datavorous/entrance-exam-dataset huggingface dataset:
If \(A=\left[\begin{array}{lll}2 & 0 & 0 \\ 2 & 2 & 0 \\ 2 & 2 & 2\end{array}\right]\), then \(\operatorname{adj}(\operatorname{adj} A)\) is equal to -

Options:
A \(8\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
B \(16\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
C \(64\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)
D None of these

The correct answer is:
\(16\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right]\)

Solution:
\(|A|=\left[\begin{array}{lll}
2 & 0 & 0 \\
2 & 2 & 0 \\
2 & 2 & 2
\end{array}\right]=(2)(2)(2)=8\)
Now adj \((\operatorname{adj} A)=|\mathrm{A}|^{3-2} \mathrm{~A}\)
\(=8\left[\begin{array}{lll}
2 & 0 & 0 \\
2 & 2 & 0 \\
2 & 2 & 2
\end{array}\right]=16\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]\)

Topic: Mathematics, Matrices, BITSAT, BITSAT 2011
If \(\alpha, \beta\) are the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\), then the roots of the equation \(\mathrm{ax}^2+\mathrm{bx}(\mathrm{x}+1)+\mathrm{c}(\mathrm{x}+1)^2=0\) are

Options:
A \(\alpha-1, \beta-1\)
B \(\alpha+1, \beta+1\)
C \(\frac{\alpha}{\alpha-1}, \frac{\beta}{\beta-1}\)
D \(\frac{\alpha}{1-\alpha}, \frac{\beta}{1-\beta}\)

The correct answer is:
\(\frac{\alpha}{1-\alpha}, \frac{\beta}{1-\beta}\)

Solution:
The second equation can be rewritten as
\(a\left(\frac{x}{x+1}\right)^2+b\left(\frac{x}{x+1}\right)+c=0\)
and hence its roots correspond to \(\frac{\mathrm{x}}{\mathrm{x}+1}=\alpha\) and \(\frac{x}{x+1}=\beta\).
Hence \(x=\frac{\alpha}{1-\alpha}\) and \(\frac{\beta}{1-\beta}\).

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2011
If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

Options:
A a straight line
B a circle
C an ellipse
D a parabola

The correct answer is:
a parabola

Solution:
The path is a parabola, because initial velocity can be resolved into two rectangular components, one along $\overrightarrow{\mathrm{E}}$ and other $\perp$ to $\overrightarrow{\mathrm{E}}$. The former decreases at a constant rate and latter is unaffected. The resultant path is therefore a parabola.

Topic: Physics, Electrostatics, BITSAT, BITSAT 2009
If $\alpha$ and $\beta$ are the roots of $x^{2}-x+1=0,$ then the equation whose roots are $\alpha^{100}$ and $\beta^{100}$ are

Options:
A $x^{2}-x+1=0$
B $x^{2}+x-1=0$
C $x^{2}-x-1=0$
D $x^{2}+x+1=0$

The correct answer is:
$x^{2}+x+1=0$

Solution:
We have $x^{3}+1 \equiv(x+1)\left(x^{2}-x+1\right)$. Therefore, $\alpha$ and $\beta$ are the complex cube roots of -1 so that we may take $\alpha=-\omega$ and $\beta=-\omega^{2},$ where $\omega \neq 1$ is a cube root of unity.

Thus $\alpha^{100}=(-\omega)^{100}=\omega$ and

$\beta^{100}=\left(-\omega^{2}\right)^{100}=\omega^{2},$ so that the required

equation is $\mathrm{x}^{2}+\mathrm{x}+1=0$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2013
If $\alpha, \beta$ are the roots of the equations $x^{2}-2 x-1=$ $0,$ then what is the value of $\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}$

Options:
A -2
B 0
C 30
D 34

The correct answer is:
34

Solution:
Since, \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x-1=0\), then

Sum of roots, \(\alpha+\beta=2\) and

product of the roots \(\alpha \beta=-1\)

