datavorous/entrance-exam-dataset huggingface dataset:
If adj $A=\left[\begin{array}{cc}a & 0 \\ -1 & b\end{array}\right]$ and $a b \neq 0$, then what is the value of $\left|A^{-1}\right| ?$

Options:
A 1
B $a b$
C $1 / \sqrt{a b}$
D $1 / a b$

The correct answer is:
1

Solution:
For $2 \times 2$ matrix,
$|A|=|\operatorname{adj} A|$
$\quad=(a b-0)=a b$
$\therefore A^{-1}=\frac{a d j A}{|A|}=\frac{1}{a b} \cdot\left(\begin{array}{cc}a & 0 \\ -1 & b\end{array}\right)$
$\left|A^{-1}\right|=\frac{1}{a b}(a b)=1$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 2)
If \( A=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{I I}\right) \\ \tan ^{-1}\left(\frac{x}{I I}\right) & \cot ^{-1}(\Pi x)\end{array}\right] \) then \( A-B \) is

Options:
A \( \frac{3}{2} \)
B \( 0 \)
C \( 21 \)
D \( \frac{1}{2} \)

The correct answer is:
\( \frac{1}{2} \)

Solution:
Given that,
\( A=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(I X) & \tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{\Pi}\right) & \cot ^{-1}(\Pi X)\end{array}\right] \)
and
\( B=\left[\begin{array}{ll}-\cos ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{\Pi}\right) & -\tan ^{-1}(\Pi x)\end{array}\right] \)
So,
\( A-B=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(\Pi x)+\cos ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{\Pi}\right)-\tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{I I}\right)-\sin ^{-1}\left(\frac{x}{I I}\right) & \cot ^{-1}(\Pi x)+\tan ^{-1}(\Pi x)\end{array}\right] \)
Since, \( \sin ^{-1} x+\cos ^{-1} x=\frac{\Pi}{2} \) and \( \tan ^{-1} x+\cot ^{-1} x=\frac{I}{2} . \) So,
\( A-B=\frac{1}{\Pi}\left[\begin{array}{ccc}\frac{I I}{2} & 0 \\ 0 & \frac{\Pi}{2}\end{array}\right]=\frac{1}{\Pi} \times \frac{\Pi}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
\( =\frac{\Pi}{2} I \times \frac{1}{\Pi}=\frac{1}{2} I \)

Topic: Mathematics, Differential Equations, KCET, KCET 2017
If \( A=\left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] \), then \( A^{2} \) equal to

Options:
A \( \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
B \( \left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right] \)
C \( \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
D \( \left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right] \)

The correct answer is:
\( \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)

Solution:
Given that \( A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
Now, \( A \cdot A=A^{2}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \)
\( =\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Therefore \( A^{2}=I \)

Topic: Mathematics, Indefinite Integration, KCET, KCET 2015
If \(A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right], P=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(X=A P A^T\), then \(A^T X^{50} A=\)

Options:
A \(\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\)
B \(\left[\begin{array}{cc}2 & 1 \\ 0 & -1\end{array}\right]\)
C \(\left[\begin{array}{cc}25 & 1 \\ 1 & -25\end{array}\right]\)
D \(\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]\)

The correct answer is:
\(\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]\)

Solution:
Given matrix \(A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\) is orthogonal matrix, because \(A A^T=I\).
\(\begin{aligned}
& \text { So, } A^T X^{50} A=A^T X^{49}\left(A P A^T\right) A=A^T X^{49} A P\left(A^T A\right) \\
& =A^T X^{49} A P \\
& =A^T X^{48}\left(A P A^T\right) A P=A^T X^{48} A P^2 \ldots \ldots \\
& \ldots \ldots =A^T A P^{50}=I P^{50}=P^{50} \\
& \because \quad P=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \Rightarrow P^2=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& \Rightarrow \quad P^3=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \ldots \ldots \\
& \Rightarrow \quad P^{50}=\left[\begin{array}{cc}
1 & 50 \\
0 & 1
\end{array}\right] \\
\end{aligned}\)
So, \(\quad A^T X^{50} A=P^{50}=\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]\)
Hence, option (4) is correct.

