datavorous/entrance-exam-dataset huggingface dataset:
The value of $\sqrt{3}$ cosec $20^{\circ}-\sec 20^{\circ}$ is equal to

Options:
A 4
B 2
C 1
D $-4$

The correct answer is:
4

Solution:
& \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}} \\ &=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \\ &=\frac{2\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)}{\sin 20^{\circ} \cos 20^{\circ}} \end{aligned}$
$=\frac{2\left(\cos 30^{\circ} \cos 20^{\circ}-\sin 30^{\circ} \sin 20^{\circ}\right)}{\left(\frac{2 \sin 20^{\circ} \cos 20^{\circ}}{2}\right)}$
$=\frac{4\left[\cos \left(30^{\circ}+20^{\circ}\right)\right]}{\sin 40^{\circ}}$
$=\frac{4 \cos 50^{\circ}}{\sin 40^{\circ}}=\frac{4 \cos \left(90^{\circ}-40^{\circ}\right)}{\sin 40^{\circ}}=\frac{4 \sin 40^{\circ}}{\sin 40^{\circ}}=4 .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 2)
The value of
$\int_{a}^{b} \frac{x^{7}+\sin x}{\cos x} d x$ where $a+b=0$ is

Options:
A $2 \mathrm{~b}-\mathrm{a} \sin (\mathrm{b}-\mathrm{a})$
B $a+3 b \cos (b-a)$
C $\sin a-(b-a) \cos b$
D 0

The correct answer is:
0

Solution:
$a+b=0 \Rightarrow a=-b$
$I=\int_{-b}^{b} \frac{x^{7}+\sin x}{\cos x}$
Using property
$\int_{-a}^{a} f(x) d x=\left\{\begin{array}{l}\int_{0}^{2 a} f(x) d x ; \text { if } f(x) \text { is even } \\ 0 & ; \text { if } f(x) \text { is odd }\end{array}\right.$
$f(x)=\frac{x^{7}+\sin x}{\cos x}$
$f(-x)=\frac{(-x)^{7}+\sin (-x)}{\cos (-x)}=\frac{-x^{7}-\sin x}{\cos x}$
$=-\left[\frac{x^{7}+\sin x}{\cos x}\right]$
$=-\mathrm{f}(\mathrm{x})$
So $\mathrm{f}(\mathrm{x})$ is odd hence
$\mathrm{I}=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 2)
The value of $\mathrm{A}$ is
Consider the function
$f(x)=\left\{\begin{array}{cc}-2 \sin x & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B & \text { if }-\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x & \text { if } x \geq \frac{\pi}{2}\end{array}\right.$
which is continuous everywhere.

Options:
A 1
B 0
C -1
D -2

The correct answer is:
-1

Solution:
$\because \mathrm{f}(\mathrm{x})=\left\{\begin{array}{ccc}-2 \sin \mathrm{x} & \text { if } & \mathrm{x} \leq-\pi / 2 \\ \mathrm{~A} \sin \mathrm{x}+\mathrm{B} & \text { if } & -\pi / 2 < \mathrm{x} < \pi / 2 \\ \cos \mathrm{x} & \text { if } & \mathrm{x} \geq \pi / 2\end{array}\right.$
$\because \mathrm{f}(\mathrm{x})$ is continuous every where :
$\therefore \lim _{x \rightarrow \pi / 2^{+}} f(x)=\lim _{x \rightarrow \pi / 2} f(x)$
$\Rightarrow \lim _{x \rightarrow \pi / 2} \cos x=\lim _{x \rightarrow \pi / 2} A \sin x+B$
$\Rightarrow A+B=0$
Also; $\lim _{x \rightarrow-\pi / 2} f(x)=\lim _{x \rightarrow-\pi / 2^{-}} f(x)$
$\Rightarrow \lim _{x \rightarrow-\pi / 2} A \sin x+B=\lim _{x \rightarrow-\pi / 2}-2 \sin x$
$\Rightarrow-A+B=2$
is from eq $(1) \&(2)$ we get,
$A=-1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 2)
The value of B is
Consider the function
$f(x)=\left\{\begin{array}{cc}-2 \sin x & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B & \text { if }-\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x & \text { if } x \geq \frac{\pi}{2}\end{array}\right.$
which is continuous everywhere.