Since, \((\alpha+\beta)=\alpha^{2}+\beta^{2}+2 \alpha \beta\)

\(\Rightarrow 4=\alpha^{2}+\beta^{2}-2\)

\(\Rightarrow \alpha^{2}+\beta^{2}=6\)

Now, \(\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}=\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}=\frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}\)

\(\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}=6^{2}\)

\(\Rightarrow \alpha^{4}+\beta^{4}+2 \alpha^{2} \beta^{2}=36\)

\(\Rightarrow \alpha^{4}+\beta^{4}+2=36\)

\(\Rightarrow \alpha^{4}+\beta^{4}=34 \ldots . .(\mathrm{i})\)

\(\Rightarrow \frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}=\frac{34}{(-1)^{2}}=34\)

[Putting value of \(\alpha^{4}+\beta^{4}=34\) from Equation (i) \(]\)

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2016
If \(\arg \left(\bar{z}_1\right)=\arg \left(z_2\right)\), then

Options:
A \(\mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)\)
B \(\mathrm{z}_2=k \mathrm{z}_1(\mathrm{k} > 0)\)
C \(\left|z_2\right|=\left|\bar{z}_1\right|\)
D None of these

The correct answer is:
\(\mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)\)

Solution:
\(\begin{aligned}
& \bar{z}_1=\frac{z_1 \bar{z}_1}{z_1}=\left|\mathrm{z}_1\right|^2 \mathrm{z}_1^{-1} \\
& \Rightarrow \arg \left(\mathrm{z}_1^{-1}\right)=\arg \left(\bar{z}_1\right) \Rightarrow \arg \left(\mathrm{z}_2\right) \\
& \Rightarrow \mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2009
If $\alpha$ be a root of the equation $4 x^2+2 x-1=0$, then the other root of the equation is

Options:
A
$4 \alpha^3+2 \alpha$

B
$4 \alpha^2-2 \alpha$

C
$4 \alpha^3-3 \alpha$

D
$4 \alpha^3+3 \alpha$

The correct answer is:

$4 \alpha^3-3 \alpha$

Solution:
$4 x^2+2 x-1=0$
$\therefore \quad \alpha+\beta=-\frac{1}{2}, \alpha \beta=-\frac{1}{4}$
Also, $4 \alpha^2+2 \alpha-1=0$ as $\alpha$ is a root and we have to prove that $\beta=4 \alpha^3-3 \alpha$.
Now, $4 \alpha^3-3 \alpha=4 \alpha^2 \cdot \alpha-3 \alpha$
$=\alpha(1-2 \alpha)-3 \alpha=-2 \alpha^2-2 \alpha$
$=-\frac{1}{2}\left[4 \alpha^2+4 \alpha\right]=-\frac{1}{2}[1-2 \alpha+4 \alpha]$
$=-\frac{1}{2}(1+2 \alpha)=-\frac{1}{2}-\alpha=\beta$
Now, $\alpha+\beta=-\frac{1}{2}$
[from Eq. (i)]
Hence, $\beta=4 \alpha^3-3 \alpha$.

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2022
If binomial coefficients of three consecutive terms of $(1+x)^{2}$ are in $H P$, then the maximum value of $\mathrm{n}$ is

Options:
A 1
B 2
C 0
D None of these

The correct answer is:
None of these

Solution:
Let the coefficients of rth, $(r+1)$ th, and (r $+$

2) th terms be in HP.

Then, $\frac{2}{{ }^{n} C_{r}}=\frac{1}{{ }^{n} C_{r-1}}+\frac{1}{{ }^{n} C_{r}+1}$

$$

\begin{array}{l}

\Rightarrow \quad 2=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}} \\

\Rightarrow \quad 2=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}+\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}} \\

\Rightarrow \mathrm{n}^{2}-4 \mathrm{nr}+4 \mathrm{r}^{2}+\mathrm{n}=0 \\

\Rightarrow(\mathrm{n}-2 \mathrm{r})^{2}+\mathrm{n}=0

\end{array}

$$

which is not possible for any value for $n$.

