datavorous/entrance-exam-dataset huggingface dataset:
What is the determinant
$\left|\begin{array}{lll}\text { bc } & a & a^{2} \\ c a & b & b^{2} \\ a b & c & c^{2}\end{array}\right|$ equal to?

Options:
A $\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|$
B $\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$
C $\left|\begin{array}{ccc}1 & a & a^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3}\end{array}\right|$
D $\left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right|$

The correct answer is:
$\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Solution:
$\left|\begin{array}{lll}b c & a & a^{2} \\ c a & b & b^{2} \\ a b & c & c^{2}\end{array}\right|$
$\mathrm{R}_{1} \rightarrow \mathrm{aR}_{1}, \mathrm{R}_{2} \rightarrow \mathrm{bR}_{2}, \mathrm{R}_{3} \rightarrow \mathrm{cR}_{3}$ and divide whole
determinant by abc
$=\frac{1}{a b c}\left|\begin{array}{ccc}a b c & a^{2} & a^{3} \\ a b c & b^{2} & b^{3} \\ a b c & c^{2} & c^{2}\end{array}\right|$
$=\frac{a b c}{a b c}\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & b^{2} & c^{2}\end{array}\right|=\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 1)
What is the determinant of the matrix $\left(\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{y}+\mathrm{z} \\ \mathrm{z} & \mathrm{z} & \mathrm{z}+\mathrm{x} \\ \mathrm{y} & \mathrm{z} & \mathrm{x}+\mathrm{y}\end{array}\right)$?

Options:
A $(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})$
B $(\mathrm{x}-\mathrm{z})(\mathrm{z}-\mathrm{x})$
C $(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})$
D $(\mathrm{z}-\mathrm{x})^{2}(\mathrm{x}+\mathrm{y}+\mathrm{z})$

The correct answer is:
$(\mathrm{z}-\mathrm{x})^{2}(\mathrm{x}+\mathrm{y}+\mathrm{z})$

Solution:
$\left|\begin{array}{ccc}x & y & y+z \\ z & x & z+x \\ y & z & x+y\end{array}\right|=\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$\left|\begin{array}{ccc}x+y+z & x+y+z & 2(x+y+z) \\ z & x & z+x \\ y & z & x+y\end{array}\right|$
$\begin{aligned} &=(x+y+z)\left|\begin{array}{ccc}1 & 1 & 2 \\ z & x & z+x \\ y & z & x+y\end{array}\right| \\ &=(x+y+z)(z+x)^{2},(\text { replacing } z \operatorname{by} x) \end{aligned}$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 2)
What is the diameter of the sphere $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-4 \mathrm{x}+6 \mathrm{y}-8 \mathrm{z}-7=0 ?$

Options:
A 4 units
B 5 units
C 6 units
D 12 units.

The correct answer is:
12 units.

Solution:
General equation of sphere is. $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{ux}+2 \mathrm{vy}+2 \mathrm{wz}+\mathrm{d}=0$
On comparing with the given equation we have $\mathrm{u}=-2, \mathrm{v}=3, \mathrm{w}=-4, \mathrm{~d}=-7$
$\begin{aligned} \text { Radius } &=\sqrt{u^{2}+v^{2}+w^{2}-d} \\ &=\sqrt{4+9+16+7}=\sqrt{36}=6 \end{aligned}$
Diameter $=2 \times 6=12$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 1)
What is the difference between a primary oocyte and a secondary oocyte?

Options:

None

Solution:
Primary oocyte a diploid $(2 n)$ cell, formed by the transformation of primordial follicles in foetal ovary when the gamete mother cell, oogonia is arrested at prophase-1 of meiosis. Secondary oocyte the hapliod (n) cell formed from primary oocyte that completes its first meiotic division, during puberty and produces the female gamete ova(n).

Topic: Biology, Human Reproduction
What is the difference between primary and secondary metabolites?

Options:
A Primary metabolites have identifiable functions while secondary metabolites do not.
B Secondary metabolites have identifiable functions while primary metabolites do not.
C Both primary and secondary metabolites have identifiable functions.
D Neither primary nor secondary metabolites have identifiable functions.

