datavorous/entrance-exam-dataset huggingface dataset:
If $\int_{\ln 2}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-1\right)^{-1} \mathrm{dx}=\ln \frac{3}{2}$, then what is the value of $\mathrm{x}$ ?

Options:
A $\mathrm{e}^{2}$
B $\frac{1}{\mathrm{e}}$
C $\ln 4$
D 1

The correct answer is:
$\ln 4$

Solution:
Let $\mathrm{I}=\int_{\ell \mathrm{n} 2}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-1\right)^{-1} \mathrm{dx}$
$=\int_{\ell^{n} 2}^{x} \frac{1}{e^{x}-1} d x$
Put $\mathrm{e}^{x}-1=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}}=\mathrm{t}+1$
$\mathrm{e}^{x} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{e}^{\mathrm{x}}}$
or, $\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{t}+1}$
when $x=\ell \mathrm{n} 2, \mathrm{t}=\mathrm{e}^{\ln 2}-1=2-1=1$
and $\mathrm{I}=\int_{1}^{\mathrm{t}} \frac{1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$
breaking into partial fractions.
$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$
and $\mathrm{I}=\int_{1}^{\mathrm{t}}\left(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}+1}\right) \mathrm{dt}=\left[\log _{\mathrm{e}}^{\mathrm{t}}-\log _{\mathrm{e}}(\mathrm{t}+1)\right]_{1}^{\mathrm{t}}$
or $\mathrm{I}=\left[\log _{\mathrm{e}} \frac{\mathrm{t}}{\mathrm{t}+1}\right]_{1}^{\mathrm{t}}=\log _{\mathrm{e}} \frac{\mathrm{t}}{\mathrm{t}+1}-\log _{\mathrm{e}} \frac{1}{2}$
$-\log _{\mathrm{e}} \frac{2 \mathrm{t}}{\mathrm{t}+1}=\log _{\mathrm{e}} \frac{3}{2}$
$\left[\right.$ Since, $\left.\int_{\ell \mathrm{n} 2}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-1\right)^{-1} \mathrm{dx}=\log _{\mathrm{e}} \frac{3}{2}\right]$
So, $\frac{2 t}{t+1}=\frac{3}{2}$
or, $4 \mathrm{t}=3 \mathrm{t}+3 \Rightarrow \mathrm{t}=3$
So, $\mathrm{e}^{\mathrm{x}}-1=3, \mathrm{e}^{\mathrm{x}}=4 \Rightarrow \mathrm{x}=\ln 4$

Topic: Mathematics, Definite Integration, NDA, NDA 2007 (Phase 1)
If \( \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right] \), then the values of \( \mathrm{x} \) and \( \mathrm{y} \) respectively are

Options:
A \( -3,-1 \)
B \( 1,3 \)
C \( 3,1 \)
D \( -1,3 \)

The correct answer is:
\( -1,3 \)

Solution:
Given that, \( \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right] \)
\( \left[\begin{array}{c}x+y \\ -x+y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right] \)
So, \( x+y=2 \rightarrow \) (1)
and \( -x+y=4 \rightarrow(2) \)
From Eqs. (1) and (2), we have \( x=-1, y=3 \)

Topic: Mathematics, Matrices, KCET, KCET 2018
If \(\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is

Options:
A 2
B 3
C 0
D 1

The correct answer is:
1

Solution:
It is given that matrix \(A=\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) have no inverse, so
\(\begin{array}{rlrl}
& |A|=0 \Rightarrow\left|\begin{array}{ccc}
1 & -1 & x \\
1 & x & 1 \\
x & -1 & 1
\end{array}\right| & =0 \\
\Rightarrow \quad 1(x+1)+1(1-x)+x\left(-1-x^2\right) & =0 \\
\Rightarrow \quad x+1+1-x-x-x^3 & =0 \\
\Rightarrow x^3+x-2=0 \Rightarrow(x-1)\left(x^2+x+2\right) & =0
\end{array}\)
Either \(x=1\) or \(x^2+x+2=0\)
But discriminant of quadratic equation \(x^2+x+2=0\) is negative so no real roots.
\(\therefore|A|=0 \Rightarrow x=1\)
Hence, option (d) is correct

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
If \( \left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0 \), then the area of the triangle whose vertices are
\[
\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right),\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \text { is }
\]