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2015
If $\mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots \ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}}$ denote the binomial

coefficientsin the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, then the value of $\mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots$

$+\left(C_{0}+C_{1}+\ldots . .+C_{n-1}\right)$

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

The correct answer is:
$\mathrm{n} \cdot 2^{\mathrm{n}-1}$

Solution:
$\begin{aligned} & \mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots \ldots \ldots \ldots \ldots \\ &+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}-1}\right) \\ &=\mathrm{n} \mathrm{C}_{0}+(\mathrm{n}-1) \mathrm{C}_{1}+(\mathrm{n}-2) \mathrm{C}_{2}+\ldots \ldots \mathrm{C}_{\mathrm{n}-1} \\ &=\mathrm{C}_{1}+2 \mathrm{C}_{2}+3 \mathrm{C}_{3}+4 \mathrm{C}_{4} \ldots \ldots \mathrm{n} \mathrm{C}_{\mathrm{n}}=\mathrm{n} .2^{\mathrm{n}-1} \end{aligned}$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2013
If $\frac{\cos A}{\cos B}=n, \frac{\sin A}{\sin B}=m$, then the value of $\left(m^{2}-n^{2}\right) \sin ^{2} B$ is

Options:
A $1+\mathrm{n}^{2}$
B $1-\mathrm{n}^{2}$
C $\mathrm{n}^{2}$
D $-\mathrm{n}^{2}$

The correct answer is:
$1-\mathrm{n}^{2}$

Solution:
$\cos A=n \cos B$ and $\sin A=m \sin B$

Squaring and adding, we get $1=n^{2} \cos ^{2} B+m^{2} \sin ^{2} B$

$$

\begin{array}{ll}

\Rightarrow & 1=n^{2}\left(1-\sin ^{2} B\right)+m^{2} \sin ^{2} B \\

\therefore & \left(m^{2}-n^{2}\right) \sin ^{2} B=1-n^{2}

\end{array}

$$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2015
If complex number $z_{1}, z_{2}$ and 0 are vertices of equilateral triangle, then $z_{1}^{2}+z_{2}^{2}-z_{1} z_{2}$ is equal to

Options:
A $0$
B $z_{1}-z_{2}$
C $z_{1}+z_{2}$
D 1

The correct answer is:
$0$

Solution:
$z_{1}, z_{2}, 0$ are vertices of an equilateral triangle, so we have $z_{1}^{2}+z_{2}^{2}+0^{2}=z_{1} z_{2}+z_{2} \cdot 0+0 \cdot z_{1}$

$$

\Rightarrow z_{1}^{2}+z_{2}^{2}=z_{1} z_{2} \Rightarrow z_{1}^{2}+z_{2}^{2}-z_{1} z_{2}=0

$$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2015
If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$,

then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to

Options:
A $2 \sin 2 \alpha$
B 4
C $4 \sin ^{2} \alpha$
D $-4 \sin ^{2} \alpha$

The correct answer is:
$4 \sin ^{2} \alpha$

Solution:
$\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$

$\Rightarrow \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}\right)=\alpha$

$\Rightarrow \cos ^{-1}\left(\frac{x y+\sqrt{4-y^{2}-4 x^{2}+x^{2} y^{2}}}{2}\right)=\alpha$

$\Rightarrow 4 x^{2}+y^{2}-4 x y \cos \alpha=4 \sin ^{2} \alpha$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2015
$\quad$ If $\cos 7 \theta=\cos \theta-\sin 4 \theta,$ then the general value

of $\theta$ is

Options:
A $\frac{n \pi}{6}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$
B $\frac{n \pi}{3}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$
C $\frac{n \pi}{4}, \frac{n \pi}{3} \pm \frac{\pi}{18}$
D $\frac{n \pi}{4}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$

