datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ then $\mathbf{A}^{100}$ :

Options:
A $2^{100} \mathrm{~A}$
B $2^{99} \mathrm{~A}$
C $2^{101} \mathrm{~A}$
D None of the above

The correct answer is:
$2^{99} \mathrm{~A}$

Solution:
Let $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

$$

\begin{array}{l}

\mathrm{A}^{2}=\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]=2\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right]=2 \mathrm{~A} \\

\mathrm{~A}^{3}=2^{2}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right], \mathrm{A}^{4}=2^{3}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right] \\

\mathrm{A}^{3}=2^{2} \mathrm{~A}, \quad \mathrm{~A}^{4}=2^{3} \mathrm{~A} \\

\therefore \mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1}\left[\begin{array}{ll}

1 & 1 \\

1 & 1

\end{array}\right] \\

\Rightarrow \mathrm{A}^{100}=2^{100-1} \mathrm{~A} \therefore \mathrm{A}^{100}=2^{99} \mathrm{~A}

\end{array}

$$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2014
If $\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}$ are in A. P., then $\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right) \left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)$ is equal to

Options:
A $\frac{4}{a c}-\frac{3}{b^{2}}$
B $\frac{b^{2}-a c}{a^{2} b^{2} c^{2}}$
C $\frac{4}{a c}-\frac{1}{b^{2}}$
D None of these

The correct answer is:
$\frac{4}{a c}-\frac{3}{b^{2}}$

Solution:
$\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}$

$\therefore\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right)\left(\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$

$=\left(\frac{2}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)\left(\frac{2}{\mathrm{c}}-\frac{1}{\mathrm{~b}}\right)=\frac{4}{\mathrm{ac}}-\frac{1}{\mathrm{~b}}\left(\frac{2}{\mathrm{a}}+\frac{2}{\mathrm{c}}\right)+\frac{1}{\mathrm{~b}^{2}}$

$=\frac{4}{\mathrm{ac}}-\frac{2}{\mathrm{~b}}\left(\frac{2}{\mathrm{~b}}\right)+\frac{1}{\mathrm{~b}^{2}}=\frac{4}{\mathrm{ac}}-\frac{3}{\mathrm{~b}^{2}}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2014
If $a^{2} x^{4}+b^{2} y^{4}=c^{4}$, then the maximum value of xy is

Options:
A $\frac{\mathrm{c}}{\sqrt{\mathrm{ab}}}$
B $\frac{c^{2}}{2 \sqrt{a b}}$
C $\frac{\mathrm{c}}{2 \sqrt{\mathrm{ab}}}$
D $\frac{c^{2}}{\sqrt{2 a b}}$

The correct answer is:
$\frac{c^{2}}{\sqrt{2 a b}}$

Solution:
If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. $\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{2}$ is maximum when

$$

a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\right.

$$

$\therefore$ maximum value of $a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}$

Maximum value of $x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2013
If \(A\) and \(B\) are \(2 \times 2\) matrices, then which of the following is true?

Options:
A \((\mathrm{A}+\mathrm{B})^2=\mathrm{A}^2+\mathrm{B}^2+2 \mathrm{AB}\)
B \((A-B)^2=A^2+B^2-2 A B\)
C \((A-B)(A+B)=A^2+A B-B A-B^2\)
D \((A-B)(A+B)=A^2-B^2\)

The correct answer is:
\((A-B)(A+B)=A^2+A B-B A-B^2\)

Solution:
For two \(2 \times 2\) matrices, \(A ~\&~ B\)
\(\begin{aligned}
& (A-B) \times(A+B) \\
& =A \times A+A \times B-B \times A-B \times B \\
& =A^2-B^2+A B-B A
\end{aligned}\)
Hence, \((A-B)(A+B)=A^2+A B-B A-B^2\)

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2009
If $A$ and $B$ are mutually exclusive events and if $\mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{13}{21}$, then $\mathrm{P}(\mathrm{A})$ is equal to