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
If \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \), then \( A^{n}=2^{k} A \), where \( K= \)

Options:
A \( 2^{n-1} \)
B n \( +1 \)
C n-1
D \( 2(n-1) \)

The correct answer is:
\( 2(n-1) \)

Solution:
Given that \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
and \( A^{n}=2^{k} A \)
So, \( A^{2}=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( A^{3}=\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]=2^{2} A \)
\( =\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]=2^{4} A \)
\( A^{4}=\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}128 & -128 \\ -128 & 128\end{array}\right]=2^{6} A \)
So, \( k=2(n-1) \)

Topic: Mathematics, Matrices, KCET, KCET 2018
If \( A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \) then \( A^{2}-5 A \) is equal to

Options:
A I
B \( -I \)
C \( 7 I \)
D $-7 I$

The correct answer is:
$-7 I$

Solution:
Given that, \( A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \)
So, \( A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \)
And
\( 5 A=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right] \)
Therefore,
\[
\begin{array}{l}
A^{2}-5 A=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right] \\
=\left[\begin{array}{cc}
-7 & 0 \\
0 & -7
\end{array}\right]=-7 I
\end{array}
\]

Topic: Mathematics, Matrices, KCET, KCET 2016
If \(A=\left(\begin{array}{cc}3 & x-1 \\ 2 x+3 & x+2\end{array}\right)\) is a symmetric matrix, then the value of \(x\) is

Options:
A 4
B 3
C -4
D -3

The correct answer is:
-4

Solution:
Hints: \(A=A^T\)
\(\begin{aligned}
& \left(\begin{array}{cc}
3 & x-1 \\
2 x+3 & x+2
\end{array}\right)=\left(\begin{array}{cc}
3 & 2 x+3 \\
x-1 & x+2
\end{array}\right) \\
& \Rightarrow x-1=2 x+3 \text { or } x=-4
\end{aligned}\)

Topic: Mathematics, Matrices, WBJEE, WBJEE 2011
If \(A=\left[\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]\) and \(A^{-1}=\left[\begin{array}{cc}a+i b & -c-i d \\ -c+i d & a-i b\end{array}\right]\). Find \(\left(a^2+b^2+c^2+d^2\right)\).

Options:
A 1
B -1
C \(i\)
D \(-i\)

The correct answer is:
1

Solution:
Given matrix
\(A=\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]\)
So, \(\quad A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]\)
\(\begin{aligned}
& =\frac{1}{a^2+b^2+c^2+d^2}\left[\begin{array}{cc}
a-i b & -c-i d \\
c-i d & a+i b
\end{array}\right] \\
& =\left[\begin{array}{cc}
a+i b & -c-i d \\
-c+i d & a-i b
\end{array}\right] \quad \text{(given)}
\end{aligned}\)
\(\therefore \quad b=0\) and \(c=d=0\)
and \(a^2+b^2+c^2+d^2=1\)

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
If \(A=\left[\begin{array}{ccc}1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{array}\right]\), then \(A^{-1}=\)

Options:
A \(A^2-2 A-4\)
B \(A^2-A-3\)
C \(\frac{1}{2}\left[A^2+A+2 I\right]\)
D \(A^2+A-2 l\)

The correct answer is:
\(A^2-A-3\)

Solution:
Characteristic equation of any square matrix \(A\) is given by
\(\begin{aligned}
& |A-\lambda I|=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\lambda\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=0 \\
& \Rightarrow \quad\left[\begin{array}{ccc}
1 & 0 & -2 \\
-2 & -1 & 2 \\
3 & 4 & 1
\end{array}\right]-\left[\begin{array}{ccc}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{array}\right]=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
1-\lambda & 0 & -2 \\
-2 & -1-\lambda & 2 \\
3 & 4 & 1-\lambda
\end{array}\right|=0 \\
& \Rightarrow(1-\lambda)[-(1+\lambda)(1-\lambda)-8]-2[-8+3(\lambda+1)]=0 \\
& \Rightarrow \quad(1-\lambda)\left[-\left(1-\lambda^2\right)-8\right]-2(-8+3 \lambda+3)=0 \\
& \Rightarrow \quad(1-\lambda)\left(\lambda^2-9\right)-2(-5+3 \lambda)=0 \\
& \Rightarrow \quad \lambda^2-9-\lambda^3+9 \lambda-6 \lambda+10=0 \\
& \Rightarrow \quad-\lambda^3+\lambda^2+3 \lambda+1=0 \\
& \Rightarrow \quad \lambda^3-\lambda^2-3 \lambda-1=0
\end{aligned}\)
According to Caley Hamilton, every square matrix satisfies its characteristics equations.
\(\begin{aligned}
& \therefore \quad A^3-A^2-3 A-I=0 \\
& \Rightarrow \quad A^{-1}\left(A^3-A^2-3 A-I\right)=0 \\
& \Rightarrow \quad A^2-A-3 I-A^{-1}=0 \\
& \Rightarrow \quad A^{-1}=A^2-A-3 I
\end{aligned}\)