The correct answer is:
$\frac{n \pi}{4}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$

Solution:
$\cos 7 \theta=\cos \theta-\sin 4 \theta$

$\Rightarrow \sin 4 \theta=\cos \theta-\cos 7 \theta$

$\Rightarrow \sin 4 \theta=2 \sin 4 \theta \sin 3 \theta$

$\Rightarrow \sin 4 \theta(1-2 \sin 3 \theta)=0$

$\therefore \sin 4 \theta=0$ or $\sin 3 \theta=\frac{1}{2}$

$\Rightarrow 4 \theta=n \pi$ or $3 \theta=n \pi+(-1)^{n} \frac{\pi}{6}$

$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2013
If cos(x-y),cosx and cos(x+y) are in HP, then cosxsec(y/2) is equal to

Options:
A ±2
B ±1/2
C ±2
D None of these

The correct answer is:
±2

Solution:
Given that, cos(x-y),cosx,cos(x+y) are in HP.

Then, cosx=2cos(x-y)cos(x+y)cos(x-y)+cos(x+y)

⇒cosx=2cos2x-sin2y2cosxcosy

⇒cos2xcosy=cos2x-sin2y

⇒cos2x(cosy-1)=-sin2y

⇒cos2x(1-cosy)=1-cos2y

⇒cos2x(1-cosy)=(1-cosy)(1+cosy)

⇒cos2x=1+cosy⇒cos2x=2cos2y/2

⇒cos2xsec2(y/2)=2

⇒cosxsec(y/2)=±2

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2017
If [ ] denotes the greatest integer less than or equal to the real number under consideration and $-1 \leq x<0 ; 0 \leq y<1 ; 1 \leq z<2,$ then the value of the determinant

is

Options:
A $[z]$
B $[y]$
C $[\mathrm{x}]$
D None of these

The correct answer is:
$[z]$

Solution:
Since, $-1 \leq x<0$ $\begin{array}{ll}\therefore \quad & {[\mathrm{x}]=-1} \\ & 0 \leq \mathrm{y}<1 \quad \therefore[\mathrm{y}]=0 \\ & 1 \leq \mathrm{z}<2 \quad \therefore[\mathrm{z}]=1 \\ \therefore \quad & \text { Given determinant }=\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2\end{array}\right|=1=[\mathrm{z}]\end{array}$

Topic: Mathematics, Determinants, JEE Main
If [ ] denotes the greatest integer less than or equal to the real number under consideration and $-1 \leq x<0 ; 0 \leq y<1 ; 1 \leq z<2,$ then the value of the determinant

is

Options:
A $[z]$
B $[y]$
C $[\mathrm{x}]$
D None of these

The correct answer is:
$[z]$

Solution:
Since, $-1 \leq x<0$ $\begin{array}{ll}\therefore \quad & {[\mathrm{x}]=-1} \\ & 0 \leq \mathrm{y}<1 \quad \therefore[\mathrm{y}]=0 \\ & 1 \leq \mathrm{z}<2 \quad \therefore[\mathrm{z}]=1 \\ \therefore \quad & \text { Given determinant }=\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2\end{array}\right|=1=[\mathrm{z}]\end{array}$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2016
If e1 and e2 are the eccentricities of a hyperbola 3x2-3y2=25 and its conjugate, then

Options:
A e12+e22=2
B e12+e22=4
C e1+e2=4
D e1+e2=2

The correct answer is:
e12+e22=4

Solution:
Given equation can be written as

x2-y2=253

∴ e1=1+b2a2

=1+1=2

The equation of conjugate hyperbola is

-x2+y2=253

∴ e2=1+b2a2=1+1=2

∴ e12+e22=22+22=4

Hence, e12+e22=4.

