datavorous/entrance-exam-dataset huggingface dataset:
If $A$ is an invertible matrix, then what is det $\left(A^{-1}\right)$ equal to?

Options:
A $\operatorname{det} A$
B $\frac{1}{\operatorname{det} A}$
C 1
D None of the above

The correct answer is:
$\frac{1}{\operatorname{det} A}$

Solution:
$\operatorname{det}\left(\mathrm{A}^{-1}\right)=\frac{1}{\operatorname{det} \mathrm{A}}$

Topic: Mathematics, Determinants, JEE Main
If A is an invertible square matrix of the order n such that Aâ 1 and adjadjA=A2n2-7n+7, then the sum of all possible values of n is

Options:
A 3
B 4
C 6
D 5

The correct answer is:
5

Solution:
We know that adjadjA=An-12
âAn-12=A2n2-7n+7
ân2-2n+1=2n2-7n+7
ân2-5n+6=0

Hence, the sum of all values of n is equal to 5

Topic: Mathematics, Determinants, JEE Main
If A is semy square matrix of order 3 and det $\mathrm{A}=5$, then what is $\operatorname{det}\left[(2 \mathrm{~A})^{-1}\right]$ equal to?

Options:
A $1 / 10$
B $2 / 5$
C $8 / 5$
D $1 / 40$

The correct answer is:
$1 / 40$

Solution:
$|A|=5 \Rightarrow|2 A|=2^{3} \times 5=40$
$\Rightarrow\left|A^{-1}\right|=\frac{1}{|A|} \Rightarrow\left|(2 A)^{-1}\right|=\frac{1}{|2 A|}=\frac{1}{40}$

Topic: Mathematics, Determinants, JEE Main
If A is the area in the first quadrant enclosed by the curve C:2x2-y+1=0, the tangent to C at the point 1, 3 and the line x+y=1, then the value of 60A is................

Options:

The correct answer is:
16

Solution:
Given,

y=2x2+1

Now tangent to above equation at 1,3 is given by,

y-3=4x-1

⇒y=4x-1

Now plotting the diagram we get,
Required area of bounded region will be,

A=∫012x2+1dx-area of △QOT -area of △PQR +area of △QRS

⇒A=23+1-12-98+940=1640

⇒60A=16

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is

Options:
A $\frac{M}{A}$
B $2MA$
C $A^2 M$
D $A M^2$

The correct answer is:
$2MA$

Solution:
Areal velocity $A=\frac{1}{2} R^2 \omega$
$\therefore \quad$ Multiplying by $M$ on both sides
$$
\begin{aligned}
M A & =\frac{1}{2} M R^2 \omega \\
M A & =\frac{1}{2} I \omega
\end{aligned}
$$
$I=$ moment of inertia.
$$
M A=\frac{1}{2} L
$$
where, $L=$ angular momentum
$$
\begin{array}{rlrl}
& =I \omega \\
\therefore & L & =2 M A
\end{array}
$$

Topic: Physics, Gravitation, JEE Main
If $\mathrm{A}$ is the domain and $\mathrm{B}$ is the range of the function
$f(x)=\left\{\begin{array}{ll}3 x-1, & x>1 \\ x^2+1, & x \leq 1\end{array}\right.$ then $A-B=$

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

The correct answer is:
$(-\infty, 1)$

Solution:
$\because f(x)= \begin{cases}3 x-1 & x>1 \\ x^2+1 & x \leq 1\end{cases}$
$\therefore$ Domain is $(-\infty, 1] \cup(1, \infty)=(-\infty, \infty)=\mathrm{A}$
When $x>1$ : $f(x)=3 x-1 \Rightarrow 2 \leq f(x) < \infty$ and $\mathrm{x} \leq 1: \mathrm{f}(\mathrm{x})=\mathrm{x}^2+1 \Rightarrow 1 \leq \mathrm{f}(\mathrm{x}) < \infty$.
$\therefore \quad$ Range $B=(1, \infty) \cup(2, \infty)=[1, \infty)$
Now, $A-B=(-\infty, \infty)-[1, \infty)=(-\infty, 1)$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2023 (18 May Shift 1)
If aα is the greatest term in the sequence an=n3n4+147, n=1, 2, 3...., then α is equal to ______