Options:
A \( \frac{1}{4} \) abo
B \( \frac{1}{8} a b d \)
C \( \frac{1}{4} \)
D \( \frac{1}{8} \)

The correct answer is:
\( \frac{1}{8} \)

Solution:
Given vertices of triangle \( \mathrm{ABC} \) are
\( A\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), B\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), C\left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right) \)
Now, \( \left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \rightarrow(1) \)
Taking L.H.S. of the Eq. (1) we have
\[
\left|\begin{array}{lll}
2 a & x_{1} & y_{1} \\
2 h & x_{2} & y_{2} \\
2 c & x_{3} & y_{3}
\end{array}\right|
\]
Now \( R_{1} \rightarrow \frac{R_{1}}{a}, R_{2} \rightarrow \frac{R_{2}}{a}, R_{3} \rightarrow \frac{R_{3}}{a} \) then \( C_{1} \rightarrow \frac{C_{1}}{2} \) we have
\( \left|\begin{array}{ccc}2 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 2 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 2 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array}\right| \begin{array}{ccc}1 & \frac{x_{1}}{a} & \frac{y_{1}}{a} \\ 1 & \frac{x_{2}}{b} & \frac{y_{2}}{b} \\ 1 & \frac{x_{3}}{c} & \frac{y_{3}}{c}\end{array} \mid \)
We know that area od triangle is given by
\( 1 \frac{x_{1}}{a} \frac{y_{1}}{a} \)

Topic: Mathematics, Basic of Mathematics, KCET, KCET 2015
If \(\left|\begin{array}{ccc}3^2+k & 4^2 & 3^2+3+k \\ 4^2+k & 5^2 & 4^2+4+k \\ 5^2+k & 6^2 & 5^2+5+k\end{array}\right|=0\), then the value of \(k\) is

Options:
A 2
B 1
C -1
D 0

The correct answer is:
1

Solution:
Breaking the given determinant into two determinants, we get
\(\begin{aligned}
& \left|\begin{array}{lll}
3^2+\mathrm{k} & 4^2 & 3^2+\mathrm{k} \\
4^2+\mathrm{k} & 5^2 & 4^2+\mathrm{k} \\
5^2+\mathrm{k} & 6^2 & 5^2+\mathrm{k}
\end{array}\right|+\left|\begin{array}{lll}
3^2+\mathrm{k} & 4^2 & 3 \\
4^2+\mathrm{k} & 5^2 & 4 \\
5^2+\mathrm{k} & 6^2 & 5
\end{array}\right|=0 \\
& \Rightarrow 0+\left|\begin{array}{ccc}
9+\mathrm{k} & 16 & 3 \\
7 & 9 & 1 \\
9 & 11 & 1
\end{array}\right|=0
\end{aligned}\)
[Applying \(R_3-R_2\) and \(R_2-R_1\) in second det.]
\(\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
9+\mathrm{k} & 16 & 3 \\
7 & 9 & 1 \\
2 & 2 & 0
\end{array}\right|=0\left[\text { Applying } \mathrm{R}_3-\mathrm{R}_2\right] \\
& \left.\Rightarrow\left|\begin{array}{ccc}
9+\mathrm{k} & 7-\mathrm{k} & 3 \\
7 & 2 & 1 \\
2 & 0 & 0
\end{array}\right|=0 \text { [Applying } \mathrm{C}_2-\mathrm{C}_1\right] \\
& \Rightarrow 2(7-\mathrm{k}-6)=0 \Rightarrow \mathrm{k}=1
\end{aligned}\)

Topic: Mathematics, Determinants, VITEEE, VITEEE 2023
If $\left|\begin{array}{ccc}x^2+x $x+1 $x-2$ 2 x^2+3 x-1 $3 x $3 x-3$ x^2+2 x+3 $2 x-1 $2 x-1\end{array}\right|=x A+B$, where $A$ and $B$ are determinants of order 3 not involving $x$, then $|A|=$

Options:
A 27
B 24
C 19
D $-8$

The correct answer is:
24

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 2)
If \( \left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right| \) then \( x \) is equal to

Options:
A \( 2 \)
B \( 04 \)
C \( 08 \)
D \( \pm 2 \sqrt{2} \)

The correct answer is:
\( \pm 2 \sqrt{2} \)

Solution:
Given that, \( \left|\begin{array}{cc}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right| \)
\( 3-x^{2}=3-8 \)
\( \Rightarrow x^{2}=8 \Rightarrow x=\pm 2 \sqrt{2} \)

Topic: Mathematics, Probability, KCET, KCET 2017
If $\lim _{x \rightarrow 0} \phi(x)=a^{2}$, where $a \neq 0$, then what is $\lim _{x \rightarrow 0} \phi\left(\frac{x}{a}\right)$ equal to?