The correct answer is:
Primary metabolites have identifiable functions while secondary metabolites do not.

Solution:
Primary metabolites have identifiable functions and play known roles in normal physiological processes, while the role or functions of all secondary metabolites in host organisms is not yet understood.

Topic: Biology, Biomolecules (B), NEET
What is the difference between the smallest five digit binary integer and the lar gest four digit binary integer?

Options:
A The smallest four digit binary integer
B The smallest one digit binary integer
C The greatest one digit binary integer
D The greatest three digit binary integer.

The correct answer is:
The greatest one digit binary integer

Solution:
The largest four digit number in binary system is 1111 which isequivalent to 15 in decimal system and smallest five digit number in binary system is 10000 which is equivalent to 16 in decimal system. Difference between numbers $=16-15=1$ Which is the greatest one digit binary integer.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
What is the difference in the roots of the equation $x^{2}-10 x+9=0 ?$

Options:
A 2
B 3
C 5
D 8

The correct answer is:
8

Solution:
$x^{2}-10 x+9=0$
$x^{2}-x-9 x+9=0$
$x(x-1)-9(x-1)=0$
$(\mathrm{x}-1)(\mathrm{x}-9)=0$
$\mathrm{x}=1,9$
Difference in roots $=9-1=8$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
What is the differentiable coefficient of $f(x)$ at $x=3$ ?
Let $f(x)$ be a function defined in $1 \leq x < \infty$ by $\quad[2014-I]$
$f(x)=\left\{\begin{array}{cc}2-x & \text { for } 1 \leq x \leq 2 \\ 3 x-x^{2} & \text { for } x>2\end{array}\right.$

Options:
A 1
B 2
C -1
D -3

The correct answer is:
-3

Solution:
$f^{\prime}(x)=\left\{\begin{array}{ccc}-1 & \text { for } & 1 \leq x \leq 2 \\ 3-2 x & \text { for } & x>2\end{array}\right.$
$f(x)$ at $x=3$
$f^{\prime}(3)=3-2(3)=3-6=-3$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 1)
What is the differential coefficient of $f(x)$ at $x=12$ ?
Consider the function $f(x)=\left\{\begin{array}{ll}x^{2}-5 & x \leq 3 \\ \sqrt{x+13} & x>3\end{array} \quad[2014-I I]\right.$

Options:
A $5 / 2$
B 5
C $1 / 5$
D $1 / 10$

The correct answer is:
$1 / 10$

Solution:
Given, $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}+13}, \mathrm{x}>3$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}+13}}$
$\Rightarrow \mathrm{f}^{\prime}(12)=\frac{1}{2 \sqrt{12+13}}=\frac{1}{2 \times 5}=\frac{1}{10}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 2)
What is the differential coefficient of $\log _{\mathrm{x}} \mathrm{x}$?

Options:
A 0
B 1
C $\frac{1}{x}$
D ${x}$

The correct answer is:
0

Solution:
Let $\mathrm{y}=\log _{\mathrm{x}} \mathrm{x}$
$\Rightarrow \mathrm{y}=1($ for $\mathrm{x}>0$ and $\mathrm{x} \neq 1)$
On differentiating both the sidew.r.t ' $x$ ', we get $\frac{\mathrm{dy}}{\mathrm{dx}}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
What is the differential equation corresponding to $\mathrm{y}^{2}-2 \mathrm{ay}+\mathrm{x}^{2}=\mathrm{a}^{2}$ by eliminating a? $\quad$

Options:
A $\left(x^{2}-2 y^{2}\right) p^{2}-4 p x y-x^{2}=0$
B $\left(x^{2}-2 y^{2}\right) p^{2}+4 p x y-x^{2}=0$
C $\left(x^{2}+2 y^{2}\right) p^{2}-4 p x y-x^{2}=0$
D $\left(x^{2}+2 y^{2}\right) p^{2}-4 p x y+x^{2}=0$
where $\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}$.

