datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following is the unit vector perpendicular to both $\overrightarrow{\mathrm{a}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ ?

Options:
A $\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$
B $\hat{\mathrm{k}}$
C $\frac{\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}$
D $\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}$

The correct answer is:
$\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$

Solution:
According to question $\vec{a}=-\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$
Then, $\vec{a} \times \vec{b}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right|$
$=\hat{i}[1+1]-\hat{j}[-1-1]+\hat{k}[1-1]$
$=2 \hat{i}+2 \hat{j}+0=2(\hat{i}+\hat{j})$
and $|\vec{a} \times \vec{b}|=\sqrt{4+4}=2 \sqrt{2}$
$\therefore$ Required unit vector $=\pm \frac{2(\hat{i}+\hat{j})}{2 \sqrt{2}}=\pm \frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 2)
Which one of the following is the unit vector
perpendicular to the vectors $4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and $-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ ?

Options:
A $\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}$
B $\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}$
C $\hat{\mathbf{k}}$
D $\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$

The correct answer is:
$\hat{\mathbf{k}}$

Solution:
Let vector $x \hat{i}+y \hat{j}+z \hat{k}$ be perpendicular to vectors
$\hat{4} \hat{i}+2 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$
Their dot product is zero.
$\therefore \quad 4 x+2 y=0$ ...(i)
and $-3 x+2 y=0 \quad$... (ii)
From eqs. (i) and (ii), $x=0, y=0$
Hence, required vector is $\hat{k}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
Which one of the following is true during ageing?

Options:
A The decreased bone mass and increased chances of fractures
B Decrease in blood urea and GFR
C Decrease in cholesterol content of cornea and lens
D Decrease in calcium content of arteries and cartilage

The correct answer is:
The decreased bone mass and increased chances of fractures

Solution:
Ageing is associated both with calcium deficiency, due to low dietary intake and decreased intestinal absorption, and with vitamin D deficiency, the secondary depletion of body stores resulting from inadequate exposure to sunlight.
The consequences of calcium deficiency might thus include not only osteoporosis, but also the decreased bone mass and increased chances of fractures.

Topic: Biology, Locomotion and Movement, NEET
Which one of the following is true?

Options:
A $\mathrm{NaOH}$ is used in the concentration of bauxite ore
B $\mathrm{NaOH}$ is a primary standard in volumetric analysis
C Manganous hydroxide is soluble in excess of $\mathrm{NaOH}$ solution
D $\mathrm{NaOH}$ solution does not react with $\mathrm{Cl}$

The correct answer is:
$\mathrm{NaOH}$ is used in the concentration of bauxite ore

Solution:
$\mathrm{NaOH}$ is used in the concentration of bauxite ore.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, KCET, KCET 2011
Which one of the following is wrong for fungi?

Options:
A They are eukaryotic
B All fungi possess a purely cellulosic cell wall
C They are heterotrophic
D They are both unicellular and multicellular

The correct answer is:
All fungi possess a purely cellulosic cell wall

Solution:
The cell wall of fungi is made up of chitin and polysaccharides. Chitin is the most abundant amino polysaccharide, a long-chain polymer of N-acetylglucosamine is a derivative of glucose occurring in nature, and is the building material that gives strength to the exoskeletons of crustaceans, insects, and the cell walls of fungi.

Topic: Biology, Biological Classification, NEET, NEET 2016 (Phase 2)
Which one of the following makes use of RNA as a template to synthesize DNA?

Options:
A Reverse transcriptase
B DNA dependant RNA polymerase
C DNA polymerase
D RNA polymerase

The correct answer is:
Reverse transcriptase

Solution:
In 1970 H. Temin and D. Baltimore independently discovered the enzyme reverse transcriptase. This enzyme uses RNA as template for the synthesis of C-DNA (Complementary DNA).
$\begin{gathered}\text { RNA } \\ \downarrow \\ \mathrm{C}-\mathrm{DNA}\end{gathered}$ Reverse transcriptase used RNA as template

Topic: Biology, Molecular Basis of Inheritance, JIPMER, JIPMER 2007
Which one of the following makes use of RNA template to synthesize DNA?

