datavorous/entrance-exam-dataset huggingface dataset:
If i ^ × a → - j ^ × i ^ + j ^ × a → - k ^ × j ^ + k ^ × a → - i ^ × k ^ = 0 and a → = x i ^ + y j ^ + z k ^ , then the value of 8x3-xy+zx is equal to

Options:

The correct answer is:
1

Solution:
i ^ × a → - j ^ × i ^ = i ^ . i ^ a → - j ^ - i ^ . a → - j ^ i ^
= a → - j ^ - i ^ . a → i ^
Therefore, a→-j^-i^.a→i^+a→-k^+j^.a→j^+a→-i^-k^.a→k^=0
⇒3a→-i^+j^+k^-a→=0
a → = 1 2 i ^ + j ^ + k ^ = x i ^ + y j ^ + z k ^
x = y = z = 1 2
8 x 3 - x y + z x = 8 x 3 - x 2 + x 2 = 8 × 1 8 = 1

Topic: Mathematics, Vector Algebra, JEE Main
If I=∫dxx2-2x+5=12tan⁡-1fx+C (where, C is the constant of integration) and f2=12, then the maximum value of y=fsin⁡x ∀x∈R is

Options:
A 4
B 2
C 0
D -1

The correct answer is:
0

Solution:
The given integral is I=∫dxx-12+4
=12tan-1⁡x-12+c
∴fx=x-12
i.e. fsin⁡x=sin⁡x-12
Hence, max fsin⁡x=0

Topic: Mathematics, Indefinite Integration, JEE Main
If I=∫dxx521+x723=kfx+c, where c is the integration constant and f1=1216, then the value of f2 is equal to

Options:
A 62316
B 6 3 2 1 6
C 2316
D 236

The correct answer is:
2316

Solution:
The given integral is I=â«dxx61+xx7213
âI=â«dxx21+1x76
Let, 1+1x=t
â-dxx2=dt
âI=â«-dtt76
=6t-16+c
=6xx+116+c
â´fx=xx+116
âf2=2316

Topic: Mathematics, Indefinite Integration, JEE Main
If I=∫dxx2⋅1+x2=fxx+C, ∀x>0 (where, C is the constant of integration) and f1=-2, then the value of f3 is

Options:

The correct answer is:
2

Solution:
Given integral is I=â«dxx31+1x2
Let 1+1x2=t2
â-2x3dx=2t dt
âI=-â«tdtt
=-x2-1x+C
âfx=-x2+1
â´f3=-3+1=2

Topic: Mathematics, Indefinite Integration, JEE Main
If $I$ is the identity matrix of order 2 and $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$, then for $n \geq 1$, mathematical induction gives

Options:
A $A^n=n A-(n-1) I$
B $A^n=n A+(n-1) I$
C $A^n=2^n A-(n+1) I$
D $A^n=2^{n-1} A-(n-1) I$

The correct answer is:
$A^n=n A-(n-1) I$

Solution:
Given $\quad A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
Now, $\quad A^2=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
$$
=\left[\begin{array}{ll}
1+0 & 1+1 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]
$$
Similarly,
$$
A^3=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]
$$
$$
A^n=\left[\begin{array}{ll}
1 & n \\
0 & 1
\end{array}\right]
$$
We have, $n A-(n-1) I$
$$
\begin{aligned}
& =\left[\begin{array}{ll}
n & n \\
0 & n
\end{array}\right]-\left[\begin{array}{cc}
n-1 & 0 \\
0 & n-1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & n \\
0 & 1
\end{array}\right]=A^n \\
A^n & =n A-(n-1) / \text { is true }
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main
If $\theta_i$, is the inversion temperature, $\theta_n$ is the neutral temperature, $\theta_c$ is the temperature of the cold junction, then