The correct answer is:
$\left(x^{2}-2 y^{2}\right) p^{2}-4 p x y-x^{2}=0$

Solution:
$\mathrm{y}^{2}-2 \mathrm{ay}+\mathrm{x}^{2}=\mathrm{a}^{2}$
$\Rightarrow y^{2}-2(y)(a)+a^{2}+x^{2}=a^{2}+a^{2}$
$\Rightarrow(y-a)^{2}+x^{2}=2 a^{2}$
Dift w.r.t x $2(\mathrm{y}-\mathrm{a}) \cdot \mathrm{y}^{1}+2 \mathrm{x}=0$
$\Rightarrow(\mathrm{y}-\mathrm{a}) \cdot \mathrm{p}=-\mathrm{x} \quad\left(\right.$ Let $\left.\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{1}=\mathrm{P}\right)$
$\Rightarrow \mathrm{y}-\mathrm{a}=\frac{-\mathrm{x}}{\mathrm{p}}$
$\Rightarrow a=y+\frac{x}{p}$
(i) $\Rightarrow\left(\frac{-x}{p}\right)^{2}+x^{2}=2\left(y+\frac{x}{p}\right)^{2}$
(from (ii) (iii))
$=2\left(\frac{(\mathrm{py}+\mathrm{x})^{2}}{\mathrm{p} 2}\right)$
$\Rightarrow \frac{x^{2}+p^{2} x^{2}}{p^{2}}=\frac{2\left(p^{2} y^{2}+x^{2}+2 p x y\right)}{p^{2}}$
$\Rightarrow x^{2}+p^{2} x^{2}-2 p^{2} y^{2}-2 x^{2}-4 p x y=0$
$\Rightarrow\left(\mathrm{x}^{2}-2 \mathrm{y}^{2}\right) \mathrm{p}^{2}-4 \mathrm{pxy}-\mathrm{x}^{2}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 1)
What is the differential equation for $y^{2}=4 a(x-a)$ ?

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

The correct answer is:
$y y^{\prime}\left(y y^{\prime}-2 x\right)+y^{2}=0$

Solution:
Given curve is $\mathrm{y}^{2}=4 a(x-a)$ On differentiating w.r.t. $\mathrm{x}$, we get $2 y y^{\prime}=4 a$
$\Rightarrow \quad a=\frac{y y^{\prime}}{2}$
On putting the value of $a$ in Eq. (i), we get
$y^{2}=4\left(\frac{y y^{\prime}}{2}\right)\left(x-\frac{y y^{\prime}}{2}\right)=y y^{\prime}\left(2 x-y y^{\prime}\right)$
$\Rightarrow \quad y y^{\prime}\left(y y^{\prime}-2 x\right)+y^{2}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 1)
What is the differential equation of all parabolas whose axes are parallel to Y-axis? $\quad$

Options:
A $\frac{d^{3} y}{d x^{3}}=0$
B $\frac{d^{2} x}{d y^{2}}=C$
C $\frac{d^{3} x}{d v^{3}}=1$
D $\frac{d^{3} y}{d r^{3}}=C$
(where $\mathrm{C}$ is a constant).

The correct answer is:
$\frac{d^{3} y}{d x^{3}}=0$

Solution:
The general equation of all parabolas where axes are parallel to $Y$ -axis, is
$$
y=\mathrm{A} x^{2}+\mathrm{B} x+\mathrm{C}
$$
where $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are arbitrary constants. On differentiating eq. (i) w.r.t. $x$, we get
$$
\frac{d y}{d x}=2 \mathrm{~A} x+\mathrm{B}
$$
(ii) On differentiating eq. (ii) w.r.t. $x$, we get
$$
\frac{d^{2} y}{d x^{2}}=2 \mathrm{~A}
$$
(iii) On differentiating eq. (iii) w.r.t. $x$, we get
$$
\frac{d^{3} y}{d x^{3}}=0
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 1)
What is the differential equation of the curve $\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}$ ?