Options:

The correct answer is:
5

Solution:
Let,

y=x3x4+147=fx

We know that, for increasing function dydx>0

So, differentiating the given function fx=x3x4+147 we get,

dydx=-x2x4-441(x4+147)2

⇒-x2x4-441(x4+147)2>0

⇒x2x4-441(x4+147)2<0

⇒x4<441

Now for maxima/minima dydx=0

⇒x4=441

⇒x=α, 4<α<5

⇒Maximum value of fx is at x=4 or x=5

f4=64403, f5=125772

∵f5>f4

⇒α=5

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If ' $a$ ' is the point of discontinuity of the function
$$
f(x)=\left\{\begin{array}{cl}
\cos 2 x, & \text { for }-\infty < x < 0 \\
e^{3 x}, & \text { for } 0 \leq x < 3 \\
x^2-4 x+3, & \text { for } 3 \leq x \leq 6 \\
\frac{\log (15 x-89)}{x-6}, & \text { for } x>6
\end{array}\right.
$$
Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9}=$

Options:
A 1
B 0
C 6
D 3

The correct answer is:
1

Solution:
We have,
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{rc}
\cos 2 x, & -\infty < x < 0 \\
e^{3 x}, & 0 \leq x < 3 \\
x^2-4 x+3, & 3 \leq x \leq 6 \\
\frac{\log (15 x-89)}{x-6}, & x>6
\end{array}\right. \\
& \because \lim _{x \rightarrow 3^{-}} e^{3 x}=e^9 \text { and } \lim _{x \rightarrow 3^{+}} x^2-4 x+3=9-12+3=0
\end{aligned}
$$
Clearly, $\quad \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$
$\therefore f(x)$ is discontinuous at $x=3$.
$$
\therefore a=3 \text {, }
$$
Now, $\lim _{x \rightarrow 3} \frac{x^2-9}{x^3-5 x^2+9 x-9}=\lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{(x-3)\left(x^2-2 x+3\right)}$
$$
=\frac{3+3}{(3)^2-2(3)+3}=\frac{2}{2}=1
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If ' $A$ ' is the reactant and ' $P$ ' is the product, which one of the following is the correct form of Nernst equation?

Options:
A $\frac{[A]}{[P]}=\exp \left(\frac{R T}{n F}\left(E-E^{\circ}\right)\right)$
B $\frac{[A]}{[P]}=\exp \left(\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$
C $\frac{[A]}{[P]}=\exp \left(-\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$
D $E=E^{\circ}-\frac{R T}{n F} \ln \frac{[A]}{[P]}$

The correct answer is:
$\frac{[A]}{[P]}=\exp \left(\frac{n F}{R T}\left(E-E^{\circ}\right)\right)$

Solution:
According to Nernst equation :
$\begin{aligned} & E=E^{\circ}-\frac{R T}{n F} \ln \frac{[\text { Product }]}{[\text { Reactant }]} \\ & \text { or, } \quad E=E^{\circ}-\frac{R T}{n F} \ln \frac{[P]}{[A]} \\ & \text { or, } \quad E-E^{\circ}=-\frac{R T}{n F} \ln \left[\frac{P}{A}\right] \\ & \text { or, } \quad\left[\frac{A}{P}\right]=e^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \text { or } \quad \frac{[A]}{[P]}=\exp ^{\left[\frac{n F}{R T}\left(E-E^{\circ}\right)\right]} \\ & \end{aligned}$
Hence, option (b) is the correct answer.

Topic: Chemistry, Electrochemistry, JEE Main
If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B are

Options:
A 29
B 92
C 32
D 29-1

The correct answer is:
29

Solution:
A={2,4,6} ; B=2,3,5
â´ AÃ B contains 3Ã3=9 elements.
Hence, the number of relations from A to B=29.

