datavorous/entrance-exam-dataset huggingface dataset:
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔCH⊖=-601.70 kJ mol-1, the magnitude of change in internal energy for the reaction is kJ. (Nearest integer)
(Given : R=8.3 J K-1 mol-1)

Options:

The correct answer is:
600

Solution:
Mgs+12O2G⟶MgOSΔHc°=-601.70KJ/Mole

ΔH°=ΔU+ΔngRT

-601.70=ΔU+-12×8.3×300×10-3

-601.7=ΔU-1.245

ΔU=-599.455KJ

ΔU=599.455KJ≈600

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (28 Jun Shift 2)
For common emitter transistor amplifier, the audio signal voltage across the collector resistance of 2 kÎ© is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kÎ© then the input signal voltage is

Options:
A 10  mV
B 20ââmV
C 30  mV
D 15  mV

The correct answer is:
20  mV

Solution:
RC=2 kΩ V0=4 V
IC=V0Rc=4 V2 kΩ=2 mA
β=ICIB=100
⇒IB=IC100=2×10-5 A
Vin=IBRi=2×10-5×1 kΩ=20 mV

Topic: Physics, Semiconductors, JEE Main
For complete combustion of ethanol,
C 2 H 5 OH l + 3 O 2 g → 2 CO 2 g + 3 H 2 O l ,
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol -1 at 2 5 ℃ . Assuming ideality the Enthalpy of combustion, Δ c H , for the reaction will be: R = 8.314 kJ mol -1

Options:
A - 1366.95 kJ mol -1
B - 1361.95 kJ mol -1
C - 1460.50 kJ mol -1
D - 1350.50 kJ mol -1

The correct answer is:
- 1366.95 kJ mol -1

Solution:
Î n g = Gaseous moles products - Gaseous moles reactants
= 2 - 3 = -1 mole
Î H = Î E + Î n g RT
= - 1364.47 - 1 Ã 8.314 Ã 1 0 - 3 Ã 2 9 8 [The value of R must be converted into kJ, so it is multiplied by 10-3]
= - 1364.47 - 2.49
= - 1366.96 kJ m -1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2014 (06 Apr)
For complete combustion of methanol

CH3OHl+32O2g→CO2g+2H2Ol the amount of heat produced as measured by bomb calorimeter is 726 kJ mol-1 at 27 °C. The enthalpy of combustion for the reaction is -x kJ mol-1, where x is
integer)
(Given : R=8.3JK-1 mol-1)

Options:

The correct answer is:
727

Solution:
CH3OHl+32O2g→CO2g+2H2Ol.

For chemical reaction : ΔH=ΔU+ΔngRT

ΔH=-726+-128.314×10-3×300

=-727.245KJ/mole

∴ x=727

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (26 Jun Shift 1)
For conversion of compound A→B, the rate constant of the reaction was found to be 4.6×10-5 Lmol -1s-1. The order of the reaction is _____ .

Options:

The correct answer is:
2

Solution:
Rate(r) =kAn

⇒k=rateAn

Unit of rate constant is related to order of reaction(n) by the relation: conc.1-ntime-1

Thus,
⇒Lmol-1⇒1-n=-1n=2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
For curves of H2and He, as the pressure increases

Options:
A PV also increases
B PV decreases
C PV remains constant
D None of the above

The correct answer is:
PV also increases

Solution:
Correct Option is : 1
PV also increases

Topic: Chemistry, States of Matter, NEET
For decolourization of 1 mole of $\mathrm{KMnO}_4$, the moles of $\mathrm{H}_2 \mathrm{O}_2$ required is

Options:
A $1 / 2$
B 3/2
C $5 / 2$
D 7/2

The correct answer is:
$5 / 2$

Solution:

Topic: Chemistry, Redox Reactions, JEE Main
For definite length of wire, if the weight used for applying tension is immersed in water, then frequency will

Options:
A become less
B become more
C remain equal
D become zero

The correct answer is:
become less

Solution:
For stretched string
n∝T
n∝Mg
When weight is dipped in water due to buoyancy force, tension decreases and hence frequency decreases

Topic: Physics, Waves and Sound, NEET
For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 Ω showing a full scale deflection for 1 mA as in figure.

(A) for voltmeter R≈50 kΩ

(B) for ammeter r≈0.2 Ω

(C) for ammeter r≈6 Ω

(D) for voltmeter R≈5 kΩ

(E) for voltmeter R≈500 Ω

Choose the correct answer from the options given below:

Options:
A A and C
B C and E
C C and D
D A and B

The correct answer is:
A and C

Solution:
The resistance is given by

R=VIg-G

⇒5010-3-54≈50 kΩ

For ammeter,

S=IgGI-Ig⇒S=10-3×54(10-1)×10-3=6 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (15 Apr Shift 1)
For determination of molecular mass, Raoult's law is applicable only to

Options:
A Dilute solutions of electrolytes
B Concentrated solutions of electrolytes
C Dilute solutions of non-electrolytes
D Concentrated solution of non-electrolytes

The correct answer is:
Dilute solutions of non-electrolytes

Solution:
Raoult's law is applicable to the dilute solutions of non-electrolytes.