Options:
A $\mathrm{a}^{2}$
B $\mathrm{a}^{-2}$
C $-\mathrm{a}^{2}$
D -a

The correct answer is:
$\mathrm{a}^{2}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \phi(x)=a^{2} \\ & \Rightarrow \lim _{x \rightarrow 0} \phi\left(\frac{x}{a}\right)=a^{2} \end{aligned}$
$a \neq 0$
[because function value is constant]

Topic: Mathematics, Limits, NDA, NDA 2016 (Phase 1)
If $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x}{x}=$ and $\lim _{x \rightarrow \infty} \frac{\cos x}{x}=m$, then which one of the
following is correct?

Options:
A $\quad l=1, \mathrm{~m}=1$
B $l=\frac{2}{\pi}, \mathrm{m}=\infty$
C $l=\frac{2}{\pi}, \mathrm{m}=0$
D $l=1, \mathrm{~m}=\infty$

The correct answer is:
$l=\frac{2}{\pi}, \mathrm{m}=0$

Solution:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x}{x}=l$ and $\lim _{x \rightarrow \infty} \frac{\cos x}{x}=m$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x}{x}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi} ; \lim _{x \rightarrow \infty} \frac{\cos x}{x}=0$
$\therefore l=\frac{2}{\pi}, \mathrm{m}=0$

Topic: Mathematics, Limits, NDA, NDA 2017 (Phase 2)
If $\lim _{\mathrm{x} \rightarrow \mathrm{a}}\left[\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right]$ exists, then which one of the following correct $?$

Options:
A Both $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ must exist
B $\lim _{x \rightarrow a} f(x)$ need not exist but $\lim _{x \rightarrow a} g(x)$ must exist
C Both $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ need not exist
D $\lim _{x \rightarrow a} f(x)$ must exist but $\lim _{x \rightarrow a} g(x)$ need not exist

The correct answer is:
Both $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ must exist

Solution:
For $\lim _{x \rightarrow a}\left[\frac{f(x)}{g(x)}\right]$ to exist, then both $\lim _{x \rightarrow a} f(x)$ and
$\lim _{x \rightarrow a} g(x)$ must exist.

Topic: Mathematics, Limits, NDA, NDA 2006 (Phase 2)
If lines represented by equation px2-qy2=0 are distinct then

Options:
A pq>0
B pq<0
C pq=0
D p+q=0

The correct answer is:
pq>0

Solution:
px2-qy2=0
Comparing with standard homogeneous equation we get,
a=p, b=-q, h=0
Lines are real and distinct if h2-ab>0
â 0+pq>0
pq>0

Topic: Mathematics, Pair of Lines, MHT CET, MHT CET 2017
If $\log _{3}\left[\log _{3}\left[\log _{3} x\right]\right]=\log _{3} 3$, then what is the value of $x$ ?

Options:
A 3
B $3^{9}$
C 27
D $3^{27}$

The correct answer is:
$3^{27}$

Solution:
Consider $\log _{3}\left[\log _{3}\left[\log _{3} x\right]\right]=\log _{3} 3$
$\Rightarrow \log _{3}\left[\log _{3} x\right]$
$\Rightarrow \log _{3} x=$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
$\mathrm{If}\left(\log _{3} \mathrm{x}\right)^{2}+\log _{3} \mathrm{x} < 2$, then which one of the following is
correct?

Options:
A $0 < \mathrm{x} < \frac{1}{9}$
B $\frac{1}{9} < \mathrm{x} < 3$
C $3 < x < \infty$
D $\frac{1}{9} \leq x \leq 3$

The correct answer is:
$\frac{1}{9} < \mathrm{x} < 3$

Solution:
Given equation is $\left(\log _{3} x\right)^{2}+\log _{3} x < 2$
$\Rightarrow\left(\log _{3} x\right)^{2}+\left(\log _{3} x\right)-2 < 0$
$\Rightarrow\left(\log _{3} x+2\right)\left(\log _{3} x-1\right) < 0$
$-2 < \log _{3} x < $
$\Rightarrow \log _{3} 3^{-2} < \log _{3} x < \log _{3} 3$
$\Rightarrow \frac{1}{9} < \mathrm{x} < 3$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2006 (Phase 1)
If $\left(\log _{3} x\right)\left(\log _{x} 2 x\right)\left(\log _{2 x} y\right)=\log _{x} x^{2}$, then what is $y$ equal
to?