Options:
A
1/7
B
4/7
C
2/7
D
5/7

The correct answer is:
2/7

Solution:
For mutually exclusive events
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $A$ and $B$ are positive acute angles satisfying $3 \cos ^{2} A+2 \cos ^{2} B=4$ and $\frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A}$

Then the value of $\mathrm{A}+2 \mathrm{~B}$ is equal to :

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{2}$
C $\frac{\pi}{3}$
D $\frac{\pi}{4}$

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

Solution:
Given, $3 \cos ^{2} \mathrm{~A}+2 \cos ^{2} \mathrm{~B}=4$

$\Rightarrow 2 \cos ^{2} \mathrm{~B}-1=4-3 \cos ^{2} \mathrm{~A}-1$

$\Rightarrow \cos 2 \mathrm{~B}=3\left(1-\cos ^{2} \mathrm{~A}\right)=3 \sin ^{2} \mathrm{~A} \ldots(1)$

and $2 \cos \mathrm{B} \sin \mathrm{B}=3 \sin \mathrm{A} \cos \mathrm{A}$

$\sin 2 \mathrm{~B}=3 \sin \mathrm{A} \cos \mathrm{A}$

Now, $\cos (\mathrm{A}+2 \mathrm{~B})=\cos \mathrm{A} \cos 2 \mathrm{~B}-\sin \mathrm{A}$

$\sin 2 \mathrm{~B}$

$=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0$

[using eqs. (1) and (2)] $\Rightarrow A+2 B=\frac{\pi}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2016
If a and bare non-zero roots of $x^{2}+a x+b=0$ then the least value of $\mathrm{x}^{2}+a \mathrm{x}+\mathrm{b}$ is

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

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

Solution:
As given a and bare the roots of the equat

$x^{2}+a x+b=0$

$\Rightarrow$ sum of roots, $a+b=-a$

$\Rightarrow \quad \mathrm{b}=-2 \mathrm{a}$

and product of roots, $a b=b$

$$

\begin{array}{l}

\Rightarrow \quad a b-b=0 \\

\Rightarrow \quad b(a-1)=0

\end{array}

$$

if $b=0$ then $a=0$

if $b \neq 0$ then $a=1$ and $b=-2$

so, the expression will be,

$f(x)=x^{2}+x-2$

$$

=x^{2}+2 \cdot \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2

$$

$\Rightarrow \mathrm{f}(\mathrm{x})=\left(\mathrm{x}+\frac{1}{2}\right)^{2}-\frac{9}{4}$

So, $\mathrm{f}(\mathrm{x})$ will be minimum, if $\left(\mathrm{x}+\frac{1}{2}\right)^{2}=0$

i.e. when $x=-\frac{1}{2}$

$\Rightarrow$ minimum value of function $=-\frac{9}{4}$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2015
If \(A\) and \(\mathrm{B}\) are mutually exclusive events and if \(\mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{13}{21}\), then \(\mathrm{P}(\mathrm{A})\) is equal to

Options:
A \(1 / 7\)
B \(4 / 7\)
C \(2 / 7\)
D \(5 / 7\)

The correct answer is:
\(2 / 7\)

Solution:
For mutually exclusive events
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7}\)

Topic: Mathematics, Probability, BITSAT, BITSAT 2011
If $a$.b $=a$.c and $a \times b=a \times c$, then correct statement is

Options:
A $\mathrm{a} \|(\mathrm{b}-\mathrm{c})$
B $a \perp(b-c)$
C $a=0$ or $b=c$
D None of these

The correct answer is:
$a=0$ or $b=c$

Solution:
$a . b .=a \cdot c \Rightarrow a \cdot(b-c)=0$
$\Rightarrow \mathrm{a}=0 \mathrm{or} \mathrm{b}-\mathrm{c}=0$ or $\mathrm{a} \perp(\mathrm{b}-\mathrm{c})$
$\Rightarrow \mathrm{a}=0$ or $\mathrm{b}=\mathrm{c}$ or $\mathrm{a} \perp(\mathrm{b}-\mathrm{c})$
Also a $x b=a x c \Rightarrow a \times(b-c)=0$
$\Rightarrow \mathrm{a}=0$ or $\mathrm{b}-\mathrm{c}=0$ or a $\|(\mathrm{b}-\mathrm{c})$
$\Rightarrow a=0$ or $b=c$ or $a \|(b-c)$
Observing to ( 1 ) and ( 2 ) we find that $\mathrm{a}=0$ or $\mathrm{b}=\mathrm{c}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2012
If $a, b$ and $c$ are real numbers then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x$$-a)=0$ are always