Options:
A DNA polymerase
B RNA polymerase
C reverse transcriptase
D DNA dependant RNA polymerase.

The correct answer is:
reverse transcriptase

Solution:
Reverse transcriptase use an RNA template and a short primer complementary to the $3^{\prime}$ end of the RNA to direct the synthesis of the first strand cDNA.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2005
Which one of the following matrices is an elementary matrix?

Options:
A $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
B $\left[\begin{array}{lll}1 & 5 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
C $\left[\begin{array}{lll}0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
D $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 2\end{array}\right]$

The correct answer is:
$\left[\begin{array}{lll}1 & 5 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

Solution:
An elementary matrix has each diagonal element $1 .$ So, option (b) is correct answer.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
Which one of the following may be the parameter of a binomial distribution?

Options:
A $\quad n p=2, n p q=4$
B $n=4, p=\frac{3}{2}$
C $\quad n=8, p=1$
D $\quad n p=10, n p q=8$

The correct answer is:
$\quad n p=10, n p q=8$

Solution:
For binomial distribution $\mathrm{p} < 1, \mathrm{q} < 1$ and $\mathrm{p}+\mathrm{q}=1$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
Which one of the following measures is determined only after the construction of cumulative frequency distribution

Options:
A Arithmetic mean
B Mode
C Median
D Geometric

The correct answer is:
Median

Solution:
After construction of cumulative frequency distribution, we can find out median easily.

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
Which one of the following measures of central tendency is. used in construction of index numbers? $\quad$

Options:
A Harmonicmean
B Geometric mean
C Median
D Mode

The correct answer is:
Geometric mean

Solution:
Geometric mean is used in construction of index numbers. option (b) is correct.

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
Which one of the following on reduction with lithium aluminium hydride yields a secondary amine?

Options:
A Methylisocyanide
B Acetamide
C Methyl cyanide
D Nitroethane

The correct answer is:
Methylisocyanide

Solution:
$\mathrm{R}-\mathrm{N} \equiv \mathrm{C}+4[\mathrm{H}] \stackrel{\mathrm{LAIH}_4}{\longrightarrow} \underset{2^{\circ} \text { amine }}{\mathrm{R}-\mathrm{NH}-\mathrm{CH}_3}$
On reaction with $\mathrm{LiAlH}_4$, Isocyanate groups gives secondary amines whereas, alkyl cyanides give $1^{\circ}$ amine on reduction.
Related Theory
LiAlH $_4$ is a nucleophilic reducing agent since the hydride transfer to the carbonyl carbon occurs prior to the coordination to the carbonyl oxygen. It reacts faster with electron deficient carbonyl groups. The reactivity of carbonyl compounds with this reagent follows the order: Aldehydes $>$ Ketones $>$ ester $>$ amide $>$ carboxylic acid.

Topic: Chemistry, Amines, NEET, NEET 2007
Which one of the following on reduction with lithium aluminium hydride yields a secondary amine ?

Options:
A Methyl isocyanide
B Acetamide
C Methyl cyanide
D Nitroethane

The correct answer is:
Methyl isocyanide

Solution:
Alkyl isocyanides on reduction with lithium aluminium hydride forms secondary amines containing methyl as one of the the alkyl groups.

Topic: Chemistry, Amines, NEET
Which one of the following operations on sets is notcorrect where B' denotes the complement of B?

Options:
A $\left(\mathrm{B}-\mathrm{A}^{\prime}\right) \cup\left(\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B})-(\mathrm{A} \cap \mathrm{B})$
B $(\mathrm{A}-\mathrm{B}) \cup(\mathrm{B}-\mathrm{A})=\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)-\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$
C $\left(B^{\prime}-A^{\prime}\right) \cap\left(A^{\prime}-B^{\prime}\right)=(B-A) \cap(A-B)$
D $\left(\mathrm{B}-\mathrm{A}^{\prime}\right) \cap\left(\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\right)=\left(\mathrm{B}-\mathrm{A}^{\prime}\right) \cup\left(\mathrm{A}^{\prime}-\mathrm{B}\right)$