Options:
A $4.5$
B 9
C 27
D 18

The correct answer is:
9

Solution:
$\left(\log _{3} x\right)\left(\log _{x} 2 x\right)\left(\log _{2 x} y\right)=\log _{x} x^{2}$
$\frac{\log x}{\log 3} \times \frac{\log 2 x}{\log x} \times \frac{\log y}{\log 2 x}=\frac{\log x^{2}}{\log x}$
$\frac{\log y}{\log 3}=\frac{2 \log x}{\log x}$
$\log y=2 \log 3$
$\log y=\log 9$
$y=9$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
If $\log _{8} \mathrm{~m}+\log _{8} \frac{1}{6}=\frac{2}{3}$, then $\mathrm{m}$ is equal to

Options:
A 24
B 18
C 12
D 4

The correct answer is:
24

Solution:
$\log _{8} \mathrm{~m}+\log _{8} \frac{1}{6}=\frac{2}{3}$
$\Rightarrow \quad \log _{8}\left(\mathrm{~m}, \frac{1}{6}\right)=\frac{2}{3}$
$\Rightarrow(8)^{\frac{2}{3}}=\mathrm{m} \cdot \frac{1}{6}$
$\Rightarrow \mathrm{m}=24$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 2)
If $\log _{10} 2, \log _{10}\left(2^{x}-1\right)$ and $\log _{10}\left(2^{x}+3\right)$ are three
consecutive terms of an A.P, then the value of $x$

Options:
A 1
B $\log _{5} 2$
C $\log _{2}$
D $\log _{10} 5$

The correct answer is:
$\log _{2}$

Solution:
$\log _{10} 2, \log _{10}\left(2^{x}-1\right)$ and $\log _{10}\left(2^{x}+3\right)$ are in A.P.
$\log _{10}\left(2^{x}-1\right)-\log _{10} 2=\log \left(2^{x}+3\right)-\log _{10}\left(2^{x}-1\right)$
$\log _{10}\left(\frac{2^{\mathrm{x}}-1}{2}\right)=\log _{10}\left(\frac{2^{\mathrm{x}}+3}{2^{\mathrm{x}}-1}\right)$
$\Rightarrow \frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1}$
Let $2^{\mathrm{x}}=\mathrm{y}$, then
$y^{2}-4 y-5=0$
$y^{2}-5 y+y-5=0$
$y(y-5)+1(y-5)=0$
$y=-1, y=5$
Therefore, $2^{\mathrm{x}}=5$ $\mathrm{x}=\log _{2} 5$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
If $\log _{10} 2, \log _{10}\left(2^{x}-1\right), \log _{10}\left(2^{x}+3\right)$ are three
consecutive terms of an AP, then which one of the following is correct?

Options:
A $x=0$
B $x=1$
C $x=\log _{2} 5$
D $x=\log _{5} 2$

The correct answer is:
$x=\log _{2} 5$

Solution:
Let $\log _{10} 2, \log _{10}\left(2^{x}-1\right)$ and $\log _{10}\left(2^{x}+3\right)$ are
in A.P
$\therefore 2 \log _{10}\left(2^{x}-1\right)=\log _{10} 2+\log _{10}\left(2^{x}+3\right)$
$\Rightarrow \log _{10}\left(2^{x}-1\right)^{2}=\log _{10} 2\left(2^{x}+3\right)$
$\Rightarrow 2^{2 x}+1-2^{x+1}=2.2^{x}+6$
$\Rightarrow a^{2}+1-2 a=2 a+6$ where $a=2^{x} .$
$\Rightarrow a^{2}-4 a-5=0$
$\Rightarrow a=5$ or $a=-1 \quad 2^{x}=5 \Rightarrow x \log _{2}=\log 5$
$\Rightarrow x=\frac{\log 5}{\log 2} \Rightarrow x=\log _{2} 5$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
If $\log _{10}(x+1)+\log _{10} 5=3$, then what is the value of $x$ ?