Options:
A
$\theta_i+\theta_c=\theta_n$

B
$\theta_{\mathrm{i}}-\theta_c=2 \theta_n$

C
$\frac{\theta_i+\theta_c}{2}=\theta_n$

D
$\theta_c-\theta_i=2 \theta_n$

The correct answer is:
$\frac{\theta_i+\theta_c}{2}=\theta_n$

Solution:
No Solution Available

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2002
If $\mathrm{I}=\lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{x}-1-\frac{\mathrm{x}^{2}}{2}}{\mathrm{x}^{2}}\right)$, then limit

Options:
A does not exist
B exists and equals 1
C exists and equals 0
D exists and equals $\frac{1}{2}$

The correct answer is:
exists and equals 0

Solution:
$I=\lim _{x \rightarrow 0} \sin \left(\frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}\right)=\lim _{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}=\lim _{x \rightarrow 0} \frac{e^{x}-1-0-x}{2 x}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{2}=0$

Topic: Mathematics, Limits, JEE Main
If I=∫ln⁡x5x2+x2ln⁡x3dx=kln⁡x3+1ln⁡x3-2+c (where c is the constant of integration), then 9k is equal to

Options:
A 4
B 2
C 6
D 10

The correct answer is:
2

Solution:
Let 1+ln⁡x3=t2
⇒I=∫t2-12tdt3t
⇒I=23∫t2-1dt
=2t39-23t+c
=29tt2-3+c
=291+ln⁡x3ln⁡x3-2+c
∴k=29⇒9k=2

Topic: Mathematics, Indefinite Integration, JEE Main
If Im,n=∫01xm-11-xn-1dx, for m,n⩾1, and ∫01xm-1+xn-11+xm+ndx=αIm,n, α∈R, then α equals ________.

Options:

The correct answer is:
1

Solution:
Im,n=∫01xm-11-xn-1dx=In,m

Now Let x=1y+1⇒dx=-1y+12dy

So, Im,n=-∫∞01y+1m-1yn-1y+1n-1dyy+12=∫0∞yn-11+ym+ndy

similarly Im,n=∫0∞ym-11+ym+ndy

Now 2Im,n=∫0∞ym-1+yn-11+ym+ndy

=∫0∞ym-1+yn-11+ym+ndy

=∫01ym-1+yn-11+ym+ndy+∫1∞ym-1+yn-11+ym+ndy⏟substitute y=1t

⇒2Im,n=∫01ym-1+yn-11+ym+ndy-∫10tn-1+tm-1tm+n-2tm+n1+tm+ndtt2

⇒ Hence 2Im,n=2∫01ym-1+yn-11+ym+ndy⇒α=1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If $I, m, n$ are real and $I \neq m$, then the roots of the equation $(I-m) x^2-5(I+m) x-2(I-m)=0$ are

Options:
A Complex
B Real and distinct
C Real and equal
D None of these

The correct answer is:
Real and distinct

Solution:
Given equation is $(I-m) x^2-5(I+m) x-2(I-m)=0$
Its discriminant $D=25(I+m)^2+8(I-m)^2$
which is positive, since $I, m, n$ are real and $I \neq m$ Hence roots are real and distinct.

Topic: Mathematics, Quadratic Equation, JEE Main
If In=∫022dx1-xn, then the value of limn→∞⁡In is equal to

Options:
A 1
B 2
C 12
D 13

The correct answer is:
2

Solution:
Ιn=∫022dx1-xn
limn→∞Ιn=limn→∞∫012dx1-xn+limn→∞∫122dx1-xn
limn→∞In=limn→∞2∫01dx1-0+0
=21-0=2

Topic: Mathematics, Definite Integration, JEE Main
If In=∫0nπmaxsin⁡x,sin-1sin⁡xdx, then I2+I4 has the value λπ22; where λ is

Options:

The correct answer is:
3

Solution:
max sinâ¡x,sin-1sinâ¡x is sinâ1â¡sinâ¡x
Hence, In=n12âÏÏ2
= nÏ24 (area of triangle)
â´ I2+I4=6Ï24

Topic: Mathematics, Definite Integration, JEE Main
If In=∫π4π2cotnxdx, then

Options:
A I2+I4,I3+I52,I4+I6 are in G.P.
B I2+I4,I3+I5,I4+I6 are in A.P.
C 1I2+I4,1I3+I5,1I4+I6 are in A.P.
D 1I2+I4,1I3+I5,1I4+I6 are in G.P.