Options:
A $x^{2} \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 y=0$
B $x^{2} \frac{d^{2} y}{d x^{2}}-y\left(\frac{d y}{d x}\right)^{2}+2=0$
C $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-\left(y \frac{d y}{d x}\right)^{2}=0$
D $\begin{array}{ll}\text { } \text { None of the above } & {}\end{array}$

The correct answer is:
$x^{2} \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 y=0$

Solution:
Given, equation is:
$\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}$
Differentiating w.r.t. $\mathrm{x}$,
$\Rightarrow \quad \frac{d y}{d x}=2 a x+b$
Differentiation g w.r.t. $\mathrm{x}$,
$\Rightarrow \frac{d^{2} y}{d x^{2}}=2 a$
$\Rightarrow \quad a=\frac{1}{2} \frac{d^{2} y}{d x^{2}}$
From (2) and (3)
$$
\mathrm{b}=\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x} \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}
$$
Putting values of $a$ and $b$ in equation (1):
$$
\begin{array}{l}
\mathrm{y}=\frac{1}{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}} \cdot \mathrm{x}^{2}+\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right) \\
\Rightarrow \quad 2 \mathrm{y}=\mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}+2 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}-2 \mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}} \\
\Rightarrow \mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}-2 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0
\end{array}
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 2)
What is the differential equation to family of parabolas having their vertices at the origin and foci on the $\mathrm{x}$ -axis?

Options:
A $y=2 x y^{\prime}$
B $\quad x=2 y y^{\prime}$
C $x y=y^{\prime}$
D $x=y y^{\prime}$

The correct answer is:
$y=2 x y^{\prime}$

Solution:
Let the equation of parabola is $y^{2}=4 a x$
On differentiating w.r.t.x, we get $2 \mathrm{yy}^{\prime}=4 \mathrm{a}$
$\Rightarrow \quad \frac{1}{2} \mathrm{yy}^{\prime}=\mathrm{a}$
put the value of 'a 'in equation (i), we get
$$
\mathrm{y}^{2}=\frac{4}{2} \mathrm{yy}^{\prime} \mathrm{x}
$$
$\Rightarrow \quad \mathrm{y}=2 \mathrm{xy}^{\prime}$

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 2)
What is the differentiation of $\log _{x} x$ with respect to $\ln x$?

Options:
A 0
B 1 $[2010-I]$
C $1 / x$
D $x$

The correct answer is:
0

Solution:
Let $u=\log _{x} x=1$ $\quad \left(\because \log _{a} a=1\right)$
Differentiate w.r.t ' $x^{\prime}$
$\frac{d u}{d x}=0$
and Let $v=\ln x \Rightarrow \frac{d v}{d x}=\frac{1}{x}$
$\therefore \quad \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
What is the distance between the lines $3 \mathrm{x}+4 \mathrm{y}=9$ and $6 x+8 y=18 ?$

Options:
A 0
B 3 units
C 9 units
D 18 units

The correct answer is:
0

Solution:
Given lines are $3 x+4 y=9 \quad \ldots(1)$
$2(3 x+4 y=9)$...(2)
Which are parallel to each other. Distance between them $=\frac{|9-9|}{\sqrt{9+16}}=\frac{0}{5}=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 1)
What is the distance between the planes $x-2 y+z-1=0$ and $-3 x+6 y-3 z+2=0 ?$

Options:
A 3 unit
B 1 unit
C 0
D None of the above

The correct answer is:
None of the above

Solution:
Given planes are $x-2 y+z=1$
and $-3 x+6 y-3 z=-2$
$=x-2 y+z=\frac{2}{3}$
Since both planes are parallel and $\mathrm{a}=1, \mathrm{~b}=-2, \mathrm{c}=1$
and $\mathrm{d}_{1}=-1, \mathrm{~d}_{2}=\frac{-2}{3}$
Distance $=\left|\frac{\mathrm{d}_{2}-\mathrm{d}_{1}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\right|$
Distance $=\left|\frac{1-\frac{2}{3}}{\sqrt{1+4+1}}\right|=\frac{1}{3 \sqrt{6}}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
What is the distance between the points which divide the line segment joining $(4,3)$ and $(5,7)$ internally and externally in the ratio $2: 3 ?$

Options:
A $\frac{12 \sqrt{17}}{5}$
B $\frac{13 \sqrt{17}}{5}$
C $\frac{\sqrt{17}}{5}$
D $\frac{6 \sqrt{17}}{5}$

