datavorous/entrance-exam-dataset huggingface dataset:
Statement I : The ferromagnetic property depends on temperature. At high temperature, ferromagnet becomes paramagnet.

Statement II : At high temperature, the domain wall area of a ferromagnetic substance increases. In the light of the above statements, choose the most appropriate answer from the options given below:

Options:
A Statement I is true but Statement II is false.
B Both Statement I and Statement II are true.
C Both Statement I and Statement II are false.
D Statement I is false but Statement II is true.

The correct answer is:
Statement I is true but Statement II is false.

Solution:
As temperature increases, domains disintegrate so ferromagnetism decreases and above curie temperature it become paramagnet.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Statement I : The law of gravitation holds good for any pair of bodies in the universe.

Statement II : The weight of any person becomes zero when the person is at the centre of the earth.

In the light of the above statements, choose the correct answer from the options given below.

Options:
A Statement 1 and statement 2 both are true.
B Statement 1 is true and statement 2 is false.
C Statement 1 and statement 2 both are false.
D Statement 1 is false and statement 2 is true.

The correct answer is:
Statement 1 and statement 2 both are true.

Solution:
Gravitational force of attraction exists in all the places in the universe. This is the force of attraction between any two substances that are having a mass of their own.

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them.

The weight of a body placed at the centre of the earth is zero. This is because acceleration due to gravity is zero at the centre and hence weight mg is zero at the centre. Thus, both statements are true.

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Statement - I : The value of the integral ∫π/6π/3dx1+tan x is equal to π6.

Statement - II : ∫abfxdx=∫abfa+b-xdx.

Options:
A Statement - I is true; Statement - II is false.
B Statement - I is false; Statement - II is true.
C Statement - I true; Statement - II is true; Statement - II is a correct explanation for Statement - I.
D Statement - I is true; Statement - II is true; Statement - II is not a correct explanation for Statement - I.

The correct answer is:
Statement - I is false; Statement - II is true.

Solution:
Statement II is a standard property of definite integrals.

For statement I : I=∫π/6π/3dx1+tan x=∫π/6π/3dx1+cot x

I=∫π/6π/3tan x1+tan xdx=∫π/6π/3sinxcosx+sinxdx=∫π/6π/3cosxcosx+sinxdx

2I=∫π/6π/3dx

2I=π3-π6

I=π12

Hence, statement I is false.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2013 (07 Apr)
Statement I : To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RL.
Statement II : To get a steady DC output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL.
In the light of the above statements, choose the most appropriate answer from the options given below :

Options:
A Both Statement I and Statement II are false
B Both Statement I and Statement II are true
C Statement I is false but Statement II is true
D Statement I is true but Statement II is false

The correct answer is:
Both Statement I and Statement II are true

Solution:
We can improve the average, DC output of the rectifier while at the same time reducing the A.C. variation of the rectified output by using smoothing capacitors to filter the output waveform. Smoothing or reservoir capacitors connected in parallel with the load across the output of the full-wave bridge rectifier circuit increase the average DC output level even higher as the capacitor acts like a storage device as shown below. Both Statement I and Statement II are true.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and N-type has excess electrons.

Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter.

In the light of above statements, choose the most appropriate answer from the options given below.

Options:
A Both Statement I and statement II are incorrect
B Statement I is incorrect but statement II is correct
C Both Statement I and statement II are correct
D Statement I is correct but statement II is incorrect

The correct answer is:
Statement I is correct but statement II is incorrect

Solution:
Doping silicon with boron leads to holes without any additional electron. This results in P-type semiconductors. Arsenic is pentavalent therefore when added with silicon it leaves one electron as a free electron. This results in N-type semiconductors.

When such P-type and N-type semiconductors are fused to make a junction, an electric field is produced from the N-side to the P-side, causing a barrier potential to develop. Because of the barrier potential, the majority of charge carriers are unable to flow through the junction, hence current is zero, unless we apply forward bias voltage. We cannot detect the current through ammeter as battery is not connected.

Hence, statement I is correct but statement II is incorrect.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Statement - I : When μ amount of an ideal gas undergoes adiabatic change from state P1,V1,T1 to state P2,V2,T2, then work done is W=μRT2-T11-γ, where γ=CPCV and R= universal gas constant.

Statement - II : In the above case, when work is done on the gas, the temperature of the gas would rise.

Options:
A Both statement-I and statement-II are true.
B Both statement-I and statement-II are false.
C Statement-I is true but statement-II is false.
D Statement-I is false but statement-II is true.