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2018
If $\frac{e^{x}+e^{5 x}}{e^{3 x}}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots$ then

the value of $2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots$ is

Options:
A $e^{2}+e^{-2}$
B $e^{4}-e^{-4}$
C $e^{t}+e^{-1}$
D 0

The correct answer is:
0

Solution:
Let $\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{5 \mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{a}^{3}+\ldots .$

$\begin{aligned} &=\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}+\frac{\mathrm{e}^{5 \mathrm{x}}}{\mathrm{e}^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\ldots . \\=& \mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{x}^{3}+\ldots . \end{aligned}$

By using $\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+{ }_{--}--$ and

$\mathrm{e}^{-\mathrm{x}}=1-\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}-\frac{\mathrm{x}^{3}}{3 !}+---$

$\mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}=2\left[1+\frac{(2 \mathrm{x})^{2}}{21}+\frac{(2 \mathrm{x})^{4}}{41}+\ldots .\right]$

$$

\begin{array}{l}

=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} a^{3}+\ldots . \\

=a_{1}=a_{3}=a_{5}=\ldots=0

\end{array}

$$

Hence, $2 \mathrm{a}_{1}+2^{3} \mathrm{a}_{3}+2^{5} \mathrm{a}_{5}+\ldots .=0$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2015
If $\mathrm{E}$ and $\mathrm{F}$ are events such that $0 < \mathrm{P}(\mathrm{F}) < 1$, then

Options:
A $P(E \mid F)+P(\bar{E} \mid F)=1$
B $P(E \mid F)+P(E \mid \bar{F})=1$
C $P(\bar{E} \mid F)+P(E \mid \bar{F})=1$
D $P(E \mid \bar{F})+P(\bar{E} \mid \bar{F})=0$

The correct answer is:
$P(E \mid F)+P(\bar{E} \mid F)=1$

Solution:
$P(E \mid F)+P(\bar{E} \mid F)$
$=\frac{P(E \cap F)+P(\bar{E} \cap F)}{P(F)}=\frac{P((E \cup \bar{E}) \cap F)}{P(F)}=\frac{P(F)}{P(F)}=1$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{K}}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by: ( $\omega$ - angular frequency) :

Options:
A $\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$
B $\omega(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{K}})$
C $\omega(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$
D $\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}}$

The correct answer is:
$\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$

Solution:
For an EM wave
$\overrightarrow{\mathrm{B}}=\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g}), \mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure are 52,394 and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively, the change in enthalpy for combustion of $\mathrm{C}_2 \mathrm{H}_4$ is equal to

Options:
A $-141.2 \mathrm{~kJ} / \mathrm{mol}$
B $-1412 \mathrm{~kJ} / \mathrm{mol}$
C $+14.2 \mathrm{~kJ} / \mathrm{mol}$
D $+1412 \mathrm{~kJ} / \mathrm{mol}$

The correct answer is:
$-1412 \mathrm{~kJ} / \mathrm{mol}$

Solution:
Enthalpy of formation of $\mathrm{C}_2 \mathrm{H}_4, \mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ are $52,-394$ and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively.
(Given)
The reaction is
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
change in enthalpy,
$(\Delta H)=\Delta H_{\text {products }}-\Delta H_{\text {reactants }}$
$=2 \times(-394)+2 \times(-286)-(52+0)$
$=-1412 \mathrm{~kJ} / \mathrm{mol}$.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2023 (Memory Based Paper 1)
If equation (10x-5)2+(10y-4)2=λ2(3x+4y-1)2 represents a hyperbola, then

Options:
A -2<λ<2
B λ>2
C λ<-2 or λ>2
D 0<λ<2

The correct answer is:
λ<-2 or λ>2

Solution:
Given, equation of hyperbola is

10x-52+10y-42=λ23x+4y-12 can be rewritten as

x-122+y-2523x+4y-15=λ2

This is of the form of PSPM=e

Where, P is any point on the hyperbola and S is a focus and M is the point of directrix.