The correct answer is:
1I2+I4,1I3+I5,1I4+I6 are in A.P.

Solution:
In=∫π/4π/2cotnxdx=∫π/4π/2cotn-2xcosec2x-1dx

=-cotn-1xn-1π/4π/2-In-2

=1n-1-In-2

⇒In+In-2=1n-1

⇒I2+I4=13

I3+I5=14

I4+I6=15

∴1I2+I4,1I3+I5,1I4+I6 are in A.P.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If In=dndxnxnln⁡x, then the value of 150I7-7I6 is equal to

Options:

The correct answer is:
14.4

Solution:
Ιn=dn-1dxn-1xn-1+nxn-1ln⁡x
⇒In=n-1!+nIn-1⇒In-nIn-1=n-1!
Putting n=7, we get,
I7-7I6=6!=720

Topic: Mathematics, Differentiation, JEE Main
If In=∫ln⁡xndx, then I10+10I9 is equal to (where C is the constant of integration)

Options:
A xlnâ¡x10+C
B 10lnâ¡x9+C
C 9lnâ¡x10+C
D xln⁡x9+C

The correct answer is:
xln⁡x10+C

Solution:
In=∫ln⁡xndx
In=xln⁡xn-∫xnln⁡xn-1xdx
Ιn=xln⁡xn-nΙn-1
In+nIn-1=xln⁡xn+C
I10+10I9=xln⁡x10+C

Topic: Mathematics, Indefinite Integration, JEE Main
If In=∫xne6xdx, then the expression 6I10+10I9 simplifies to (where, c is the constant of integration)

Options:
A x10e5x+c
B x10e6x+c
C x9e5x+c
D x10e10x+c

The correct answer is:
x10e6x+c

Solution:
Using integration by parts in In, we get,
In=∫xn⏟Ie6x⏟IIdx
=xne6x6-∫nxn-1e6x6dx
⇒6In=xne6x-nIn-1+c
⇒6In+nIn-1=xne6x+c
Put n=10,
6I10+10I9=x10e6x+c

Topic: Mathematics, Indefinite Integration, JEE Main
If $I_n=\int x^n \cdot e^{c x} d x$ for $n \geq 1, \quad$ then $c \cdot I_n+n \cdot I_{n-1}$ is equal to

Options:
A $x^n e^{c x}$
B $x^n$
C $e^{c x}$
D $x^n+e^{c x}$

The correct answer is:
$x^n e^{c x}$

Solution:
Given that,
$$
\begin{aligned}
& I_n=\int x^n \cdot e^{c x} d x \\
& I_n=\frac{e^{c x}}{c} \cdot x^n-\int \frac{e^{c x}}{c} \cdot n x^{n-1} d x \\
& \Rightarrow \quad I_n=\frac{e^{c x} \cdot x^n}{c}-\frac{n}{c} I_{n-1} \\
&
\end{aligned}
$$
$\Rightarrow \quad c I_n+n I_{n-1}=e^{c x} \cdot x^n$

Topic: Mathematics, Indefinite Integration, JEE Main
If I=∫sin⁡x3sin⁡x+cos⁡x+2dx and J=∫cos⁡x3sin⁡x+cos⁡x+2dx, then 3J-I is equal to (where C is the constant of integration)

Options:
A x+ln⁡3sin⁡x+cos⁡x+2+c
B x-ln⁡3sin⁡x+cos⁡x+2+c
C ln⁡3sin⁡x+cos⁡x+2+c
D 2x+ln⁡3sin⁡x+cos⁡x+2+c

The correct answer is:
ln⁡3sin⁡x+cos⁡x+2+c

Solution:
3J-I=∫3cosx-sinx3sinx+cosx+2dx
Let 3sinx+cosx+2=t
⇒3cos⁡x-sin⁡xdx=dt
So, 3J-I=∫dtt=ln⁡t+C