Options:
A 1
B 0
C -1
D -2

The correct answer is:
1

Solution:
Put the value of A in eq. (2) we get:
$\Rightarrow \mathrm{B}=1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 2)
The value of $[\mathrm{C}(7,0)+\mathrm{C}(7,1)]+[\mathrm{C}(7,1)+\mathrm{C}(7,2)]+\ldots .+$
$[\mathrm{C}(7,6)+\mathrm{C}(7,7)]$ is

Options:
A 254
B 255
C 256
D 257

The correct answer is:
254

Solution:
$\left(7 \mathrm{c}_{0}+7 \mathrm{c}_{1}\right)+\left(7 \mathrm{c}_{1}+7 \mathrm{c}_{2}\right)+\ldots+\left(7 \mathrm{c}_{6}+7 \mathrm{c}_{7}\right)$
We know, $\mathrm{n}_{\mathrm{c}_{\mathrm{r}}}+\mathrm{n}_{\mathrm{c}_{\mathrm{r}-1}}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}$
$=8_{\mathrm{c}_{1}}+8_{\mathrm{c}_{2}}+\ldots . .+8_{\mathrm{c}_{7}}$
$=\left(8_{\mathrm{c}_{0}}+8_{\mathrm{c}_{1}}+8_{\mathrm{c}_{2}}+\ldots . .+8_{\mathrm{c}_{7}}+8_{\mathrm{c}_{8}}\right)-\left(8_{\mathrm{c}_{0}}+8_{\mathrm{c}_{8}}\right)$
$=2^{8}-(1+1)$
$\left[\right.$ Since,$\left.n_{c_{0}}+n_{c_{1}}+n_{c_{2}}+\ldots+n_{c_{n}}=2^{n}\right]$
$=256-2$
$=254$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2017 (Phase 1)
The value of determinant \( \left|\begin{array}{ccc}a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c\end{array}\right| \) is

Options:
A \( a^{3}+b^{3}+c^{3} \)
B 3abc
C \( a^{3}+b^{3}+c^{3}-3 a b c \)
D None of the above

The correct answer is:
None of the above

Solution:
Given that, \( \left|\begin{array}{ccc}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| \)
\( C_{3} \rightarrow C_{2}+C_{3} \)
\( \left|\begin{array}{lll}a-b & b+c & a+b+c \\ b-c & c+a & a+b+c \\ c-a & a+b & a+b+c\end{array}\right| \)
\( =a+b+c)\left|\begin{array}{ccc}a-b & b+c & 1 \\ b-c & c+a & 1 \\ c-a & a+b & 1\end{array}\right| \)
\( =(a+b+c)((c+a)(c-a)-(a+b)(b-c) \)
\( -(b+c)(c-a)-(a+b)(a-b) \)
\( +(b+c)(b-c)-(c+a)(a-b)) \)
\( =(a+b+c)\left(c^{2}-a^{2}-a b+a c-b^{2}+b c\right) \)
\( -b c+a b-c^{2}+a c-a^{2}+b^{2} \)
\( \left.+b^{2}-c^{2}-a c+b c-a^{2}+a b\right) \)
\( =(a+b+c)\left(-c^{2}-3 a^{2}+a b+a c+b^{2}+b c\right) \)
\( =a\left(-c^{2}-3 a^{2}+a b+a c+b^{2}+b c\right) \)
\( +b\left(-c^{2}-3 a^{2}+a b+a c+b^{2}+b c\right) \)
\( +c\left(-c^{2}-3 a^{2}+a b+a c+b^{2}+b c\right) \)
\( =-a c^{2}-3 a^{3}+a^{2} b+a^{2} c+a b^{2}+a b c \)
\( -b c^{2}-3 a^{2} b+a b^{2}+a b c+b^{3}+b^{2} c \)
\( -c^{3}-3 a^{2} c+a b c+a c^{2}+b^{2} c+b c^{2} \)

