datavorous/entrance-exam-dataset huggingface dataset:
If the coefficient of $(2 r+1)$ th term and $(r+2)$ th term in the expansion of $(1+x)^{43}$ are equal, then $r$ is equal to :

Options:
A 12
B 14
C 16
D 18

The correct answer is:
14

Solution:
Given that,
Coefficient of $(2 r+1)$ th term $=$ co-efficient of $(r+2)$ th term
$\Rightarrow \quad{ }^{43} C_{2 r}={ }^{43} C_{r+1}$
$\Rightarrow \quad 2 r+(r+1)=43$
or $\quad 2 r=r+1$
$\Rightarrow \quad r=14$ or $r=1$
thus $r=14$

Topic: Mathematics, Binomial Theorem, JEE Main
If the coefficient of friction between the wedge A and block B shown in the figure is μ, then the maximum possible horizontal acceleration of A for which B doesn't slip is [angle of inclination of wedge = 45°]

Options:
A μg
B g1+μ1-μ
C gμ
D g1-μ1+μ

The correct answer is:
g1+μ1-μ

Solution:
FBD of block B w.r.t. wedge A, for maximum 'a':

Perpendicular to wedge:
Î£fy=(mgcosÎ¸+masinÎ¸-N)=0
and Î£fx=mgsinÎ¸+Î¼N-macosÎ¸=0 (for maximum a)
â mgsinÎ¸+Î¼(mgcosÎ¸+masinÎ¸)-macosÎ¸=0
â a=(gsinÎ¸+Î¼gcosÎ¸)cosÎ¸-Î¼sinÎ¸

for Î¸=45Â°

a=gtan45Â°+Î¼cot45Â°-Î¼;a=g1+Î¼1-Î¼

Topic: Physics, Laws of Motion, JEE Main
If the coefficient of the three successive terms in the binomial expansion of 1+xn are in the ratio 1:7:42, then the first of these terms in the expansion is

Options:
A 9th
B 6th
C 8th
D 7th

The correct answer is:
7th

Solution:
Let coefficient of first term is Crn

CrnCr+1n=17

⇒ r+1n-r=17

⇒ 7r+7=n-r

n-8r=7 …1

Also, Cr+1nCr+2n=742=16

⇒r+2n-r-1=16

⇒6r+12=n-r-1

n-7r=13 …2

Solving 1 and 2, we get r=6

Hence, 7th term is the answer.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2015 (10 Apr Online)
If the coefficient of x6 in the expansion of 2+x33+x25+x3 is K, then the value of K100 is

Options:

The correct answer is:
3.12

Solution:
x+23x+32x+53=x8+x72+2+2+3+3+5+5+5+x6K+.......
Clearly, K is equal to the sum taken two at a time of 2,2,2,3,3,5,5,5.
⇒K=2+2+2+3+3+5+5+52-22+22+22+32+32+52+52+522
=272-4+4+4+9+9+25+25+252
=729-1052=6242=312
⇒K100=3.12

Topic: Mathematics, Binomial Theorem, JEE Main
If the co-efficient of x7 and x8 in the expansion of 2+x3n are equal, then the value of n is equal to :

Options:

The correct answer is:
55

Solution:
We have, the co-efficient of x7 and x8 in the expansion of 2+x3n are equal.

Therefore, C7n2n-7137=C8n2n-8138

⇒n-7=48

⇒n=55.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If the coefficient of x7 in ax-1bx213 and the coefficient of x-5 in ax+1bx213 are equal, then a4b4 is equal to:

Options:
A 11
B 44
C 22
D 33.

The correct answer is:
22

Solution:
Given,

The coefficient of x7 in the expansion of ax2-1bx13 is equal to the coefficient of x-5 in ax+1bx213.