Hence, 3J-I=ln⁡3sin⁡x+cos⁡x+2+c

Topic: Mathematics, Indefinite Integration, JEE Main
If I=â«tanâ¡x+cotâ¡xdx, then I equals to (where, C is an arbitrary constant)

Options:
A 2sin-1⁡sin⁡x+cos⁡x+C
B 2cos-1⁡sin⁡x-cos⁡x+C
C 2sin-1⁡sin⁡x-cos⁡x+C
D 2cos-1⁡sin⁡x+cos⁡x+C

The correct answer is:
2sin-1⁡sin⁡x-cos⁡x+C

Solution:
I=∫sin⁡x+cos⁡xcos⁡xsin⁡xdx
Put sin⁡x-cos⁡x=t , so that sin⁡x+cos⁡xdx=dt and 1 – 2sin⁡xcos⁡x=t2
∴ I=2 ∫dt1-t2=2sin-1⁡t+C
=2sin-1⁡sin⁡x-cos⁡x+C

Topic: Mathematics, Indefinite Integration, JEE Main
If I=â«x3-1x5+x4+x+1dx =14lnâ¡fx-lnâ¡gx+c (where, c is the constant of integration) and f0=g0=1, then the value of f1âg1 is equal to

Options:

The correct answer is:
4

Solution:
The given integral is I=∫x3-1x+1x4+1dx
=∫x3+x4-x4-1x+1x4+1dx
=∫x3x+1-x4+1x+1x4+1dx
=∫x3x4+1dx-∫dxx+1
=14∫4x3dxx4+1-∫dxx+1
=14ln⁡x4+1-ln⁡1+x+c

Hence, fx=x4+1, gx=x+1
∴f1·g1=2×2=4

Topic: Mathematics, Indefinite Integration, JEE Main
If I=∫x5+1x+1dx=fx+c and f0=0 (where, c is the constant of integration), then

Options:
A f1>1
B 0<f1<1
C f1=1
D f1>2

The correct answer is:
0<f1<1

Solution:
The given integral is
I=∫x+1x4-x3+x2-x+1x+1dx
=∫x4-x3+x2-x+1dx
=x55-x44+x33-x22+x+c
∴fx=x55-x44+x33-x22+x
∴f1=15-14+13-12+1
=1-120-16=4760
⇒0<f1<1

Topic: Mathematics, Indefinite Integration, JEE Main
If Ix=∫esin2xcosx sin2x-sinxdx and I0=1, then Iπ3 is equal to

Options:
A -12e34
B 12e34
C -e34
D e34

The correct answer is:
12e34

Solution:
Given,

Ix=∫esin2xcosx sin2x-sinxdx

⇒Ix=∫esin2xcosx-12 cosxsin2x dx

Now let, sin2x=t⇒sin2x dx=dt

Ix=∫et1-t-121-tdt

Now comparing with,

Ix=∫etft+f'tdt=etft+c

Here, ft=1-t & f't=-121-t

⇒Ix=et1-t+c=esin2x·cosx+c

Now using, I0=1⇒c=0

Hence, the value of Iπ3=12e34

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If \( i z^{3}+z^{2}-z+i=0 \), then \( |z| \) is equal to

Options:

The correct answer is:
1

Solution:
Given, iz3+z2−z+i=0
z3+1iz2-zi+1=0
⇒ z3-iz2+iz+1=0
⇒ z2z-i+iz-i=0
⇒ z-iz2+i=0
⇒z=i or z2=−i
Hence, |z|=1

Topic: Mathematics, Complex Number, JEE Main
If \( I=\int_{1}^{4}(\{x\})^{[x]} d x \) then \( \frac{24}{13} I \) is equal to

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 2 \)

Solution:
I=∫12x+∫23x2+∫23x3
=∫01x+x2+x3 = 12+13+14 = 6+4+312
=1312

Topic: Mathematics, Definite Integration, JEE Main
If Imiz+2z+i=-1 represents part of a circle with radius r units, then the value of 4r2 is (where, z∈C, z≠i, Imz represents the imaginary part of z and i2=-1)