Options:
A 199
B 200
C 290
D 300

The correct answer is:
199

Solution:
Let $\log _{10}(x+1)+\log _{10} 5=3$
$\Rightarrow \log _{10} 5(x+1)=3 \quad(\because \log m+\log n=\log m n)$
초 $5(x+1)=10$
$\Rightarrow(x+1)=\frac{1000}{5}=200$
$\Rightarrow x=200-1=199$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
If $\log _{a}(a b)=x$, then what is $\log _{h}(a b)$ equal to?

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

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

Solution:
$\log _{a} a b=x$
$\log _{a} a+\log _{a} b=x$
$\frac{1}{\log _{b} a}=x-1$
$\log _{b} a=\frac{1}{x-1}$...(i)
$\log _{b} a b=\log _{b} a+\log _{b} b$
$\quad=\frac{1}{x-1}+1 \quad($ From $(1))$
$=\frac{1+x-1}{x-1}$
$\log _{b} a b=\frac{x}{x-1}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 1)
If $\log _{k} x \log _{5} k=3$, then what is $x$ equal to?

Options:
A $k^{5}$
B $5 k^{3}$
C $24^{2}$
D 125

The correct answer is:
125

Solution:
$\Rightarrow \log y=\log 3^{2} \Rightarrow y=3^{2} \Rightarrow y=9$
Given, $\log _{5} k \log _{k} x=3$
$\frac{\log k}{\log 5} \cdot \frac{\log x}{\log k}=3 \Rightarrow \frac{\log x}{\log 5}=3$
$\Rightarrow \log x=3 \log 5 \Rightarrow \log x=\log 5^{3}$
$\Rightarrow x=5^{3} \Rightarrow x=125$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $\left(\log _{x} x\right)\left(\log _{3} 2 x\right)\left(\log _{2 x} y\right)=\log _{x} x^{2}$, then what is the value
of $y$ ?

Options:
A $9 / 2$
B 9
C 27
D 18

The correct answer is:
9

Solution:
$\left(\log _{x} x\right)\left(\log _{3} 2 x\right)\left(\log _{2 x} y\right)=\log _{x} x^{2}$
$\left.\Rightarrow 1\left(\log _{3} 2 x\right)\left(\log _{2 x} y\right)=2\left(\because \log _{x} x^{2}\right)=2 \log _{x} x\right)$
$\Rightarrow\left(\frac{\log 2 x}{\log 3}\right)\left(\frac{\log y}{\log 2 x}\right)=2$
$\Rightarrow \frac{\log y}{\log 3}=2 \Rightarrow \log y=2 \log 3$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $\mathrm{m}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $\mathrm{n}=\left[\begin{array}{ll}0 & 1 \\ -1 & 0\end{array}\right]$, then what is the value of the determinant of $m \cos \theta-n \sin \theta ?$

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

The correct answer is:
1

Solution:
$\mathrm{m} \cos \theta-\mathrm{n} \sin \theta=\left[\begin{array}{cc}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\right]-\left[\begin{array}{cc}0 & \sin \theta \\ -\sin \theta & 0\end{array}\right]$
$=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
det. $(\mathrm{m} \cos \theta-\mathrm{n} \sin \theta)=\cos ^{2} \theta-\left(-\sin ^{2} \theta\right)$
$=\cos ^{2} \theta+\sin ^{2} \theta=1$

Topic: Mathematics, Matrices, NDA, NDA 2016 (Phase 2)
If $\mathrm{m}$ and $\mathrm{n}$ are integers, then what is the value of
$\int_{0}^{\pi} \sin m x \sin n x d x$, if $m \neq n ?$

Options:
A 0
B $\frac{1}{m+n}$
C $\frac{1}{\mathrm{~m}-\mathrm{n}} \quad$
D $\mathrm{m} \quad\mathrm{}$

The correct answer is:
0

Solution:
The given integral
$\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=\int_{0}^{\pi} \sin m(\pi-x) \cdot \sin n(\pi-x) d x$
$=\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
So, $\int_{0}^{\pi} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \sin m x \cdot \sin n x d x$
$=2 \int_{0}^{\pi / 2} \frac{1}{2}[\cos (m x-n x)-\cos (m x+n x)] d x$
$=\int_{0}^{\pi / 2}[\cos (m-n) x-\cos (m+n) x] d x$
$=\left[\frac{\sin (m-n) x}{m-n}-\frac{\sin (m+n) x}{m+n}\right]_{0}^{\pi / 2}=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2006 (Phase 1)
If $m$ and $n$ are the roots of the equation $(x+p)(x+q)-k=0$, then the roots of the equation $(x-m)(x-n)+k=0$ are