We know that, the general term Tr+1 in the expansion a+bn is

Tr+1=Crnan-rbr

Applying to ax-1bx213, we get

Tr+1=Cr13ax13-r-1bx2r

⇒Tr+1=-1r×Cr13a13-rx13-3rb-r

Therefore, 13-3r=7⇒r=2 for coefficient of x7.

Thus,

T3=C213a11b2

Similarly, applying to ax+1bx213, we get

Tr+1=Cr13ax13-r1bx2r

⇒Tr+1=Cr13a13-rx13-3rb-r

Therefore, 13-3r=-5 for coefficient of x-5

⇒r=6

So,

T7=C613a7b-6

Hence, applying the given condition we get

C213a11b2=C613a7b-6

⇒a4b4=C613C213

⇒a4b4=13!7!·6!×2!·11!13!

⇒a4b4=11×10×9×86×5×4×3

⇒a4b4=22

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If the coefficient of $x^7$ in $\left[a x^2+\left(\frac{1}{b x}\right)\right]^{11}$ equals the coefficient of $x^{-7}$ in $\left[a x^2-\left(\frac{1}{b x}\right)\right]^{11}$, then $a$ and $b$ satisfy the relation

Options:
A
$a-b=1$

B
$a+b=1$

C
$\frac{a}{b}=1$

D
$a b=1$

The correct answer is:
$a b=1$

Solution:
$$
\begin{aligned}
& T_{r+1} \text { in the expansion }\left[a x^2+\frac{1}{b x}\right]^{11}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{b x}\right)^r \\
& ={ }^{11} C_r(a)^{11-r}(b)^{-r}(x)^{22-2 r-r} \\
& \Rightarrow 22-3 r=7 \Rightarrow r=5 \\
& \therefore \text { coefficient of } x^7={ }^{11} C_5(a)^6(b)^{-5} ....(i)
\end{aligned}
$$
Again $T_{r+1}$ in the expansion $\left[a x-\frac{1}{b x^2}\right]^{11}={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{b x^2}\right)^r$ $={ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{11-\mathrm{r}}(-1)^{\mathrm{r}} \times(\mathrm{b})^{-\mathrm{r}}(\mathrm{x})^{-2 \mathrm{r}}(\mathrm{x})^{11-\mathrm{r}}$
Now $11-3 r=-7 \Rightarrow 3 r=18 \Rightarrow r=6$
$$
\begin{aligned}
& \therefore \text { coefficient of } x^{-7}={ }^{11} C_6 a^5 \times 1 \times(b)-6 \\
& \Rightarrow{ }^{11} C_5(a)^6(b)^{-5}={ }^{11} C_6 a^5 \times(b)^{-6} \\
& \Rightarrow a b=1
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If the coefficient of $x^7$ in $\left[a x^2+\left(\frac{1}{b x}\right)\right]^{11}$ equals the coefficient of $x^{-7}$ in $\left[a x^2-\left(\frac{1}{b x}\right)\right]^{11}$, then $a$ and $b$ satisfy the relation

Options:
A
$a-b=1$

B
$a+b=1$

C
$\frac{a}{b}=1$

D
$a b=1$

The correct answer is:
$a b=1$

Solution:
$$
\begin{aligned}
& T_{r+1} \text { in the expansion }\left[a x^2+\frac{1}{b x}\right]^{11}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{b x}\right)^r \\
& ={ }^{11} C_r(a)^{11-r}(b)^{-r}(x)^{22-2 r-r} \\
& \Rightarrow 22-3 r=7 \Rightarrow r=5 \\
& \therefore \text { coefficient of } x^7={ }^{11} C_5(a)^6(b)^{-5} ....(i)
\end{aligned}
$$
Again $T_{r+1}$ in the expansion $\left[a x-\frac{1}{b x^2}\right]^{11}={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{b x^2}\right)^r$ $={ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{11-\mathrm{r}}(-1)^{\mathrm{r}} \times(\mathrm{b})^{-\mathrm{r}}(\mathrm{x})^{-2 \mathrm{r}}(\mathrm{x})^{11-\mathrm{r}}$
Now $11-3 r=-7 \Rightarrow 3 r=18 \Rightarrow r=6$
$$
\begin{aligned}
& \therefore \text { coefficient of } x^{-7}={ }^{11} C_6 a^5 \times 1 \times(b)-6 \\
& \Rightarrow{ }^{11} C_5(a)^6(b)^{-5}={ }^{11} C_6 a^5 \times(b)^{-6} \\
& \Rightarrow a b=1
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2005
If the coefficient of x 7 in ax 2 + 1 bx 1 1 is equal to the coefficient of x-7 in ax - 1 bx 2 1 1 , then a and b satisfy the relation