Here, λ2>1⇒λ>2 ∵e>1

⇒ λ<-2 or λ>2

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2017
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right) .$ Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-

Options:
A $\pi / 2 \mathrm{x}_{0}$
B $2 \pi x_{0}$
C $\pi$
D $\pi x_{0}$

The correct answer is:
$\pi x_{0}$

Solution:
$y=x_{0} \cos 2 \pi\left(\right.$ nt $\left.-\frac{x}{\lambda}\right)$

$\mathrm{y}=\mathrm{x}_{0} \cos \frac{2 \pi}{\lambda}(\mathrm{vt}-\mathrm{x}) \quad[\because \mathrm{v}=\mathrm{n} \lambda]$

$\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\mathrm{x}_{0} \times \frac{2 \pi}{\lambda} \mathrm{v}=2 \mathrm{v}($ given $) \therefore \lambda=\pi \mathrm{x}_{0}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2013
If f and g are differentiable function in 0,1 satisfying f0=2=g1, g0=0 and f1=6, then for some c∈0,1

Options:
A 2f'c=gc
B 2f'c=3g'c
C f'c=g'c
D f'c=2g'c

The correct answer is:
f'c=2g'c

Solution:
Let

hx=fx-2gx

h0=f0-2g0

h0=2-20 ∵f0=2,g0=0

h0=2

h1=f1-2g1

h(1)=6-2(2) ∵f1=6,g1=2

h1=2

h0=h1

h'x=f'x-2g'x

By Rolle's theorem h'c=0

∴ f'c-2g'c=0,f'c=2g'c

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2019
If $\mathrm{f}$ is an even function and $\mathrm{g}$ is an odd function, then the function fog is

Options:
A an even function
B an odd function
C neither even nor odd
D a periodic function

The correct answer is:
an even function

Solution:
We have, $f o g(-x)=f[g(-x)]=f[-g(x)]$ $(\because g$ is odd $)$
$\begin{array}{l}
=\mathrm{f}[\mathrm{g}(\mathrm{x})] \quad(\because \mathrm{fis} \text { even }) \\
=\operatorname{fog}(\mathrm{x}) \forall \mathrm{x} \in \mathrm{R}
\end{array}$
$\therefore$ fog is an even function.

Topic: Mathematics, Functions, BITSAT, BITSAT 2012
If $\mathrm{f}(\mathrm{x})=\sqrt{1+\cos ^2\left(\mathrm{x}^2\right)}$, then the value of $f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)$

Options:
A $\frac{\sqrt{\pi}}{6}$
B $-\sqrt{\frac{\pi}{6}}$
C $\frac{1}{\sqrt{6}}$
D $\frac{\pi}{\sqrt{6}}$

The correct answer is:
$-\sqrt{\frac{\pi}{6}}$

Solution:
On differentiating (i) w.r.t.x, we get
$f^{\prime}(x)=\frac{-2 \sin x^2 \cos x^2}{\sqrt{1+\cos ^2 x^2}}(x)$
Put, $x=\frac{\sqrt{\pi}}{2}$ in (ii), we get
$\begin{aligned}
& \mathrm{f}^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin 2\left(\frac{\pi}{4}\right)}{\sqrt{1+\frac{1}{2}}} \\
& =-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin \frac{\pi}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{\pi}{6}}
\end{aligned}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $f(x)=\frac{x}{\sqrt{1+x^2}}$, then (fof of) $(x)$ is

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

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

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

Topic: Mathematics, Functions, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{a}^{2}-1}{\mathrm{a}^{2}+1} \mathrm{x}^{3}-3 \mathrm{x}+5$ is a decreasing
function of $\mathrm{x}$ in $\mathbf{R}$, then the set of possible values of a (independent of $\mathrm{x}$ ) is

Options:
A $(1, \infty)$
B $(-\infty,-1)$
C [-1,1]
D None of these

The correct answer is:
[-1,1]

Solution:
$f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3$
$\mathrm{f}^{\prime}(\mathrm{x})<0$ for all $\mathrm{x}$ if $\mathrm{a}^{2}-1 \leq 0 \Rightarrow-1 \leq \mathrm{a} \leq 1$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2012
If $f(x)=\cos x, g(x)=\cos 2 x, h(x)=\cos 3 x$ and $I(x)$ $=\tan x$, then which of the following option is correct?

