datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=\left\{\begin{array}{cl}\frac{x \log \cos x}{\log \left(1+x^{2}\right)} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$ then $f(x)$ is

Options:
A continuous as well as differentiable at $\mathrm{x}=0$
B continuous but not differentiable at $x=0$
C differentiable but not continuous at $\mathrm{x}=0$
D neither continuous nor differentiable at $\mathrm{x}=0$

The correct answer is:
continuous as well as differentiable at $\mathrm{x}=0$

Solution:
We have,

$$

\begin{aligned}

\mathrm{Lf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h} \log \cosh }{-\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)} \\

=& \lim _{\mathrm{h} \rightarrow 0} \frac{\log \cosh }{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text { form }\right)

\end{aligned}

$$

$\begin{aligned} &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{~h} /\left(1+\mathrm{h}^{2}\right)}=-1 / 2 \\ \mathrm{Rf}^{\prime}(0) &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \log \cosh }{\mathrm{h} \log \left(1+\mathrm{h}^{2}\right)} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\log \cos \mathrm{h}}{\log \left(1+\mathrm{h}^{2}\right)}\left(\frac{0}{0} \text { form }\right) \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{-\tan \mathrm{h}}{2 \mathrm{~h} /\left(1+\mathrm{h}^{2}\right)}=\frac{-1}{2} \end{aligned}$

Since $\mathrm{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0),$ therefore $\mathrm{f}(\mathrm{x})$ is dif-

ferentiable at $x=0$

Since differentiability $\Rightarrow$ continuity, there-fore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2016
If $f(x)=\left\{\begin{array}{l}x^2-1,0 < x < 2 \\ 2 x+3,2 \leq x < 3\end{array}\right.$, then the quadratic equation whose roots are $\lim _{x \rightarrow 2^{-}} f(x)$ and $\lim _{x \rightarrow 2^{+}} f(x)$ is

Options:
A $x^2-6 x+9=0$
B $x^2-7 x+8=0$
C $x^2-14 x+49=0$
D $x^2-10 x+21=0$

The correct answer is:
$x^2-10 x+21=0$

Solution:
Since, $f(x)= \begin{cases}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^2-1\right)$
$\begin{aligned}
& =\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right] \\
& =\lim _{h \rightarrow 0}\left[h^2-4 h+3\right]=0-0+3=3
\end{aligned}$
and $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)$
$=\lim _{h \rightarrow 0}[2(2+h)+3]=\lim _{h \rightarrow 0}[4+2 h+3]=7$
Hence the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7) x+(3+7)=0 \Rightarrow x^2$ $-10 x+21=0$

Topic: Mathematics, Limits, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $f(x)=\left\{\begin{array}{cl}\frac{x^{2}+3 x-10}{x^{2}+2 x-15} & \text { , when } x \neq-5 \\ \text { a }, & \text { when } x=-5\end{array}\right.$

is continuous at $x=-5$, then the value of 'a' will be

Options:
A $3 / 2$
B $7 / 8$
C $8 / 7$
D $2 / 3$

The correct answer is:
$7 / 8$

Solution:
$\lim _{x \rightarrow-5} f(x)=\frac{(x-2)(x+5)}{(x+5)(x-3)}=\frac{-7}{-8}=\frac{7}{8}$

Topic: Mathematics, Limits, BITSAT, BITSAT 2013
If $f(x)=x^{\alpha} \log x$ and $f(0)=0,$ then the value of $\alpha$ for which Rolle's theorem can be applied in [0,1] is