Options:
A ab=1
B a b = 1
C a+b=1
D a−b=1

The correct answer is:
ab=1

Solution:
In the expansion of ax 2 + 1 bx 1 1
T r + 1 = 1 1 C r ax 2 1 1 - r 1 bx r
= 1 1 C r a 1 1 - r b r Â· x 2 2 - 3 r
For the coefficient of x7, put 22-3r = 7
âr=5
â´ T 6 = 1 1 C 5 a 6 b 5 Â· x 7
â´ Coefficient of x7 in the expansion of ax 2 + 1 bx 1 1 is 1 1 C 5 a 6 b 5
Similarly, the coefficient of x-7 in the expansion axâ 1 bx 2 11 is 11C6a5b6
Now, 1 1 C 5 a 6 b 5 = 1 1 C 6 a 5 b 6
âab=1

Topic: Mathematics, Binomial Theorem, JEE Main
If the co-efficient of x9 in αx3+1βx11and the co-efficient of x-9 in αx-1βx311 are equal, then (αβ)2 is equal to

Options:

The correct answer is:
1

Solution:
Given,

The coefficient of x9 in αx3+1βx11and the coefficient of x-9 in αx-1βx311 are equal,

Now the rth term of αx3+1βx11 is given by =Cr11·α11-rβr×x311-r-r

So, for x9, 33-4r=9⇒r=6

Hence, coefficient of x9 in αx3+1βx11=C611·α5β6

Similarly, for coefficient of x-9 in αx-1βx311=-C511·α6β5

Now given both coefficient are equal,

So, 11C6·α5β6=-11C5·α6β5

⇒1β=-α

⇒αβ=-1

⇒(αβ)2=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (29 Jan Shift 1)
If the coefficient of x10 in the binomial expansion of x514+5x1360 is 5kl, where l,k∈N and l is coprime to 5, then k is equal to ______.

Options:

The correct answer is:
5

Solution:
Tr+1=Cr60x51460-r5x13r

For x10, 60-r2-r3=10⇒ r=24

Hence coefficient of x10 will be C246053

Now, exponent of 5 in C2460 will be the power of 5 in 60!24!36! i.e. 51454·58=52

i.e. k=5

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If the coefficient of x15 in the expansion of ax3+1bx1315 is equal to the coefficient of x-15 in the expansion of ax13-1bx315, where a and b are positive real numbers, then for each such ordered pair a,b:

Options:
A a=b
B ab=1
C a=3b
D ab=3

The correct answer is:
ab=1

Solution:
(r+1)th term in the expansion of ax3+1bx1315 is

Tr+1=Cr15ax315-r1bx1/3r

⇒Tr+1=Cr15a15-rx45-10r31br

For the coefficient of x15 in ax3+1bx1/315:

45-10r3=15

⇒30=10r3

⇒r=9

Coefficient of x15=C915a6b-9

(r+1)th term in the expansion of ax1/3-1bx315 is:

Tr+1=Cr15ax1/315-r-1bx3r

For the coefficient of x-15 in ax1/3-1bx315 is:

5-r3-3r=-15

⇒10r3=20

⇒r=6

Coefficient of x-15 =C615a9×b-6

Hence,

C915a6b-9=C615a9×b-6

⇒a9b6=a6b9

⇒a3b3=1⇒ab=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If the coefficients of $r$ th, $(r+1)$ th and $(r+2)$ th terms in the binomial expansion of $(1+$ $y)^m$ are in A.P., then $m$ and $r$ satisfy the equation

Options:
A
$m^2-m(4 r-1)+4 r^2-2=0$

B
$m^2-m(4 r+1)+4 r^2+2=0$

C
$m^2-m(4 r+1)+4 r^2-2=0$

D
$m^2-m(4 r-1)+4 r^2+2=0$

The correct answer is:
$m^2-m(4 r+1)+4 r^2-2=0$

Solution:
Given ${ }^m C,{ }^m C,{ }^m C_{r+1}$ are in A.P.
$$
\begin{aligned}
& 2{ }^m C_r={ }^m C_{r-1}+{ }^m C_{r+1} \\
& \Rightarrow 2=\frac{{ }^m C_{r-1}}{{ }^m C_r}+\frac{{ }^m C_{r+1}}{{ }^m C_r} \\
& =\frac{r}{m-r+1}+\frac{m-r}{r+1} \\
& \Rightarrow m^2-m(4 r+1)+4 r^2-2=0
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2005
If the coefficients of three consecutive terms in the expansion of 1+xn are in the ratio 1:7:42, then the value of n is equal to

Options:
A 49
B 50
C 55
D 56

The correct answer is:
55

Solution:
For Tr+1= nCrxr, coefficient â nCr
nCr-1: nCr: nCr+1=1:7:42
â nCr nCr-1=71 and nCr+1 nCr=427
ân-r+1r=7 and n-rr+1=6
ân-r+1=7r and n-r=6r+6
ân=8r-1 and n=7r+6
â8r-1=7r+6âr=7
ân=55

Topic: Mathematics, Binomial Theorem, JEE Main
if the coefficients of three consecutive terms in the expansion of 1+xn are the ratio 1 : 5 :20 then the coefficient of the fourth term is

Options:
A 2436
B 5481
C 1827
D 3654

The correct answer is:
3654

Solution:
Given,

Cr-1n:Crn:Cr+1n=1:5:20

Now using the formula CrnCr-1n=n-r+1r we get,

.CrnCr-1n=n-r+1r=51

⇒n-6r+1=0 .........1

And Cr+1nCrn=n-rr+1=205

⇒n-5r-4=0 ..........2

Solving above equations we get,

n=29 & r=5

So, the coefficient of forth term will be C3n=C329=29×28×273×2×1=3654

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If the coefficients of variation of two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively, then their arithmetic means are respectively

Options:
A $35,22.85$
B $32,25.85$
C $35,28.25$
D $35,25.25$

The correct answer is:
$35,22.85$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, JEE Main
If the coefficients of x2 and x3, are both zero, in the expansion of the expression 1+ax+bx21-3x15, in powers of x , then the ordered pair a,b is equal to

Options:
A 28,315
B -21,714
C 28,861
D -54,315

The correct answer is:
28,315

Solution:
1+ax+bx21-3x15=1+ax+bx2C015-15C1 3x + 15C2 3x2-15C33x3+…-15C153x15

∴ Coefficient of x2 is: b.15C0-3.a15C1+32..15C2=0

(Given)
⇒b-45a+945=0......i

Coefficient of x3 is:

-b.3.15C1+9a.15C2-27.15C3=0

⇒ -45b+9a×152×141-27×153×142×131=0

⇒ -b+21a-273=0

⇒ b-21a+273=0 .......ii

From i and ii

a=28 and b=315

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If the coefficients of x-2 and x-4, in the expansion of x13+12x1318,x>0, are m and n respectively, then mn is equal to