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
If \( A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right] \) and \( A+A^{T}=I \),
where \( \mathrm{I} \) is the unit matrix of \( 2 \times 2 \& \mathrm{~A}^{\mathrm{T}} \) is the transpose of \( \mathrm{A} \), then the value of \( \theta \) is equal to

Options:
A \( \frac{\pi}{6} \)
B \( \frac{\pi}{3} \)
C \( \pi \)
D \( \frac{3 \pi}{2} \)

The correct answer is:
\( \frac{\pi}{6} \)

Solution:
Given that, $\mathrm{A}=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$
So, $\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]$
$=2 \cos 2 \theta\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
Given that, $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{I}$ so
$\Rightarrow(2 \cos 2 \theta) I=I$
$\Rightarrow 2 \cos 2 \theta=1 \Rightarrow \cos 2 \theta=\frac{1}{2}$
$\Rightarrow 2 \theta=\frac{\Pi}{3} \Rightarrow \theta=\frac{\Pi}{6}$

Topic: Mathematics, Matrices, KCET, KCET 2016
If \( A=\left[\begin{array}{ll}1 & 3 \\ 4 & 2\end{array}\right], B=\left[\begin{array}{cc}2 & -1 \\ 1 & 2\end{array}\right] \) then \( \left|A B B^{\prime}\right|= \)

Options:
A \( -250 \)
B \( 250 \)
C \( 50 \)
D \( 100 \)

The correct answer is:
\( -250 \)

Solution:
(A)
\[
\begin{array}{l}
A=\left[\begin{array}{ll}
1 & 3 \\
4 & 2
\end{array}\right], B=\left[\begin{array}{cc}
2 & -1 \\
1 & 2
\end{array}\right] \\
B^{\prime}=\left[\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right] \\
\left|A B B^{\prime}\right|=|A||B|\left|B^{\prime}\right| \\
=(2-12) \times(4+1) \times(4+1) \\
=-10 \times 5 \times 5=-250
\end{array}
\]

Topic: Mathematics, Determinants, KCET, KCET 2019
If \(A=\left[\begin{array}{ll}3 & 2 \\ 0 & 1\end{array}\right]\), then \(\left(A^{-1}\right)^3\) is equal to

Options:
A \(\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]\)
B \(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & 27\end{array}\right]\)
C \(\frac{1}{27}\left[\begin{array}{ll}1 & -26 \\ 0 & -27\end{array}\right]\)
D \(\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \\ 0 & -27\end{array}\right]\)

The correct answer is:
\(\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]\)

Solution:
\(|\mathrm{A}|=\left|\begin{array}{ll}
3 & 2 \\
0 & 1
\end{array}\right|=3-0=3 \neq 0 \quad \therefore \mathrm{A}^{-1} \text { exists }\)
Adj. \(A=\left[\begin{array}{cc}1 & 0 \\ -2 & 3\end{array}\right]^{\prime}=\left[\begin{array}{cc}1 & -2 \\ 0 & 3\end{array}\right]\)
\(\therefore \quad \mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\)
\(\begin{aligned}
& \therefore\left(A^{-1}\right)^3=\frac{1}{27}\left[\begin{array}{cc}
1 & -2 \\
1 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right] \\
& =\frac{1}{27}\left[\begin{array}{cc}
1 & -8 \\
0 & 9
\end{array}\right]\left[\begin{array}{cc}
1 & -2 \\
0 & 3
\end{array}\right]=\frac{1}{27}\left[\begin{array}{cc}
1 & -26 \\
0 & 27
\end{array}\right]
\end{aligned}\)

Topic: Mathematics, Matrices, VITEEE, VITEEE 2022
If \(A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4\end{array}\right]\) and \(B=A^3\), then \(B^{-1}=\)

Options:
A \(\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -4\end{array}\right]\)
B \(\left[\begin{array}{ccc}-27 & 0 & 0 \\ 0 & -125 & 0 \\ 0 & 0 & -64\end{array}\right]\)
C \(\left[\begin{array}{ccc}\frac{1}{27} & 0 & 0 \\ 0 & \frac{1}{125} & 0 \\ 0 & 0 & \frac{1}{64}\end{array}\right]\)
D \(\left[\begin{array}{ccc}\frac{-1}{27} & 0 & 0 \\ 0 & \frac{-1}{125} & 0 \\ 0 & 0 & \frac{-1}{64}\end{array}\right]\)