Options:
A $P$ and $q$
B $\frac{1}{n}$ and $\frac{1}{ }$
C $-p$ and $-q$
D $p+q$ and $p-q$

The correct answer is:
$-p$ and $-q$

Solution:
Here $\mathrm{m}$ and $\mathrm{n}$ are the roots of equation. $(\mathrm{x}+\mathrm{p})(\mathrm{x}+\mathrm{q})-\mathrm{k}=0$
$\mathrm{x}^{2}+\mathrm{x}(\mathrm{p}+\mathrm{q})+\mathrm{p} \mathrm{q}-\mathrm{k}=0$...(i)
If $\mathrm{m}$ and $\mathrm{n}$ are the roots of equation, then $(\mathrm{x}-\mathrm{m})(\mathrm{x}-\mathrm{n})=0$
$\mathrm{x}^{2}-(\mathrm{m}+\mathrm{n}) \mathrm{x}+\mathrm{mn}=0$...(ii)
Now equation (i) should be equal to equation (ii), $(m+n)=-(p+q) a n d m n=p q-k$
Now, we have to find roots of $(x-m)(x-n)+k=0$ $x^{2}-(m+n) x+m n+k=0$
$x^{2}+(p+q) x+(p q-k)+k=0$
$\mathrm{x}^{2}+(\mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{pq}=0$
$\mathrm{x}^{2}+\mathrm{px}+\mathrm{qx}+\mathrm{pq}=0$
$x(x+p)+q(x+p)=0$
$x+q=0$ or $x+p=0$
$\mathrm{x}=-\mathrm{q}$ and $\mathrm{x}=-\mathrm{p}$
Option $(\mathrm{c})$ is correct

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 1)
If $\mathrm{m}$ is the geometric mean of
$\left(\frac{y}{z}\right)^{\log (y z)},\left(\frac{z}{x}\right)^{\log (z x)}$ and $\left(\frac{x}{y}\right)^{\log (x y)}$
then what is the value of $\mathrm{m}$ ?

Options:
A 1
B 3
C 6
D 9

The correct answer is:
1

Solution:
Three terms are
$G_{1}=\left(\frac{y}{z}\right)^{\log y z} G_{2}=\left(\frac{z}{x}\right)^{\log z x} G_{3}=\left(\frac{x}{y}\right)^{\log x y}$
Geometric mean of three terms is
$m=\sqrt[3]{G_{1} G_{2} G_{3}}$ ...(i)
$\begin{aligned} G_{1} G_{2} G_{3}=&\left(\frac{y}{z}\right)^{\log y z} \cdot\left(\frac{z}{x}\right)^{\log z x} \cdot\left(\frac{x}{y}\right)^{\log x y} \\ &=\frac{y^{\log y} \cdot y^{\log z}}{z^{\log y} \cdot z^{\log z}} \times \frac{z^{\log z} \cdot z^{\log x}}{x^{\log z} \cdot x^{\log x}} \times \frac{x^{\log x} \cdot x^{\log y}}{y^{\log x} \cdot y^{\log y}} \\ &=\left(\frac{y}{x}\right)^{\log z} \cdot\left(\frac{z}{y}\right)^{\log x} \cdot\left(\frac{x}{z}\right)^{\log y} \end{aligned}$
Taking log both sides $\begin{aligned} \log G_{1} G_{2} G_{3}=& \log \left[\left(\frac{y}{x}\right)^{\log z}\right]+\log \left[\left(\frac{z}{y}\right)^{\log x}\right] \\ &+\log \left[\left(\frac{x}{z}\right)^{\log y}\right] \\=& \log z \log y-\log z \log x+\log x \log z \\ &-\log x \log y+\log y \log x-\log y \log z \end{aligned}$
$\log G_{1} G_{2} G_{3}=0$
$G_{1} G_{2} G_{3}=\mathrm{e}^{0}=1$
Hence $m=\sqrt[3]{G_{1} G_{2} G_{3}}=(1)^{\frac{1}{3}}$
$m=1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 1)
If $\mathrm{m}, \mathrm{n}, \mathrm{r}$ are three vectors, $\theta$ is the angle between the
vectors $\overrightarrow{\mathrm{m}}$ and $\overrightarrow{\mathrm{n}}$, what is $\mathrm{mnr} \cos \theta$ equal to ?