Options:
A real
B imaginary
C positive
D negative

The correct answer is:
real

Solution:
Given equation is $(x-a)(x-b)+(x-b)$ $(x-c)+(x-c)(x-a)=0$

$\Rightarrow 3 x^{2}-2(b+a+c) x+a b+b c+c a=0$

Now, here $A=3, B=-2(a+b+c)$

$C=a b+b c+c a$

$\therefore D=\sqrt{B^{2}-4 A C}$

$=\sqrt{(-2(a+b+c))^{2}-4(3)(a b+b c+c a)}$

$=\sqrt{4(a+b+c)^{2}-12(a b+b c+c a)}$

$=2 \sqrt{a^{2}+b^{2}+c^{2}-a b-b c-c a}$

$=2 \sqrt{\frac{1}{2}\left\{(a-b)^{2}-(b-c)^{2}+(c-a)^{2}\right\}} \geq 0$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2016
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$ then

Options:
A $\tan A \tan B+\tan B \tan C+\tan C \tan A=1$
B $\cot A+\cot B+\cot C=\cot A \cot B \cot C$
C $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}=1+4 \sin \mathrm{A} \sin \mathrm{B} \sin C$
D All three are correct

The correct answer is:
All three are correct

Solution:
$\mathrm{B}+\mathrm{C}=\frac{\pi}{2}-\mathrm{A} \Rightarrow \tan (\mathrm{B}+\mathrm{C})=\cot \mathrm{A}$

$\Rightarrow \frac{\tan \mathrm{B}+\tan \mathrm{C}}{1-\tan \mathrm{B} \tan \mathrm{C}}=\cot \mathrm{A}$

$\Rightarrow \tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan \mathrm{A}=1$

$\Rightarrow \tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}(\cot \mathrm{C}+\cot \mathrm{A}+\cot \mathrm{B})=1$

$\Rightarrow \cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}=\cot \mathrm{A} \cot \mathrm{B} \cot \mathrm{C}$

Again $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}$

$=2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2 \mathrm{C}$

$=2 \cos \left(\frac{\pi}{2}-C\right) \cos (A-B)+1-2 \sin ^{2} C$

$=2 \sin C[\cos (A-B)-\sin C]+1$

$=2 \sin \mathrm{C}\left[\cos (\mathrm{A}-\mathrm{B})-\sin \frac{\pi}{2}-(\mathrm{A}+\mathrm{B})\right]+1$

$\begin{array}{l}

=2 \sin C[\cos (A-B)-\cos (A+B)]+1 \\

=4 \sin A \sin B \sin C+1

\end{array}$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2021
If a, b, c are in GP and loga-log2b, log2b-log3c and log3c-loga are in AP, then a, b and c are the lengths of the sides of a triangle, which is

Options:
A equilateral
B right angled
C acute-angled
D obtuse angled

The correct answer is:
obtuse angled

Solution:
Given, a, b, c are in GP.

∴ b2=ac

and 2log2b-log3c=loga-log2b+log3c-loga

⇒b2=ac and 2b=3c

∵ a+b=5a3>c, b+c=10a9>a

and c+a=13a9>b

∴ a, b, c are the sides of a triangle.

Also, a is the greatest side

∴ cosA=b2+c2-a22bc=-2948<0

∴ ∆ABC is an obtuse angled triangle.

Hence, option (d) is correct.

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2018
If a→,b→,c→ are non-coplaner vectors such that b→×c→=a→; c→×a→=b→; a→×b→=c→, then which of the following is not TRUE?