The correct answer is:
\(\left[\begin{array}{ccc}\frac{1}{27} & 0 & 0 \\ 0 & \frac{1}{125} & 0 \\ 0 & 0 & \frac{1}{64}\end{array}\right]\)

Solution:
The given matrix \(A=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4\end{array}\right]\) and
\(\begin{aligned}
B & =A^3=\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right] \\
& =\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 25 & 0 \\
0 & 0 & 16
\end{array}\right]\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{ccc}
27 & 0 & 0 \\
0 & 125 & 0 \\
0 & 0 & 64
\end{array}\right] \\
\therefore & \\
B^{-1} & =\frac{1}{27 \times 125 \times 64}\left[\begin{array}{ccc}
125 \times 64 & 0 & 0 \\
0 & 27 \times 64 & 0 \\
0 & 0 & 527 \times 125
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\frac{1}{27} & 0 & 0 \\
0 & \frac{1}{125} & 0 \\
0 & 0 & \frac{1}{64}
\end{array}\right]
\end{aligned}\)
Hence, option (c) is correct.

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
If all permutations of the letters of the word 'LAGAN' are arranged as in dictionary, then what is therank of'NAAGL'?

Options:
A 48 th word
B 49 th word
C 50th word
D 51st word

The correct answer is:
49 th word

Solution:
Starting with the letter A and arranging the other four letters, there are 24 words. There are the first 24 words. Then starting with $\mathrm{G}$ that comes next in dictionaryorder and arranging $\mathrm{A}, \mathrm{A}, \mathrm{L}, \mathrm{N}$ in different ways, there are
$\frac{4 !}{2 !}=12$ words. Next the 37 th word starts with $\mathrm{L}$, that
comes next in dictionary order there are 12 words starting with $\mathrm{L}$. This accounts up to the 48 words. The 49 th word is 'NAAGL'

Topic: Mathematics, Permutation Combination, NDA, NDA 2006 (Phase 1)
If all the natural numbers between 1 and 20 are multiplied by 3 , then what is the variance of the resulting series?

Options:
A $99.75$
B $199.75$
C $299.25$
D $399.25$

The correct answer is:
$299.25$

Solution:
The series is $1,2,3, \ldots \ldots .20$ Variance $(\sigma)=\frac{\Sigma x^{2}}{n}-\Sigma(\bar{x})^{2}$
$=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^{2}$
$=\frac{(n+1)}{12}(n-1)$
$=\frac{n^{2}-1}{12}=\frac{(20)^{2}-1}{12}=\frac{399}{12}=\frac{133}{4}=33.25$
- Numbers are multiplied by 3,
wiance $(\sigma)=33.25 \times 9=299,25$

Topic: Mathematics, Statistics, NDA, NDA 2019 (Phase 1)
If all the seven letters of the word LEADING are permuted in all possible ways and the words thus formed are arranged as in the dictionary order, then the word in 2017th place is

Options:
A ELIGDAN
B ELNADGI
C ELINADG
D ELNDAGI

The correct answer is:
ELNADGI

Solution:
First we will arrange the letters of the word LEADING in the alphabetical order.

A, D, E, G, I, L, N

Number of words staring with letter A=6!=720

Number of words staring with letter D=6!=720

Number of words staring with letter E=6!=720

Total words are 720+720+720=2160>2017.

Hence, the required word will start from the letter E and before E we have total 720+720=1440 words.

Now,

Number of words staring with letters EA=5!=120

Number of words staring with letters ED=5!=120

Number of words staring with letters EG=5!=120

Number of words staring with letters EI=5!=120

Total words=1440+120+120+120+120=1920

Number of words staring with letters EL=5!=120

Total words

=1920+120=2040>2017

Hence, the required word will start from letters EL. And, words before EL are 1920.

Now,

Number of words staring with letters ELA=4!=24

Number of words staring with letters ELD=4!=24

Number of words staring with letters ELG=4!=24

Total words

=1920+24+24+24=1992

Now,

Number of words staring with letters ELI=4!=24

Now, total words=1992+24=2016

Number of words staring with letters ELN=4!=24

Total words =2016+24>2017

Hence, required word is ELNADGI.