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\left[\begin{array}{ccc}k / 2 & 0 & 0 \\ 0 & l / 3 & 0 \\ 0 & 0 & m / 4\end{array}\right]$ and $A^{-1}=\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$, then $k+l+m=$

Options:
A 1
B 9
C 14
D 29

The correct answer is:
29

Solution:
Given,
$$
A=\left[\begin{array}{ccc}
k / 2 & 0 & 0 \\
0 & l / 3 & 0 \\
0 & 0 & m / 4
\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{ccc}
1 / 2 & 0 & 0 \\
0 & 1 / 3 & 0 \\
0 & 0 & 1 / 4
\end{array}\right]
$$
Since, $\quad A A^{-1}=I$
$$
\begin{aligned}
& \Rightarrow \quad\left[\begin{array}{ccc}
\frac{k}{2} & 0 & 0 \\
0 & \frac{l}{3} & 0 \\
0 & 0 & \frac{m}{4}
\end{array}\right]\left[\begin{array}{lll}
\frac{1}{2} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{4}
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{ccc}
\frac{k}{4} & 0 & 0 \\
0 & \frac{l}{9} & 0 \\
0 & 0 & \frac{m}{16}
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$$
On comparing, we get
$$
\begin{aligned}
& & \frac{k}{4} & =1, \frac{l}{9}=1 \text { and } \frac{m}{16}=1 \\
& & k & =4, l=9, m=16 \\
\therefore & & k+l+m & =4+9+16=29
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main
If a ladder weighing $250 \mathrm{~N}$ is placed against a smooth vertical wall having coefficient of friction between it and floor is 0.3 , then what is the maximum force of friction available at the point of contact between the ladder and the floor

Options:
A $75 \mathrm{~N}$
B $50 \mathrm{~N}$
C $35 \mathrm{~N}$
D $25 \mathrm{~N}$

The correct answer is:
$75 \mathrm{~N}$

Solution:
$F=\mu R=0.3 \times 250=75 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
If a LED is forward biased, then

Options:
A electrons from the n - type side cross the p - n junction and recombine with holes in the p - type side
B electrons and holes neutralise each other in depletion region
C at junction electrons and holes remain at rest
D none of these

The correct answer is:
electrons from the n - type side cross the p - n junction and recombine with holes in the p - type side

Solution:
Forward bias causes a force on the electrons pushing them from the N side toward the P side. With forward bias, the depletion region is narrow enough that electrons can cross the junction into the p-type material.

Topic: Physics, Semiconductors, NEET
If a lens is cut into two pieces perpendicular to the principal axis and only one part is used, the intensity of the image

Options:
A Remains same
B $\frac{1}{2}$ times
C 2 times
D Infinite

The correct answer is:
Remains same

Solution:
Since light transmitting area is same, there is no effect on intensity.

Topic: Physics, Ray Optics, JEE Main
If a=limn→∞∑k=1n2nn2+k2 and fx=1-cosx1+cosx,x∈0,1, then:

Options:
A 22fa2=f'a2
B fa2f'a2=2
C 2fa2=f'a2
D fa2=2f'a2

The correct answer is:
2fa2=f'a2

Solution:
Given, a=limn→∞∑k=1n2nn2+k2

⇒a=1n∑k=1n21+kn2=∫0121+x2dx

⇒a=2tan-1x01

⇒a=2π4-0=π2

Now, fx=1-cosx1+cosx

⇒fx=tanx2;x∈0,1

So, fa2=fπ4

⇒fπ4=2-1

And f'π4=12sec2π8=22+1

So, f'π4=2fπ4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (26 Jul Shift 1)
If $a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$, then

Options:
A $a=b$
B $2 \mathrm{a}=3 \mathrm{~b}$
C $a=2 b$
D $3 \mathrm{a}=2 \mathrm{~b}$

The correct answer is:
$2 \mathrm{a}=3 \mathrm{~b}$

Solution:
$$
\begin{aligned}
& a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)}{2(n)(n)}=\lim _{n \rightarrow \infty}\left(\frac{1}{2}\right)\left(\frac{n+1}{n}\right)=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2} \\
& b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots .+n^2}{n^3} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=\lim _{n \rightarrow \infty} \frac{2 n^2+3 n+1}{6 n^2} \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{6}\right)\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
Thus $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{1}{3} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}$

Topic: Mathematics, Limits, JEE Main
If a=limx→01+1+x4-2x4 and b=limx→0sin2x2-1+cosx, then the value of ab3 is :