The correct answer is:
$\left(\mathrm{B}-\mathrm{A}^{\prime}\right) \cap\left(\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\right)=\left(\mathrm{B}-\mathrm{A}^{\prime}\right) \cup\left(\mathrm{A}^{\prime}-\mathrm{B}\right)$

Solution:
Let there be two sets $\mathrm{A}$ and $\mathrm{B}$ and universal set of $\mathrm{A}$ and $\mathrm{B}$, be $\mathrm{U}$
Then drawing these sets on a Venn-diagram, four regions are created as shown in the figure
$\mathrm{B}=$ regions $\mathrm{b}, \mathrm{c}$
$\mathbf{B}^{\prime}$ =regions a, d $\mathrm{A}=$ regions $\mathrm{a}, \mathrm{b}$
$\mathrm{A}^{\prime}=$ regions $\mathrm{c}, \mathrm{d}$
$\mathrm{A}-\mathrm{B}=$ region a
$B-A=$ region $c$ $\mathrm{B}^{\prime}-\mathrm{A}^{\prime}$ =region a
$\mathrm{A}^{\prime}-\mathrm{B}^{\prime}=$ region $\mathrm{c}$
$B-A^{\prime}=$ region $b$ $\mathrm{A}^{\prime}-\mathrm{B}=$ region $\mathrm{d}$
From these we check the operations given in the choice choices (a), (b) and (c) are correct
(d) $\mathrm{LHS}=\mathrm{region} \mathrm{a} \cap \mathrm{c}=\phi$
$\mathrm{RHS}=$ region b $\cup$ region $\mathrm{d}=\mathrm{b}, \mathrm{d}$
So, for (d) LHS $\neq$ RHS

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Which one of the following options is correct?

Options:
A $\sin ^{2} 30^{\circ}, \sin ^{2} 45^{\circ}, \sin ^{2} 60^{\circ}$ are in GP
B $\cos ^{2} 30^{\circ}, \cos ^{2} 45^{\circ}, \cos ^{2} 60^{\circ}$ are in GP
C $\cot ^{2} 30^{\circ}, \cot ^{2} 45^{\circ}, \cot ^{2} 60^{\circ}$ are in GP
D $\tan ^{2} 30^{\circ}, \tan ^{2} 45^{\circ}, \tan ^{2} 60^{\circ}$ are in GP

The correct answer is:
$\tan ^{2} 30^{\circ}, \tan ^{2} 45^{\circ}, \tan ^{2} 60^{\circ}$ are in GP

Solution:
Three numbers a, b and c will be in G.P. if $b^{2}=a c$. Only option (d) i.e. $\tan ^{2} 30^{\circ}, \tan ^{2} 45^{\circ}$ and $\tan ^{2} 60^{\circ}$ are
in GP.
$\because \quad \tan ^{2} 30^{\circ}=\frac{1}{3}$
$\tan ^{2} 45^{\circ}=1$
and $\tan ^{2} 60^{\circ}=3$
$\tan ^{2} 30^{\circ}, \tan ^{2} 45^{\circ}$ and $\tan ^{2} 60^{\circ}$ are in G.P.

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 1)
Which one of the following pairs is not correctly matched?

Options:
A $\sin 2 \pi$ : $\sin (-2 \pi)$
B $\tan 45^{\circ}: \tan \left(-315^{\circ}\right)$
C $\cot \left(\tan ^{-1} 0.5\right)$ : $\tan \left(\cos ^{-1} 0.5\right)$
D $\tan 420^{\circ} : \tan \left(-60^{\circ}\right)$

The correct answer is:
$\tan 420^{\circ} : \tan \left(-60^{\circ}\right)$

Solution:
Fourth pair is not correct matched explained below $\tan 420^{\circ}=\tan (360+60)=\tan 60^{\circ}$
$\tan 60^{\circ} \neq \tan \left(-60^{\circ}\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 1)
Which one of the following pairs of hormones are the examples of those that can easily pass through the cell membrane of the target cell and bind to a receptor inside it (mostly in the nucleus)?