The correct answer is:
Both statement-I and statement-II are true.

Solution:
The work done in an adiabatic process is given by,

Wadiabatic =nRTf-Ti1-γ→ statement I is true.

For adiabatic process ∆Q=0. From the first law of thermodynamics,

∆Q=W+ΔU

0=W+ΔU

ΔU=-W

If work is done on the gas, i.e. work is negative

∴ΔU is positive.

∴ The temperature will increase. Hence, statement II is also true.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Statement II The condition on $a$ and $\mathrm{b}$ for which two distinct chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=$ 2 passing through $(a,-b)$ are bisected by the line $x+y=b$ is $a^2+6 a b-7 b^2\gt0$ Statement II Equation of chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose mid-point $\left(x_1, y_1\right)$ is $T=S_1$

Options:
A Statement I is true. Statement II is true: Statement II is a correct explanation for statement I
B Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I
C Statement I is true. Statement II is flase
D Statement I is false, Statement II is true

The correct answer is:
Statement I is true. Statement II is true: Statement II is a correct explanation for statement I

Solution:
Let $(\lambda, b-\lambda)$ is a point on the line $x+y=b$, then equation of chord whose mid-point $(\lambda, b-\lambda)$ is $T=S_1$ or
$\begin{array}{ll}\frac{\lambda x}{2 a^2}+\frac{(b-\lambda) y}{2 b^2}-1=\frac{\lambda^2}{2 a^2}+\frac{(b-\lambda)^2}{2 b^2}-1(a,-b) \text { lies on Eq. (i), then } \\ \Rightarrow \frac{\lambda a}{2 a^2}-\frac{b(b-\lambda)}{2 b^2}=\frac{\lambda^2}{2 a^2}+\frac{(b-\lambda)^2}{2 b^2} \\ \Rightarrow \lambda^2\left(a^2+b^2\right)-a b \lambda(3 a+b)+2 a^2 b^2=0\end{array}$
$\Rightarrow a^2 b^2(3 a+b)^2-8 a^2 b^2\left(a^2+b^2\right) \gt0$
$\Rightarrow a^2+6 a b-7 b^2 \gt0$
For two distinct chords $D\gt0$
Hence, both Statements are true and Statement II is correct explanation for Statement I.

Topic: Mathematics, Ellipse, JEE Main
Statement (P⇒Q)∧(R⇒Q) is logically equivalent to

Options:
A P⇒R∨Q⇒R
B P∧R⇒Q
C P⇒R∧Q⇒R
D P∨R⇒Q

The correct answer is:
P∨R⇒Q

Solution:
Given that (P⇒Q)∧(R⇒Q)

We know that P⇒Q≡~P∨Q

⇒~P∨Q∧~R∨Q

⇒~P∧~R∨Q

⇒~P∨R∨Q

⇒P∨R⇒Q

Hence this is the correct option.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (06 Apr Shift 1)
STATEMENT-1 : Average power consumed in an any ac circuit is always equal to average power consumed by resistors in the circuit.
STATEMENT-2 : Average power consumed in an any ac by capacitor and inductor is zero.

Options:
A Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C Statement-1 is True, Statement-2 is False
D Statement-1 is False, Statement-2 is True

The correct answer is:
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Solution:
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Topic: Physics, Alternating Current, JEE Main
Statement-1: $\sim(p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$.
Statement-2 : $\sim(p \leftrightarrow \sim q)$ is a tautology.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2009
Statement-1: $\sum_{r=0}^n(r+1)^n C_r=(n+2) 2^{n-1}$
Statement -2: $\quad \sum_{r=0}^n(r+1)^n C_r x^r=(1+x)^n+n x(1+x)^{n-1}$.

Options:
A
Statement $-1$ is false, Statement $-2$ is true

B
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$.

D
Statement $-1$ is true, Statement $-2$ is false.