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

The correct answer is:
$1 / 2$

Solution:
For Rolle's theorem in $[\mathrm{a}, \mathrm{b}], \mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b}),$
$\operatorname{In}[0,1] \Rightarrow \mathrm{f}(0)=\mathrm{f}(1)=0$
$\because$ the function has to be continuous in [0,1] $\Rightarrow f(0)=\lim _{x \rightarrow 0^{+}} f(x)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{\alpha} \log x=0 \Rightarrow \lim _{x \rightarrow 0} \frac{\log x}{x^{-\alpha}}=0$
Applying L.H. Rule $\lim _{x \rightarrow 0} \frac{1 / x}{-\alpha x^{-\alpha-1}}=0$
$\Rightarrow \lim _{x \rightarrow 0} \frac{-x^{\alpha}}{\alpha}=0 \Rightarrow \alpha>0$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2012
If force $(F)$, length $(L)$ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

Options:
A $\left[\mathrm{FL}^{-1} \mathrm{~T}^{2}\right]$
B $\left[\mathrm{FL}^{-1} \mathrm{~T}^{-2}\right]$
C $\left[\mathrm{FL}^{-1} \mathrm{~T}^{-1}\right]$
D $\left[\mathrm{FL}^{2} \mathrm{~T}^{2}\right]$

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

Solution:
Let $\mathrm{m}=\mathrm{KF}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}$

Substituting the dimensions of

$[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right],[\mathrm{L}]=[\mathrm{L}]$ and $[\mathrm{T}]=[\mathrm{T}]$ and comparing both side, we get $\mathrm{m}=\mathrm{FL}^{-1} \mathrm{~T}^{2}$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2021
If \(\frac{1-i \alpha}{1+i \alpha}=A+i B\), then \(A^2+B^2\) equals to

Options:
A 1
B \(\alpha^2\)
C -1
D \(-\alpha^2\)

The correct answer is:
1

Solution:
\(\begin{aligned}
& A+i B=\frac{1-i \alpha}{1+i \alpha} \Rightarrow A-i B=\frac{1+i \alpha}{1-i \alpha} \\
& \Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1 \\
& \Rightarrow A^2+B^2=1
\end{aligned}\)

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2010
If \(\frac{2 x+3}{5} < \frac{4 x-1}{2}\), then \(x\) lies in the interval

Options:
A \(\left[0, \frac{11}{16}\right)\)
B \(\left[\frac{11}{16}, \infty\right)\)
C \(\left(0, \frac{11}{16}\right)\)
D \(\left(\frac{11}{16}, \infty\right)\)

The correct answer is:
\(\left(\frac{11}{16}, \infty\right)\)

Solution:
\(\begin{aligned}
& \frac{2 x+3}{5} < \frac{4 x-1}{2} \Rightarrow-16 x < -11 \\
& \Rightarrow 16 x > 11 \Rightarrow x > \frac{11}{16} \\
& \text { Hence, } x \in\left(\frac{11}{16}, \infty\right)
\end{aligned}\)

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2009
If \(f(t)=\frac{1-t}{1+t}\), then \(f^{\prime}(1 / t)\) is equal to

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

The correct answer is:
\(\frac{-2 t^2}{(t+1)^2}\)

Solution:
\(\begin{aligned}
& f^{\prime}(t)=\frac{d}{d t}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t) \times(1)}{(1+t)^2} \\
& =\frac{-1-t-1+t}{(1+t)^2}=\frac{-2}{(1+t)^2}
\end{aligned}\)
\(f^{\prime}[1 / t]=\frac{-2}{\left(1+\frac{1}{t}\right)^2}=\frac{-2 t^2}{(t+1)^2}\)

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2009
If \(f(x)\) is a function that is odd and even simultaneously, then \(\mathrm{f}(3)-\mathrm{f}(2)\) is equal to

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

The correct answer is:
0

Solution:
\(\mathrm{f}(\mathrm{x})=0 ~\forall~ \mathrm{x} \in \mathrm{R} \Rightarrow \mathrm{f}(3)-\mathrm{f}(2)=0\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2011
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is

Options:
A \(\mathrm{R}\)
B \(\mathrm{R}-\{-1\}\)
C \((-\infty,-1)\)
D \((-1, \infty)\)

The correct answer is:
\(\mathrm{R}-\{-1\}\)

