datavorous/entrance-exam-dataset huggingface dataset:
If \(M(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\); \(M(\beta)=\left[\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]\) then \([M(\alpha) M(\beta)]^{-1}\) is equal to -

Options:
A \(\mathrm{M}(\beta) \mathrm{M}(\alpha)\)
B \(\mathrm{M}(-\alpha) \mathrm{M}(-\beta)\)
C \(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)
D \(-\mathrm{M}(\beta) \mathrm{M}(\alpha)\)

The correct answer is:
\(\mathrm{M}(-\beta) \mathrm{M}(-\alpha)\)

Solution:
\(\begin{aligned}
& \left.[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(\beta)^{-1} \mathrm{M}(\alpha)^{-1}\right] \\
& \operatorname{Now~M(\alpha )^{-1}}=\left[\begin{array}{ccc}
\cos \alpha & \sin \alpha & 0 \\
-\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (-\alpha) & -\sin (-\alpha) & 0 \\
\sin (-\alpha) & \cos (-\alpha) & 0 \\
0 & 0 & 1
\end{array}\right]=\mathrm{M}(-\alpha) \\
& M(\beta)^{-1}=\left[\begin{array}{ccc}
\cos \beta & 0 & -\sin \beta \\
0 & 1 & 0 \\
\sin \beta & 0 & \cos \beta
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (-\beta) & 0 & \sin (-\beta) \\
0 & 1 & 0 \\
-\sin (-\beta) & 0 & \cos (-\beta)
\end{array}\right]=\mathrm{M}(-\beta) \\
& {[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(-\beta) \mathrm{M}(-\alpha)} \\
&
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2009
If \(\mathrm{a} > 0, \mathrm{~b} > 0, \mathrm{c} > 0\) are respectively the \(\mathrm{pth}, \mathrm{q}\) th, rth terms of GP., then the value of the determinant \(\left|\begin{array}{lll}\log \mathrm{a} & \mathrm{p} & 1 \\ \log \mathrm{b} & \mathrm{q} & 1 \\ \log \mathrm{c} & \mathrm{r} & 1\end{array}\right|\) is

Options:
A 0
B 1
C -1
D None of these

The correct answer is:
0

Solution:
Let \(A\) be the 1 st term and \(\mathrm{R}\) the common ratio of G.P., then
\(\begin{aligned}
& a=T_p=A R^{p-1} \\
& \therefore \log \mathrm{a}=\log \mathrm{A}+(\mathrm{p}-1) \log \mathrm{R}
\end{aligned}\)
Similarly, \(\log \mathrm{b}=\log \mathrm{A}+(\mathrm{q}-1) \log \mathrm{R}\)
and \(\log \mathrm{c}=\log \mathrm{A}+(\mathrm{r}-1) \log \mathrm{R}\)
\(\therefore \Delta=\left|\begin{array}{lll}
\log A+(p-1) \log R & p & 1 \\
\log A+(q-1) \log R & q & 1 \\
\log A+(r-1) \log R & r & 1
\end{array}\right|\)
Split into two determinants and in the first take \(\log A\) common and in the second take \(\log \mathrm{R}\) common
\(\Delta=\log A\left|\begin{array}{lll}
1 & p & 1 \\
1 & q & 1 \\
1 & r & 1
\end{array}\right|+\log R\left|\begin{array}{lll}
p-1 & p & 1 \\
q-1 & q & 1 \\
r-1 & r & 1
\end{array}\right|\)
Apply \(\mathrm{C}_1 \rightarrow \mathrm{C}_2 \rightarrow \mathrm{C}_3\) in the second
\(\Delta=0+\log R\left|\begin{array}{lll}
0 & p & 1 \\
0 & q & 1 \\
0 & r & 1
\end{array}\right|=0\)

Topic: Mathematics, Determinants, BITSAT, BITSAT 2009
If \(\mathrm{I}_{\mathrm{m}}=\int_1^{\mathrm{e}}(\ln \mathrm{x})^{\mathrm{m}} \mathrm{dx}\), where \(\mathrm{m} \in \mathrm{N}\), then \(I_{10}+10 I_9\) is equal to -