Topic: Mathematics, Determinants, KCET, KCET 2018
The value of E M + / M o of four metals A, B, C and D are -1.2 V, 0.6 V, 0.85 V and -0.76 V, respectively. The order of tendency of their unipositive cations getting deposited at the cathode during electrolysis of the respective salts is

Options:
A B>D>C>A.
B A>C>B>D.
C C>B>D>A.
D D>A>B>C.

The correct answer is:
C>B>D>A.

Solution:
During electrolysis, the species having higher reduction potential is preferably reduced at the cathode.

As per the data given in the question, the standard reduction potential of the metals are in the order-

E C ∘ > E B ∘ > E D ∘ > E A ∘

Hence, C>B>D>A will be the preference of deposition of these metals on the cathode.

Topic: Chemistry, Electrochemistry, NEET
The value of $i^{2 n}+i^{2 n+1}+i^{2 n+2}+i^{2 n+3}$, where $i$
$=\sqrt{-1}$, is

Options:
A 0
B 1
C i
D -i

The correct answer is:
0

Solution:
$\begin{aligned} & \mathrm{i}^{2 \mathrm{n}}+\mathrm{i}^{2 \mathrm{n}}+1+\mathrm{i}^{2 \mathrm{n}+2}+\mathrm{i}^{2 \mathrm{n}} \cdot+{ }^{3} \\ &=\mathrm{i}^{2 \mathrm{n}}+\mathrm{i}^{2 \mathrm{n}} \cdot \mathrm{i}+\mathrm{i}^{2 \mathrm{n}} \cdot \mathrm{i}^{2}+{ }^{\mathrm{i} 2 \mathrm{n}} \cdot \mathrm{i}^{3} \\ &=\mathrm{i}^{2 \mathrm{n}}\left(1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}\right)\left[\text { since }, \mathrm{i}^{2}=-1, \mathrm{i}^{3}=\mathrm{i}^{2} \mathrm{i}=-\mathrm{i}\right] \\ &=\mathrm{i}^{2 \mathrm{n}}(1+\mathrm{i}-1-\mathrm{i}) \\ &=\mathrm{i}^{2 \mathrm{n}}(0) \\ &=0 \end{aligned}$

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 1)
The value of $k$ for which the lines $2 x+3 y+a=0$ and $5 x+k y+a=0$ represent family of parallel lines is

Options:
A 3
B $4.5$
C $7.5$
D 15

The correct answer is:
$7.5$

Solution:
$2 \mathrm{x}+3 \mathrm{y}+\mathrm{a}=0$ or $\mathrm{y}=\frac{-2}{3} \mathrm{x}-\frac{\mathrm{a}}{3}$
Slope $=\frac{-2}{3}$
$5 x+k y+a=0$ or $y=\frac{-5}{k} x \frac{-a}{5}$
Slope $=\frac{-5}{\mathrm{k}}$
lines are parallel $\frac{-2}{3}=\frac{-5}{\mathrm{k}}$
$\mathrm{k}=\frac{15}{2}=7.5$

Topic: Mathematics, Pair of Lines, NDA, NDA 2013 (Phase 2)
The value of $\mathrm{k}$ which makes $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl}\sin x & x \neq 0 \\ k & x=0\end{array}\right.$ continuous at $\mathrm{x}=0$, is

Options:
A 2
B 1
C -1
D 0

The correct answer is:
0

Solution:
$f(x)=\left\{\begin{array}{cl}\sin x, & x \neq 0 \\ k, & x=0\end{array}\right.$
Given, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \sin x=k$
$\Rightarrow \mathrm{k}=0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2019 (Phase 1)
The value of $\frac{1}{\log _{3} e}+\frac{1}{\log _{3} e^{2}}+\frac{1}{\log _{3} e^{4}}+\ldots$ up to infinite
terms is