The correct answer is:
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

Solution:
$$
\begin{aligned}
& \sum_{r=0}^n(r+1){ }^n C_r=\sum_{r=0}^n r{ }^n C_r+{ }^n C_r \\
& =\sum_{r=0}^n r \frac{n}{r}{ }^{n-1} C_{r-1}+\sum_{r=0}^n{ }^n C_r=n 2^{n-1}+2^n \\
& =2^{n-1}(n+2)
\end{aligned}
$$
Statement $-1$ is true
$$
\begin{aligned}
& \sum(r+1)^n C_r x^r=\sum r{ }^n C_r x^r+\sum{ }^n C_r x^r \\
& =n \sum_{r=0}^n{ }^{n-1} C_{r-1} x^r+\sum_{r=0}^n{ }^n C_r x^r=n x(1+x)^{n-1}+(1+x)^n
\end{aligned}
$$
Substituting $x=1$
$$
\sum(r+1)^n C_r=n 2^{n-1}+2^n
$$
Hence Statement $-2$ is also true and is a correct explanation of Statement $-1$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2008
Statement-1: The equation $x \log x=2-x$ is satisfied by at least one value of $x$ lying between 1 and 2.
Statement-2: The function $f(x)=x \log x$ is an increasing function in $[1,2]$ and $g(x)=2-x$ is a decreasing function in $[1,2]$ and the graphs represented by these functions intersect at a point in $[1,2]$

Options:
A
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is true; Statement-2 is true; Statement-2 is not correct explanation for Statement-1.

C
Statement-1 1 is false, Statement- 2 is true.

D
Statement-1 1 is true, Statement- 2 is false.

The correct answer is:
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Solution:
$f(x)=x \log x, f(1)=0, f(2)=4$
$g(x)=2-x, g(1)=1, g(2)=0$
$\log 10>\log 4 \Rightarrow 1>\log 4$
Thus statement $-1$ and 2 both are true and statement-2 is a correct explanation of statement 1 .

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (09 Apr Online)
Statement-1: The function $x^2\left(e^x+e^{-x}\right)$ is increasing for all $x>0$.
Statement-2: The functions $x^2 e^x$ and $x^2 e^{-x}$ are increasing for all $x>0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b)$.

Options:
A
Statement-1 is false; Statement-2 is true.

B
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

C
Statement-1 is true; Statement-2 is false.

D
Statement-1is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.

The correct answer is:
Statement-1 is true; Statement-2 is false.

Solution:
Let $y=x^2 \cdot e^{-x}$
For increasing function,
$$
\begin{aligned}
& \frac{d y}{d x}>0 \Rightarrow x\left[(2-x) e^{-x}\right]>0 \\
& \because x>0, \therefore(2-x) e^{-x}>0 \\
& \Rightarrow(2-x) \frac{1}{e^x}>0 \\
& \text { For } 0 < x < 2,(2-x) < 0 \\
& \therefore \frac{1}{e^x} < 0, \text { but it is not possible }
\end{aligned}
$$
Hence the statement- 2 is false.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (22 Apr Online)
Statement-1: The line $x-2 y=2$ meets the parabola, $y^2+2 x=0$ only at the point $(-2,-2)$.
Statement-2: The line $y=m x-\frac{1}{2 m}(m \neq 0)$ is tangent to the parabola, $y^2=-2 x$ at the point $\left(-\frac{1}{2 m^2},-\frac{1}{m}\right)$

Options:
A
Statement-1 is true; Statement- 2 is false.

B
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.

C
Statement-1 is false; Statement-2 is true.

D
Statement-1 a true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

The correct answer is:
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.

Solution:
Both statements are true and statement-2 is the correct explanation of statement-1
$\therefore \quad$ The straight line $y=m x+\frac{a}{m}$ is always a tangent to the parabola $y^2=4 a x$ for any value of $m$.
The co-ordinates of point of contact
$$
\left(\frac{a}{m^2}, \frac{2 a}{m}\right)
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2013 (22 Apr Online)
Statement-1: The number of common solutions of the trigonometric equations $2 \sin ^2 \theta$ $-\cos 2 \theta=0$ and $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is two.
Statement-2: The number of solutions of the equation, $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0, \pi]$ is two.

Options:
A
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.

B
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

C
Statement-1 is false; Statement-2 is true.

D
Statement-1 is true; Statement-2 is false.