Options:
A \(\mathrm{e}^{10}\)
B \(\frac{\mathrm{e}^{10}}{10}\)
C \(\mathrm{e}\)
D \(\mathrm{e}-1\)

The correct answer is:
\(\mathrm{e}\)

Solution:
\(\begin{aligned}
& \mathrm{I}_{10}=\int_1^{\mathrm{e}} 1 \cdot(\ln \mathrm{x})^{10} \mathrm{dx}=\left[(\ln \mathrm{x})^{10} \mathrm{x}\right]_1^{\mathrm{e}} \\
& -\int_1^{\mathrm{e}} 10(\ln \mathrm{x})^9 \cdot \frac{1}{\mathrm{x}} \cdot \mathrm{xdx}=\mathrm{e}-0-10 \int_1^{\mathrm{e}}(\ln \mathrm{x})^9 \mathrm{dx} \\
& =\mathrm{e}-10 \mathrm{I}_9+10 \mathrm{I}_9=\mathrm{e}
\end{aligned}\)

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2010
If \(\mathrm{m}\) arithmetic means are inserted between 1 and 31 so that the ratio of the \(7^{\text {th }}\) and \((\mathrm{m}-1)^{\text {th }}\) means is \(5: 9\), then find the value of \(m\).

Options:
A 14
B 24
C 10
D 20

The correct answer is:
14

Solution:
Let the means be \(x_1, x_2, \ldots \ldots x_m\) so that \(1, \mathrm{x}_1, \mathrm{x}_2, \ldots . \mathrm{x}_{\mathrm{m}}, 31\) is an A.P. of \((\mathrm{m}+2)\) terms.
Now, \(31=\mathrm{T}_{\mathrm{m}+2}=\mathrm{a}+(\mathrm{m}+1) \mathrm{d}=1+(\mathrm{m}+1) \mathrm{d}\)
\(\therefore \mathrm{d}=\frac{30}{\mathrm{~m}+1}\) Given : \(\frac{\mathrm{x}_7}{\mathrm{x}_{\mathrm{m}-1}}=\frac{5}{9}\)
\(\begin{aligned}
& \therefore \frac{\mathrm{T}_8}{\mathrm{~T}_{\mathrm{m}}}=\frac{\mathrm{a}+7 \mathrm{~d}}{\mathrm{a}+(\mathrm{m}-1) \mathrm{d}}=\frac{5}{9} \\
& \Rightarrow 9 \mathrm{a}+63 \mathrm{~d}=5 \mathrm{a}+(5 \mathrm{~m}-5) \mathrm{d} \\
& \Rightarrow 4.1=(5 \mathrm{~m}-68) \frac{30}{\mathrm{~m}+1} \\
& \Rightarrow 2 \mathrm{~m}+2=75 \mathrm{~m}-1020 \Rightarrow 73 \mathrm{~m}=1022 \\
& \therefore \quad \mathrm{m}=\frac{1022}{73}=14
\end{aligned}\)

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2010
If \(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{2}{3}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}\) and \(\mathrm{P}(\mathrm{A})=\frac{1}{3}\) then

Options:
A A and \(\mathrm{B}\) are independent events
B A and \(\mathrm{B}\) are disjoint events
C A and \(\mathrm{B}\) are dependent events
D None of these

The correct answer is:
A and \(\mathrm{B}\) are independent events

Solution:
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AB})\)
\(\Rightarrow \frac{2}{3}=\frac{1}{3}+\mathrm{P}(\mathrm{B})-\frac{1}{6} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{1}{2}\)
Now, \(\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A})~ \mathrm{P}(\mathrm{B}), \mathrm{A}\) and \(\mathrm{B}\) are independent events.

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2010
If \(\mathrm{p}^{\text {th }}, \mathrm{q}^{\text {th }}\) and \(\mathrm{r}^{\text {th }}\) terms of H.P. are \(\mathrm{u}, \mathrm{v}, \mathrm{w}\) respectively, then find the value of the expression \((q-r) v w+(r-p) w u+(p-q) u v\).