Options:
A $\log _{e} 9$
B 0
C 1
D $\log _{e} 3$

The correct answer is:
$\log _{e} 9$

Solution:
$\frac{1}{\log _{3} \mathrm{e}}+\frac{1}{\log _{3} \mathrm{e}^{2}}+\frac{1}{\log _{3} \mathrm{e}^{4}}+\ldots .$
$=\frac{1}{\log _{3} \mathrm{e}}+\frac{1}{2 \log _{3} \mathrm{e}}+\frac{1}{4 \log _{3} \mathrm{e}}+\ldots .\left(\right.$ Since, $\left.\log _{\mathrm{a}} \mathrm{b}^{\mathrm{m}}=\mathrm{m} \log \mathrm{a} \mathrm{b}\right)$
$=\log _{\mathrm{e}} 3+\frac{\log _{\mathrm{e}} 3}{2}+\frac{\log _{\mathrm{e}} 3}{4}+\ldots \cdot\left(\right.$ Since $\left.\log _{\mathrm{a}} \mathrm{b}=\frac{1}{\log _{\mathrm{b}} \mathrm{a}}\right)$
$\quad=\log _{\mathrm{e}} 3\left(1+\frac{1}{2}+\frac{1}{4}+\ldots .\right)$
$=\log _{\mathrm{e}} 3\left(\frac{1}{1-\frac{1}{2}}\right)\left(\because 1, \frac{1}{2}, \frac{1}{4} \ldots .\right.$ is G.P. with $\left.\mathrm{a}=1, \mathrm{r}=\frac{1}{2}\right)$
$=\log _{\mathrm{e}} 3(2)=2 \log _{\mathrm{e}} 3=\log _{\mathrm{e}} 3^{2}=\log _{\mathrm{e}} 9$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
The value of \(\left|\begin{array}{cc}\log _5 729 & \log _3 5 \\ \log _5 27 & \log _9 25\end{array}\right|\) \(\times\left|\begin{array}{ll}\log _3 5 & \log _{27} 5 \\ \log _5 9 & \log _5 9\end{array}\right|\) is

Options:
A 1
B 6
C \(\log _5 9\)
D \(\left(\log _3 5\right) \times\left(\log _5 81\right)\)

The correct answer is:
\(\left(\log _3 5\right) \times\left(\log _5 81\right)\)

Solution:
\(\begin{aligned}
& \left|\begin{array}{ll}
\log _5 729 & \log _3 5 \\
\log _5 27 & \log _9 25
\end{array}\right| \times\left|\begin{array}{ll}
\log _3 5 & \log _{27} 5 \\
\log _5 9 & \log _5 9
\end{array}\right| \\
& =\left|\begin{array}{ll}
6 \log _5 3 & \log _3 5 \\
3 \log _5 3 & \log _3 5
\end{array}\right| \times\left|\begin{array}{ll}
\log _3 5 & \frac{1}{3} \log _3 5 \\
2 \log _5 3 & 2 \log _5 3
\end{array}\right| \\
& =3 \log _5 3 \log _3 5\left|\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right| \times 2 \log _3 5 \log _5 3\left|\begin{array}{cc}
1 & 1 / 3 \\
1 & 1
\end{array}\right| \\
& =6[2-1] \times\left[1-\frac{1}{3}\right]=4
\end{aligned}\)
\(\because\) Hence the options
\(\log _3 5 \times \log _5 81=\left(\log _3 5\right) \times\left(4 \log _5 3\right)=4\)
Hence, option (d) is correct.