Options:
A 27
B 182
C 54
D 45

The correct answer is:
182

Solution:
We have, x13+12x1318,x>0

General term, Tr+1=18Crx1318-r12x13r=18Cr x6-2r312r
Now,6-2r3=-2⇒r=12 and, 6-2r3=-4⇒r=15
Hence, mn=Coefficient of x-2Coefficient of x-4=C1218212C1518215

=18!6!.12!18!3!.15!×23

=15×14×136×5×4×23

=182

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2016 (10 Apr Online)
If the coefficients of x3 and x4 in the expansion of 1+ax+bx21-2x18 in powers of x are both zero, then a,b is equal to

Options:
A 1 4 , 2 7 2 3
B 1 6 , 2 7 2 3
C 1 6 , 2 5 1 3
D 1 4 , 2 5 1 3

The correct answer is:
1 6 , 2 7 2 3

Solution:
We have, 1+ax+bx21-2x18=a0+a1x+a2x2+.......
=1+ax+bx21-18C1·2x+18C2·22x2+.....18C182x18
⇒a3=-18C3×8+18C2×4a-36b
⇒a4=18C4×16-18C3×8a+18C2×4×b
102a-6b=1088 ...i

64a-6b=480 ...ii

Solution of the above two equation gives.

a=16
b = 2 7 2 3

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2014 (06 Apr)
If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression, then the maximum value of $n$ is:

Options:
A 7
B 21
C 28
D 14

The correct answer is:
14

Solution:
$\begin{aligned} & \text { Coeff. of } x^4={ }^n C_4 \\ & \text { Coeff. of } x^5={ }^n C_5 \\ & \text { Coeff. of } x^6={ }^n C_6 \\ & { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P \\ & 2 C_5={ }^n C_4+{ }^n C_6 \\ & 2=\frac{{ }^n C_4}{{ }^n C_5}+\frac{{ }^n C_6}{{ }^n C_5} \quad\left\{\frac{{ }^n C_r}{{ }^n C_r}=\frac{n-r+1}{r}\right\} \\ & 2=\frac{5}{n-4}+\frac{n-5}{6} \\ & 12(n-4)=30+n^2-9 n+20 \\ & n^2-21 n+98=0 \\ & (n-14)(n-7)=0 \\ & n_{\max }=14 \quad n_{\min }=7\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (04 Apr Shift 2)
If the coefficients of x7 in ax2+12bx11 and x-7 in ax-13bx211 are equal, then

Options:
A 729ab=32
B 32ab=729
C 64ab=243
D 243ab=64

The correct answer is:
729ab=32

Solution:
The coefficient of x7 in ax2+12bx11 and ax-13bx211 are equal, then

Tr+1=11Crax211-r12bxr

=11Cra11-r12brx22-3r

Now solving 22-3r=7⇒r=5

Coefficient of x-7 in ax-13bx211

Tr+1=11Crax11-r-13bx2r

=11Cra11-r-13brx11-3r

Now solving 11-3r=-7⇒r=6

Now equating coefficient of x7 and x-7 we get,

∴ 11C5a612b5=11C6 a5-13b6

⇒36ab=32

⇒729ab=32

Hence this is the correct option.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If the coefficients of x7 in x2+1bx11 and x-7 in x-1bx211,b≠0, are equal, then the value of b is equal to:

Options:
A 2
B -1
C 1
D -2

The correct answer is:
1

Solution:
Coefficient of x7 in x2+1bx11

=Cr11x211-r·1bxr

=Cr11x22-3r·1 br

So, 22-3r=7

⇒r=5

So, the term is C511·1 b5·x7

Coefficient of x-7 in x-1bx211

=Cr11x11-r·-1 bx2r

=Cr11x11-3r·-1rbr

So, 11-3r=-7 ∴ r=6

So, the term is C611·1 b6x-7

Hence, C511·1 b5=C611·1 b6

Since b≠0 ∴ b=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If the coefficients of x and x2 in (1+x)p(1-x)q are 4 and -5 respectively, then 2p+3q is equal to

Options:
A 60
B 69
C 66
D 63

The correct answer is:
63

Solution:
Given that The coefficient of x and x2 in 1+xp1-xq are 4 and -5.