Options:

The correct answer is:
2.25

Solution:
Let, z=x+iy
Imix+yi+2x+iy+i=-1
⇒Im-y-2+ixx+iy+1=-1
⇒Im-y-2+ixx-iy+1x2+y+12=-1
⇒x2+y2-y-2=-x2-y2-2y-1
⇒2x2+2y2+y-1=0
⇒x2+y2+y2-12=0
⇒ the radius r of the circle is 142+12=34

Hence, 4r2=4×916=94=2.25

Topic: Mathematics, Complex Number, JEE Main
If Imiz-2z-i=-1, (where zâC, zâ i) represents part of a circle, then the radius (in units) of the circle is (here, Im(z) represents the imaginary part of the complex number z)

Options:

The correct answer is:
0.75

Solution:
Let, z=x+iy
Imix+yi-2x+iy-i=-1
⇒Im-y+2+ixx+iy-1=-1
⇒Im-y+2+ixx-iy-1x2+y-12=-1
⇒x2+y2+y-2=-x2-y2+2y-1
⇒2x2+2y2-y-1=0
⇒x2+y2-y2-12=0
⇒ radius of the circle is 142+12=34
= 0.75

Topic: Mathematics, Complex Number, JEE Main
If in 4f the number of radial nodes and angular nodes are X and Y respectively. Find the sum of X + Y?

Options:

The correct answer is:
3

Solution:
If 4f n = 4, l = 3
Radial nodes X=n-l-1=4-3-1=0
Angular nodes Y=0=3
X+Y=0+3=3

Topic: Chemistry, Structure of Atom, JEE Main
If in a circular coil $A$ of radius $R$, current I is flowing and in another coil $B$ of radius $2 R$ a current $2 \mathrm{l}$ is flowing, then the ratio of the magnetic fields $\mathrm{B}_{\mathrm{A}}$ and $\mathrm{B}_{\mathrm{B}}$, produced by them will be

Options:
A
1

B
2

C
1/2

D
4

The correct answer is:
1

Solution:
In coil A, B $=\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{I}}{\mathrm{R}} . \quad \therefore \mathrm{B} \alpha \frac{\mathrm{I}}{\mathrm{R}}$; Hence, $\frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{I}_1}{\mathrm{R}_1} \cdot \frac{\mathrm{R}_2}{\mathrm{I}_2}=\frac{2}{2}=1$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2002
If in a class there are 200 students in which 120 take Mathematics, 90 take Physics, 60 take Chemistry, 50 take Mathematics & Physics, 50 take Mathematics & Chemistry, 43 take Physics & Chemistry and 38 take Mathematics, Physics & Chemistry, then the number of students who have taken exactly one subject is

Options:
A 42
B 56
C 270
D 98

The correct answer is:
98

Solution:
Let, MâMathematics, PâPhysics, CâChemistry
Given that total students =200
nM=120, nP=90, nC=60
nMâ©P=50, nMâ©C=50, nPâ©C=43
nMâ©Pâ©C=38
Required number of students taking exactly one subject is
nM+nP+nC-2nMâ©P-2nPâ©C-2nMâ©C+3nMâ©Pâ©C
=120+90+60-250-250-243+338
=98

Topic: Mathematics, Sets and Relations, JEE Main
If in a container neither mass and nor heat exchange occurs then it constitutes a

Options:
A Closed system
B Open system
C Isolated system
D Imaginary system

The correct answer is:
Isolated system

Solution:
An isolated system neither shows exchange of heat nor matter with surroundings.