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
If all the seven letters of the word LEADING are permuted in all possible ways and the words thus formed are arranged as in the dictionary order, then the word in $2017^{\text {th }}$ place is

Options:
A ELIGDAN
B ELNADGI
C ELINADG
D ELNDAGI

The correct answer is:
ELNADGI

Solution:
When letters of LEADING are arranged in ascending order then letters are A, D, E, G, I, L, N Now, words starts with $\mathrm{A}$ is $6 !=720$, ie. $A$........
Similarly, words starts with $D=6 !=720$
Now, words starts with $E$ is also $6 !=720$.
But total number is 2160 that is more than 2017th place. Hence, word will start with $E$.
Now, words starts with
$$
\begin{aligned}
& E A \ldots \ldots .=5 !=120 \\
& E D \ldots \ldots . .=5 !=120 \\
& E G \ldots \ldots . .=5 !=120 \\
& E I \ldots \ldots . .=5 !=120
\end{aligned}
$$
Here, total words till now is $720+480=1920$
Now, words start with
$$
\begin{aligned}
& \text { ELA } \ldots \ldots \ldots=4 !=24 \\
& E L D \ldots \ldots .=4 !=24 \\
& E L G \ldots \ldots .=4 !=24 \\
& E L I \ldots \ldots . .=4 !=24
\end{aligned}
$$
Now, total words is 2016.
Next word is 2017th word and the word is ELNADGI.

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 1)
If AM of numbers $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots . \mathrm{x}_{\mathrm{n}}$ is $\mu$, then what is the $\mathrm{AM}$ of the numbers which are increased by $1,2,3, \ldots \mathrm{n}$ respectively?

Options:
A $\mu+\left(\frac{\mathrm{n}+1}{2}\right) \quad$
B $\mu \quad[2007-I I]$
C $\mu+\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
D $\mu-\left(\frac{\mathrm{n}+1)}{2}\right)$

The correct answer is:
$\mu+\left(\frac{\mathrm{n}+1}{2}\right) \quad$

Solution:
Since, AM of number $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots \mathrm{x}_{\mathrm{n}}$ is $\mu$
$\mathrm{n} \mu=\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots \mathrm{x}_{\mathrm{n}}$
Sum of new numbers
$=\left(x_{1}+1\right)+\left(x_{2}+2\right)+\left(x_{3}+3\right)+\ldots+\left(x_{n}+n\right)$
$=\left(x_{1}+x_{2}+\ldots+x_{n}\right)+(1+2+3+\ldots+n)$
$=n \mu+\frac{n(n+1)}{2}$
$\mathrm{AM}=\mu+\frac{(\mathrm{n}+1)}{2}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
If an angle $\mathrm{B}$ is complement of an angle $\mathrm{A}$, what are the greatest and least values of $\cos A$ cos $B$ respectively ?

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

The correct answer is:
\frac{1}{2},-\frac{1}{2}

Solution:
Since, A and B are complementary angles, then A $+\mathrm{B}=90^{\circ}$
Now, $\cos A \cos B=\cos A \cos \left(90^{\circ}-\mathrm{A}\right)$
$=\cos \mathrm{A} \sin \mathrm{A}=\frac{1}{2} \sin 2 \mathrm{~A}$
Since, $-1 \leq \sin 2 \mathrm{~A} \leq 1$
Hence, $-\frac{1}{2} \leq \frac{1}{2} \sin 2 \mathrm{~A} \leq \frac{1}{2}$.
Thus, greatest and least values of $\cos \mathrm{A} \cos \mathrm{B}$ are $\frac{1}{2}$ and $-\frac{1}{2}$.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
If an angle $\alpha$ is divided into two parts $A$ and $B$ such that $A-$ $B=x$ and $\tan A: \tan B=2: 1$, then what is $\sin x$ equal to?