Solution:
Let \(f(x)=y\). Then \(\frac{1-x}{1+x}=y\)
\(\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}\)
Thus, \(f^{-1}(x)=\frac{1-x}{1+x}\)
Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\).
Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2011
If \(f(x)=\frac{x}{\sqrt{1+x^2}}\), then (fof of \()(x)\) is

Options:
A \(\frac{3 x}{\sqrt{1+x^2}}\)
B \(\frac{\mathrm{x}}{\sqrt{1+3 \mathrm{x}^2}}\)
C \(\frac{3 x}{\sqrt{1-x^2}}\)
D None of these

The correct answer is:
\(\frac{\mathrm{x}}{\sqrt{1+3 \mathrm{x}^2}}\)

Solution:
\(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\)
\((f o f)(x)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}\)
\((fofof)(x)=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2010
If \(f(x)=\left\{\begin{array}{l}x ; \text { when } x \text { is rational } \\ 1-x ; \text { when } x \text { is irrational }\end{array}\right.\), then fof \((x)\) is given as

Options:
A 1
B \(\mathrm{x}\)
C \(1+\mathrm{x}\)
D None of these

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

Solution:
\(\begin{aligned}
& f o f(x)=\left\{\begin{array}{l}
f(x) ; \text { when } f(x) \text { is rational } \\
1-f(x) \text {; when } f(x) \text { is irrational }
\end{array}\right. \\
& =\left\{\begin{array}{l}
\mathrm{x} ; \quad \text { when } \mathrm{x} \text { is rational } \\
1-(1-\mathrm{x}) ; \text { when } \mathrm{x} \text { is irrational }
\end{array}=\mathrm{x}\right. \\
&
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2011
If $g$ is the inverse of function $f$ and $f^{\prime}(x)=\sin x$, then $\mathrm{g}^{\prime}(\mathrm{x})$ is equal to

Options:
A $\operatorname{cosec}\{\mathrm{g}(\mathrm{x})\}$
B $\sin \{g(x)\}$
C $\frac{1}{\sin \{g(x)\}}$
D None of these

The correct answer is:
$\frac{1}{\sin \{g(x)\}}$

Solution:
Since, $g$ is the inverse of function $f$. Therefore, $\mathrm{g}(\mathrm{x})=\mathrm{f}^{-1}(\mathrm{x})$

$\Rightarrow \mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{x}$

$\Rightarrow$ fog $(\mathrm{x})=\mathrm{x},$ for all $\mathrm{x}$

Differentiate both side,w.r.tx

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\{\operatorname{fog}(\mathrm{x})\}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}),$ for all $\mathrm{x}$

$\Rightarrow \mathrm{f}^{\prime}[\mathrm{g}(\mathrm{x})] \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$

$\Rightarrow \sin \{\mathrm{g}(\mathrm{x})\} \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$

(By defn of $\left.f^{\prime}(x)\right)$

$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\sin \{\mathrm{g}(\mathrm{x})\}}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2015
If $g(x)=x^2+x-2$ and $\frac{1}{2} g o f(x)=2 x^2-5 x+2$ then $f(x)$ is equal to

Options:
A
$2 x-3$

B
$2 x+3$

C
$2 x^2+3 x+1$

D
$2 x^2-3 x+1$

The correct answer is:

$2 x-3$

Solution:
We have, $\frac{1}{2} go f(x)=2 x^2-5 x+2$
$\begin{aligned}
& \Rightarrow g(f(x))=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-2=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm(4 x-5)}{2}=2 x-3,-2 x+2
\end{aligned}$
Hence, $f(x)=2 x-3$

Topic: Mathematics, Functions, BITSAT, BITSAT 2022
If h is the altitude of a parallelopiped determined by the vectors a^, b^, c^ and the base is taken to be the parallelogram determined by a^ and b^ where a^=i^+j^+k^, b^=2i^+4j^-k^ and c^=i^+j^+3k^, then the value of 19h2 is

Options:
A 19
B 16
C 8
D None of these

The correct answer is:
8

Solution:
Volume of the parallelopiped =a^ b^ c^

⇒ (area of the base parallelogram) ×h=a^ b^ c^

⇒a^×b^h=a^ b^ c^

⇒-5i^+3j^+2k^h=11124-1113

⇒38h=4⇒h=438

⇒38h2=42⇒19h2=8

Hence, option (c) is correct.