Topic: Chemistry, Thermodynamics (C), JEE Main
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

Options:
A $25.5$
B $24.0$
C $22.0$
D $20.5$

The correct answer is:
$24.0$

Solution:
$\begin{aligned} \text { Mode } & =3 \text { Median }-2 \text { Mean }=3(22)-2(21) \\ & =66-42=24 . \end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

Options:
A $25.5$
B $24.0$
C $22.0$
D $20.5$

The correct answer is:
$24.0$

Solution:
$\begin{aligned} \text { Mode } & =3 \text { Median }-2 \text { Mean }=3(22)-2(21) \\ & =66-42=24 .\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

Options:
A
22.0

B
20.5

C
25.5

D
24.0

The correct answer is:
24.0

Solution:
Mode $+2$ Mean $=3$ Median
$$
\Rightarrow \text { Mode }=3 \times 22-2 \times 21=66-42=24 \text {. }
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2005
If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Options:
A 7
B 4
C 5
D 6

The correct answer is:
6

Solution:
For the given GP, n=64, Sn=7So ; where So→Sum of odd terms

⇒a+ar+ar2+ar3+...+ar63=7a+ar2+ar4…..+ar62

⇒a1-r641-r=7a1-r641-r2

⇒1-r64=71-r641+r

⇒1=71+r

⇒1+r=7

⇒r=6

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If in a nuclear fusion process, the masses of the fusing nuclei are \( m_{1} \) and \( m_{2} \) and the mass of the resultant nucleus is \( m_{3} \),
then

Options:
A \( m_{3}=m_{1}+m_{2} \)
B \( m_{3}=\left|m_{1} m_{2}\right| \)
C \( m_{3} < \left(m_{1}+m_{2}\right) \)
D \( m_{3}>\left(m_{1}+m_{2}\right) \)

The correct answer is:
\( m_{3} < \left(m_{1}+m_{2}\right) \)

Solution:
In a nuclear fusion, when two light nuclei of different masses are combined to form a stable nucleus, then some mass is lost and appears in the form of energy called the mass defect. So, the mass of resultant nucleus is always less than the sum of masses of fusing nuclei, i.e., m3< (m1 + m2).

Topic: Physics, Nuclear Physics, JEE Main
If in a $\mathrm{p-n}$ junction diode, a square input signal of $10 \mathrm{~V}$ is applied as shown
Then the output signal across $R_{\mathrm{L}}$ will be

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2007
If in a parallelogram $\mathrm{ABDC}$, the coordinates of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively (1,2),(3,4) and $(2,5),$ then the equation of the diagonal $\mathrm{AD}$ is :

Options:
A $5 x-3 y+1=0$
B $5 x+3 y-11=0$
C $3 x-5 y+7=0$
D $3 x+5 y-13=0$

The correct answer is:
$5 x-3 y+1=0$

Solution:
Since, in parallelogram mid points of both diagonals

coinsides.

$\therefore$ mid-point of $A D=$ mid-point of $B C$

$\left(\frac{x_{1}+1}{2}, \frac{y_{1}+2}{2}\right)=\left(\frac{3+2}{2}, \frac{4+5}{2}\right)$

$\therefore \quad\left(x_{1}, y_{1}\right)=(4,7)$

Then, equation of $A D$ is,

$y-7=\frac{2-7}{1-4}(x-4)$

$y-7=\frac{5}{3}(x-4)$

$3 y-21=5 x-20$

$5 x-3 y+1=0$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If in a plano-convex lens, the radius of curvature of the convex surface is $10 \mathrm{~cm}$ and the focal length of the lens is $30 \mathrm{~cm}$, then the refractive index of the material of lens will be

Options:
A 1.5
B 1.66
C 1.33
D 3

The correct answer is:
1.33

Solution:
$f=\frac{R}{(\mu-1)} \Rightarrow 30=\frac{10}{(\mu-1)} \Rightarrow \mu=1.33$.

Topic: Physics, Ray Optics, JEE Main
If in a regular polygon the number of diagonals is 54, then the number of sides of this polygon is:

Options:
A 12
B 10
C 6
D 9

The correct answer is:
12

Solution:
Number of diagonals of an ′n′ sided polygon is
nC2 -n=54
nn-11.2-n=54
n2-n2-n=54
n2 -3n = 108
n2-3n-108=0
n-12n+9=0
n=12

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (11 Apr Online)
If in a resonance tube, oil of density higher than that of water is used, then the resonance frequency would

Options:
A Increase
B Decrease
C Slightly increase
D Remain same

The correct answer is:
Remain same

Solution:
In a resonance tube, water works as a reflector and the resonance frequency is independent of the substance (liquid) which is filled in the tube.