Options:
A Thyroxine, Insulin
B Somatostatin, oxytocin
C Cortisol, Thyroxine
D Insulin, glucagon

The correct answer is:
Cortisol, Thyroxine

Solution:
Hormonal action depends on whether the hormone is Lipid or water soluble. Lipid soluble hormones can cross the target cell membrane easily, get attached to the mobile receptors within the cytoplasm and reach the nucleus to perform their functions. Examples are cortisol, thyroxine, estrogen, progesterone etc. However water soluble hormones are proteinaceous and these include insulin, oxytocin etc. These cannot cross the cell membrane easily. These get attached to the static receptors on the cell membrane. This attachment triggers the formation of secondary messengers that help these hormones to act on cells. Estrogen and Progesterone are steroid hormones that means these are lipid soluble hormones. Hence these attach to receptors within the cytoplasm.

Topic: Biology, Chemical Coordination and Integration, NEET
Which one of the following pairs of plant structures has haploid number of chromosomes?

Options:
A Megaspore mother cell and antipodal cells
B Egg cell and antipodal cells
C Nucellus and antipodal cells
D Egg nucleus and secondary nucleus

The correct answer is:
Egg cell and antipodal cells

Solution:
In angiosperms a single diploid megaspore mother cell matures within an ovule. Through meiosis it gives rise to four megaspores (haploid). In most plants only one of these megaspores survive. This functional megaspore divides three times by mitosis and produces eight haploid nuclei enclosed within a 7-celled embryo sac. One nucleus is located near the opening of embryo sac in egg cell. Two nuclei are located in a single cell in the midde of embryo sac and are called polar nuclei, two nuclei are contained in cells called synergids that flank the egg cell and other three nuclei are resided in the cells called antipodals, located at the end of embryo sac, opposite the egg cell.
Nucellus is centrally located mass of thin-walled parenchymatous cells covered by integuments in ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2008 (Screening)
Which one of the following pairs of plant structures has haploid number of chromosomes?

Options:
A Nucellus and antipodal cells
B Egg nucleus and secondary nucleus
C Megaspore mother cell and antipodal cells
D Egg cell and antipodal cells

The correct answer is:
Egg cell and antipodal cells

Solution:
The egg cell is one of the haploid cell of egg apparatus present in the micropylar end of embryo sac in seeded plants. It is also called oosphere and represents the single female gamete of the embryo sac. Antipodal cells are the haploid cells, usually three in number, present in embryo sac at the opposite end of micropyle. Their function is unknown and at fertilization, they may disintegrate or multiply and enlarge.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2008
Which one of the following pairs of plants structures has a haploid number of chromosomes?

Options:
A Nucellus and antipodal cells
B Egg nucleus and secondary nucleus
C Megaspore mother cell and antipodal cells
D Egg cell and antipodal cells

The correct answer is:
Egg cell and antipodal cells

Solution:
The nucleus undergoes three divisions to produce 8 haploid nuclei which get organised into three groups: egg apparatus towards the micropylar end, two middle polar nuclei or secondary nucleus and three antipodal cells towards the chalazal end.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Which one of the following planes contains the z-axis?

Options:
A $\quad x-z=0$
B $\mathrm{z}+\mathrm{y}=0$
C $3 x+2 y=0$
D $\quad 3 x+2 z=0 \quad$

The correct answer is:
$3 x+2 y=0$

Solution:
Theequation of planewhich contains z-axis is $3 \mathrm{x}+2 \mathrm{y}=0$ as $\mathrm{z}$ is absent in this equation.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 1)
Which one of the following planes is normal to the plane $3 x+y+z=5 ? \quad$

Options:
A $x+2 y+z-6$
B $x-2 y+z-6$
C $x+2 y-z=6$
D $x-2 y-z=6$

The correct answer is:
$x-2 y-z=6$

Solution:
Direction cosines of the normal to the plane $3 x+y+z=5$ are $3,1,1$
Direction cosines of the normal to the plane $x-2 y-z=6$ are $1,-2,-1$
Sum of the product of direction cosines $=3 \times 1+1 \times(-2)+1 \times(-1)=0$
Hence, normals to the two planes are perpendicular to each other. Therefore two planes are alsoperpendicular.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
Which one of the following plants shows vexillary aestivation and diadelphous stamens?