Options:
A 2
B 0
C 4
D 8

The correct answer is:
0

Solution:
Let H.P. be \(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{a}+\mathrm{d}}+\frac{1}{\mathrm{a}+2 \mathrm{~d}}+\ldots \ldots\)
\(\begin{aligned}
& \therefore \mathrm{u}=\frac{1}{\mathrm{a}+(\mathrm{p}-1) \mathrm{d}}, \mathrm{v}=\frac{1}{\mathrm{a}+(\mathrm{q}-1) \mathrm{d}}, \\
& \mathrm{w}=\frac{1}{\mathrm{a}+(\mathrm{r}-1) \mathrm{d}} \Rightarrow \mathrm{a}+(\mathrm{p}-1) \mathrm{d}=\frac{1}{\mathrm{u}} \\
& \mathrm{a}+(\mathrm{q}-1) \mathrm{d}=\frac{1}{\mathrm{v}}, \mathrm{a}+(\mathrm{r}-1) \mathrm{d}=\frac{1}{\mathrm{w}} \\
& \Rightarrow(\mathrm{q}-\mathrm{r})\{\mathrm{a}+(\mathrm{p}-1) \mathrm{d}\}+(\mathrm{r}-\mathrm{p})\{\mathrm{a}+(\mathrm{q}-1) \mathrm{d}\}+\ldots \ldots \\
& =\frac{1}{\mathrm{u}}(\mathrm{q}-\mathrm{r})+\frac{1}{\mathrm{v}}(\mathrm{r}-\mathrm{p})+\ldots . \Rightarrow(q-r) \mathrm{vw}+\ldots . . = 0
\end{aligned}\)

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2011
If mean of a poisson distribution of a random variable \(X\) is 2 , then the value of \(P(X > 1.5)\) is

Options:
A \(\frac{3}{e^2}\)
B \(\frac{3}{e}\)
C \(1-\frac{3}{e}\)
D \(1-\frac{3}{e^2}\)

The correct answer is:
\(1-\frac{3}{e^2}\)

Solution:
Since, \(\mathrm{P}(\mathrm{X}=\mathrm{r})=\frac{e^{-\lambda} \lambda^r}{r !}\) (where \(\lambda=\) mean)
\(\begin{aligned}
\therefore \quad & \mathrm{P}(\mathrm{X}=\mathrm{r} > 1.5)=\mathrm{P}(2)+\mathrm{P}(3)+\ldots \infty \\
& =1-\mathrm{P}(\mathrm{X}=\mathrm{r} \leq 1)=1-\mathrm{P}(0)-\mathrm{P}(1) \\
& =1-\left(e^{-2}+\frac{e^{-2} \times 2}{1 !}\right)=1-\frac{3}{e^2}
\end{aligned}\)

Topic: Mathematics, Probability, BITSAT, BITSAT 2010
If $n(\mathrm{~A})=1000, n(\mathrm{~B})=500$ and if $n(\mathrm{~A} \cap \mathrm{B}) \geq 1$ and $n(\mathrm{~A} \cup \mathrm{B})=p$, then

Options:
A $500 \leq p \leq 1000$
B $1001 \leq p \leq 1498$
C $1000 \leq p \leq 1498$
D $1000 \leq p \leq 1499$

The correct answer is:
$1000 \leq p \leq 1499$

Solution:
$\mathrm{n}(\mathrm{A})=1000, \mathrm{n}(\mathrm{B})=500, \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \geq 1$, $n(A \cup B)=p ; n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\mathrm{p}=1000+500-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \Rightarrow 1 \leq \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq 500$
Hence $p \leq 1499$ and $p \geq 1000 \Rightarrow 1000 \leq p \leq 1499$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $n$ is a positive integer, then $2.4^{2 n+1}+3^{3 n+1}$ is divisible by:

Options:
A
2
B
7
C
11
D
27

The correct answer is:
11

Solution:
Let $P(n)=2 \cdot 4^{2 n+1}+3^{3 n+1}$
Then $P(1)=2.4^3+3^4=209$, which is divisible by 11 but not divisible by 2,7 or 27 .
Further, let $P(k)=2.4^{2 k+1}+3^{3 k+1}$ is divisible by 11 , that is,
$2.4^{2 k+1}+3^{3 k+1}=11 q$ for some integer q. Now
$\begin{aligned}
\mathrm{P}(\mathrm{k}+1) & =2 \cdot 4^{2 \mathrm{k}+3}+3^{3 \mathrm{k}+4} \\
& =2 \cdot 4^{2 \mathrm{k}+1} \cdot 4^2+3^{3 \mathrm{k}+1} \cdot 3^3 \\
& =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+27 \cdot 3^{3 \mathrm{k}+1} \\
& =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+(16+11) \cdot 3^{3 \mathrm{k}+1} \\
& =16\left[2 \cdot 4^{2 \mathrm{k}+1}+3^{3 \mathrm{k}+1}\right]+11 \cdot 3^{3 \mathrm{k}+1} \\
& =16 \cdot 11 \mathrm{q}+11 \cdot 3^{3 \mathrm{k}+1} \\
& =11\left(16 \mathrm{q}+3^{3 \mathrm{k}+1}\right)=11 \mathrm{~m}
\end{aligned}$
where $m=16 q+3^{3 k+1}$ is another integer.
$\therefore \mathrm{P}(\mathrm{k}+1)$ is divisible by 11 .

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If on combining two charged bodies, the current does not flow then

Options:
A charge is equal on both
B capacitance is equal on both
C potential is equal on both
D resistance is equal on both

The correct answer is:
potential is equal on both

Solution:
No Solution available

Topic: Physics, Electrostatics, BITSAT, BITSAT 2009
If one root of the equation $x^2+p x+12=0$ is 4, while the equation $x^2+p x+q=0$. has equal roots, then the value of ' $q$ ' is

Options:
A 4
B 12
C 3
D $\frac{49}{4}$

The correct answer is:
$\frac{49}{4}$

Solution:
4 is a root of $x^2+p x+12=0$
$\Rightarrow 16+4 p+12=0 \Rightarrow p=-7$
Now, the equation $x^2+p x+q=0$ has equal roots.
$\therefore p^2-4 q=0 \Rightarrow q=\frac{p^2}{4}=\frac{49}{4}$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If one strand of DNA has sequence ATCGTATG, the sequence in the complementary strand would be

Options:
A TAGCTTAC
B TCACATAC
C TAGCATAC
D TACGATAC

The correct answer is:
TAGCATAC

Solution:
Due to size and geometries of bases, the only possible pairing in DNA are between G (guanine) and C (cytosine) through three H-bonds and between A (adenine) and T (thymine) through two H-bonds.

Thus, the complementary strand of DNA will be

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2019
If $P_{1}$ and $P_{2}$ be the length of perpendiculars from the origin upon the straight lines $x \sec \theta+y \operatorname{cosec} \theta$ $=a$ and $x \cos \theta-y \sin \theta=a \cos 2 \theta$ respectively,

then the value of $4 \mathrm{P}_{1}^{2}+\mathrm{P}_{2}^{2}$

Options:
A $a^{2}$
B $2 a^{2}$
C $a^{2} / 2$
D $3 a^{2}$

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

Solution:
We have $P_{1}=$ length of perpendicular from (0,0) on $x \sec \theta+y \operatorname{cosec} \theta=a$

$\begin{aligned} \text { i.e. } P_{1} &=\left|\frac{a}{\sqrt{\sec ^{2} \theta+\cos e c^{2} \theta}}\right|=|a \sin \theta \cos \theta| \\ &=\left|\frac{a}{2} \sin 2 \theta\right| \text { or } 2 P_{1}=|a \sin 2 \theta| \end{aligned}$

$\mathrm{P}_{2}=$ Length of the perpendicular from (0,0) on $x \cos \theta-y \sin \theta=a \cos 2 \theta$

$P_{2}=\left|\frac{a \cos 2 \theta}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\right|=|a \cos 2 \theta|$

Now, $4 P_{1}^{2}+P_{2}^{2}=a^{2} \sin ^{2} 2 \theta+a^{2} \cos ^{2} 2 \theta=a^{2}$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2014
If p:4 is an even prime number, q:6 is a divisor of 12 and r: the HCF of 4 and 6 is 2, then which of the following is correct?

Options:
A p∧q
B p∨q∧~r
C ~q∧r∨p
D ~p∨q∧r

The correct answer is:
~p∨q∧r

Solution:
Since, p:4 is an even prime number,

q:6 is a divisor of 12 and

r: the HCF of 4 and 6 is 2.

So, ~p∨q∧r is correct.