⇒1+xp1-xq=1+px+pp-12!x2+....1-qx+qq-12!x2-....

Now coefficient of x from the above expansion will be

p-q which is equal to 4

⇒p-q=4.

Similarly coefficient of x2 is -5.

⇒pp-12+qq-12-pq=-5

⇒p2-2pq+q22-p+q2=-5

⇒p-q22-p+q2=-5

⇒162+5=p+q2

⇒p+q=26 and p-q=4

On solving the above equations we get,

p=15 and q=11.

⇒2p+3q=215+311=63

Therefore, the required answer is 63.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If the coefficients of x and x2 in the expansion of 1+xp1-xq,p,q≤15, are -3 and -5 respectively, then the coefficient of x3 is equal to ______.

Options:

The correct answer is:
23

Solution:
Given coefficient of x in expansion of 1+xp1-xq is -3

⇒ C0qC1p-C0pC1q=-3

⇒p-q=-3 ⋯1

Now coefficients of x2 in expansion of 1+xp1-xq

-C1qPC1+C0qC2P+C0PC2q=-5

⇒-pq+pp-12+qq-12=-5 ⋯2

Solving 1 and 2

p=8,q=11

Coefficient of x3 in expansion of 1+xp1-xq will be,

-C3q+C3p+Cp1qC2-Cp2qC1

=-C311+C38+C18C211-C28C111

=23

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (26 Jul Shift 1)
If the collision frequency of hydrogen molecules in a closed chamber at $27^{\circ} \mathrm{C}$ is $\mathrm{Z}$, then the collision frequency of the same system at $127^{\circ} \mathrm{C}$ is :

Options:
A $\frac{\sqrt{3}}{2} \mathrm{z}$
B $\frac{2}{\sqrt{3}} \mathrm{Z}$
C $\frac{3}{4} \mathrm{Z}$
D $\frac{4}{3} \mathrm{Z}$

The correct answer is:
$\frac{2}{\sqrt{3}} \mathrm{Z}$

Solution:
Assuming mean free path constant.
$\begin{aligned} & \mathrm{f} \propto \mathrm{V} \propto \sqrt{\mathrm{T}} \\ & \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}=\sqrt{\frac{300}{400}} \\ & \mathrm{f}_2=\sqrt{\frac{4}{3}}=\mathrm{f}_1=\frac{2}{\sqrt{3}} \mathrm{Z}\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If the colours of salts of transition elements are due to the presence of unpaired electrons in the transition metal ions, which of the following ions will be colourless in aqueous solution

Options:
A $T i^{3+}$
B $T i^{4+}$
C $\mathrm{Fe}^{2+}$
D $\mathrm{Fe}^{3+}$

The correct answer is:
$T i^{4+}$

Solution:
$T i^{+4} \rightarrow 3 d^0 4 s^0 \quad \therefore$ no unpaired $\mathrm{e}^{-}$

Topic: Chemistry, d and f Block Elements, JEE Main
If the common tangents of x2+y2=r2 and x216+y29=1 form a square, then the area (in sq. units) of the square is

Options:
A 50
B 100
C 25
D 40

The correct answer is:
50

Solution:
Let, the equation of tangent in slope form for the circle be y=mx±r1+m2

and the equation of tangent in slope form for the ellipse be y=mx±16m2+9

Now, for common tangets,
y=mx±r1+m2 is same as y=mx±16m2+9
⇒r2+r2m2=16m2+9⇒r2-16m2+r2-9=0

Since, here m1m2=-1
⇒2r2=25⇒r=52

So, the side of the square =2r=52
⇒ area of the square is 50 sq. units

Topic: Mathematics, Ellipse, JEE Main
If the common tangents to the parabola, x2=4y and the circle, x2+y2=4 intersect at the point P, then the distance of P from the origin (units), is:

Options:
A 23+22
B 3+22
C 2+1
D 22+1

The correct answer is:
22+1

Solution:
Let y=mx+c be the common tangent.

