datavorous/entrance-exam-dataset huggingface dataset:
What is $\lim _{x \rightarrow 0} \frac{\mathrm{a}^{\mathrm{X}}-\mathrm{b}^{\mathrm{X}}}{\mathrm{x}}$ ?

Options:
A $\log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)$
B $\log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$
C ab
D $\log (\mathrm{ab})$

The correct answer is:
$\log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)$

Solution:
Let $\mathrm{L}=\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{a}^{\mathrm{x}}-\mathrm{b}^{\mathrm{x}}}{\mathrm{x}}$
This is of $\frac{0}{0}$ form so, $L$ 'hospital's rule is applicable.
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{a}^{\mathrm{x}} \log \mathrm{a}-\mathrm{b}^{\mathrm{x}} \log \mathrm{b}}{1}$ (byL'Hospital's rule)
$=\log a-\log b=\log \frac{a}{b}$

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}$ equal to ?

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

The correct answer is:
$\frac{1}{\pi}$

Solution:
$\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 2)
What is $\lim _{x \rightarrow 0} \frac{\log _{5}(1+x)}{x}$ equal to ?

Options:
A 1
B $\log _{5} \mathrm{e}$
C $\log _{e} 5$
D 5

The correct answer is:
$\log _{5} \mathrm{e}$

Solution:
Given equation $\lim _{x \rightarrow 0} \frac{\log _{5}(1+x)}{x}$
$=\lim _{x \rightarrow 0} \frac{\log _{\mathrm{e}}(1+\mathrm{x})}{\mathrm{x} \log _{\mathrm{e}} 5} \quad\left[\because \log _{\mathrm{x}} \mathrm{y}=\frac{\log _{\mathrm{e}} \mathrm{y}}{\log _{\mathrm{e}} \mathrm{x}}\right]$
$=\frac{1}{\log _{\mathrm{e}} 5} \lim _{\mathrm{x} \rightarrow 0} \frac{\log _{\mathrm{e}}(1+\mathrm{x})}{\mathrm{x}}=\log _{5} \mathrm{e}$
$\left[\because \lim _{\mathrm{x} \rightarrow 0} \frac{\log _{\mathrm{e}}(1+\mathrm{x})}{\mathrm{x}}=1, \log _{\mathrm{x}} \mathrm{y}=\frac{1}{\log _{\mathrm{y}} \mathrm{x}}\right]$

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 2)
What is $\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-4}$ equal to?

Options:
A 0
B $\frac{1}{4}$
C $\frac{1}{2}$
D 1

The correct answer is:
$\frac{1}{4}$

Solution:
Let $\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{x-2}{(x-2)(x+2)}$
$=\lim _{x \rightarrow 2} \frac{1}{x+2}=\frac{1}{2+2}=\frac{1}{4}$

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 2)
What is $\lim _{x \rightarrow 2} \frac{2-x}{x^{3}-8}$ equal to ?

Options:
A $\frac{1}{8}$
B $-\frac{1}{8}$
C $\frac{1}{12}$
D $-\frac{1}{12}$

The correct answer is:
$-\frac{1}{12}$

Solution:
$\lim _{x \rightarrow 2} \frac{2-x}{x^{3}-8}=\lim _{x \rightarrow 2} \frac{2-x}{(x-2)\left(x^{2}+2 x+4\right)}$
$=\lim _{x \rightarrow 2} \frac{-1}{x^{2}+2 x+4}$
Putting $\mathrm{x}=2$, we get
$\frac{-1}{2^{2}+4+4}=-\frac{1}{12}$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 2)
What is $\lim _{\mathrm{x} \rightarrow-2}\left(\frac{\mathrm{x}+2}{\mathrm{x}^{3}+8}\right)$ equal to?