Hence, option (d) is correct.

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2018
If P=abcbcacab, abc=1, PTP=I then the value of a3+b3+c3 is

Options:
A 2
B 1
C 0
D 5

The correct answer is:
2

Solution:
Given, P=abcbcacab

∴ PT=abcbcacab

Also given PTP=I

∴ PTP=I=1

or abcbcacab2=1⇒abcbcacab=±1

or 3abc-a3-b3-c3=±1

or 3∓1=a3+b3+c3

∴ a3+b3+c3=2 or 4

Topic: Mathematics, Matrices, BITSAT, BITSAT 2018
If $p \neq a, q \neq b, r \neq c$ and the system of equations
$\begin{aligned}
& p x+a y+a z=0 \\
& b x+q y+b z=0 \\
& c x+c y+r z=0
\end{aligned}$
has a non-trivial solution, then the value of
$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

Options:
A
$1$

B
$2$

C
$\frac{1}{2}$

D
$0$

The correct answer is:

$2$

Solution:
As the given system of equations has a non-trivial solution.
$\Delta=\left|\begin{array}{lll}p & a & a \\ b & q & b \\ c & c & r\end{array}\right|=0$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$
$\Delta=\left|\begin{array}{ccc}p & a-p & a-p \\ b & q-b & 0 \\ c & 0 & r-c\end{array}\right|=0$
Expanding along $C_3$, we get
$(a-p)\left|\begin{array}{cc}b & q-b \\ c & 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p & a-p \\ b & q-b\end{array}\right|=0$
$\begin{aligned}
& \Rightarrow(a-p)(-c)(q-b)+(r-c) \\
& \{p(q-b)-b(a-p)\}=0 \\
& \Rightarrow \\
& (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0
\end{aligned}$
Dividing by $(p-a)(q-b)(r-c)$, we get
$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2022
If p and p' denote the lengths of the perpendicular from a focus and the centre of an ellipse with semi-major axis of length a, respectively, on a tangent to the ellipse and r denotes the focal distance of the point, then

Options:
A ap=rp'
B rp=ap'
C ap=rp'+1
D ap'+rp=1

The correct answer is:
ap=rp'

Solution:
Let the ellipse be x2a2+y2b2=1

and let xcosθa+ysinθb=1 ...(i)

be a tangent to it at point (acosθ,bsinθ). Then, p= length of the perpendicular from S(ae, 0) on Eq. (i)

p=ecosθ-1cos2θa2+sin2θb2

p'= length of the perpendicular from O(0,0) on Eq. (i)

⇒ p'=1cos2θa2+sin2θb2

and r=aecosθ-a

Clearly, rp'=ap

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2017
If $\mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{4}{5}$, then $\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=$

Options:
A $\frac{1}{5}$
B $\frac{4}{5}$
C $\frac{1}{2}$
D 1

The correct answer is:
1

Solution:
$P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$
$P(A \cap B)=P(A \mid B) P(B)=\frac{1}{2} \cdot \frac{3}{5}=\frac{3}{10}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\mathrm{P}(\mathrm{A})=\frac{4}{5}-\frac{3}{10}=\frac{1}{2} . \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=\frac{1}{2}$
We know, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})$
[as $A \cap B$ and $A^{\prime} \cap B$ are mutually exclusive events]
$\Rightarrow \quad \frac{3}{10}+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}$
$\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}$
Now, $\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B})-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$
$=\frac{1}{2}+\frac{3}{5}-\frac{3}{10}=\frac{5+6-3}{10}=\frac{4}{5}$
$\mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\frac{4}{5}=\frac{1}{5}$
$\therefore \mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\frac{1}{5}+\frac{4}{5}=1$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0,$ then the value of

$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}$ is

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

The correct answer is:
2

Solution:
$\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$

Apply $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},$ we get

$\left|\begin{array}{ccc}\mathrm{x} & 0 & -\mathrm{z} \\ 0 & \mathrm{y} & -\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$

$\Rightarrow \mathrm{x}[\mathrm{yr}+\mathrm{z}(\mathrm{q}-\mathrm{y})]-\mathrm{z}[0-\mathrm{y}(\mathrm{p}-\mathrm{x})]=0$

[Expansion along first row $]$

$\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2014
If: \(p\) Raju is tall and \(q\) : Raju is intelligent, then the symbolic statement \(\sim p \vee q\) means

Options:
A Raju is not tall or he is intelligent.
B Raju is tall or he is intelligent
C Raju is not tall and he is intelligent
D Raju is not tall implies he is intelligent

The correct answer is:
Raju is not tall or he is intelligent.