Then c2=41+m2 ...(1) (condition of tangency for circle)

Solving with x2=4y, we get x2=4mx+c

i.e. x2-4mx-4c=0 ...(1)

Being a tangent, (1) must have equal roots.

⇒-4m2=41-4c⇒m2=-c ...(2)

From 1 & 2, c2=4-4c & c<0

⇒c2+4c-4=0⇒c=-22-2 (As c<0, so c≠22-2).

So, both tangents have common y intercept and thus intersect at P0,-2-2.

Thus, distance of P from origin is 22+1 units.

Topic: Mathematics, Circle, JEE Main, JEE Main 2017 (08 Apr Online)
If the complex number ω=x+iy (∀x,y∈R and i2=-1) satisfy the equation ω3=8i, then the maximum value of y is

Options:
A 1
B 32
C 12
D 2

The correct answer is:
1

Solution:
ω=2⋅i13=2cos⁡2nπ+π2+isin⁡2nπ+π213
=2cos⁡2nπ+π23+isin⁡2nπ+π23
Put n=1
=2cos⁡5π6+isin⁡5π6=-3+i

Hence, the maximum value of y is 1

Topic: Mathematics, Complex Number, JEE Main
If the complex numbers sinâ¡x+icosâ¡2x and cosâ¡x-isinâ¡2x are conjugate of each other, then the number of values of x in the interval 0,2Ï is equal to (where, i2=-1)

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
sinâ¡x=cosâ¡x and cosâ¡2x=sinâ¡2x
âsinâ¡x=cosâ¡x & 2cos2x-1=2sinâ¡xcosâ¡x
âsinâ¡x=cosâ¡x & 2cos2x-1=2cos2x
âsinâ¡x=cosâ¡x & -1=0 (Which is impossible)
Hence, no value of x exists

Topic: Mathematics, Complex Number, JEE Main
If the concentration of glucose C6H12O6 in blood is 0.72 gL-1, the molarity of glucose in blood is -×10-3M (Nearest integer):
[Given : Atomic mass of C=12,H=1, O=16 u]

Options:

The correct answer is:
4

Solution:
Glucose=C(gm/ℓ)M(gm/mol)=0.72180=4×10-3M

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the concentration of \( \left[\mathrm{NH}_{4}^{+}\right] \)in a solution having \( 0.02 \mathrm{M} \mathrm{NH}_{3} \) and \( 0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} \) is \( \mathrm{a} \times 10^{-6} \mathrm{M} \), determine a .
\[
\left[\mathrm{k}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\right]
\]

Options:

The correct answer is:
36

Solution:
NH4OH⇋NH4++OH-
kb=NH4+ OH-NH4OH
⇒ 1.8×10-5=NH4+ 0.010.02
{As OH- will mainly come from CaOH2 only}
⇒NH4+=36×10-6=a×10-6
⇒ a=36

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the concentration of [NH4+] in a solution having 0.02 M NH3 and 0.005 M CaOH2 is aÃ10-6 M , determine a . [kbNH3=1.8Ã10-5]

Options:

The correct answer is:
36

Solution:
NH4OH⇋NH4++OH-
kb=NH4+ OH-NH4OH
⇒ 1.8×10-5=NH4+ 0.010.02
{As OH- will mainly come from CaOH2 only}
⇒NH4+=36×10-6=a×10-6
⇒ a=36

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the concentration of $\mathrm{OH}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH})_3(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$ is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by

Options:
A 4 times
B 8 times
C 16 times
D 64 times

The correct answer is:
64 times

Solution:
To maintain the constant value of $\mathrm{K}_c$, the concentration of $\mathrm{Fe}^{3+}$ ion will increase by 64 times.