The correct answer is:
$\frac{12 \sqrt{17}}{5}$

Solution:
Let $\mathrm{A}(\mathrm{x}, \mathrm{y})$ be the point that divides $(4,3)$ and $(5,7)$ internally is ratio $2: 3$
$\mathrm{A}=(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m} \mathrm{x}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\right)$
$=\left(\frac{10+12}{2+3}, \frac{14+9}{2+3}\right)=\left(\frac{22}{5}, \frac{23}{5}\right)$
Let $\mathrm{B}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be point that divides $(4,3)$ and $(5,7)$ externally in ratio $2: 3$
$\mathrm{B}=\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)=\left(\frac{\mathrm{m} \mathrm{x}_{2}-\mathrm{nx}_{1}}{\mathrm{~m}-\mathrm{n}}, \frac{\mathrm{m} \mathrm{y}_{2}-\mathrm{ny}_{1}}{\mathrm{~m}-\mathrm{n}}\right)$
$=\left(\frac{10-12}{2-3}, \frac{14-9}{2-3}\right)=(2,-5)$
Distance between $\mathrm{A}$ and $\mathrm{B}=\sqrt{\left(\mathrm{x}^{\prime}-\mathrm{x}\right)^{2}+\left(\mathrm{y}^{\prime}-\mathrm{y}\right)^{2}}$
$=\sqrt{\left(2-\frac{22}{5}\right)^{2}+\left(-5-\frac{23}{5}\right)^{2}}$
$=\sqrt{\frac{(-12)^{2}}{25}+\frac{(-48)^{2}}{25}}$
$=\frac{12}{5} \sqrt{1+4^{2}}=\frac{12}{5} \sqrt{17}$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
What is the distance between the straight lines $3 \mathrm{x}+4 \mathrm{y}=9$ and $6 x+8 y=15 ?$

Options:
A $\frac{3}{2}$
B $\frac{3}{10}$
C 6
D 5

The correct answer is:
$\frac{3}{10}$

Solution:
Given lines, $L_{1}=3 x+4 y-9=0$
$\mathrm{L}_{2}=6 \mathrm{x}+8 \mathrm{y}-15=0 \Rightarrow 3 \mathrm{x}+4 \mathrm{y}-\frac{15}{2}=0$
Observe that the coefficients of $\mathrm{x}$ and $\mathrm{y}$ are same. $\therefore \mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are parallel lines.
Distance between parallel lines $=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}$
$=\frac{\left|-9+\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}}$
$=\frac{\left|\frac{-18+15}{2}\right|}{\sqrt{25}}=\frac{3}{10}$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
What is the distance of the line $2 x+y+2 z=3$ from the origin ?

Options:
A 1 units
B $1.5$ units
C 2 units
D $2.5$ units

The correct answer is:
1 units

Solution:
$\mathrm{d}=\left|\frac{2 \times 0+0 \times 1+2 \times 0-3}{\sqrt{2^{2}+1^{2}+2^{2}}}\right|=\frac{3}{3}=1$ unit

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 2)
What is the distance of the origin from the plane $2 x+6 y-3 z+7=0 ?

Options:
A 1
B 2
C 3
D 6

The correct answer is:
1

Solution:
Required distance $=\frac{|2(0)+6(0)-3(0)+7|}{\sqrt{(2)^{2}+(6)^{2}+(-3)^{2}}}$
$=\frac{|7|}{\sqrt{4+36+9}}=\left|\frac{7}{7}\right|=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 2)
What is the distance of the point $(1,2,0)$ from yz-plane is:

Options:
A 1 unit
B 2 units
C 3 units
D 4 units

The correct answer is:
1 unit

Solution:
Equation of plane is $x=0$ $\therefore \quad$ Required distance from $(1,2,0)$ is
$=\frac{1.1+2.0+0.0-0}{\sqrt{1^{2}+0^{2}+0^{2}}}=\frac{1}{\sqrt{1}}=1$ unit

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What is the distance of the point $(2,3,4)$ from the plane $3 \mathrm{x}$ $6 y+2 z+11=? \quad$

Options:
A 1 unit
B 2 unit
C 3 unit
D 4 units

The correct answer is:
1 unit

Solution:
Distance $=\left|\frac{3 \times 2-6 \times 3+2 \times 4+11}{\sqrt{3^{2}+(-6)^{2}+(2)^{2}}}\right|$
$=\left|\frac{6-18+8+11}{\sqrt{49}}\right|=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 2)
What is the eccentricity of an ellipse, if its latusrectum is equal toone-half of its minor axis?