Options:
A |a→|-|b→|=0
B |a→|=|b→|=|c→|=2
C a→b→c→=1
D |a→||b→||c→|=1

The correct answer is:
|a→|=|b→|=|c→|=2

Solution:
Given,

a→×b→=c→

b→×c→=a→

c→×a→=b→

(b→×c→)·a→=a→·a→

(c→×a→)·b→=b→·b→

a→2-b→2=(b→×c→)·a→-(c→×a→)·b→

=[c→a→b→]-[b→c→a→]=0

∴ |a→|2-|b→|2=0

|a→|-|b→|=0

⇒|a→|=|b→|

We know that,

a→×b→ b→×c→ c→×a→=[a→b→c→]2

a→b→c→=a→b→c→2

⇒a→b→c→=1

|a→||b→||c→|=1

∴ |a→|=|b→|=|c→|≠2

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2019
If $a, b, c$ are positive numbers, then least value of $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is

Options:
A 1
B 6
C 9
D None

The correct answer is:
9

Solution:
$\mathrm{AM} \geq \mathrm{HM}$

$\Rightarrow \frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$\Rightarrow(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
If a, b, c are three natural numbers in AP and a + $\mathrm{b}+\mathrm{c}=21$ then the possible number of values of the ordered triplet $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ is

Options:
A 15
B 14
C 13
D None of these

The correct answer is:
13

Solution:
$\quad$ Let $a=b-d$ and $c=b+d$, then $a+b+c=21$ $\Rightarrow \mathrm{b}=7$

So, the equation is $\mathrm{a}+\mathrm{c}=14$

$\therefore$ No. of solution $=$ coeff. of $x^{14}$ in $\left(x+x^{2}+\ldots .\right)$ $={ }^{13} \mathrm{C}_{12}=13$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2020
If $\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},$ where $\overrightarrow{0}$ is null vector, then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is :

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

The correct answer is:
$-\frac{3}{2}$

Solution:
We have $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

$|\vec{a}+\vec{b}+\vec{c}|=0 \Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}=0$

$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}$

$+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$

$\Rightarrow \vec{a} \cdot \vec{b}+\bar{b} \cdot \vec{c}+\bar{c} \cdot \bar{a}=-\frac{3}{2}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2013
If $a, c, b$ are in G.P., then the area of the triangle formed by the lines $a x+b y+c=0$ with the coordinates axes is equal to

Options:
A 1
B 2
C $\frac{1}{2}$
D None of these

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

Solution:
Area of the triangle $=\frac{1}{2}(\mathrm{x}$ intercept $) \times(\mathrm{y}$ intercept $)$

$=\frac{1}{2}\left(-\frac{\mathrm{c}}{\mathrm{a}}\right)\left(-\frac{\mathrm{c}}{\mathrm{b}}\right)=\frac{1}{2} \frac{\mathrm{c}^{2}}{\mathrm{ab}}=\frac{1}{2} \mathrm{unit}$

$\left[\because \mathrm{a}, \mathrm{c}, \mathrm{b}\right.$ are in $\left.\mathrm{G} . \mathrm{P} . \Rightarrow \mathrm{c}^{2}=\mathrm{ab}\right]$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
If a chord of the circle $x^{2}+y^{2}=8$ makes equal intercets of length a on the coordinate axes, then

Options:
A $|\mathrm{a}|<8$
B $|\mathrm{a}|<4 \sqrt{2}$
C $|\mathrm{a}|<4$
D $|\mathrm{a}|>4$

The correct answer is:
$|\mathrm{a}|<4$

Solution:
Since the chord makes equal intercepts of length a on the coordinate axes. So, its equation can be written as $\mathrm{x} \pm \mathrm{y}=\pm \mathrm{a}$. This line meets the given circle at two distinct points. So, length of the perpendicular from the centre (0,0) of the given circle must be less than the radius.

i.e. $\left|\pm \frac{\mathrm{a}}{\sqrt{2}}\right|<\sqrt{8} \Rightarrow \mathrm{a}^{2}<16 \Rightarrow|\mathrm{a}|<4$