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
The value of $\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

Options:
A 0
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
$\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$\sin \theta=\frac{3}{5}$
$\tan \theta=\frac{3}{4}$
$=\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{ \left.1-\frac{3}{4} \cdot \frac{1}{7}\right)}\right.$
$=\tan ^{-1}\left(\frac{25}{25}\right)=\tan ^{-1}(1)=\frac{\pi}{4} .$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2017 (Phase 2)
The value of $\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\sin ^{2} 20^{\circ}+\ldots$
$+\sin ^{2} 90^{\circ}$ is

Options:
A $\overline{7}$
B 8
C 9
D $\frac{19}{2}$

The correct answer is:
$\frac{19}{2}$

Solution:
$\sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots \ldots .+\ldots \ldots \ldots .+\sin ^{2} 75^{\circ}+$
$\sin ^{2} 80^{\circ}+\sin ^{2} 85^{\circ}+\sin ^{2} 90^{\circ}$
$\Rightarrow \sin ^{2} 5^{\circ}+\sin ^{2} 10^{\circ}+\sin ^{2} 15^{\circ}+\ldots \ldots . \sin ^{2}\left(90-15^{\circ}\right)+$
$\sin ^{2}(90-10)+\sin ^{2}(90-5)+1$
$\Rightarrow \sin ^{2} 5+\sin ^{2} 10+\sin ^{2} 15^{\circ}+\ldots \ldots .+\cos ^{2} 15^{\circ}+$
$\cos ^{2} 10^{\circ}+\cos ^{2} 5^{\circ}+1$
$\Rightarrow(1+1+1+\ldots .8$ times $)+\sin ^{2} 45^{\circ}+1$
$\Rightarrow 8+\frac{1}{2}+1=\frac{19}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
The value of $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)$ is

Options:
A $-\frac{7}{17}$
B $\frac{5}{16}$
C $\frac{5}{4}$
D $\frac{7}{17}$

The correct answer is:
$-\frac{7}{17}$

Solution:
$\quad 2 \tan ^{-1}\left(\frac{1}{5}\right)=\tan ^{-1}\left[\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right]$
$=\tan ^{-1}\left[\frac{10}{24}\right]$
$=\tan ^{-1}\left(\frac{5}{12}\right)$
Let $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\mathrm{x}$
$\Rightarrow \quad \tan \left[\tan ^{-1}\left(\frac{5}{12}\right)-\frac{\pi}{4}\right]=\mathrm{x}$
$\Rightarrow \tan ^{-1}\left(\frac{5}{12}\right)-\frac{\pi}{4}=\tan ^{-1} \mathrm{x}$
$\Rightarrow \tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)=\tan ^{-1} \mathrm{x}$
$\Rightarrow \quad \mathrm{x}=\frac{-7 / 12}{17 / 12}=-7 / 17$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 2)
The value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is equal to

Options:
A $-1$
B 0
C 1
D 4

The correct answer is:
4

Solution:
$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$
$=\tan 9^{\circ}-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right)+\tan \left(90^{\circ}-9^{\circ}\right)$
$=\tan 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}+\cot 9^{\circ}$
$=\tan 9^{\circ}+\cot 9^{\circ}-\left(\tan 27^{\circ}+\cot 27^{\circ}\right)$
We know, $\tan \theta+\cot \theta=\frac{1}{\sin \theta \cos \theta}=\frac{2}{2 \sin \theta \cos \theta}=\frac{2}{\sin 2 \theta}$
$\begin{aligned} & \therefore \tan 9^{\circ}+\cot 9^{\circ}-\left(\tan 27^{\circ}+\cot 27^{\circ}\right) \\ &=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=2\left(\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}\right) \\ & \text { We also know, } \sin \mathrm{C}-\sin \mathrm{D} \\ &=2 \cos \frac{(\mathrm{C}+\mathrm{D})}{2} \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right) . \\ & \therefore 2\left(\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}\right)=2 \cdot 2 \cdot \frac{\cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}} \\ &=2 \cdot 2 \cdot \frac{\sin 54^{\circ} \cdot \sin 18^{\circ}}{\sin 18^{\circ} \cdot \sin 54^{\circ}}=4 \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 2)
The value of the determinant $\left|\begin{array}{ccc}1-\alpha & \alpha-\alpha^{2} & \alpha^{2} \\ 1-\beta & \beta-\beta^{2} & \beta^{2} \\ 1-\gamma & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$ is equal to