The correct answer is:
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

Solution:
$2 \sin ^2 \theta-\cos 2 \theta=0$
$\Rightarrow 2 \sin ^2 \theta-\left(1-2 \sin ^2 \theta\right)=0$
$\Rightarrow 2 \sin ^2 \theta-1+2 \sin ^2 \theta=0$
$\Rightarrow 4 \sin ^2 \theta=1 \Rightarrow \sin \theta=\pm \frac{1}{2}$
$\therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}, \theta \in[0,2 \pi]$
$\therefore \theta=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$
$$
\begin{aligned}
& \text { Now } 2 \cos ^2 \theta-3 \sin \theta=0 \\
\Rightarrow & 2\left(1-\sin ^2 \theta\right)-3 \sin \theta=0 \\
\Rightarrow & -2 \sin ^2 \theta-3 \sin \theta+2=0 \\
\Rightarrow & -2 \sin ^2 \theta-4 \sin \theta+\sin \theta+2=0 \\
\Rightarrow & 2 \sin ^2 \theta-\sin \theta+4 \sin \theta-2=0 \\
\Rightarrow & \sin \theta(2 \sin \theta-1)+2(2 \sin \theta-1)=0 \\
\Rightarrow & \sin \theta=\frac{1}{2},-2
\end{aligned}
$$
But $\sin \theta=-2$, is not possible
$$
\therefore \quad \sin \theta=\frac{1}{2}, \Rightarrow \theta=\frac{\pi}{6}, \frac{5 \pi}{6}
$$
Hence, there are two common solution, there each of the statement-1 and 2 are true but statement- 2 is not a correct explanation for statement-1.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2013 (22 Apr Online)
Statement-1: The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5$.
Statement-2: The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4)$.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is false

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

Solution:
$$
A(3,1,6) ; B=(1,3,4)
$$
Mid-point of $A B=(2,2,5)$ lies on the plane.
and d.r's of $A B=(2,-2,2)$
d.r's Of normal to plane $=(1,-1,1)$.
$\mathrm{AB}$ is perpendicular bisector
$\therefore \mathrm{A}$ is image of $\mathrm{B}$
Statement-2 is correct but it is not correct explanation.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2010
Statement-1: The slope of the tangent at any point $\mathrm{P}$ on a parabola, whose axis is the axis of $x$ and vertex is at the origin, is inversely proportional to the ordinate of the point $\mathrm{P}$.
Statement-2: The system of parabolas $y^2=4 a x$ satisfies a differential equation of degree 1 and order 1.

Options:
A
Statement-1 is true; Statement- 2 is true; Statement-2 is a correct explanation for statement-1.

B
Statement-1 is true; Statement-2 is true; Statement- 2 is not a correct explanation for statement-1.

C
Statement-1 is true; Statement- 2 is false.

D
Statement-1 is false; Statement- 2 is true.

The correct answer is:
Statement-1 is true; Statement-2 is true; Statement- 2 is not a correct explanation for statement-1.

Solution:
- Statement $-1: y^2=\pm 4 a x$
$$
\Rightarrow \frac{d y}{d x}=\pm 2 a \cdot \frac{1}{y} \Rightarrow \frac{d y}{d x} \propto \frac{1}{y}
$$
Statement $-2: y^2=4 a x \Rightarrow 2 y \frac{d y}{d x}=4 a$ Thus both statements are true but statement-2 is not a correct explanation for statement-1.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2013 (09 Apr Online)
Statement-1: The statement $A \rightarrow(B \rightarrow A)$ is equivalent to $A \rightarrow(A \vee B)$.
Statement-2: The statement $\sim[(\mathrm{A} \wedge \mathrm{B}) \rightarrow$ $(\sim \mathrm{A} \vee \mathrm{B})]$ is a Tautology.

Options:
A
Statement- 1 is false; Statement- 2 is true.

B
Statement-1 is true; Statement-2 is true; Statement- 2 is not correct explanation for Statement-1.

C
Statement-1 is true; Statement- 2 is false.

D
Statement-1 is true; Statement-2 is true; Statement- 2 is the correct explanation for Statement-1.

The correct answer is:
Statement-1 is true; Statement- 2 is false.

Solution:

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2013 (09 Apr Online)
Statement-1: The system of linear equations
$$
\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x-(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}
$$
has a non-trivial solution for only one value of $\alpha$
lying in the interval $\left(0, \frac{\pi}{2}\right)$.
Statement-2: The equation in $\alpha$
$$
\left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0
$$
has only one solution lying in the interval
$\left(0, \frac{\pi}{2}\right)$

Options:
A
Statement-1 is true, Statement-2 is true, Statement-2 is not correct explantion for Statement-1.

B
Statement-1 is true, Statement-2 is true, Statement-2 is a correct explantion for Statement-1.

C
Statement-1 is true, Statement- 2 is false.

D
Statememt-1 is false, Statement-2 is true.

The correct answer is:
Statement-1 is true, Statement- 2 is false.