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2018
If hydrogen atom, an electron jumps from bigger orbit to smaller orbit so that radius of smaller orbit is one-fourth of radius of bigger orbit. If speed of electron in bigger orbit was $\mathrm{v}$, then speed in smaller orbit is

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

The correct answer is:
$2 v$

Solution:
Radius of the orbit, $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^{2}$

$$

\begin{aligned}

& \frac{\mathrm{r}_{\mathrm{nbig}}}{\mathrm{r}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {big }}^{2}}{\mathrm{n}_{\text {small }}^{2}}=\frac{4}{1} \\

\Rightarrow & \frac{\mathrm{n}_{\text {big }}}{\mathrm{n}_{\text {small }}}=2 \\

\Rightarrow & \frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2}

\end{aligned}

$$

Velocity of electron in $\mathrm{n}^{\text {th }}$ orbit

$$

\begin{array}{c}

\quad \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\

\frac{\mathrm{v}_{\mathrm{n} \mathrm{big}}}{\mathrm{v}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2} \\

\Rightarrow \mathrm{v}_{\mathrm{n} \text { small }}=2\left(\mathrm{v}_{\mathrm{n} \mathrm{big}}\right)=2 \mathrm{v}

\end{array}

$$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2014
If in a $\triangle A B C, 2 b^2=a^2+c^2$, then $\frac{\sin 3B}{\sin B}$ is equal to

Options:
A
$\frac{c^2-a^2}{2 c a}$

B
$\frac{c^2-a^2}{c a}$

C
$\frac{\left(c^2-a^2\right)^2}{(c a)^2}$

D
$\left[\frac{c^2-a^2}{2 c a}\right]^2$

The correct answer is:

$\left[\frac{c^2-a^2}{2 c a}\right]^2$

Solution:
$\frac{\sin 3 B}{\sin B}=\frac{3 \sin B-4 \sin ^3 B}{\sin B}$
$=3-4 \sin ^2 B$
$=3-4\left(1-\cos ^2 B\right)$
$=-1+\frac{4\left(a^2+c^2-b^2\right)^2}{4(a c)^2}$
$=-1+\frac{\left(\frac{a^2+c^2}{2}\right)^2}{(a c)^2}$
$=-1+\frac{\left(a^2+c^2\right)^2}{4(a c)^2}$
$=(\frac{c^2-a^2}{2 a c})^2$

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2022
If in a binomial distribution $\mathrm{n}=4, \mathrm{P}(\mathrm{X}=0)=\frac{16}{81}$, then $\mathrm{P}(\mathrm{X}=4)$ equals

Options:
A $\frac{1}{16}$
B $\frac{1}{81}$
C $\frac{1}{27}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{81}$

Solution:
Given $n=4$ and $P(X=0)=\frac{16}{81}$

Let $\mathrm{p}$ be the probability of success and $\mathrm{q}$ that of failure in a trial.

$$

\text { Then, } \mathrm{P}(\mathrm{X}=0)={ }^{4} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{4}=\frac{16}{81}

$$

$\Rightarrow \quad \mathrm{q}^{4}=\left(\frac{2}{3}\right)^{4}$

$\left(\because{ }^{n} C_{0}=1\right)$

$\Rightarrow \quad \mathrm{q}=\frac{2}{3} \Rightarrow \mathrm{p}=\frac{1}{3}$

$\therefore \quad \mathrm{P}(\mathrm{X}=4)={ }^{4} \mathrm{C}_{4} \mathrm{p}^{4} \mathrm{q}^{0}=\mathrm{p}^{4}=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2015
If in the expansion of \(\left(2^x+\frac{1}{4^x}\right)^n, T_3=7 T_2\) and sum of the binomial coefficients of second and third terms is 36 , then the value of \(x\) is -