Options:
A $1 / 4$
B $-1 / 4$
C $1 / 12$
D $-1 / 12$

The correct answer is:
$1 / 12$

Solution:
$\lim _{x \rightarrow-2}\left(\frac{x+2}{x^{3}+8}\right)=\lim _{x \rightarrow-2} \frac{x+2}{(x)^{3}+(2)^{3}}$
$=\lim _{x \rightarrow-2} \frac{x+2}{(x+2)\left(x^{2}-2 x+4\right)}$
$=\frac{1}{(-2)^{2}-2(-2)+4}=\frac{1}{12}$

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 1)
What is $\lim _{x \rightarrow 3} f(x)$ equal to ?
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 2
B 4
C 5
D 13

The correct answer is:
4

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x}^{2}-5 & , \quad \mathrm{x} \leq 3 \\ \sqrt{\mathrm{x}+13} & , \quad \mathrm{x}>3\end{array}\right.$
To find $\lim _{x \rightarrow 3} f(x)$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})$
$=\lim _{\mathrm{x} \rightarrow 3^{-}}\left(\mathrm{x}^{2}-5\right)=\lim _{\mathrm{x} \rightarrow(3-\mathrm{h})}\left[(3-\mathrm{h})^{2}-5\right]$
$=\lim _{\mathrm{h} \rightarrow 0}\left(9-6 \mathrm{~h}+\mathrm{h}^{2}-5\right)=4$
$\mathrm{RH}=\lim _{\mathrm{x} \rightarrow 3^{+}}(\sqrt{\mathrm{x}+13})$
$=\lim _{\mathrm{x} \rightarrow(3+\mathrm{h})}(\sqrt{3+\mathrm{h}+13})=\lim _{\mathrm{h} \rightarrow 0}(\sqrt{16+\mathrm{h}})=4$
$\because \lim _{\mathrm{x} \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 3^{+}} \mathrm{f}(\mathrm{x})=4$
$\therefore \lim _{\mathrm{x} \rightarrow 3} \mathrm{f}(\mathrm{x})=4$

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 2)
What is $\lim _{x \rightarrow \infty}\left(\frac{x}{3+x}\right)^{3 x}$ equal to?

Options:
A $e$
B $e^{3}$
C $e^{-9}$
D $e^{9}$

The correct answer is:
$e^{-9}$

Solution:
Consider $\lim _{x \rightarrow \infty}\left(\frac{x}{3+x}\right)^{3 x}=\lim _{x \rightarrow \infty}\left(\frac{1}{1+\frac{3}{x}}\right)^{3 x}$
$=\lim _{x \rightarrow \infty}\left(1+\frac{3}{x}\right)^{-3 x}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{3}{x}\right)^{x}\right]^{-3}$
$=\left[e^{3}\right]^{-3}=e^{-9} \quad\left(\because \lim _{n \rightarrow \infty}\left(1+\frac{\lambda}{n}\right)^{n}=e^{\lambda}\right)$

Topic: Mathematics, Limits, NDA, NDA 2009 (Phase 2)
What is $\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}$ to?

Options:
A $-\frac{1}{2}$
B $-\frac{1}{3}$
C $-2$
D $-3$

The correct answer is:
$-3$

Solution:
$\quad \lim _{x \rightarrow \frac{\pi}{6}} \frac{(\sin x+1)(2 \sin x-1)}{(\sin
x-1)(2 \sin x-1)}=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 2)
What is $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right)$ equal to?

Options:
A $\frac{1}{2}$
B 1
C 2
D 0

The correct answer is:
$\frac{1}{2}$

Solution:
Consider, $\lim _{x \rightarrow \infty}\left[\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right]$
$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right)\left(\sqrt{a^{2} x^{2}+a x+1}+\sqrt{a^{2} x^{2}+1}\right)}{\left(\sqrt{a^{2} x^{2}+a x+1}+\sqrt{a^{2} x^{2}+1}\right)}$
$=\lim _{x \rightarrow \infty}\left[\frac{a^{2} x^{2}+a x+1-a^{2} x^{2}-1}{\sqrt{a^{2} x^{2}+a x+1}+\sqrt{a^{2} x^{2}+1}}\right]$
$=\lim _{x \rightarrow \infty} \frac{a}{\sqrt{a^{2}+\frac{a}{x}+\frac{1}{x^{2}}}+\sqrt{a^{2}+\frac{1}{x^{2}}}}$
$=\frac{a}{\sqrt{a^{2}}+\sqrt{a^{2}}}=\frac{a}{2 a}=\frac{1}{2}$

Topic: Mathematics, Limits, NDA, NDA 2011 (Phase 1)
What is $\lim _{x \rightarrow o} \frac{\sin ^{2} a x}{b x}(\mathrm{a}, \mathrm{b}$ are constants) equal to?

