlinef_{i} & 2 & 4 & 7 & 12 & 8 & 4 & 3 \\
\hline
\end{array}

Options:
A 17
B 18
C 19
D 20

The correct answer is:
19

Solution:
Given table can be rewritten \begin{array}{|ccc|}
\hlinex_{i} & f_{i} & f_{i} x_{i} \\
\hline 6 & 2 & 12 \\
10 & 4 & 40 \\
14 & 7 & 98 \\
18 & 12 & 216 \\
24 & 8 & 192 \\
28 & 4 & 112 \\
30 & 3 & 90 \\
\hline Total & 40 & 760 \\
\hline
\end{array} So, required mean $=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{760}{40}=19$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
Which one of the following is the plane containing the line
$\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{5}$ and parallel to z-axis?

Options:
A $2 x-3 y=0$
B $5 x-2 z=0$
C $5 y-3 z=0$
D $\quad 3 x-2 y=0 \quad$

The correct answer is:
$\quad 3 x-2 y=0 \quad$

Solution:
The equation of the line is
$\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{5}=r$
where $\mathrm{r}$ is a constant. Any point on this line, is given by $x=2 r+2, y=3 r+2$ and $z=5 r+4$
Since, a plane that is parallel to z-axis will have no zco-ordinate, $z=0$
$\mathrm{z}=0 \Rightarrow 5 \mathrm{r}+4=0$ or, $\mathrm{r}=\frac{-4}{5}$
Putting this value of $\mathrm{r}$ for $\mathrm{x}$ and $\mathrm{y}$ co-ordinates.
$\mathrm{x}=2 \mathrm{r}+2=2 \times\left(-\frac{4}{5}\right)+2$
or, $5 \mathrm{x}=-8+10=2$
$\mathrm{x}=\frac{2}{5}$, or $\frac{2}{\mathrm{x}}=5$
Similarly, $\mathrm{y}=3 \mathrm{r}+3=3 \times\left(-\frac{4}{5}\right)+3$
or, $5 \mathrm{y}=-12+15=3$
$\mathrm{y}=\frac{3}{5} \Rightarrow \frac{3}{\mathrm{y}}=5$
From equations (1) and (2)
$\frac{2}{x}=\frac{3}{y} \Rightarrow 3 x-2 y=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 1)