Solution:
$$
\begin{aligned}
& \Rightarrow \alpha=\frac{64 x}{3 \times 16 x}=\frac{4}{3} \\
& \Delta_1=\left|\begin{array}{ccc}
1 & \sin \alpha & \cos \alpha \\
1 & \cos \alpha & \sin \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & \sin \alpha-\cos \alpha & \cos \alpha-\sin \alpha \\
0 & \cos \alpha+\sin \alpha & \sin \alpha-\cos \alpha \\
1 & -\sin \alpha & \cos \alpha
\end{array}\right| \\
& =(\sin \alpha-\cos \alpha)^2-\left(\cos ^2 \alpha-\sin ^2 \alpha\right) \\
& =\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha \\
& =2 \sin ^2 \alpha-2 \sin \alpha \cdot \cos \alpha \\
& =2 \sin ^2 \alpha(\sin \alpha-\cos \alpha)
\end{aligned}
$$
Now, $\sin \alpha-\cos \alpha=0$ for only
$$
\begin{aligned}
\alpha & =\frac{\pi}{4} \text { in }\left(0, \frac{\pi}{2}\right) \\
\therefore \quad \Delta_1 & =2(\sin \alpha) \times 0=0,
\end{aligned}
$$
since value of $\sin \alpha$ is finite for $\alpha \in\left(0, \frac{\pi}{2}\right)$
Hence non-trivivial solution for only one value of $\alpha$ in $\left(0, \frac{\pi}{2}\right)$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
\cos \alpha & \sin \alpha & \cos \alpha \\
\sin \alpha & \cos \alpha & \sin \alpha \\
\cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
0 & \sin \alpha & \cos \alpha \\
0 & \cos \alpha & \sin \alpha \\
2 \cos \alpha & -\sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow 2 \cos \alpha\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\
& \therefore \cos \alpha=0 \text { or } \sin ^2 \alpha-\cos ^2 \alpha=0
\end{aligned}
$$
But $\cos \alpha=0$ not possible for any value of
$$
\begin{aligned}
& \alpha \in\left(0, \frac{\pi}{2}\right) \\
& \therefore \sin ^2 \alpha-\cos ^2 \alpha=0 \Rightarrow \sin \alpha=-\cos \alpha,
\end{aligned}
$$
which is also not possible for any value of
$$
\alpha \in\left(0, \frac{\pi}{2}\right)
$$
Hence, there is no solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2013 (23 Apr Online)
Statement-1 : The variance of first $n$ even natural numbers is $\frac{n^2-1}{4}$
Statement-2 : The sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$ and the sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2 n+1)}{6}$

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is false, Statement-2 is true

Solution:
Statement-2 is true
Statement-1:
Sum of $n$ even natural numbers $=n(n+1)$
$$
\begin{aligned}
& \text { Mean }(\bar{x})=\frac{n(n+1)}{n}=n+1 \\
& \text { Variance }=\left[\frac{1}{n} \sum\left(x_1\right)^2\right]-(\bar{x})^2=\frac{1}{n}\left[2^2+4^2+\ldots .+(2 n)^2\right]-(n+1)^2 \\
& =\frac{1}{n} 2^2\left[1^2+2^2+\ldots .+n^2\right]-(n+1)^2=\frac{4}{n} \frac{n(n+1)(2 n+1)}{6}-(n+1)^2 \\
& =\frac{(n+1)[2(2 n+1)-3(n+1)]}{3}=\frac{(n+1)[4 n+2-3 n-3]}{3}=\frac{(n+1)(n-1)}{3}=\frac{n^2-1}{3}
\end{aligned}
$$
$\therefore$ Statement 1 is false.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2009
Statement-1: Two longitudinal waves given by equations: \( y_{1}(x, t)=2 a \sin (\omega t-k x) \) and \( y_{2}(x, t)=a \sin (2 \omega t-2 k x) \) will
have equal intensity.
Statement-2: Intensity of waves of given frequency in same medium is proportional to square of amplitude.

Options:
A Statement- \( 1 \) is true, statement- \( 2 \) true; statement- \( 2 \) is the correct explanation of statement- \( 1 \)
B Statement- \( 1 \) is true, statement- \( 2 \) true; statement- \( 2 \) is not correct explanation of statement- \( 1 \)
C Statement- \( 1 \) is true, statement- \( 2 \) is false
D Statement- \( 1 \) is false, statement- \( 2 \) is true

The correct answer is:
Statement- \( 1 \) is true, statement- \( 2 \) true; statement- \( 2 \) is not correct explanation of statement- \( 1 \)

Solution:
I ∝ ω2A2

Topic: Physics, Waves and Sound, JEE Main