Options:
A 0
B a
C $\mathrm{a} / \mathrm{b}$
D Does not exist

The correct answer is:
0

Solution:
Consider $\lim _{x \rightarrow 0} \frac{\sin ^{2} a x}{b x}$
Multiply and divide by $a^{2} x^{2}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin ^{2} a x}{b x}=\lim _{x \rightarrow 0}\left[\frac{\sin ^{2} a x}{b x} \times \frac{a^{2} x^{2}}{a^{2} x^{2}}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\sin ^{2} a x}{a^{2} x^{2}} \times \frac{a^{2} x^{2}}{b x}\right]$
$=\lim _{x \rightarrow 0}\left[\left(\frac{\sin a x}{a x}\right)^{2} \times \frac{a^{2} x}{b}\right]$
$=\lim _{x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)^{2} \cdot \lim _{x \rightarrow 0} \frac{a^{2} x}{b}$
$=1 \times 0=0 \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)$

Topic: Mathematics, Limits, NDA, NDA 2009 (Phase 1)
What is $\int \ln \left(x^{2}\right) d x$ equal to?

Options:
A $2 x \ln (x)-2 x+c$
B $\frac{2}{x}+c$
C $2 x \ln (x)+c$
D $\frac{2 \ln (x)}{x}-2 x+c$

The correct answer is:
$2 x \ln (x)-2 x+c$

Solution:
$\int \ln \left(x^{2}\right) \cdot d x=2 \cdot \int \ln x \cdot d x$
$=2 \int 1 \cdot \ln x d x=2\left[\ln x_{-} x-\int \frac{1}{x} x \cdot d x\right.$
$=2(x, \ln x-x)+c=2 x \ln x-2 x+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2019 (Phase 1)
What is $\log _{81} 243$ equal to?

Options:
A $0.75$
B $1.25$
C $1.5$
D 3

The correct answer is:
$1.25$

Solution:
$\log _{81} 243=\log _{3} 53^{6}=\frac{\log 3^{6}}{\log 3^{5}}=\frac{6}{5}=1.25$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
What is
$\log \left(a+\sqrt{a^{2}+1}\right)+\log \left(\frac{1}{a+\sqrt{a^{2}+1}}\right)$ equal to?

Options:
A 1
B 0
C 2
D $\frac{1}{2}$

The correct answer is:
0

Solution:
Let $\log \left(a+\sqrt{a^{2}+1}\right)+\log \left(\frac{1}{a+\sqrt{a^{2}+1}}\right)$
$=\log \left(a+\sqrt{a^{2}+1}\right)+\log 1-\log \left(a+\sqrt{a^{2}+1}\right)$
$=\log \left(a+\sqrt{a^{2}+1}\right)-\log \left(a+\sqrt{a^{2}+1}\right)$
$=0$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
What is $\int \log (x+1) d x$ is equal to?

Options:
A $x \log (x+1)-x+c$
B $(x+1) \log (x+1)-x+c$
C $\frac{1}{x+1}+c$
D $\frac{\log (x+1)}{x+1}+c$

The correct answer is:
$(x+1) \log (x+1)-x+c$

Solution:
Let $\mathrm{I}=\int \log (\mathrm{x}+1) \mathrm{d} \mathrm{x}$
Let $x+1=t$ $\Rightarrow \mathrm{dx}=\mathrm{dt}$
$\Rightarrow \quad \mathrm{I}=\int 1 . \log \mathrm{tdt}$
Integrating by parts, taking log tas first function $\Rightarrow \mathrm{I}=\mathrm{t} \log \mathrm{t}-\int \frac{1}{\mathrm{t}} \cdot \mathrm{tdt}+\mathrm{c}_{1}$
$=\mathrm{t} \log \mathrm{t}-\int 1 \mathrm{dt}+\mathrm{c}_{1}=\mathrm{t} \log \mathrm{t}-\mathrm{t}+\mathrm{c} 1$
$=(x+1) \log (x+1)-x-1+c_{1}$
$=(x+1) \log (x+1)-x+c \quad\left[\because c=c_{1}-1\right]$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2008 (Phase 1)
What is maximum value of $\sin ^{2} \mathrm{x} ?$

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

The correct answer is:
1

Solution:
$-1 < \sin x < 1 \Rightarrow 1 < \sin ^{2} x < 1$
Hence, maximum value of $\sin ^{2} x=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is meant by tertiary structure of proteins?