Options:
A 36
B 32
C 25
D 30

The correct answer is:
32

Solution:
Given: a=limx→01+1+x4-2x4

⇒a=limx→01+1+x4-2x4×1+1+x4+21+1+x4+2

⇒a=limx→01+1+x4-2x41+1+x4+2

⇒a=limx→01+x4-1x41+1+0+2

⇒a=limx→01+x4-1x41+1+2

⇒a=122limx→01+x4-1x4

⇒a=122limx→01+x4-1x4×1+x4+11+x4+1

⇒a=122limx→01+x4-1x41+x4+1

⇒a=122limx→0x4x41+0+1

⇒a=122limx→0x4x42

⇒a=142limx→0x4x4

⇒a=142

Also, b=limx→0sin2x2-1+cosx

⇒b=limx→01-cos2x(2+1+cosx)2-(1+cosx)

⇒b=limx→01-cosx1+cosx(2+1+cosx)1-cosx

⇒b=limx→0(1+cosx)(2+1+cosx)

⇒b=(1+cos0)(2+1+cos0)

⇒b=(1+1)(2+1+1)

⇒b=(2)(22)

⇒b=42

⇒ab3=142×(42)3

⇒ab3=32

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (27 Jan Shift 1)
If a line along a chord of the circle 4x2+4y2+120x+675=0, passes through the point (-30,0) and is tangent to the parabola y2=30x, then the length of this chord is:

Options:
A 5
B 7
C 35
D 53

The correct answer is:
35

Solution:
We have,

4x2+4y2+120x+675=0

⇒x2+y2+30x+6754=0

Centre≡-15,0 and radius=225-6754=152 units

Let the chord cuts the circle at A & B and passes through P-30,0

Equation of tangent to parabola y2=30x is

y=mx+304m ...i

It passes through -30,0, so

0=-30m+304m

⇒4m2=1

⇒m=±12

For m=12, we have

y=x2+15

⇒x-2y+30=0

Length of OB=-15+0+305=35

Length of BC=2254-45⇒BC=352

Length of chord AC=2×352=35.

Similarly for m=-12, length of chord AC=35.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If a line in the space makes angle \( \alpha, \beta \) and \( \gamma \) with the coordinate axes, then
\( \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma \) equals

Options:
A \( -1 \)
B \( 0 \)
C \( 1 \)
D \( 2 \)

The correct answer is:
\( 1 \)

Solution:
cos⁡2α+cos⁡2β+cos⁡2γ+sin2⁡α+sin2β+sin2γ
=cos 2α-sin2α+cos 2β-sin2β
+cos 2γ-sin2γ+sin2⁡α+sin2β+sin2γ
=cos2⁡α+cos2β+cos2γ
=1

Topic: Mathematics, Vector Algebra, JEE Main
If a line in the space makes angle $\alpha, \beta$ and $\gamma$ with the coordinate axes, then
$\begin{aligned}
\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+\sin ^2 \alpha & +\sin ^2 \beta \\
& +\sin ^2 \gamma \text { equals }
\end{aligned}$

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
$\begin{aligned} & \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+\sin ^2 \alpha+\sin ^2 \beta \\ & \quad+\sin ^2 \gamma \\ & =\left(\cos ^2 \alpha-\sin ^2 \alpha\right)+\left(\cos ^2 \beta-\sin ^2 \beta\right) \\ & \quad+\left(\cos ^2 \gamma-\sin ^2 \gamma\right)+\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma \\ & =\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma \\ & =1\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a line intercepted between the coordinate axes is trisected at a point $\mathrm{A}(4,3)$, which is nearer to $x$-axis, then its equation is:

Options:
A
$4 x-3 y=7$

B
$3 \mathrm{x}+2 \mathrm{y}=18$

C
$3 x+8 y=36$

D
$x+3 y=13$

The correct answer is:
$3 \mathrm{x}+2 \mathrm{y}=18$

Solution:
$$
\text { A divides } \mathrm{CB} \text { in } 2: 1
$$
$$
\begin{aligned}
&\Rightarrow 4=\left(\frac{1 \times 0+2 \times a}{1+2}\right)=\frac{2 a}{3} \\
&\Rightarrow a=6 \Rightarrow \text { coordinate of B is B }(6,0) \\
&3=\left(\frac{1 \times b+2 \times 0}{1+2}\right)=\frac{b}{3} \\
&\Rightarrow b=9 \text { and } \mathrm{C}(0,9)
\end{aligned}
$$
Slope of line passing through $(6,0),(0,9)$
slope, $m=\frac{9}{-6}=-\frac{3}{2}$
Equation of line $y-0=\frac{-3}{2}(x-6)$
$$
\begin{aligned}
&2 y=-3 x+18 \\
&3 x+2 y=18
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (12 Apr Online)
If a line L is perpendicular to the line 5x-y=1, and the area of the triangle formed by the line L and the coordinate axes is 5 sq units, then the distance of the line L from the line x+5y=0 is