Topic: Mathematics, Circle, BITSAT, BITSAT 2020
If a circular plate is heated uniformly, its area expands \(3 c\) times as fast as its radius, then the value of \(c\) when the radius is 6 units, is

Options:
A \(4 \pi\)
B \(2 \pi\)
C \(6 \pi\)
D \(3 \pi\)

The correct answer is:
\(4 \pi\)

Solution:
Let \(A\) sq. units in the area measure when the radius is \(r\) units.
their \(A=\pi r^2\)
Differentiate both side w.r.t ' \(t\) '
\(\frac{d A}{d t}=2 \pi r \frac{d r}{d t}\) ...(i))
We have, \(\frac{d A}{d t}=3 c \frac{d r}{d t}\)
From eqn (i), we get
\(3 c \cdot \frac{d r}{d t}=2 \pi r \cdot \frac{d r}{d t} \Rightarrow 3 c=2 \pi r\)
Now, \(c=\frac{2}{3} \pi(6)=4 \pi\) when \(r=6\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2010
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R, break into N small droplets each of radius r, then decrease (drop) in temperature Q (given, specific heat of liquid drop=S and surface tension =T )

Options:
A 3TρS1r-1R
B -2TρS1r-1R
C 2RρS1R-1r
D 3TρS1R-1r

The correct answer is:
3TρS1R-1r

Solution:
since, volume remains unchanged, during this phenomenon, so

43πR3=N×43πr3

N=R3r3

Now, change in surface area =4πR2-N4πr2

=4πR2-Nr2

Energy released (ΔU)=T× change in surface area

=T×4πR2-Nr2

Here, all this energy released is al the cost of lowering the temperature and mass of the big drop of liquid =43πR2ρ

Now, change in temperature, Δθ=ΔUms

=T×4πR2-Nr243πR3ρS

=3TμS1R-Nr2R3

=3TρS1R-R3×r2r3×R3

=3TρS1R-1r

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2018
If a flagstaff of 6 metres high placed on the top of a tower throws a shadow of \(2 \sqrt{3}\) metres along the ground, then the angle (in degrees) that the sun makes with the ground is

Options:
A \(60^{\circ}\)
B \(80^{\circ}\)
C \(75^{\circ}\)
D None of these

The correct answer is:
\(60^{\circ}\)

Solution:
Accordingly,
\(\tan \theta=\frac{h}{x}=\frac{h+6}{x+2 \sqrt{3}}=\frac{6}{2 \sqrt{3}} \Rightarrow \theta=60^{\circ}\)
[Since the triangles \(A E C\) and \(B D C\) are similar] \(h \cot \alpha=(h-100) \cot \beta\)
\(\therefore h=\frac{100 \cot \beta}{\cot \beta-\cot \alpha}\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2010
If a function f:R→R satisfy the equation f(x+y)=f(x)+f(y),∀x,y and the function f(x) is continuous at x=0, then

Options:
A f(x) is continuous for all positive real values of x
B f(x) is continuous for all x
C f(x)=0 for all x
D None of the above

The correct answer is:
f(x) is continuous for all x

Solution:
∵ limx→0fx=f0

Let a be any point

Now, at x=a,limx→afx=limh→0fa+h

=limh→0fa+limh→0fh

=fa+f0=fa+0=fa

∴ fx is continuous at x=a, where a is any arbitrary point.

Hence, fx is continuous for all x.

Topic: Mathematics, Functions, BITSAT, BITSAT 2017
If a function $f(x)$ is given by

$f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}$

$+\frac{x}{(2 x+1)(3 x+1)}+\ldots, \infty$, then at $x=0, f(x)$

Options:
A has no limit
B is not continuous
C is continuous but not differentiable
D is differentiable

The correct answer is:
is not continuous

Solution:
Let

$f(x)=\frac{x}{1+x}+\frac{x}{(x+1)(2 x+1)}+\frac{x}{(2 x+1)(3 x+1)}+\ldots \infty$