Options:
A $f(x)$ and $g(x)$ are strictly decreasing in $(0, \pi /$ 2)
B $h(x)$ is neither increasing nor decreasing in $(0, \pi / 2)$
C $\mathrm{I}(\mathrm{x})$ is strictly increasing in $(0, \pi / 2)$
D
All are correct

The correct answer is:
All are correct

Solution:
(a) Let $f(x)=\cos x$, then $f^{\prime}(x)=-\sin x$.
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(b) Let $f(x)=\cos 2 x \Rightarrow f^{\prime}(x)=-2 \sin x 2 x$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$
Because $\sin 2 x$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
(c) Let $f^{\prime}(x)=\cos 3 x$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-3 \sin 3 \mathrm{x}$. In Interval $\left(0, \frac{\pi}{3}\right), \mathrm{f}^{\prime}(\mathrm{x}) < 0$
Because $\sin 3 \mathrm{x}$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $\mathrm{f}(\mathrm{x})$ irstrictly decreasing on $\left(0, \frac{\pi}{3}\right)$.
When $\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$, then $\mathrm{f}^{\prime}(\mathrm{x})>0$
Because $\sin 3 x$ will lie in the third quadrant. Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$
(d) Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x$.
In Interval $\mathrm{x} \in\left(0, \frac{\pi}{2}\right), \mathrm{f}^{\prime}(\mathrm{x})>0$
Therefore, $f(x)$ is not strictly decreasing on
$\left(0, \frac{\pi}{2}\right)$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If fx=ex,x≤01-x,x>0, then

Options:
A fx is differentiable at x=0
B fx is continuous at x=0,1
C fx is differentiable at x=1
D None of the above

The correct answer is:
fx is continuous at x=0,1

Solution:
fx=ex,x≤01-x,0<x≤1x-1,x>1

RHD: f'0=limh→0f0+h-f0h=limh→0-h-1h=-1

LHD: f'0=limh→0f0-h-f0-h=limh→0e-h-1-h=1

So, it is not differentiable at x=0.

Similarly, it is not differentiable at x=1 but it is continuous at x=0 and 1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=ex,x≤01-x,x>0, then

Options:
A fx is differentiable at x=0
B fx is continuous at x=0,1
C fx is differentiable at x=1
D None of the above

The correct answer is:
fx is continuous at x=0,1

Solution:
fx=ex,x≤01-x,0<x≤1x-1,x>1

RHD: f'0=limh→0f0+h-f0h=limh→0-h-1h=-1

LHD: f'0=limh→0f0-h-f0-h=limh→0e-h-1-h=1

So, it is not differentiable at x=0.

Similarly, it is not differentiable at x=1 but it is continuous at x=0 and 1.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2017
If f'x=fx+∫01fxdx,f0=1, then fx=

Options:
A 2ex3-e+1-e3-e
B ex3-e+1+e1-e
C 3ex2-e+1+e1-e
D 3ex2-e+1-e3+e

The correct answer is:
2ex3-e+1-e3-e

Solution:
Given,

f'x=fx+∫01fxdx,f0=1 ...(i)

⇒ f''x=f'x+0

⇒ f''xf'x=1

⇒ ∫f''xf'xdx=∫dx

⇒ logf'x=x+C

⇒ f'x=Aex

⇒ fx=Aex+K

f0=A+K=1 ...(ii)

∴ Aex=Aex+K+∫01Aex+kdx

⇒ k+Aex+Kx01=0

⇒ k+Ae-A+K=0

⇒ A(e-1)+2k=0 ...(iii)

From Eqs. (ii) and (iii), we get

A=23-e,k=1-e3-e

∴ fx=2ex3-e+1-e3-e

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2019