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

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

Solution:
Since, Latusrectum of an ellipse $=\frac{2 b^{2}}{a}$ and minor axis $=2 \mathrm{~b}$
$b=\frac{2 b^{2}}{a} \Rightarrow a=2 b$
Also, $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Ellipse, NDA, NDA 2009 (Phase 1)
What is the eccentricity of rectangular hyperbola?

Options:
A $\sqrt{2}$
B $\sqrt{3}$
C $\sqrt{5}$
D $\sqrt{6}$

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

Solution:
Here, $b^{2}=a^{2}\left(e^{2}-1\right)$
For rectangular hyperbola : $a=b$ $\Rightarrow b^{2}=b^{2}\left(e^{2}-1\right)$
$\Rightarrow e^{2}-1=1$
$\Rightarrow e^{2}=2 \Rightarrow e=\pm \sqrt{2}$
For hyper bola, $\mathrm{e}>1$
Hence, $\mathrm{e}=\sqrt{2}$

Topic: Mathematics, Hyperbola, NDA, NDA 2016 (Phase 2)
What is the eccentricity of the conic $4 x^{2}+9 y^{2}=144 ?$

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

The correct answer is:
$\frac{\sqrt{5}}{3} \quad$

Solution:
Given equation can be written as
$\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$
Thisis an ellipse.
$\Rightarrow a^{2}=36, b^{2}=16$
$\quad e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{16}{36}}=\sqrt{\frac{20}{36}}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}$

Topic: Mathematics, Ellipse, NDA, NDA 2012 (Phase 1)
What is the effluent from the primary settling tank used for?

Options:
A Primary treatment
B Secondary treatment
C Physical removal of particles
D Increase in pollution potential

The correct answer is:
Secondary treatment

Solution:
The effluent from the primary settling tank is taken for secondary treatment.

Topic: Biology, Microbes in Human Welfare, NEET
What is the electrode potential (in V ) of the following electrode at $25^{\circ} \mathrm{C}$ ?
$\mathrm{Ni}^{2+}(0.1 \mathrm{M}) \mid \mathrm{Ni}(s)$
(Standard reaction potential of $\mathrm{Ni}^{2+} \mid \mathrm{Ni}$ is $-0.25 \mathrm{~V}, \frac{2.303 R T}{F}=0.06$ )

Options:
A -0.28 V$
B -0.34 V$
C -0.82 V$
D -0.22 V$

The correct answer is:
-0.28 V$

Solution:
$\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log \left[\frac{[\mathrm{Ni}]}{\left[\mathrm{Ni}^{2+}\right]}\right] \\ & =-0.25-\frac{0.06}{2} \log \left(\frac{1}{0.1}\right) \quad(\because n=2) \\ & =-0.25-0.03 \times 1 \\ E_{\text {cell }} & =-0.28 \mathrm{~V}\end{aligned}$

Topic: Chemistry, Electrochemistry, JIPMER, JIPMER 2009
What is the enthalpy change that accompanies the melting of one mole of a solid substance in its standard state called?

Options:
A Standard enthalpy of sublimation
B Standard enthalpy of fusion or molar enthalpy of fusion
C Standard enthalpy of vaporization
D Standard enthalpy of condensation

The correct answer is:
Standard enthalpy of fusion or molar enthalpy of fusion

Solution:
The enthalpy change that accompanies the melting of one mole of a solid substance in its standard state is called the standard enthalpy of fusion or molar enthalpy of fusion.