Topic: Physics, Waves and Sound, NEET
If in a triangle $A B C, \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos \mathrm{A}$ is equal to

Options:
A
$5 / 7$

B
$1 / 5$

C
$35 / 19$

D
$19 / 35$

The correct answer is:
$1 / 5$

Solution:
Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $A B C$. $\Rightarrow b+c=11 K, c+a=12 K, a+b=13 K$ On solving these, we get $a=7 K, b=6 K, c=5 K$
Now we know,
$$
\begin{aligned}
& \cos A=\frac{b^2+c^2-a^2}{2 b c} \\
& =\frac{36 K^2+25 K^2-49 K^2}{2(6 K) 5 K}=\frac{1}{5}
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2012 (12 May Online)
If in a triangle $\mathrm{ABC} \mathrm{a} \cos ^2\left(\frac{\mathrm{C}}{2}\right)+\mathrm{c}^2 \cos ^2\left(\frac{\mathrm{A}}{2}\right)=\frac{3 \mathrm{~b}}{2}$, then the sides $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$

Options:
A
satisfy $a+b=c$

B
are in A.P.

C
are in G.P.

D
are in H.P.

The correct answer is:
are in A.P.

Solution:
If $a \cos ^2\left(\frac{\mathrm{C}}{2}\right)+\cos ^2\left(\frac{\mathrm{A}}{2}\right)=\frac{3 \mathrm{~b}}{2}$
$\mathrm{a}[\cos \mathrm{C}+1]+\mathrm{c}[\cos \mathrm{A}+1]=3 \mathrm{~b}$
$(\mathrm{a}+\mathrm{c})+(\mathrm{a} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{B})=3 \mathrm{~b}$
$\mathrm{a}+\mathrm{c}+\mathrm{b}=3 \mathrm{~b}$ or $\mathrm{a}+\mathrm{c}=2 \mathrm{~b}$ or $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2003
If in a triangle $\mathrm{ABC}$, the altitudes from the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ on opposite sides are in H.P., then $\sin A, \sin B, \sin C$ are in

Options:
A
G.P.

B
A.P.

C
Arithmetic - Geometric Progression

D
H.P.

The correct answer is:
A.P.

Solution:
$\Delta=\frac{1}{2} p_1 a=\frac{1}{2} p_2 b=\frac{1}{2} p_3 b$
$p_1, p_2, p_3$ are in H.P.
$\Rightarrow \frac{2 \Delta}{a}, \frac{2 \Delta}{b}, \frac{2 \Delta}{c}$ are in H.P.
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in H.P
$\Rightarrow a, b, c$ are in A.P.
$\Rightarrow \sin A, \sin B, \sin C$ are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2005
If in compound microscope $m_1$ and $m_2$ be the linear magnification of the objective lens and eye lens respectively, then magnifying power of the compound microscope will be

Options:
A $m_1-m_2$
B $\sqrt{m_1+m_2}$
C $\left(m_1+m_2\right) / 2$
D $m_1 \times m_2$

The correct answer is:
$m_1 \times m_2$

Solution:
Magnification of a compound microscope is given by $m=-\frac{v_o}{u_o} \times \frac{D}{u_e} \Rightarrow|m|=m_o \times m_e$.

Topic: Physics, Ray Optics, JEE Main
If in I2Cl6 the number of covalent bonds and coordinate bonds are m and n respectively. Find the sum of m + n.

Options:

The correct answer is:
8

Solution:
It contains 6 covalent bonds and 2 coordinate bonds.
i.e., m = 6, n = 2
m + n = 6 + 2 = 8

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
If in nuclear fission, a piece of uranium of mass 6.0 g is lost, the energy obtained (in kWh) is nÃ107. Find the value of n.