Options:
A Pisum sativum
B Allium cepa
C Solanum nigrum
D Colchicum autumnale

The correct answer is:
Pisum sativum

Solution:
Inflorescence: racemose

Flower: bisexual, zygomorphic

Calyx: sepals five, gamosepalous; valvate/imbricate aestivation

Corolla: petals five, polypetalous, papilionaceous, consisting of a posterior standard, two lateral wings, two anterior ones forming a keel (enclosing stamens and pistil), vexillary aestivation.

Androecium: ten, diadelphous, anther dithecous

Gynoecium: ovary superior, mono carpellary, unilocular with many ovules, style single.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2022 (Phase 1)
Which one of the following points lies inside a circle of radius 6 and centre at $(3,5) ?$

Options:
A $(-2,-1)$
B $(0,1)$
C $(-1,-2)$
D $(2,-1)$

The correct answer is:
$(0,1)$

Solution:
Radius of circle $=6$, centre $=(3,5)$ $\therefore \quad$ Equation of circle is $\begin{aligned} & \mathrm{S} \equiv(\mathrm{x}-3)^{2}+(\mathrm{y}-5)^{2}=(6)^{2} \\ \Rightarrow \quad \mathrm{S} & \equiv(\mathrm{x}-3)^{2}+(\mathrm{y}-5)^{2}-36 \end{aligned}$
Now, consider all the four options.
(a) $(-2,-1)$
Put it in $\mathrm{S}$ $\mathrm{S} \equiv(-2-3)^{2}+(-1-5)^{2}-36=25+36-36=25>0$
$\Rightarrow(-2,-1)$ is outside the circle. (b) $(0,1)$
$\mathrm{S} \equiv(0-3)^{2}+(1-5)^{2}-36$
$=9+16-36=25-36=-11 < 0$
Hence, $(0,1)$ lies inside a circle.

Topic: Mathematics, Circle, NDA, NDA 2013 (Phase 1)
Which one of the following points lies on the plane $2 x+3 y-6 z=21 ?

Options:
A $(3,2,2)$
B $(3,7,1)$
C $(1,2,3)$
D $(2,1,-1)$

The correct answer is:
$(3,7,1)$

Solution:
Only point $(3,7,1)$ satisfy the equation of plane
$2 x+3 y-6 z=21$
Hence, $(3,7,1)$ lies on the plane

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 1)
Which one of the following points lies outside the ellipse $\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1$ ?

Options:
A $(\mathrm{a}, 0)$
B $(0, b)$
C $(-a, 0)$
D $(a, b)$

The correct answer is:
$(a, b)$

Solution:
The equation of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1=0$
The point for which $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}-1>0$ is outsideellipse
Since, at $(a, 0), 1+0-1=0$
It lies on the ellipse. $\operatorname{At}(0, b), 0+1-1=0$
It lies on the ellipse.
$\operatorname{At}(-a, 0), 1+0-1=0$
It lies on the ellipse. $\mathrm{At}(\mathrm{a}, \mathrm{b}), 1+1-1>0$
So, the point $(a, b)$ lies outside the ellipse

Topic: Mathematics, Ellipse, NDA, NDA 2006 (Phase 2)
Which one of the following points on the line $2 \mathrm{x}-3 \mathrm{y}=5$ is equidistant from $(1,2)$ and $(3,4) ?$

Options:
A $(7,3)$
B $(4,1)$
C $(1,-1)$
D $(-2,-3)$

The correct answer is:
$(4,1)$

Solution:
Let point $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ be equidistant from point $\mathrm{A}(1,2)$ and $\mathrm{B}(3,4)$.
$\begin{array}{ll}\therefore & \mathrm{PA}=\mathrm{PB} \\ \Rightarrow & \mathrm{PA}^{2}=\mathrm{PB}^{2} \\ \Rightarrow & \left(1-\mathrm{x}_{1}\right)^{2}+\left(2-\mathrm{y}_{1}\right)^{2}=\left(3-\mathrm{x}_{1}\right)^{2}+\left(4-\mathrm{y}_{1}\right)^{2} \\ \Rightarrow & 1+\mathrm{x}_{1}^{2}-2 \mathrm{x}_{1}+4+\mathrm{y}_{1}^{2}-4 \mathrm{y}_{1} \\ & =9+\mathrm{x}_{1}^{2}-6 \mathrm{x}_{1}+16+\mathrm{y}_{1}^{2}-8 \mathrm{y}_{1} \\ \Rightarrow & 4 \mathrm{x}_{1}+4 \mathrm{y}_{1}=20 \\ \Rightarrow & \mathrm{x}_{1}+\mathrm{y}_{1}=5 ...(i)\end{array}$
As $P\left(x_{1}, y_{1}\right)$ lies on $2 x-3 y=5$
$2 x_{1}-3 y_{1}=5$...(ii)
On solving Eqs. (1) and (2),
we get $\mathrm{x}_{1}=4$ and $\mathrm{y}_{1}=1$
Coordinates of $\mathrm{P}$ are $(4,1)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 2)
Which one of the following real valued functions is never zero?