Options:
A \(-1 / 3\)
B \(-1 / 2\)
C \(1 / 3\)
D \(1 / 2\)

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

Solution:
\(\begin{aligned}
& { }^n C_1+{ }^n C_2=36 \Rightarrow n=8 \\
& T_3=7 \mathrm{~T}_2 \Rightarrow\left(2^{\mathrm{x}}\right)^3=1 / 2 \\
& 3 \mathrm{x}=-1 \Rightarrow \mathrm{x}=-\frac{1}{3}
\end{aligned}\)

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2009
If \(\int \frac{2 \cos x-\sin x+\lambda}{\cos x+\sin x-2} d x\) \(=A \ln |\cos x+\sin x-2|+B x+C\).
Then the ordered triplet \(A, B, \lambda\) is -

Options:
A \(\left(\frac{1}{2}, \frac{3}{2},-1\right)\)
B \(\left(\frac{3}{2}, \frac{1}{2},-1\right)\)
C \(\left(\frac{1}{2},-1, \frac{3}{2}\right)\)
D \(\left(\frac{3}{2},-1, \frac{1}{2}\right)\)

The correct answer is:
\(\left(\frac{3}{2}, \frac{1}{2},-1\right)\)

Solution:
\(\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A} \ln |\cos \mathrm{x}+\sin \mathrm{x}-2|+\mathrm{Bx}+\mathrm{C}) \\
&= \mathrm{A} \frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}-2}+\mathrm{B} \\
&= \frac{\mathrm{A} \cos \mathrm{x}-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x}-2 \mathrm{~B}}{\cos \mathrm{x}+\sin \mathrm{x}-2} \\
& \therefore 2=\mathrm{A}+\mathrm{B} \text { or }-1=-\mathrm{A}+\mathrm{B} ; \lambda=-2 \mathrm{~B} \\
& \therefore \mathrm{A}=3 / 2, \mathrm{~B}=1 / 2, \lambda=-1
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2009
If \(\int_0^n[x] \mathrm{dx}=66\), then \(n=\)

Options:
A 24
B 9
C 12
D 7

The correct answer is:
12

Solution:
\(\begin{aligned}
& \int_0^n[x] d x=\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\ldots \ldots+ \\
& \int_{n-1}^n(n-1) d x \\
& =1+2+3+\ldots .+(n-1)=\frac{(n-1) n}{2}=66 \\
& \Rightarrow n(n-1)=132 \Rightarrow n=12
\end{aligned}\)

Topic: Mathematics, Pair of Lines, BITSAT, BITSAT 2009
If $\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ is square root of identity matrix of order 2 then -

Options:
A $1+\alpha^{2}+\beta \gamma=0$
B $1+\alpha^{2}-\beta \gamma=0$
C $1-\alpha^{2}+\beta \gamma=0$
D $\alpha^{2}+\beta \gamma=1$

The correct answer is:
$\alpha^{2}+\beta \gamma=1$

Solution:
$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]=\sqrt{\mathrm{I}_{2}}$;

$\left[\begin{array}{lc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow \alpha^{2}+\beta \gamma=1$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2014
If θ is the angle between the vectors 4i^-k^ and i^+j^+k^, then sinθ+cosθ equals to

Options:
A 0
B 12
C 1
D 2

The correct answer is:
1

Solution:
Let

a→=i^+j^+k^

b→=4i^-k^

Then,

a→=12+12+12=3

b→=42+02+-42=42

Now,

cosθ=a→·b→a→b→

⇒cosθ=i^+j^+k^·4i^-4k^a→b→

⇒cosθ=4-4a→b→=0

Hence, sinθ=1-cos2θ=1

Therefore, sinθ+cosθ=1

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2018
If $\omega$ is the complex cube root of unity, then the value of $\omega+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots\right)}$ is