Options:

The correct answer is:
1.25

Solution:
E=Δmc2
=0.5×10-3×3×1082
=4.5×1013
E=4.5×10133.6×106
=1.25×107 kWh

Topic: Physics, Nuclear Physics, JEE Main
If in the circuit, power dissipation is $150 \mathrm{~W}$, then $\mathrm{R}$ is

Options:
A
$2 \Omega$

B
$6 \Omega$

C
$5 \Omega$

D
$4 \Omega$

The correct answer is:
$6 \Omega$

Solution:
$W=\frac{V^2}{R_{\text {net }}} ; 150=\frac{(15)^2}{R}+\frac{(15)^2}{2} \Rightarrow R=6 \Omega$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2002
If in the expansion of $(1+x)^{21}$, the coefficients of $x^r$ and $x^{r+1}$ be equal, then $r$ is equal to

Options:
A $9$
B $10$
C $11$
D $12$

The correct answer is:
$10$

Solution:
$T_{r+1}={ }^{21} C_r(1)^{21-r}(x)^r={ }^{21} C_r$
$\therefore$ Coefficient of $x^r={ }^{21} C_r$ and coefficient of $x^{r+1}={ }^{21} C_{r+1}$
So, we must have ${ }^{21} C_r={ }^{21} C_{r+1} \Rightarrow r=10$

Topic: Mathematics, Binomial Theorem, JEE Main
If in the expansion of (1+x)m(1–x)n the coefficients of x and x2 are –2 and –6 respectively, then the value of m+n is

Options:
A 12
B 14
C 16
D 18

The correct answer is:
16

Solution:
(1+x)m(1âx)n=1+mx+mmâ12x2+...
1ânx+nnâ12x2+...
â m- n=-2 and
m(mâ1)2+n(nâ1)2âmn=â6
â m=7 and n=9

Topic: Mathematics, Binomial Theorem, JEE Main
If in the expansion of 1+xm1-xn, the coefficients of x and x2 are 3 and â6 respectively, then the value of m is m,nâN

Options:
A 6
B 9
C 12
D 24

The correct answer is:
12

Solution:
Given expansion =1+mC1x+mC2x2+....1-nC1x+nC2x2-....
Coefficient of x2=-nC1+mC1=m-n=3
Coefficient of x2=nC2-mC1mC1+mC2=nn-12-mn+mm-12
⇒ nn-12-mn+mm-12=-6
n=mâ3
⇒ m-3m-42+mm-12-mm-3=-6
⇒m2-7m+12+m2-m-2m2+6m=-12
⇒-2m=-24⇒m=12

Topic: Mathematics, Binomial Theorem, JEE Main
If in the expansion of $(a-2 b)^{\mathrm{n}}$, the sum of the 5 th and 6 th term is zero, then the value of $\frac{\mathrm{a}}{\mathrm{b}}$ is

Options:
A $\frac{\mathrm{n}-4}{5}$
B $\frac{2(n-4)}{5}$
C $\frac{5}{n-4}$
D $\frac{5}{2(n-4)}$

The correct answer is:
$\frac{2(n-4)}{5}$

Solution:
$$
\text { Hints: } \begin{gathered}
(a-2 b)^{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(a)^{\mathrm{n}-\mathrm{r}}(-2 \mathrm{~b})^{\mathrm{r}} \\
\mathrm{t}_5+\mathrm{t}_6=0
\end{gathered}
$$
$$
\begin{aligned}
& \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_4(\mathrm{a})^{\mathrm{n}-4}(-2 \mathrm{~b})^4+{ }^{\mathrm{n}} \mathrm{C}_5(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5=0 \Rightarrow \frac{\mathrm{n} !}{4 !(\mathrm{n}-4) !} \mathrm{a}^{\mathrm{n}-4}(-2 \mathrm{~b})^4=-\frac{\mathrm{n} !}{5 !(\mathrm{n}-5) !}(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5 \\
& \Rightarrow \frac{1}{(\mathrm{n}-4)} \times \mathrm{a}=\frac{-1}{5}(-2 \mathrm{~b}) \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{2(\mathrm{n}-4)}{5}
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main