Solution:
\(\sim \mathrm{p} \vee \mathrm{q}:\) Raju is not tall or he is intelligent.

Topic: Mathematics, Statistics, BITSAT, BITSAT 2009
If $P$ represents radiation pressure, c represents speed of light and $\mathrm{Q}$ represents radiation energy striking a unit area per second, the non-zero integers $x, y$ and $z$ such that $P^{x} Q^{y} c^{z}$ is dimensionless, are.

Options:
A $x=1, y=1, z=-1$
B $x=1, y=-1, z=1$
C $\quad x=-1, y=1, z=1$
D $x=1, y=1, z=1$

The correct answer is:
$x=1, y=-1, z=1$

Solution:
No Solution available

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2010
If $\sum_{r=0}^{n} \frac{r+2}{r+1}^{n} C_{r}=\frac{2^{8}-1}{6},$ then $n=$

Options:
A 8
B 4
C 6
D 5

The correct answer is:
5

Solution:
$\sum_{r=0}^{n} \frac{r+2}{r+1}{ }^{n} C_{r}=\frac{2^{8}-1}{6}$

$$

\begin{aligned}

& \Rightarrow \sum_{r=0}^{n}\left[1+\frac{1}{r+1}\right]{ }^{n} C_{r}=\frac{2^{8}-1}{6} \\

& \Rightarrow 2^{n}+\sum_{r=0}^{n} \frac{1}{n+1} \cdot{ }^{n+1} C_{r+1}=\frac{2^{8}-1}{6} \\

\Rightarrow 2^{n}+& \frac{2^{n+1}-1}{n+1}=\frac{2^{8}-1}{6} \Rightarrow \frac{2^{n}(n+3)-1}{n+1}=\frac{2^{8}-1}{6} \\

\Rightarrow & \frac{2^{n}(n+1+2)-1}{n+1}=\frac{2^{5}(6+2)-1}{6}

\end{aligned}

$$

Comparing we get $\mathrm{n}+1=6 \Rightarrow \mathrm{n}=5$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2016
If ∑r=1ntr=nn+1n+2n+38, where tr denotes the rth term of a series, then limn→∞∑r=1n1tr is

Options:
A 18
B 14
C 12
D 1

The correct answer is:
12

Solution:
Given, ∑r=1ntr=nn+1n+2n+38=Sn (say)

∴ ∑r=1n-1tr=n-1nn+1n+28=Sn-1

Now, tn=Sn-Sn-1=nn+1n+22

∴ limn→∞∑r=1n1tr=limn→∞∑r=1n2nn+1n+2

=limn→∞∑r=1n1nn+1-1n+1n+2

=-limn→∞∑r=1n1n+1n+2-1nn+1

=-limn→∞1n+1n+2-12

=-0-12=12

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2018
If $\vec{r}$ and $\vec{s}$ arenon-zero constant vectors and the scalar $b$ is chosen such that $|\vec{r}+b \vec{s}|$ is minimum, then the value of $|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}$ is equal to

Options:
A $2|\vec{r}|^{2}$
B $|\vec{r}|^{2} / 2$
C $3|\vec{r}|^{2}$
D $|\vec{r}|^{2}$

The correct answer is:
$|\vec{r}|^{2}$

Solution:
For minimum value $|\vec{r}+b \vec{s}|=0$. Let $\vec{r}$ and $\vec{s}$ are anti-parallel so $b \vec{s}=-\vec{r}$

$\therefore|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2020
If \(|r| > \) land \(x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots\) to \(\infty\),
\(\mathrm{y}=\mathrm{b}-\frac{\mathrm{b}}{\mathrm{r}}+\frac{\mathrm{b}}{\mathrm{r}^2}-\ldots . \text { to } \infty\)
and \(z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots\) to \(\infty\), then \(\frac{x y}{z}=\)