Topic: Chemistry, Thermodynamics (C), NEET
What is the equation of a circle, whose centre lies on the $\mathrm{x}$ -axis at a distance $\mathrm{h}$ from the origin and the circle passes through the origin?

Options:
A $x^{2}+y^{2}-2 h x=0$
B $x^{2}+y^{2}-2 h x+h^{2}=0$
C $x^{2}+y^{2}+2 h x y=0$
D $x^{2}+y^{2}-h^{2}=0$

The correct answer is:
$x^{2}+y^{2}-2 h x=0$

Solution:
Centre of the circle is $(\mathrm{h}, 0)$ and circle passes through the origin. In the general equation of circle:
$x^{2}+y^{2}+2 g x+2 f x+c=0$
$\mathrm{g}=-\mathrm{h}$ and $\mathrm{f}=0$
so, $x^{2}+y^{2}-2 h x+0+c$
$=x^{2}+y^{2}-2 h x+c=0$
since circle passes through origin $(0,0)$ $0+0-0+\mathrm{c}=0 \Rightarrow \mathrm{c}=0$
and equation (i) radius to $x^{2}+y^{2}-2 h x=0$

Topic: Mathematics, Circle, NDA, NDA 2007 (Phase 1)
What is the equation of a curve passing through $(0,1)$ and whose differential equation is given by $d y=y \tan x d x$ ?

Options:
A $y=\cos x$
B $y=\sin x$
C $y=\sec x$
D $y=\operatorname{cosec} x$

The correct answer is:
$y=\sec x$

Solution:
$\mathrm{dy}=\mathrm{y} \tan \mathrm{x} \mathrm{d} \mathrm{x}$
$\begin{aligned} \int \frac{\mathrm{dy}}{\mathrm{y}} &=\int \tan \mathrm{x} \mathrm{d} \mathrm{x} \\ \log |\mathrm{y}| &=\log |\sec \mathrm{x}|+\log |\mathrm{c}| \\ \log |\mathrm{y}| &=\log |\mathrm{c} \sec \mathrm{x}| \\ \mathrm{y} &=\operatorname{csec} \mathrm{x} \\ \text { at } \mathrm{x} &=0, \mathrm{y}=1 \\ \mathrm{y} &=\mathrm{c} \end{aligned}$
Solution is given by
$\mathrm{y}=\sec x$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2014 (Phase 1)
What is the equation of a line parallel to $\mathrm{x}$ -axis at a distance of 5 units below x-axis?

Options:
A $x=5$
B $x=-5$
C $y=5$
D $y=-5$

The correct answer is:
$y=-5$

Solution:
Equation of a line parallel to $\mathrm{x}$ -axis at a distance of 5 units below $\mathrm{x}$ -axis is $y=-5$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 2)
What is the equation of a plane through the $x$ -axis and passing through the point $(1,2,3)$ ?

Options:
A $x+y+z=6$
B $x=1$
C $y+z=5$
D $z+y=1$

The correct answer is:
$x=1$

Solution:
The equation of plane which is passing through $x$ -axis
is
$x=a$
This plane also passes through $(1,2,3)$ By putting $x=1$ in eqn of plane, we get $a=1$ Hence $x=1$ is required equation of plane.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 1)
What is the equation of a straight line which passes through $(3,4)$ and sum of whose $x$ and $y$ intercepts is 14 ?

Options:
A $4 x+3 y=24$
B $x+y=14$
C $\quad 4 x-3 y=0$
D $3 x+4 y=25$

The correct answer is:
$4 x+3 y=24$

Solution:
Equation of a straight line in intercept form is
$\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{3}{a}+\frac{4}{b}=1$
Given, $a+b=14$ On solving (i) and (ii) we get
$\frac{3}{a}+\frac{4}{14-a}=1 \Rightarrow a^{2}-13 a+42=0$
$\Rightarrow(a-7)(a-6)=0$
$\Rightarrow \mathrm{a}=6$ and $\mathrm{b}=8$
or $\mathrm{a}=7$ and $\mathrm{b}=7$.
- Required eqns are $4 \mathrm{x}+3 \mathrm{y}=24$ or $\mathrm{x}+\mathrm{y}=1$.

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)