Options:

None

Solution:
Tertiary structure of protein : When the individual peptide chains of secondary structure of protein are further extensively coiled and folded into sphere like shapes with the hydrogen bonds between the amino and carboxyl group and various other kinds of bonds cross linking on chain to another they form tertiary structure.
The ability of proteins to carry out specific reactions is the result of their primary, secondary and tertiary structure.

Topic: Biology, Biomolecules
What is meant by the term "Tyndall effect"?

Options:
A The scattering of light by colloidal particles, making the beam visible.
B The irreversible aggregation of colloidal particles leading to the formation of a precipitate.
C The process of removing small dissolved molecules from a colloid by diffusion through a semipermeable membrane.
D The migration of charged colloidal particles under the influence of an electric field.

The correct answer is:
The scattering of light by colloidal particles, making the beam visible.

Solution:
Correct Option is : (A)
The scattering of light by colloidal particles, making the beam visible.

Topic: Chemistry, Surface Chemistry, NEET
What is modulus of $\frac{1}{1+3 i}-\frac{1}{1-3 i} ?$

Options:
A $\frac{3}{5}$
B $\frac{9}{25}$
C $\frac{3}{25}$
D $\frac{5}{3}$

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

Solution:
Let $z=\frac{1}{1+3 i}-\frac{1}{1-3 i}$
$=\frac{(1-3 i)-(1+3 i)}{(1+3 i)(1-3 i)}=\frac{-6 i}{(1)^{2}-(3 i)^{2}}=-\frac{6 i}{10}=-\frac{3}{5} i$
$\therefore \quad|z|=\sqrt{(0)^{2}+\left(-\frac{3}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 1)
What is $\frac{(\mathrm{n}+2) !+(\mathrm{n}+1)(\mathrm{n}-1) !}{(\mathrm{n}+1)(\mathrm{n}-1) !}$ equal to ?

Options:
A 1
B Always an odd integer
C A perfect square
D None of the above

The correct answer is:
A perfect square

Solution:
Given expression is:
$\frac{(\mathrm{n}+2) !+(\mathrm{n}+1) !(\mathrm{n}-1) !}{(\mathrm{n}+1) !(\mathrm{n}-1) !}=\mathrm{x}$ (let)
$\Rightarrow \quad x=\frac{(n+2)(n+1) n(n-1) !+(n+1)(n-1) !}{(n+1)(n-1) !}$
$=(n+2) n+1=n^{2}+2 n+1=(n+1)^{2}$
Which is a perfect square.

Topic: Mathematics, Permutation Combination, NDA, NDA 2007 (Phase 2)
What is $\mathrm{O}_{0}^{1} \frac{\tan ^{-1}}{1+\mathrm{x}^{2}} \mathrm{dx}$ equal to?

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{8}$
C $\frac{\pi^{2}}{8}$
D $\frac{\pi^{2}}{32}$

The correct answer is:
$\frac{\pi^{2}}{32}$

Solution:
Let $\mathrm{I}=\dot{\mathrm{O}}_{0}^{1} \frac{\tan ^{-1}}{1+\mathrm{x}^{2}} \mathrm{dx}$
$\operatorname{Put} \tan ^{-1} \mathrm{x}=\mathrm{t}$
$\frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}=\mathrm{dt}$
$\mathrm{x}=0, \Rightarrow \mathrm{t}=0$
$\mathrm{x}=1, \Rightarrow \mathrm{t}=\pi / 4$
$\therefore \mathrm{I}=\int_{0}^{\pi / 4} \mathrm{tdt}=\left.\frac{\mathrm{t}^{2}}{2}\right|_{0} ^{\pi / 4}=\frac{\pi^{2}}{32}$

Topic: Mathematics, Definite Integration, NDA, NDA 2012 (Phase 1)
What is one of the square roots of $3+4 i$, where $i=\sqrt{-1}$ ?

Options:
A $2+i$
B $2-i$
C $-2+i$
D $-3-i$

The correct answer is:
$2+i$

Solution:
$3+4 \mathrm{i}=2^{2}+\mathrm{i}^{2}+2 \times 2 \times \mathrm{i}=(2+\mathrm{i})^{2}$
$\sqrt{3+4 i}=2+i$

Topic: Mathematics, Complex Number, NDA, NDA 2013 (Phase 2)
What is one of the values of $\sqrt{\mathrm{i}}+\sqrt{-\mathrm{i}}$ ?