Options:
A $(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}})(\overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{r}} / \mathrm{r}))$
B $(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}})(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{r}})$
C $(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{r}})(\overrightarrow{\mathrm{n}} \cdot(\overrightarrow{\mathrm{n}} / \mathrm{n}))$
D $(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}) \overrightarrow{\mathrm{r}}$

The correct answer is:
$(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}) \overrightarrow{\mathrm{r}}$

Solution:
Given that $\overrightarrow{\mathrm{m}}, \overrightarrow{\mathrm{n}}$ and $\overrightarrow{\mathrm{r}}$ are three vectors and $\theta$ is the
angle between $\overrightarrow{\mathrm{m}}$ and $\overrightarrow{\mathrm{n}}$,
We get, $\cos \theta=\frac{\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{m}} \| \overrightarrow{\mathrm{n}}|}$
$\Rightarrow m n \cos \theta=\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}$
[where, $|\overrightarrow{\mathrm{m}}|=\mathrm{m}$ and $|\overrightarrow{\mathrm{n}}|=\mathrm{n}]$
$\Rightarrow \mathrm{mn} \cos \theta=(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}) \frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}$
$\Rightarrow \mathrm{mn} \cos \theta=(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}) \frac{\overrightarrow{\mathrm{r}}}{\mathrm{r}}$
[where $|\overrightarrow{\mathrm{r}}|=\mathrm{r}]$
$\Rightarrow \operatorname{mn} r \cos \theta=(\overrightarrow{\mathrm{m}} \cdot \overrightarrow{\mathrm{n}}) \overrightarrow{\mathrm{r}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
If mass $(\mathrm{M})$, distance $(\mathrm{L})$ and time $(\mathrm{T})$ are fundamental quantities, then find dimension of torque.

Options:
A $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
B $\left[\mathrm{MLT}^{-2}\right]$
C $[M L T]$
D $\left[M L^2 T\right]$

The correct answer is:
$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

Solution:
Dimension of torque
$\begin{aligned}
\tau & =\mathrm{F} \times \mathrm{r}=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}$

Topic: Physics, Units and Dimensions, JIPMER, JIPMER 2019
If \(\mathrm{A}=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) \\ \sin ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) & \cot ^{-1}(\pi \mathrm{x})\end{array}\right] \mathrm{B}=\left[\begin{array}{cc}-\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) \\ \sin ^{-1}\left(\frac{\mathrm{x}}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]\) then \(\mathrm{A}-\mathrm{B}\) equal to

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

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

Solution:
We have, \(A=\left[\begin{array}{ll}\frac{1}{\pi} \sin ^{-1}(x \pi) & \frac{1}{\pi} \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \frac{1}{\pi} \sin ^{-1}\left(\frac{x}{\pi}\right) & \frac{1}{\pi} \cot ^{-1}(\pi x)\end{array}\right] B=\left[\begin{array}{cc}-\frac{1}{\pi} \cos ^{-1}(x \pi) & \frac{1}{\pi} \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \frac{1}{\pi} \sin ^{-1}\left(\frac{x}{\pi}\right) & -\frac{1}{\pi} \tan ^{-1}(\pi x)\end{array}\right]\)
\(\therefore A-B=\left[\begin{array}{ll}
\frac{1}{\pi}\left(\sin ^{-1} x \pi+\cos ^{-1} x \pi\right) & \frac{1}{\pi}\left(\tan ^{-1} \frac{x}{\pi}-\tan ^{-1} \frac{x}{\pi}\right) \\
\frac{1}{\pi}\left(\sin ^{-1} \frac{x}{\pi}-\sin ^{-1} \frac{x}{\pi}\right) & \frac{1}{\pi}\left(\cot ^{-1} \pi x\right)+\tan ^{-1} \pi x
\end{array}\right]\)
\(\left[\begin{array}{cc}
\frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \\
0 & \frac{1}{\pi} \cdot \frac{\pi}{2}
\end{array}\right]\left[\begin{array}{l}
\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\
\text { and } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}
\end{array}\right]\)
\(=\frac{1}{2}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\frac{1}{2} \mathrm{I}\)

Topic: Mathematics, Matrices, KCET, KCET 2016