Options:
A \(\frac{\mathrm{ab}}{\mathrm{c}}\)
B \(\frac{\mathrm{ac}}{\mathrm{b}}\)
C \(\frac{\mathrm{bc}}{\mathrm{a}}\)
D 1

The correct answer is:
\(\frac{\mathrm{ab}}{\mathrm{c}}\)

Solution:
Since \(|\mathrm{r}| > 1, \frac{1}{|\mathrm{r}|} < 1\)
\(\therefore \mathrm{x}=\frac{\mathrm{a}}{1-\frac{1}{\mathrm{r}}}=\frac{\mathrm{ar}}{\mathrm{r}-1}\)
Similarly, \(y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{b r}{r+1}\) and
\(\begin{gathered}
z=\frac{c}{1-\frac{1}{r^2}}=\frac{cr^2}{r^2-1} \quad ...(1)\\
\therefore \quad x y=\frac{a r}{r-1} \times \frac{b r}{r+1}=\frac{a b r^2}{r^2-1} \quad ...(2)
\end{gathered}\)
Dividing (2) by (1), we get
\(\frac{\mathrm{xy}}{\mathrm{z}}=\frac{\mathrm{abr}^2}{\mathrm{r}^2-1} \times \frac{\mathrm{r}^2-1}{\mathrm{cr}^2}=\frac{\mathrm{ab}}{\mathrm{c}}\)

Topic: Mathematics, Circle, BITSAT, BITSAT 2009
If $\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}=\mathrm{A} x+\mathrm{B} \log \sin (x-\alpha)+\mathrm{C}$, then value of $(A, B)$ is

Options:
A $(-\cos \alpha, \sin \alpha)$
B $(\cos \alpha, \sin \alpha)$
C $(-\sin \alpha, \cos \alpha)$
D $(\sin \alpha, \cos \alpha)$

The correct answer is:
$(\cos \alpha, \sin \alpha)$

Solution:
$\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{dx}=\int \frac{\sin (x-\alpha+\alpha}{\sin (x-\alpha)} \mathrm{dx}$

$=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}$

$=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} \mathrm{d} \mathrm{x}$

$=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+\mathrm{C}$

$\therefore \mathrm{A}=\cos \alpha, \mathrm{B}=\sin \alpha$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2021
If sample A contains 100 observations $101,102, \ldots$ 200 and sample B contains 100 obsections $151,152,$ \ldots 250, then ratio of variance $\mathrm{v}_{\mathrm{A}} / \mathrm{v}_{\mathrm{B}}=$

Options:
A 1
B $\frac{9}{4}$
C $\frac{4}{9}$
D $\frac{2}{3}$

The correct answer is:
1

Solution:
$\because \quad \delta_{\mathrm{x}}^{2}=\frac{\Sigma \mathrm{d}^{2} \mathrm{i}}{\mathrm{n}}$
But both $A$ and $B$ have 100 observations, then both the sample $A$ and $B$ have same standard deviation and the same variance.
Hence, $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=1$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2012
If $\sec ^2 \theta=\frac{4}{3}$, then the general value of $\theta$ is

Options:
A $2 \mathrm{n} \pi \pm \frac{\pi}{6}$
B $\mathrm{n} \pi \pm \frac{\pi}{6}$
C $2 \mathrm{n} \pi \pm \frac{\pi}{3}$
D $\mathrm{n} \pi \pm \frac{\pi}{3}$

The correct answer is:
$\mathrm{n} \pi \pm \frac{\pi}{6}$

Solution:
We have $\sec ^2 \theta=\frac{4}{3} \Rightarrow \cos ^2 \theta=\frac{3}{4}$
$\Rightarrow \cos ^2 \theta=\cos ^2\left(\frac{\pi}{6}\right) \Rightarrow$
$\theta=n \pi \pm \frac{\pi}{6} \ldots \ldots \ldots$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$,

then what is the value of $x$ ?

Options:
A $a / b$
B $a b$
C $b / a$
D $\frac{a-b}{1+a b}$

The correct answer is:
$\frac{a-b}{1+a b}$

Solution:
Given,

$\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

$\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x$

$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x$

$\Rightarrow \quad \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x$

$\Rightarrow \quad x=\frac{a-b}{1+a b}$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2016