Options:
A $\sqrt{2}$
B 0
C $\pm \frac{1+\mathrm{i}}{\sqrt{2}}$
D $\pm \frac{1-\mathrm{i}}{\sqrt{2}}$

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

Solution:
$\sqrt{i}=\sqrt{\frac{2 i}{2}}=\sqrt{\frac{1+2 i-1}{2}}$
$=\sqrt{\frac{1+2 i+i^{2}}{2}}=\sqrt{\frac{(1+i)^{2}}{2}}=\frac{1+i}{\sqrt{2}}$
$\sqrt{-i}=\sqrt{-\frac{2 i}{2}}$
$=\sqrt{\frac{1-2 i-1}{2}}=\sqrt{\frac{1-2 i+i^{2}}{2}}=\sqrt{\frac{(1-i)^{2}}{2}}=\frac{(1-i)}{\sqrt{2}}$
So, value of $\sqrt{i}+\sqrt{-i}$
$=\frac{(1+i)+(1-i)}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 1)
What is oogenesis? Give a brief account of oogenesis.

Options:

None

Solution:
The process of formation of a mature female gamete (ovum) is called oogenesis. It occurs in the ovaries of female reproductive system. Oogenesis is a discontinuous process it begins before birth, stops in midprocess \& only resumes after menarch. It occurs in three phases:
Multiplicative phase (formation of oogonia mitotically from the primary germ cells), Growth phase (growth of oogonia into primary oocyte) \& Maturation phase (formation of mature ova from primary oocyte through meiosis). Maturation phase produces two haploid cells Larger one called secondary oocyte \& the smaller one called polar bodies (1st polar body). Meiosis II of secondary oocyte results in the formation of functional egg or ovum
and a second polar body. The first polar body may also divide to form two polar bodies of equal sizes which do not take part in reproduction \& ultimately degenerates. First maturation division may be completed in the ovaries just prior to ovulation but second one (Final) is completed outside the ovary after fertilization. Secondary oocyte is female gamete in which the 1st meiotic division is completed \& second meiotic division (Metaphase stage) has begin. The egg is released at secondary oocyte stage under the effect of LH.

Topic: Biology, Human Reproduction
What is periderm? How does periderm formation take place in the dicot stems?

Options:

None

Solution:
As the stem continues to increase in girth due to activity of vascular cambium, the outer cortical and epidermis layers get broken and need to be replaced to provide new protective cell layers. Hence, sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is made of thinwalled rectangular cells. Its cuts off cells on both side. The outer cell differentiate into cork or phellem while inner cells differentiate into secondary cortex or phelloderm. Phelloderm, phellem and phelloderm are collectively called as periderm.

Topic: Biology, Anatomy of Flowering Plants
What is primary productivity? Give brief description of factors that affect primary productivity.

Options:

None

Solution:
Primary productivity of an ecosystem is the amount of energy fixed or biomass synthesized by primary producers or green plants per unit area per unit time during photosynthesis.
Factors affecting primary productivity are -
- Plant species inhabiting a particular area
- Sunlight
- Temperature
- Soil water
- Nutrients
In deserts, sunlight is abundant but water is scarce or nutrients are lacking. Therefore, in such areas, water \& nutrients supply become the limiting factors.

Topic: Biology, Ecosystem
What is $\sum_{r=0}^{1}{ }^{n+r} C_{n}$ equal to?

Options:
A $n+3 C_{n+1}$
B $n+3 C_{n}$
C $n+2 C_{n}$
D $n+2 C_{n+1}$

The correct answer is:
$n+2 C_{n+1}$

Solution:
$\sum_{\mathrm{r}=0}^{1} \sum^{\mathrm{n}+\mathrm{r}} \mathrm{C}_{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}$
$=1+\frac{(\mathrm{n}+1) !}{(\mathrm{n}+1-\mathrm{n}) ! \mathrm{n} !}=1+\frac{(\mathrm{n}+1)(\mathrm{n} !)}{\mathrm{n} !}$
$=1+\mathrm{n}+1=\mathrm{n}+2$
${ }^{n+2} C_{n+1}=\frac{(n+2) !}{(n+2-n-1) !(n+1) !}$
$=\frac{(\mathrm{n}+2)(\mathrm{n}+1) !}{(\mathrm{n}+1) !}=\mathrm{n}+2$
$\mathrm{OR}$
$\sum_{\mathrm{r}=0}^{1}{ }^{\mathrm{n}+\mathrm{r}} \mathrm{C}_{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}$
$=1+(\mathrm{n}+1)=\mathrm{n}+2$
Now,
${ }^{n+2} C_{1}=\frac{(n+2) !}{1 !(n+2-1) !}=\frac{(n+2)(n+1) !}{(n+1) !}=(n+2)$
$\therefore$ Option (d) is correct.