Options:
A -1
B 1
C $-\mathrm{i}$
D i

The correct answer is:
-1

Solution:
Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots .$

Which can be written as

$$

\frac{3^{0}}{2^{1}}+\frac{3^{1}}{2^{3}}+\frac{3^{2}}{2^{5}}+\frac{3^{3}}{2^{7}}+\ldots .

$$ $=\frac{1}{2}\left[1+\frac{3}{2^{2}}+\frac{3^{2}}{2^{4}}+\frac{3^{3}}{2^{6}}+\ldots .\right]$

Since $\left(1+\frac{3}{2^{2}}+\frac{3^{2}}{2^{4}}+_{--}-\right)$ is a G.P.

therfore by sum of infinite G.P, we have

$$

=\frac{1}{2}\left[\frac{1}{1-\frac{3}{2^{2}}}\right]=\frac{1}{2}\left[\frac{1}{1-\frac{3}{4}}\right]=2

$$

$\therefore \quad$ Given expression

$$

\begin{array}{l}

=-1 \\

{\left[\because 1+\omega+\omega^{2}=0\right]}

\end{array}

$$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2015
If $k \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq K,$ then $-$

Options:
A $\mathrm{k}=0, \mathrm{~K}=\pi$
B $\mathrm{k}=0, \mathrm{~K}=\pi / 2$
C $\mathrm{k}=\pi / 2, \mathrm{~K}=\pi$
D None of these

The correct answer is:
None of these

Solution:
$\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}+\tan ^{-1} x$
Since domain of the function $\mathrm{x} \in[-1,1]$ $\therefore-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4}$
Hence, $\mathrm{k}=\frac{\pi}{4}$ and $\mathrm{K}=\frac{3 \pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2012
If limn→∞an-1+n21+n=b, where a is finite number, then

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

The correct answer is:
b=1

Solution:
limn→∞an(1+n)-1+n21+n

=limn→∞(a-1)n2+an-1n+1=∞,

If a-1≠0 limit does not exist and if a-1=0, then

limn→∞an-1n+1=a=b

⇒ a=b=1

Topic: Mathematics, Limits, BITSAT, BITSAT 2017
If $\lim _{x \rightarrow} x \sin \left(\frac{1}{x}\right)=A$ and $\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=B$,

then which one of the following is correct?

Options:
A $A=1$ and $B=0$
B $\quad A=0$ and $B=1$
C $A=0$ and $B=0$
D $A=1$ and $B=1$

The correct answer is:
$A=1$ and $B=0$

Solution:
As given,

$$

\mathrm{A}=\lim _{\mathrm{x} \rightarrow \infty} \mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)=\lim _{\mathrm{x} \rightarrow \infty} \frac{\sin \left(\frac{1}{\mathrm{x}}\right)}{\left(\frac{1}{\mathrm{x}}\right)}

$$

Let $\mathrm{t}=\frac{1}{\mathrm{x}}$ when $\mathrm{x} \rightarrow \alpha, \mathrm{t} \rightarrow 0$

$$

\begin{aligned}

\Rightarrow \mathrm{A} &=\lim _{t \rightarrow \infty} \frac{\sin \mathrm{t}}{\mathrm{t}}=1 \\

&\left[\because \lim _{\mathrm{t} \rightarrow 0} \frac{\sin \mathrm{x}}{\mathrm{x}}=1\right]