Options:
A $(\alpha-\beta)(\beta-\gamma)(\alpha-\gamma)$
B $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
C $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$
D ${0}$

The correct answer is:
$(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$

Solution:
$\left|\begin{array}{lll}1-\alpha & \alpha-\alpha^{2} & \alpha^{2} \\ 1-\beta & \beta-\beta^{2} & \beta^{2} \\ 1-\gamma & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$
$\mathrm{c}_{1} \rightarrow \mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha-\alpha^{2} & \alpha^{2} \\ 1 & \beta-\beta^{2} & \beta^{2} \\ 1 & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$
$\mathrm{c}_{2} \rightarrow \mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha & \alpha^{2} \\ 1 & \beta & \beta^{2} \\ 1 & \gamma & \gamma^{2}\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 2)
The value of the determinant $\left|\begin{array}{lll}\mathrm{m} & \mathrm{n} & \mathrm{p} \\ \mathrm{p} & \mathrm{m} & \mathrm{n} \\ \mathrm{n} & \mathrm{p} & \mathrm{m}\end{array}\right|$

Options:
A is a perfect cube
B is a perfect square
C has linear factor
D is zero

The correct answer is:
has linear factor

Solution:
Consider $\left|\begin{array}{ccc}\mathrm{m} & \mathrm{n} & \mathrm{p} \\ \mathrm{p} & \mathrm{m} & \mathrm{n} \\ \mathrm{n} & \mathrm{p} & \mathrm{m}\end{array}\right|$
$\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$
$=\left|\begin{array}{ccc}\mathrm{m}+\mathrm{n}+\mathrm{p} & \mathrm{n} & \mathrm{p} \\ \mathrm{p}+\mathrm{m}+\mathrm{n} & \mathrm{m} & \mathrm{n} \\ \mathrm{n}+\mathrm{p}+\mathrm{m} & \mathrm{p} & \mathrm{m}\end{array}\right|$
Take $\mathrm{m}+\mathrm{n}+\mathrm{p}$ common from $\mathrm{C}_{1}$. $=(\mathrm{m}+\mathrm{n}+\mathrm{p})\left|\begin{array}{ccc}1 & \mathrm{n} & \mathrm{p} \\ 1 & \mathrm{~m} & \mathrm{n} \\ 1 & \mathrm{p} & \mathrm{m}\end{array}\right|$
$=(m+n+p)\left[\left(m^{2}+n^{2}+p^{2}\right)-m n-n p-p m\right]$
Hence, value of the determinant has linear factor.

Topic: Mathematics, Determinants, NDA, NDA 2013 (Phase 1)
The value of the determinant $\left|\begin{array}{lll}x^{2} & 1 & y^{2}+z^{2} \\ y^{2} & 1 & z^{2}+x^{2} \\ z^{2} & 1 & x^{2}+y^{2}\end{array}\right|$ is:

Options:
A 0
B $x^{2}+y^{2}+z^{2}$
C $x^{2}+y^{2}+z^{2}+1$
D None of the above

The correct answer is:
0

Solution:
Let $\Delta=\left|\begin{array}{ccc}x^{2} & 1 & y^{2}+z^{2} \\ y^{2} & 1 & z^{2}+x^{2} \\ z^{2} & 1 & x^{2}+y^{2}\end{array}\right|$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3}$
$\Delta=\left|\begin{array}{ccc}\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} & 1 & \mathrm{y}^{2}+\mathrm{z}^{2} \\ \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} & 1 & \mathrm{z}^{2}+\mathrm{x}^{2} \\ \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} & 1 & \mathrm{x}^{2}+\mathrm{y}^{2}\end{array}\right|$
$\Delta=\left(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}\right)\left|\begin{array}{ccc}1 & 1 & \mathrm{y}^{2}+\mathrm{z}^{2} \\ 1 & 1 & \mathrm{z}^{2}+\mathrm{x}^{2} \\ 1 & 1 & \mathrm{x}^{2}+\mathrm{y}^{2}\end{array}\right|$
$\Delta=0\left[\because \mathrm{C}_{1} \& \mathrm{C}_{2}\right.$ are identical $]$