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2008 (Mains)
If the conductivity of mercury at 0°C is 1.07×106 S m-1 and the resistance of a cell containing mercury is 0.243Ω, then the cell constant of the cell is x×104 m-1. The value of x is ______ . (Nearest integer)

Options:

The correct answer is:
26

Solution:
k=1.07×106Sm-1, R=0.243Ω

G=1R=10.243Ω-1

k=G×G*

G*=kG=1.07×10610.243=26×104 m-1

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
If the conjugate of $(x+i y)(1-2 i)$ be $1+i$, then

Options:
A $x=\frac{1}{5}$
B $y=\frac{3}{5}$
C $x+i y=\frac{1-i}{1-2 i}$
D $x-i y=\frac{1-i}{1+2 i}$

The correct answer is:
$x+i y=\frac{1-i}{1-2 i}$

Solution:
Given that $\bar{(x+i y)(1-2 i)}=1+i$
$\Rightarrow \quad x-i y=\frac{1+i}{1+2 i} \Rightarrow x+i y=\frac{1-i}{1-2 i}$

Topic: Mathematics, Complex Number, JEE Main
If the constant forces $2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is

Options:
A 15 unit
B 9 unit
C $-15$ unit
D $-9$ unit

The correct answer is:
$-15$ unit

Solution:
Resultant force,
$$
\overrightarrow{\mathbf{F}}=(2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=\hat{\mathbf{i}}-3 \mathbf{j}+5 \hat{\mathbf{k}}
$$
Displacement,
$$
\begin{aligned}
\overrightarrow{\mathbf{d}}=\overrightarrow{\mathbf{A B}} &=(6 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})-(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\
&=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\
W &=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}}=(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
&=2-12-5=-15 \text { unit }
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If the constant term, in binomial expansion of 2xr+1x210 is 180, then r is equal to ____________.

Options:

The correct answer is:
8

Solution:
The general term in the binomial expansion of a+bn is CRnan-RbR.

Thus, the general term in the expansion of 2xr+1x210 is =CR102xr10-R1x2R

=CR10210-Rxr10-Rx-2R

Given, the constant term in the expansion is 180

⇒CR10·210-R=180 …1

And (10-R)r-2R=0

⇒r=2R10-R

⇒r=2(R-10)10-R+2010-R

⇒r=-2+2010-R …2

Since R is positive integer less than or equal to 10 and r is also an integer, hence, the value of R is such that, 10-R must divide 20.

This is possible, when the value of 10-R can be one from the divisors of 20 i.e. 10-R can be from 1, 2, 4 or 5 and consequently, R can be 9, 8, 6 or 5 respectively.

But for R=9, 6 & 5 equation 1 is not satisfied, hence R=8

as CR10210-R=180

⇒r=2×810-8=8.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the constant term, in binomial expansion of \( \left(2 x^{r}+\frac{1}{x^{2}}\right)^{10} \) is \( 180 \), then \( r \) is equal to

Options:

The correct answer is:
8

Solution:
The general term in the binomial expansion of a+bn is CRnan-RbR.

Thus, the general term in the expansion of 2xr+1x210 is =CR102xr10-R1x2R

=CR10210-Rxr10-Rx-2R

Given, the constant term in the expansion is 180

⇒CR10·210-R=180 …1

And (10-R)r-2R=0

⇒r=2R10-R

⇒r=2(R-10)10-R+2010-R

⇒r=-2+2010-R …2

Since R is positive integer less than or equal to 10 and r is also an integer, hence, the value of R is such that, 10-R must divide 20.

This is possible, when the value of 10-R can be one from the divisors of 20 i.e. 10-R can be from 1, 2, 4 or 5 and consequently, R can be 9, 8, 6 or 5 respectively.

But for R=9, 6 & 5 equation 1 is not satisfied, hence R=8

as CR10210-R=180

⇒r=2×810-8=8.

Topic: Mathematics, Binomial Theorem, JEE Main