\end{aligned}

$$

and $\mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)$

$$

\Rightarrow \mathrm{B}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \cdot \lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{1}{\mathrm{x}}\right)

$$

$$

\Rightarrow \mathrm{B}=0

$$

$\therefore \quad A=1$ and $B=0$ is correct

Topic: Mathematics, Limits, BITSAT, BITSAT 2015
If $\log$ a, $\log \mathrm{b}$, and $\log \mathrm{c}$ are in A.P. and also $\log$ a $-\log 2 b, \log 2 b-\log 3 c, \log 3 c-\log$ a are in A.P.,

then

Options:
A $\mathrm{a}, \mathrm{b}, \mathrm{c},$ are in $\mathrm{H.P.}$
B $a, 2 b, 3 c$ are in A.P.
C $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the sides of a triangle
D none of the above

The correct answer is:
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the sides of a triangle

Solution:
\(\log \mathrm{a}, \log \mathrm{b}, \log \mathrm{c}\) are in A.P.

\(\begin{aligned}

&\Rightarrow 2 \log \mathrm{b}=\log \mathrm{a}+\log \mathrm{c} \\

&\Rightarrow \log \mathrm{b}^{2}=\log (\mathrm{ac}) \\

&\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are in G.P. } \\

&\log \mathrm{a}-\log 2 \mathrm{~b}, \log 2 \mathrm{~b}-\log 3 \mathrm{c}, \log 3 \mathrm{c}-\log \text { a are in A.P. }

\end{aligned}\)

\(\Rightarrow 2(\log 2 b-\log 3 c)=(\log a-\log 2 b)+(\log 3 c-\log a)\)

\(\Rightarrow 3 \log 2 \mathrm{~b}=3 \log 3 \mathrm{c} \Rightarrow 2 \mathrm{~b}=3 \mathrm{c}\)

Now, \(b^{2}=a c \Rightarrow b^{2}=a \cdot \frac{2 b}{3} \Rightarrow b=\frac{2 a}{3}, c=\frac{4 a}{9}\)

\(\begin{aligned}

&\text { i.e., } a=a, b=\frac{2 a}{3}, c=\frac{4 a}{9} \\

&\Rightarrow a: b: c=1: \frac{2}{3}: \frac{4}{9}=9: 6: 4

\end{aligned}\)

Since, sum of any two is greater than the \(3^{\mathrm{rd}}, \mathrm{a}, \mathrm{b}, \mathrm{c}\) form a triangle.

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2016
If $m$ is magnetic moment and $B$ is the magnetic field, then the torque is given by

Options:
A $\vec{m} \cdot \vec{B}$
B $\frac{|\vec{m}|}{|\vec{B}|}$
C $\vec{m} \times \vec{B}$
D $|\vec{m}| \cdot|\vec{B}|$

The correct answer is:
$\vec{m} \times \vec{B}$

Solution:
$\vec{\tau}=\vec{m} \times \vec{B}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2012
If $M$ is magnetic moment and $B$ is the magnetic field, then the torque is given by

Options:
A $\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}$
B $\frac{|\overrightarrow{\mathrm{M}}|}{|\overrightarrow{\mathrm{B}}|}$
C $\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$
D $|\overrightarrow{\mathrm{M}} \| \overrightarrow{\mathrm{B}}|$

The correct answer is:
$\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$

Solution:
Torque, $\tau=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2011
If $\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1$ touches the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ then

Options:
A $\mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$
B $\mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}+1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}+1}$
C $\mathrm{m}^{2}=\frac{\mathrm{n}^{2}+1}{\mathrm{n}^{2}}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}+1}{\mathrm{~m}^{2}}$
D $\mathrm{m}^{2}=\frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}-1}{\mathrm{~m}^{2}}$

The correct answer is:
$\mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$

Solution:
The line is $y=-\frac{\text { nb }}{\text { ma }} x-n b$. It will touch the ellipse if $(-n b)^{2}=a^{2}\left(-\frac{n b}{m a}\right)^{2}+b^{2}$

$\left[\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}\right]$

$\Rightarrow n^{2}=\frac{n^{2}}{m^{2}}+1$

$\Rightarrow \mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2020