Options:
A 7 1 3 units
B 7 5 units
C 5 1 3 units
D 5 7 units

The correct answer is:
5 1 3 units

Solution:
The equation of a line perpendicular to the line ax+by+c=0 is bx-ay+k=0, k∈R.

Given, line L is perpendicular to 5x-y=1, hence the equation of L is x+5y+k=0

For finding x-Intercept, put y=0

⇒x+k=0, ⇒x=-k

∴ x-intercept=-k

Similarly, for finding y- Intercept, put x=0

⇒5y+k=0, ⇒y=-k5

∴ y-intercept=-k5

Given, the area formed by the line L with the coordinate axes is 5 sq units.

Hence, the area of Δ=12×-k-k5=5

⇒k2=50

⇒k=±52

Hence, the equation of the line is L: x⁡+5y⁡±52=0

Now, we know that the distance between the parallel lines ax+by+c=0 and ax+by+c1=0 is c-c1a2+b2

Thus, the distance between the lines x+5y=0 and x+5y⁡±52=0 is

d=±52-012+52

⇒d=5226

⇒d=5213×2=513 units.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (19 Apr Online)
If a line makes an angle of $\frac{\pi}{4}$ with the positive directions of each of $x$-axis and $y$-axis, then the angle that the line makes with the positive direction of the $z-$axis is

Options:
A
$\frac{\pi}{6}$

B
$\frac{\pi}{3}$

C
$\frac{\pi}{4}$

D
$\frac{\pi}{2}$

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

Solution:
$\mathrm{l}=\cos \frac{\pi}{4}, m=\cos \frac{\pi}{4}$
we know $\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2=1$
$\begin{aligned}
& \frac{1}{2}+\frac{1}{2}+n^2=1 \\
& \Rightarrow n=0
\end{aligned}$
Hence angle with positive direction of $\mathrm{z-}$ axis is $\frac{\pi}{2}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2007
If a line makes angles \(\alpha, \beta, \gamma\) with the positive directions of \(X, Y\) and \(Z\)-axes respectively, then the value of \(\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=\)

Options:
A 1
B 2
C 3
D -1

The correct answer is:
2

Solution:
Direction cosines will be \((\cos \alpha, \cos \beta, \cos \gamma)\)
So, \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\)
\(\Rightarrow \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2\)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a line passing through -2,1,α and 4,1,2 is perpendicular to the vector 3i^-4j^+5k^ and parallel to the plane containing vectors i^+2βk^ and 2βj^+αk^ ∀β≠0, then 10α+β is equal to

Options:
A 53
B 54
C 55
D 56

The correct answer is:
53

Solution:
Let, coordinates of A,B are respectively -2,1,Î±,4,1,2 respectively
ABâ=6i^+2-Î±k^
âµABâ is perpendicular to 3i^-4j^+5k^
â18+52-Î±=0âÎ±=285
Also, ABâ,i^+2Î²k^,2Î²j^+Î±k^ are coplanar
â602-Î±102Î²02Î²Î±=0
â2Î²12Î²+Î±-2=0âÎ²=0 or 12Î²=2-Î±
12Î²=2-285
12Î²=-185âÎ²=-310

Topic: Mathematics, Vector Algebra, JEE Main
If a line y=mx+c, is a tangent to the circle x-32+y2=1, and it is perpendicular to a line L1, where L1 is the tangent to the circle x2+y2=1, at the point 12, 12, then

Options:
A c2-7c+6=0
B c2+7c+6=0
C c2+6c+7=0
D c2-6c+7=0

The correct answer is:
c2+6c+7=0

Solution:
Slope of tangent to x2+y2=1 at 12,12
x2+y2=1
2x+2yy'=0
y'=-xy=-1
Given that, y=mx+c, is perpendicular to L1,
So, m=1 ⇒y=x+c

Now distance of 3,0 from y=x+c is
c+32=1
c2+6c+9=2
c2+6c+7=0.