Options:
A 3 \sin \alpha
B (2 \sin \alpha) / 3 & \text
C $(\sin \alpha) / 3$
D $2 \sin \alpha$

The correct answer is:
$(\sin \alpha) / 3$

Solution:
Since, angle $\alpha$ is divided into two parts $\mathrm{A}$ and $\mathrm{B}$. $\therefore \quad \alpha=\mathrm{A}+\mathrm{B}$
and $x=\mathrm{A}-\mathrm{B}$ (given)
On solving (1) and (2) we get,
$\mathrm{A}=\frac{\alpha+x}{2}, \mathrm{~B}=\frac{\alpha-x}{2}$
Now, consider $\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{2}{1}$
$\Rightarrow \frac{\tan \left(\frac{\alpha+x}{2}\right)}{\tan \left(\frac{\alpha-x}{2}\right)}=\frac{2}{1}$
$\Rightarrow \frac{\sin \left(\frac{\alpha+x}{2}\right) \cos \left(\frac{\alpha-x}{2}\right)}{\cos \left(\frac{\alpha+x}{2}\right) \sin \left(\frac{\alpha-x}{2}\right)}=\frac{2}{1}$
Multiply and divide by 2 ,
$\Rightarrow \frac{2 \sin \left(\frac{\alpha+x}{2}\right) \cos \left(\frac{\alpha-x}{2}\right)}{2 \cos \left(\frac{\alpha+x}{2}\right) \sin \left(\frac{\alpha-x}{2}\right)}=2$
$\Rightarrow \frac{\sin \alpha+\sin x}{\sin \alpha-\sin x}=2$
$\Rightarrow \sin \alpha+\sin x=2 \sin \alpha-2 \sin x$
$\Rightarrow 3 \sin x=\sin \alpha$
$\Rightarrow \sin x=\frac{\sin \alpha}{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 1)
If an individual wants to view diakinesis, which of these would be suitable?

Options:
A
B
C
D

The correct answer is:

Solution:
Flower buds are the reproductive structures that perform meiosis. Onion flower buds are commonly used to observe all the phases of meiosis with acetocarmine or acetoorcein staining method.

Topic: Biology, Cell Cycle and Cell Division, NEET
If an infinite GP has the first term $\mathrm{x}$ and the sum 5 , then which of the following is correct?

Options:
A $x < -10$
B $-10 < x < 0$
C $0 < x < 10$
D $x>10$

The correct answer is:
$0 < x < 10$

Solution:
Sum of an infinite G.P. with first term $x$ and common ratio $r( < 1)$ is
$$
s=\frac{x}{1-r}
$$
From question $s=5$
$$
\text { then, } \begin{aligned}
5 &=\frac{x}{1-r} \\
1-r &=\frac{x}{5} \\
r &=1-\frac{x}{5} \\
\text { For, }-1 & < 1-\frac{x}{5} < 1 \\
-2 & < -\frac{x}{5} < 0 \\
0 & < x < 10
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
If $\alpha$ and $\beta(\neq 0)$ are the rootsof the quadratice quation $x^{2}+\alpha x$ $-\beta=0$, then the quadratic expression $-x^{2}+\alpha x+\beta$ where $x \in$ $\begin{array}{ll}\text { R has }\end{array}$

Options:
A Least value $-\frac{1}{4}$
B Least value $-\frac{9}{4}$
C Greatest value $\frac{1}{4}$
D Greatest value $\frac{9}{4}$

The correct answer is:
Greatest value $\frac{9}{4}$

Solution:
$\alpha+\beta=-\alpha, \alpha \beta=-\beta$
$\Rightarrow \alpha \beta+\beta=0$
$\Rightarrow(\alpha+1) \beta=0$
$\Rightarrow \alpha=-1(\beta \neq 0)$
$\Rightarrow(2 \alpha+\beta)=0$
$\Rightarrow \beta=2$
$\therefore-x^{2}+\alpha x+\beta=-x^{2}-x+2$
Greatest value $=-\frac{1+8}{-4}=\frac{9}{4}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 2)
If $\alpha$ and $\beta$ are different complex numbers with $|\alpha|=1$, then
what is $\left|\frac{\alpha-\beta}{1-\alpha \beta}\right|$ equal to?

Options:
A $|\beta|$
B 2
C 1
D 0

The correct answer is:
1

Solution:
We know, it $|\alpha|=1$ $\Rightarrow \alpha \bar{\alpha}=1$
$\therefore\left|\frac{\alpha-\beta}{1-\alpha \bar{\beta}}\right|=\left|\frac{\alpha-\beta}{\alpha \cdot \bar{\alpha}-\alpha \bar{\beta}}\right| \quad($ from $(1))$
$=\left|\frac{\alpha-\beta}{\alpha(\bar{\alpha}-\bar{\beta})}\right|$
$=\frac{|\alpha-\beta|}{|\alpha||\alpha / \beta|} \quad(\because$ sin ce $|\bar{z}|=|z|)$
$=\frac{1}{|\alpha|}=\frac{1}{1}=1$

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)