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 2)
The value of the infinite product $6^{\frac{1}{2}} \times 6^{\frac{1}{2}} \times 6^{\frac{3}{8}} \times 6^{\frac{1}{4}} \times \ldots$ is

Options:
A 6
B 36
C 216
D $\infty$

The correct answer is:
36

Solution:
$\mathrm{X}=6^{1 / 2} \times 6^{1 / 2} \times 6^{3 / 8} \times 6 / 16 \times \ldots \ldots \ldots \ldots . . \infty .$
$\left.=6^{\left(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16} \ldots \ldots \ldots \ldots \ldots 00\right.}\right) .$
Let $\mathrm{S}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\ldots \ldots \ldots \ldots \infty$
$\frac{\frac{1}{2} \mathrm{~S}=+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\ldots \ldots \ldots . \infty}{\left(1-\frac{1}{2}\right) \mathrm{S}=\frac{1}{2}+\frac{(2-1)}{4}+\frac{(3-2)}{8}+\frac{(4-3)}{16}+\ldots \ldots \ldots \ldots \infty}$
$\frac{\mathrm{S}}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots \ldots \ldots \infty$
$\frac{\mathrm{S}}{2}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$
$\mathrm{~S}=2 \therefore \mathrm{x}=6^{2}=36$

Topic: Mathematics, Sequences and Series, NDA, NDA 2015 (Phase 2)
The value of the product $6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots$ up to
infinite terms is $\quad[2017-I I]$

Options:
A 6
B 36
C 216
D 512

The correct answer is:
6

Solution:
$6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots .$ upto infinite terms.
$=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots . \infty}=6$
Sum of $\infty$ terms of G.P. is $S_{\infty}=\frac{a}{1-r} .$ Here $a=\frac{1}{2}, r=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$
Sum of $\infty$ terms ofG.P. is $S_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}} .$ Here $\mathrm{a}=\frac{1}{2}, \mathrm{r}=\frac{1}{2}$
$\therefore 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \infty}=6^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=6^{\frac{1}{2}}=6^{1}=6$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
The value of the sum $\sum_{\mathrm{n}=1}^{13}\left(\mathrm{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}\right)$ where $\mathrm{i}=\sqrt{-1}$ is:

Options:
A i
B -i
C 0
D i-1

The correct answer is:
i-1

Solution:
$\quad \sum_{n=1}^{13}\left[i^{n}+i^{n+1}\right]=\sum_{n=1}^{13} i^{n}[1+i]$
$=(1+i)\left[i+i^{2}+i^{3} \ldots . i^{13}\right]=\frac{(1+i)}{(1-i)} i\left[1-i^{13}\right]$
$=\frac{(-1+i)\left(1-i^{13}\right)}{(1-i)}=\frac{-1+i^{13}+i-i^{14}}{(1-i)}$
$\quad=\frac{-1+\left(i^{2}\right)^{6} \cdot i+i-\left(i^{2}\right)^{7}}{(1-i)}=\frac{2 i+2 i^{2}}{1-i^{2}}=(i-1)$

Topic: Mathematics, Complex Number, NDA, NDA 2012 (Phase 2)
The value of the term independent of $\mathrm{x}$ in the expansion of
$\left(\mathrm{x}^{2}-\frac{1}{\mathrm{x}}\right)^{9}$ is:

Options:
A 9
B 18
C 48
D 84

The correct answer is:
84

Solution:
$\left(x^{2}-\frac{1}{x}\right)^{9}$
$$
\begin{array}{l}
\mathrm{t}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(\frac{-1}{\mathrm{x}}\right)^{\mathrm{r}} \\
{ }^{9} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{18-2 \mathrm{r}} \cdot(-1)^{\mathrm{r}} \cdot x^{-\mathrm{r}}
\end{array}
$$
$={ }^{9} C_{r}(x)^{18-3 r}(-1)^{r}$
Term will be independent of $\mathrm{x}$ when
$$
\begin{array}{l}
18-3 \mathrm{r}=0 \\
\mathrm{r}=6
\end{array}
$$
Put $\mathrm{r}=6$, in $[1]$
$$
\mathrm{t}_{7}={ }^{9} \mathrm{C}_{6}(-1)^{6}=\frac{9 !}{6 ! 3 !}=84
$$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2012 (Phase 1)
The value ofthe determinant $\left|\begin{array}{rr|l}\cos ^{2} \frac{\theta}{2} & \sin ^{2} \frac{\theta}{2} \\ \sin ^{2} \frac{\theta}{2} & \cos ^{2} \frac{\theta}{2}\end{array}\right|$ for all values

Options:
A 1
B $\cos \theta$
C $\sin \theta$
D $\cos 2 \theta$

The correct answer is:
$\cos \theta$

Solution:
$\left|\begin{array}{rr}\cos ^{2} \frac{\theta}{2} & \sin ^{2} \frac{\theta}{2} \\ \sin ^{2} \frac{\theta}{2} & \cos ^{2} \frac{\theta}{2}\end{array}\right|=\left(\cos ^{4} \frac{\theta}{2}-\sin ^{4} \frac{\theta}{2}\right)$
$=\left(\cos ^{2} \frac{\theta}{2}+\sin ^{2} \frac{\theta}{2}\right)\left(\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}\right)$
$=(1)\left(\cos 2\left(\frac{\theta}{2}\right)\right)$
$=\cos \theta .$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 2)
The variance of 20 observations is 5 . If each observation is multiplied by 3 , then what is the new variance of the resulting observations? $\quad$

Options:
A 5
B 10
C 15
D 45

The correct answer is:
45

Solution:
New variance $=\mathrm{k}^{2}$ (old variance) $=3^{2} \times 5$
$=9 \times 5=45$

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
The variance of 20 observations is $5 .$ If each observation multiplied by 2, then what is the new varianve ofrhe obseryation:

Options:
A 5
B 10
C 20
D 40

The correct answer is:
20

Solution:
Let $x_{1}, x_{2}, \ldots, x_{20}$ be the given observations. Given, $\frac{1}{20} \sum_{i=1}^{20}\left(x_{i}-\bar{x}\right)^{?}=5$
To find variance of $2 \mathrm{x}_{1}, 2 \mathrm{x}_{2}, 2 \mathrm{x}_{3}, \ldots ., 2 \mathrm{x}_{20}$
Let $\bar{x}$ denotes the mean of new observation,
Clenrly, $\bar{x}=\frac{\sum_{\mathrm{i}=1}^{20} 2 \mathrm{x}_{\mathrm{i}}}{20}=\frac{2 \sum_{\mathrm{i}=1}^{20} \mathrm{x}_{\mathrm{i}}}{20}=2 \overline{\mathrm{x}}$
Now, variance of new observation
$=\frac{1}{20} \sum_{i=1}^{20}\left(2 x_{i}-\bar{x}\right)^{2}=\frac{1}{20} \sum_{i=1}^{20}\left(2 x_{i}-2 \bar{x}\right)^{2}$
$=\frac{1}{20} \sum_{i=1}^{20} 4\left(x_{i}-\bar{x}\right)^{2}=4\left(\frac{1}{20} \sum_{i=1}^{20}\left(x_{i}-\bar{x}\right)^{2}\right)=4 \times 5=20$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 2)