$=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{x}{[(r-1) x+1](r x+1)}$

$=\lim _{x \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}}\left[\frac{\mathrm{x}}{[(\mathrm{r}-1) \mathrm{x}+1]}-\frac{1}{\mathrm{rx}+1}\right]$

$=\lim _{n \rightarrow \infty}\left[1-\frac{1}{n x+1}\right]=1$

For $x=0,$ we have $f(x)=0$

Thus, we have $f(x)=\left\{\begin{array}{ll}1, & x \neq 0 \\ 0, & x=0\end{array}\right.$

Clearly, $\lim _{x \rightarrow 0^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x}) \neq \mathrm{f}(0)$

So, $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2015
If a→=i^+2j^+3k^, b→=-i^+2j^+k^, c→=3i^+j^ and a→+pb→ is normal to c→, then p is equal to

Options:
A 0
B 1
C 5
D 3

The correct answer is:
5

Solution:
a→+pb→ is normal to c→

∴ a→+pb→·c→=0

1-pi^+2+2pj^+3+pk^·3i^+j^=0

3-3p+2+2p=0⇒p=5

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2019
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\vec{j}-\vec{k}$, then $\vec{b}$ is

Options:
A $\hat{i}-\hat{j}+\hat{k}$
B $2 \hat{j}-\hat{k}$
C $2 \hat{i}$
D $\hat{i}$

The correct answer is:
$\hat{i}$

Solution:
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\vec{j}-\vec{k}$,
Let $\bar{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \widehat{k}$
Now, $\hat{j}-\widehat{k}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 1 & 1 \\ b_2 & b_2 & b_3\end{array}\right|$
$\Rightarrow b_3-b_2=0, b_1-b_3=1, b_2-b_1=-1$
$\Rightarrow b_3=b_2, b_1=b_2+1$
Also, $\bar{a} \cdot \bar{b}=1$
$\Rightarrow b_1+b_2+b_3=1$
$\Rightarrow 3 b_2+1=1$
$\Rightarrow b_2=0$
$\Rightarrow b_1=1, b_3=0$
Thus, $\bar{b}=\hat{i}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $\vec{a}$ is a vector of magnitude 50 , collinear with the vector $\vec{b}=6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}$ and makes an acute angle with the positive direction of $Z$ - axis, then $\vec{a}$ is equal to

Options:
A $-24 \hat{i}+32 \hat{j}+30 \hat{k}$
B $24 \hat{i}-32 \hat{j}-30 \hat{k}$
C $-12 \hat{i}+16 \hat{j}-15 \hat{k}$
D $12 \hat{i}-16 \hat{j}-15 \hat{k}$

The correct answer is:
$-24 \hat{i}+32 \hat{j}+30 \hat{k}$

Solution:
Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e.,

$\begin{array}{l}

\vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right) \\

\Rightarrow|a|=|m| \sqrt{36+64+\frac{225}{4}} \\

\Rightarrow 50=\frac{25}{2}|m| \\

\Rightarrow|m|=4 \\

\Rightarrow m=\pm 4

\end{array}$

Since, a makes an acute angle with the positive direction of $Z$ - axis, so its $z$ component must be positive and hence, $m$ must be $-4$.

$\therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)=-24 \hat{i}+32 \hat{j}+30 \hat{k}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2021
If A is an 3×3 non- singular matrix, such that AA'=A'A and B=A-1A', then BB' equals to

Options:
A B-1'
B I+B
C I
D B-1

The correct answer is:
I

Solution:
Given,

AA'=A'A

and B=A-1A'

BB'=A-1A'A-1A''

=A-1A'A''A-1'

=A-1A'AA'-1

∵ A''=A,A-1'=A'-1

=A-1A'AA'-1

=A-1AA'A'-1

=A-1AA-1A'

=I·I=I

Topic: Mathematics, Matrices, BITSAT, BITSAT 2019
If a=log23,b=log25 and c=log72, then log14063 in terms of a,b,c is

Options:
A 2ac+12c+abc+1
B 2ac+12a+c+a
C 2ac+12c+ab+a
D None of these

The correct answer is:
None of these

Solution:
We have,

log14063=log22×5×7(3×3×7)