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If a live earthworm is pricked with a needle on its outer surface without damaging its gut, the fluid that comes out is :

Options:
A Slimy mucus
B excretory fluid
C Coelomic fluid
D haemolymph

The correct answer is:
Coelomic fluid

Solution:
No Solution Available

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2009 (Mains)
If a magnet is plugged into a coil, then the magnitude of induced emf does not depend upon

Options:
A The number of turns in the coil
B The medium of the core of the coil
C The strength of the magnet
D The resistance of the coil

The correct answer is:
The resistance of the coil

Solution:
The magnitude of induced emf is given by, e=Ndϕdt=NBdsdt
Which depends upon the number of turns (N), magnetic field intensity (B) which depends upon the medium, rate of change of area dsdt which depends upon insertion speed of magnet. It is independent of resistance of the coil.

Topic: Physics, Electromagnetic Induction, JEE Main
If $\mathbf{a}$ makes an acute angle with $\mathbf{b}, \mathbf{r} \cdot \mathbf{a}=0$ and $\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}$, then $\mathbf{r}=$

Options:
A $\mathbf{a} \times \mathbf{c}-\mathbf{b}$
B $c \times a$
C $c-\left(\frac{c \cdot a}{b \cdot a}\right) b$
D $c+\left(\frac{c \cdot a}{b \cdot a}\right) b$

The correct answer is:
$c-\left(\frac{c \cdot a}{b \cdot a}\right) b$

Solution:
Given,
and
$$
\begin{aligned}
\mathbf{r} \cdot \mathbf{a} & =0 \\
\mathbf{r} \times \mathbf{b} & =\mathbf{c} \times \mathbf{b}
\end{aligned}
$$
So,
$$
\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}
$$
$$
\begin{aligned}
& \mathbf{r}(\mathbf{a} \cdot \mathbf{b})-\mathbf{b}(\mathbf{a} \cdot \mathbf{r})=\mathbf{c}(\mathbf{a} \cdot \mathbf{b})-\mathbf{b}(\mathbf{a} \cdot \mathbf{c}) \\
& \Rightarrow \quad \mathbf{r}(\mathbf{a} \cdot \mathbf{b})=\mathbf{c}(\mathbf{a} \cdot \mathbf{b})-\mathbf{b}(\mathbf{a} \cdot \mathbf{c}) \\
& \Rightarrow \quad \mathbf{r}=\mathbf{c}-\mathbf{b}\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{b}}\right) \\
&
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If a man of mass M jumps to the ground from a height h and his centre of mass moves a distance x in the time taken by him to hit the ground, then the average force acting on him is

Options:
A Mghx
B Mgxh
C Mghx2
D Mgxh2

The correct answer is:
Mghx

Solution:
Here, F=-∆u∆x=--Mghx=Mghx

Topic: Physics, Work Power Energy, JEE Main
If a matrix $A$ is such that $3 A^3+2 A^2+5 A+I=0$, then its inverse is

Options:
A $-\left(3 A^2+2 A+5 I\right)$
B $3 A^2+2 A+5 I$
C $3 A^2-2 A-51$
D None of these

The correct answer is:
$-\left(3 A^2+2 A+5 I\right)$

Solution:
$3 A^3+2 A^2+5 A+I=0 \Rightarrow I=-3 A^3-2 A^2-5 A$
$\Rightarrow I A^{-1}=-3 A^2-2 A-5 I$
$A^{-1}=-\left(3 A^2+2 A+5 I\right)$

Topic: Mathematics, Matrices, JEE Main
If a message signal of frequency fm is amplitude modulated with a carrier signal of frequency fc and radiated through an antenna, the wavelength of the corresponding signal in air is

Options:
A cfc-fm
B cfm
C cfc
D cfc+fm

The correct answer is:
cfc

Solution:
Given frequency of massage signal =fm frequency of carrier signal =fc
the wavelength of the corresponding signal in air is ⇒λ=cfc