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
What is $\sum_{r=0}^{n} C(n, r)$ equal to ?

Options:
A $2^{n}-1$
B $n$
C $n l$
D $2^{n}$

The correct answer is:
$2^{n}$

Solution:
We know that, $(1+\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1}^{\mathrm{x}}+{ }^{\mathrm{n}} \mathrm{C}_{2}^{\mathrm{x}^{2}}+\ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}$
For $\mathrm{x}=1,(1+1)^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}$
$\therefore \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{c}(\mathrm{n}, \mathrm{r})=2^{\mathrm{n}}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2013 (Phase 2)
What is referred to as satellite chromosome?

Options:

None

Solution:
Additional constriction or secondary constriction at the chromosomal ends as distal part of the arm formed by chromatin thread are known satellite chromosomees. These constriction gives appearance of an outgrowth or a small fragment.
These are also known as (sat) chromosomes or marker chromosomes. Chromosomes $13,14,15,16,21$ and 21 satellite chromosomes.

Topic: Biology, Cell - The Unit of Life
What is $\left(\frac{\sec 18^{\circ}}{\sec 144^{\circ}}+\frac{\operatorname{cosec} 18^{\circ}}{\operatorname{cosec} 144^{\circ}}\right)$ equal to?

Options:
A $\sec 18^{\circ}$
B $\operatorname{cosec} 18^{\circ}$
C $-\sec 18^{\circ}$
D $-\operatorname{cosec} 18^{\circ}$

The correct answer is:
$\sec 18^{\circ}$

Solution:
$\begin{aligned} \text { Consider }\left(\frac{\sec 18^{\circ}}{\sec 144^{\circ}}+\frac{\cos e c 18^{\circ}}{\cos e c 144^{\circ}}\right) \\=& \frac{\sec 18^{\circ}}{\sec \left(180^{\circ}-36^{\circ}\right)}+\frac{\cos e c 18^{\circ}}{\cos e c\left(180^{\circ}-36^{\circ}\right)} \\=&-\frac{\sec 18^{\circ}}{\sec 36^{\circ}}+\frac{\cos e c 18^{\circ}}{\cos e c 36^{\circ}}(\because \sin , \operatorname{cosec} \operatorname{are}+\mathrm{ve}\\ & \left.\text { in } 2^{\text {nd }} \text { quadrant and sec is }-\text { ve in } 2^{\text {nd }} \text { quadrant }\right) \\=& \frac{\sin 36^{\circ}}{\sin 18^{\circ}}-\frac{\cos 36^{\circ}}{\cos 18^{\circ}}=\frac{\sin 36^{\circ} \cos 18^{\circ}-\cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 18^{\circ}} \\=& \frac{\sin \left(36^{\circ}-18^{\circ}\right)}{\sin 18^{\circ} \cos 18^{\circ}}=\frac{\sin 18^{\circ}}{\sin 18^{\circ} \cos 18^{\circ}}=\sec 18^{\circ} \\ &\left(\because \frac{1}{\cos x}=\sec x\right) \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
What is $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$ equal to?

Options:
A 0
B 1
C 2
D 4

The correct answer is:
4

Solution:
$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$
$=\frac{2 \times 2\left[\frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ}\right]}{2 \sin 10^{\circ} \cos 10^{\circ}}$
$=\frac{4\left(\cos 60^{\circ} \cos 10^{\circ}-\sin 60^{\circ} \sin 10^{\circ}\right)}{\sin 20^{\circ}}$
$=\frac{4 \cdot \cos \left(60^{\circ}+10^{\circ}\right)}{\sin 20^{\circ}}=4 \cdot \frac{\cos 70^{\circ}}{\sin 20^{\circ}}=4 \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}}=4$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
What is $\int \frac{d \theta}{\sin ^{2} \theta+2 \cos ^{2} \theta-1}$ equal to?