Options:
A Polynomial function
B Trigonometric function
C Logarithmic function
D Exponential function

The correct answer is:
Exponential function

Solution:
When $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$
$\mathrm{f}(\mathrm{x}) \neq 0, \forall \mathrm{x} \in \mathrm{R}$
$\Rightarrow$ An exponential function is never zero

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
Which one of the following sets has all elements as odd positive integers?

Options:
A $\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R} 4 \mathrm{x}^{3}-8 \mathrm{x}^{2}+19 \mathrm{x}-12=0\right\}$
B $\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R} 4 \mathrm{x}^{3}-9 \mathrm{x}^{2}+23 \mathrm{x}-15=0\right\}$
C $\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R} \mathrm{x}^{3}-7 \mathrm{x}^{2}+14 \mathrm{x}-8=0\right\}$
D $\mathrm{S}=\left\{\mathrm{x} \in \mathrm{E} \mathrm{x}^{3}-12 \mathrm{x}^{2}+44 \mathrm{x}-48=0\right\}$

The correct answer is:
$\mathrm{S}=\left\{\mathrm{x} \in \mathrm{R} 4 \mathrm{x}^{3}-9 \mathrm{x}^{2}+23 \mathrm{x}-15=0\right\}$

Solution:
We take option (a) : $x^{3}-8 x^{2}+19 x-12=0$ $\Rightarrow(x-1)\left(x^{2}-7 x+12\right)=0$
$\Rightarrow(x-1)(x-3)(x-4)=0$
$\Rightarrow x=1,3,4$
Thus, it is nota setofelements as odd positive integers
(b) $x^{3}-9 x^{2}+23 x-15=0$
$\Rightarrow(x-1)\left(x^{2}-8 x+15\right)=0$
$\Rightarrow(x-1)(x-3)(x-5)=0$
$\Rightarrow \mathrm{x}=1,3,5$
Thus, $\mathrm{S}$ will be a set of elements as odd positive integers.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Which one of the following statement is correct?

Options:
A The derivative of a function $\mathrm{f}(\mathrm{x})$ at a point will exist if there is one tangent to the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at that point and the tangent is parallel to y-axis
B The derivative of a function $\mathrm{f}(\mathrm{x})$ at a point will exist if there is one tangent to the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at that point and the tangent must be parallel to $\mathrm{x}$ -axis
C The derivative of a function $\mathrm{f}(\mathrm{x})$ at a point will exist if there is one and only one tangent to the curve $\mathrm{y}=\mathrm{f}[\mathrm{x}]$ at that point and the tangent is not parallel to y-axis
D None of the above

The correct answer is:
The derivative of a function $\mathrm{f}(\mathrm{x})$ at a point will exist if there is one and only one tangent to the curve $\mathrm{y}=\mathrm{f}[\mathrm{x}]$ at that point and the tangent is not parallel to y-axis

Solution:
It is obvious

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 1)
Which one of the following statement is not correct?

Options:
A Median divides distributions into two equal subgroups
B The third quartile is the same as the 75 th percentile
C The 5 th decile is the same as the 50 th percentile
D The 50 th decile is the same as the 5 th percentile

The correct answer is:
The 50 th decile is the same as the 5 th percentile

Solution:
$50_{\text {th }}$ decile $=\frac{50}{10}=5$ and $5_{\text {th }}$ percentile $=\frac{5}{100}$ $5 \neq \frac{5}{100}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)