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If a metallic circular plate of radius $50 \mathrm{~cm}$ is heated so that its radius increases at the rate of $1 \mathrm{~mm}$ per hour, then the rate at which, the area of the plate increases (in $\mathrm{cm}^2 /$ hour) is

Options:
A
$5 \pi$

B
$10 \pi$

C
$100 \pi$

D
$50 \pi$

The correct answer is:
$10 \pi$

Solution:
Let $A=\pi \mathrm{r}^2$ be area of metalic circular plate of $r=50 \mathrm{~cm}$.
Also, given $\frac{d r}{d t}=1 \mathrm{~mm}=\frac{1}{10} \mathrm{~cm}$
$$
\begin{aligned}
& \therefore A=\pi r^2 \\
& \Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}=2 \pi \cdot 50 \cdot \frac{1}{10}=10 \pi
\end{aligned}
$$
Hence, area of plate increases in $10 \pi \mathrm{cm}^2 /$ hour.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (26 May Online)
If a molecule \( \mathrm{MX}_{3} \) has zero dipole moment, the sigma bonding orbitals used by \( \mathrm{M} \) (atomic number \( < 21 \) ) are

Options:
A Pure \( \mathrm{p} \)
B sp hybridised
C \( \mathrm{sp}^{2} \) hybridised
D \( \mathrm{sp}^{3} \) hybridised

The correct answer is:
\( \mathrm{sp}^{2} \) hybridised

Solution:
The atomic number of the element M is given as less than 21. The electronic configuration for an atom with an atomic number of 21 will be [Ar]3s23p1. This implies that the element M with less than 21 electrons does not have d-orbital electrons.

If the molecule MX3 is non-polar in nature, and contains no d-orbitals, then it must have three sigma (single) bonds. The hybridisation for three sigma bonds is sp2. Also, the fact that the molecule has zero dipole moment confirms that the molecule has a symmetrical geometry of triagonal planar arrangement, which is possible only for an sp2hybridisation around M.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
If $a_n=\int_0^{\frac{\pi}{2}} \frac{\sin ^2 n x}{\sin x} d x$, then $a_2-a_1, a_3-a_2, a_4-a_3$ are in

Options:
A Arithmetic progression
B Geometric progression
C Harmonic progression
D Arithmetico-geometric progression

The correct answer is:
Harmonic progression

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Definite Integration, JEE Main
If An=â«0nÏsinâ¡xdx, ânâN, then âr=110Ar is equal to

Options:
A 100
B 110
C 55
D 105

The correct answer is:
110

Solution:
As the period of y=sin⁡x is π, hence

An=n∫0πsin⁡xdx
=n∫0πsin⁡xdx
=n-cos⁡x0π
=2n
∴∑r=110Ar=∑r=1102r
=21+2+....+10
=2×10×112=110

Topic: Mathematics, Definite Integration, JEE Main
If $A_n=\left[\begin{array}{cc}1-n & n \\ n & 1-n\end{array}\right]$, then $\left|A_1\right|+\left|A_2\right|+\ldots\left|A_{2021}\right|=$

Options:
A -2021
B $-(2021)^2$
C $(2021)^2$
D 4042

The correct answer is:
$-(2021)^2$

Solution:
Given, $A_n=\left[\begin{array}{cc}1-n & n \\ n & 1-n\end{array}\right]$ and
$$
\begin{aligned}
& \left|A_n\right|=\left[\begin{array}{cc}
1-n & n \\
n & 1-n
\end{array}\right] \\
& =(1-n)^2-n^2=1+n^2-2 n-n^2 \\
& \left|A_n\right|=1-2 n \\
& \text { Now. }\left|A_1\right|+\left|A_2\right|+\left|A_3\right|+\ldots .+\left|A_{2021}\right| \\
& =(1-2)+(1-4)+(1-6)+\ldots+(1-4042) \\
& =(1+1+\ldots .+1)-(2+4+6+\ldots .+4042) \\
& =2021-\left[\frac{2021}{2}(2+4042)\right] \\
& =2021-\frac{2021}{2} \times 4044=2021-2021 \times 2022 \\
& =2021(1-2022)=-(2021)^2 \\
&
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main