Options:
A $\tan \theta+c$
B $\cot \theta+c$
C $\frac{1}{2} \tan \theta+c$
D $\frac{1}{2} \cot \theta+c$

The correct answer is:
$\tan \theta+c$

Solution:
Let $I=\int \frac{d \theta}{\sin ^{2} \theta+2 \cos ^{2} \theta-1}$
$=\int \frac{d \theta}{1-\cos ^{2} \theta+2 \cos ^{2} \theta-1}=\int \frac{d \theta}{\cos ^{2} \theta}$
$=\int \sec ^{2} \theta d \theta=\tan \theta+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2008 (Phase 2)
What is $\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equal to?

Options:
A $\tan x+\cot x+\mathrm{c}$
B $\tan x-\cot x+\mathrm{c}$
C $(\tan x+\cot x)^{2}+\mathrm{c}$
D $(\tan x-\cot x)^{2}+c$
where $c$ is the constant of integration.

The correct answer is:
$\tan x-\cot x+\mathrm{c}$

Solution:
Let $I=\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=\int \frac{\left(\sin ^{2} x+\cos ^{2} x\right) d x}{\sin ^{2} x \cos ^{2} x}$
$=\int\left[\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right] d x$
$=\int\left[\frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x}\right] d x$
$=\int\left(\sec ^{2} x+\operatorname{cosec}^{2} x\right) d x$
$=\tan x-\cot x+c$ where $c$ is constant of Integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2011 (Phase 2)
What is $\left[\frac{\sin \frac{\pi}{6}+i\left(1-\cos \frac{\pi}{6}\right)}{\sin \frac{\pi}{6}-i\left(1-\cos \frac{\pi}{6}\right)}\right]^{3}$
where $i=\sqrt{-1}$, equal to?

Options:
A 1
B -1
C i
D -i

The correct answer is:
i

Solution:
$\left[\frac{\sin \frac{\pi}{6}+i\left(1-\cos \frac{\pi}{6}\right)}{\sin \frac{\pi}{6}-i\left(1-\cos \frac{\pi}{6}\right)}\right]^{3}$
$=\left[\frac{2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}+i\left(2 \sin ^{2} \frac{\pi}{12}\right)}{2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}-i\left(2 \sin ^{2} \frac{\pi}{12}\right)}\right]^{3}$
$=\left[\frac{\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}}{\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}}\right]^{3}=\left(\frac{e^{i \frac{\pi}{12}}}{e^{-i \frac{\pi}{12}}}\right)^{3}$
$=\left(e^{i \frac{\pi}{6}}\right)^{3}=e^{i \times 3 \times \frac{\pi}{6}}=e^{i \frac{\pi}{2}}=\cos \frac{\pi}{2}+\mathrm{i} \sin \frac{\pi}{2}=\mathrm{i}$
$\therefore$ Option (c) is correct.

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 1)
What is $\frac{\cos 7 x-\cos 3 x}{\sin 7 x-2 \sin 5 x+\sin 3 x}$ equal to ?

Options:
A $\tan \mathrm{x}$
B $\cot x$
C $\tan 2 \mathrm{x}$
D $\cot 2 \mathrm{x}$

The correct answer is:
$\cot x$

Solution:
$\frac{\cos 7 x-\cos 3 x}{\sin 7 x-2 \sin 5 x+\sin 3 x}$
$=\frac{-2 \sin \frac{7 x+3 x}{2} \cdot \sin \frac{7 x-3 x}{2}}{2 \sin \frac{7 x+3 x}{2} \cos \frac{7 x-3 x}{2}-2 \sin 5 x}$
$\quad\left(\because \sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right)\right.$
and $\left.\cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right)$
$=\frac{-2 \sin 5 x \cdot \sin 2 x}{2 \sin 5 x \cos 2 x-2 \sin 5 x}$
$=\frac{-2 \sin 5 x \cdot \sin 2 x}{-2 \sin 5 x[1-\cos 2 x]}$
$=\frac{\sin 2 x}{1-1+2 \sin ^{2} x} \quad\left(\because \cos 2 x=1-2 \sin ^{2} x\right)$
$=\frac{2 \sin x \cos x}{2 \sin ^{2} x}=\cot x$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)