=log2(3×3×7)log222×5×7=log23+log23+log272log22+log25+log27

=2a+1c2+b+1c=2ac+12c+bc+1

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2017
If a parallelogram is cut by two sets of m lines parallel to its sides, then the number of parallelogram thus formed, is

Options:
A C2m×C2m
B 2C2m+2
C C2m+22
D None of these

The correct answer is:
C2m+22

Solution:
The two sets of m parallel lines along with two sets of two parallel lines of the given parallelogram will form two sets of m+2 parallel lines. Each parallelogram is formed by choosing two parallel lines from each of the above

∴ Total number of parallelograms

=C2m+2×C2m+2=C2m+22

Hence, option (c) is correct.

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2018
If $A=\{x: x$ is a multiple of 4$\}$ and, $B=\{x: x$ is a multiple of 6$\}$, then $A \cap B$ consists of multiples of

Options:
A
16

B
12

C
88

D
4

The correct answer is:

12

Solution:
Let $x \in A \cap B$. Then,
$x \in A$ and $x \in B$
$\Rightarrow x$ is a multiple of 4 and $x$ is a multiple of 6
$\Rightarrow x$ is a multiple of 4 and 6 both
$\therefore x$ is a multiple of 12

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2022
If \(A=\{1,2,3,4,5\}\), then the number of proper subsets of \(A\) is

Options:
A 31
B 38
C 48
D 54

The correct answer is:
31

Solution:
Number of proper subsets of \(A=2^n-1\) Given: \(A=\{1,2,3,4,5\}\)
Here \(n=5\)
\(\therefore\) no. of proper subsets \(=2^5-1\)

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2009
If \(A=\left[\begin{array}{ll}1 & 3 \\ 3 & 2 \\ 2 & 5\end{array}\right]\) and \(B=\left[\begin{array}{cc}-1 & -2 \\ 0 & 5 \\ 3 & 1\end{array}\right]\) and \(\mathrm{A}+\mathrm{B}-\mathrm{D}=0\) (zero matrix), then \(\mathrm{D}\) matrix will be -

Options:
A \(\left[\begin{array}{ll}0 & 2 \\ 3 & 7 \\ 6 & 5\end{array}\right]\)
B \(\left[\begin{array}{ll}0 & 2 \\ 3 & 7 \\ 5 & 6\end{array}\right]\)
C \(\left[\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}\right]\)
D \(\left[\begin{array}{cc}0 & -2 \\ -3 & -7 \\ -5 & -6\end{array}\right]\)

The correct answer is:
\(\left[\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}\right]\)

Solution:
Let \(D=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]\)
\(\begin{aligned} & \therefore A+B-C=\left[\begin{array}{ll}1 & 3 \\ 3 & 2 \\ 2 & 5\end{array}\right]+\left[\begin{array}{cc}-1 & -2 \\ 0 & 5 \\ 3 & 1\end{array}\right]-\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \\ \mathrm{e} & \mathrm{f}\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}1-1-\mathrm{a} & 3-2-\mathrm{b} \\ 3+0-\mathrm{c} & 2+5-\mathrm{d} \\ 2+3-\mathrm{e} & 5+1-\mathrm{f}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}\)
\(\begin{array}{ll}-\mathrm{a}=0 \Rightarrow \mathrm{a}=0,1-\mathrm{b}=0 \Rightarrow \mathrm{b}=1, \\ 3-\mathrm{c}=0 \Rightarrow \mathrm{c}=3, & 7-\mathrm{d}=0 \Rightarrow \mathrm{d}=7, \\ 5-\mathrm{e}=0 \Rightarrow \mathrm{e}=5, & 6-\mathrm{f}=0 \Rightarrow \mathrm{f}=6,\end{array}\)
\(\therefore D=\left[\begin{array}{ll}0 & 1 \\ 3 & 7 \\ 5 & 6\end{array}\right]\)

Topic: Mathematics, Determinants, BITSAT, BITSAT 2010