For which,
P solvent - P solution P solvent = X solute

Raoult's law states that partial pressure exerted by vapours of any liquid is directly proportional to the mole fraction of that liquid.
Besides this, Raoult's law is also applicable to ideal solutions of two volatile liquids.
For which,
Psolution=PAo+PBo-PAo. XB

Topic: Chemistry, Solutions, JEE Main
For different values of $\alpha$, the locus of the point of intersection of the two straight lines $\sqrt{3} x-y-4 \sqrt{3} \alpha=0$ and $\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0$ is

Options:
A a hyperbola with eccentricity 2
B an ellipse with eccentricity $\sqrt{\frac{2}{3}}$
C a hyperbola with eccentricity $\sqrt{\frac{19}{16}}$
D an ellipse with eccentricity $\frac{3}{4}$

The correct answer is:
a hyperbola with eccentricity 2

Solution:
Hints : $\sqrt{3} x-y=4 \sqrt{3} \alpha$
(1) $\sqrt{3} x+y=\frac{4 \sqrt{3}}{\alpha}$
(1) $x(2) \Rightarrow 3 x^2-y^2=48 \Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$
$$
\mathrm{e}=\sqrt{\frac{48+16}{16}}=2
$$

Topic: Mathematics, Straight Lines, JEE Main
For each $n \in \mathbf{N}$, let $A_n=\{(n+1) k / k \in \mathbf{N}\}$ and $X=\bigcup_{n \in \mathbf{N}} A_n \cdot A$ mapping $f: X \rightarrow N$ defined by $f(x)=x, \forall x \in \mathbf{X}$, is

Options:
A one-one and onto
B one-one but not onto
C onto but not one-one
D neither one-one nor onto

The correct answer is:
one-one but not onto

Solution:
Given that, $A_n=\{(n+1) k \mid k \in \mathbf{N}\}$
If $n=1, A_1=\{2 k \mid k \in \mathbf{N}\}$
for, $k=1,2,3 \ldots A_1=\{2,4,6,8, \ldots\}$
If $\quad n=2, A_2=\{3 k \mid k \in \mathbf{N}\}$
for $\quad k=1,2,3, \ldots \quad A_2=\{3,6,9, \ldots\}$
If $\quad n=3, A_3=\{4 k \mid k \in \mathbf{N}\}$
for $\quad k=1,2,3, \ldots A_3=\{4,8,12, \ldots\}$
Now, $X=\bigcup_{n \in \mathbf{N}} A_n \quad$ or $X=A_1 \cup A_2 \cup A_3 \ldots$
or $\quad X=\{2,3,4,5,6, \ldots\}$
Here, $f: X \rightarrow N$ and $f(x)=x \forall x \in \mathbf{X}$
$\because f(x)$ is linear function i.e., it is one-one function but there is no value of $x$ in domain where $f(x)=1$
$\therefore f(x)$ is not onto function.

Topic: Mathematics, Functions, JEE Main
For each $n \in \mathbf{N}$, let $A_n=\{(n+1) k / k \in \mathbf{N}\}$ and $X=\bigcup_{n \in \mathbf{N}} A_n \cdot A$ mapping $f: X \rightarrow N$ defined by $f(x)=x, \forall x \in \mathbf{X}$, is

Options:
A one-one and onto
B one-one but not onto
C onto but not one-one
D neither one-one nor onto

The correct answer is:
one-one but not onto

Solution:
Given that, $A_n=\{(n+1) k \mid k \in \mathbf{N}\}$
If $n=1, A_1=\{2 k \mid k \in \mathbf{N}\}$
for, $k=1,2,3 \ldots A_1=\{2,4,6,8, \ldots\}$
If $\quad n=2, A_2=\{3 k \mid k \in \mathbf{N}\}$
for $\quad k=1,2,3, \ldots \quad A_2=\{3,6,9, \ldots\}$
If $\quad n=3, A_3=\{4 k \mid k \in \mathbf{N}\}$
for $\quad k=1,2,3, \ldots A_3=\{4,8,12, \ldots\}$
Now, $X=\bigcup_{n \in \mathbf{N}} A_n \quad$ or $X=A_1 \cup A_2 \cup A_3 \ldots$
or $\quad X=\{2,3,4,5,6, \ldots\}$
Here, $f: X \rightarrow N$ and $f(x)=x \forall x \in \mathbf{X}$
$\because f(x)$ is linear function i.e., it is one-one function but there is no value of $x$ in domain where $f(x)=1$
$\therefore f(x)$ is not onto function.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
For each non-zeroreal number $\mathrm{x}$, let $f(x)=\frac{x}{|x|}$. The range of $f$ is

Options:
A a null set
B a set consisting of only one element
C a set consisting of two elements
D a set consisting of infinitely many elements

The correct answer is:
a set consisting of two elements

Solution:
For a non-zero real number $\mathrm{X}$
$f(x)=\frac{x}{|x|}$
domain $f(x)=\frac{x}{|x|}$ range
for any value of $\mathrm{x}$ in domain a set consisting of two dementsi.e. $(-1,+1)$ $\therefore$ Option (c) is correct.

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 1)
For each positive integer $\mathrm{n}$, define $\mathrm{f}_{\mathrm{n}}(\mathrm{x})=$ minimum $\left(\frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n} !}, \frac{(1-\mathrm{x})^{\mathrm{n}}}{\mathrm{n} !}\right)$, for $0 \leq \mathrm{x} \leq 1$. Let $\mathrm{I}_{\mathrm{n}}=\int_{0}^{1} \mathrm{f}_{\mathrm{n}}(\mathrm{x}) \mathrm{dx}, \mathrm{n} \geq 1$. Then $I_{n}=\sum_{n=1}^{\infty} I_{n}$ is equal to-

Options:
A $2 \sqrt{\mathrm{e}}-3$
B $2 \sqrt{\mathrm{e}}-2$
C $2 \sqrt{\mathrm{e}}-1$
D $2 \sqrt{\mathrm{e}}$

The correct answer is:
$2 \sqrt{\mathrm{e}}-3$

Solution:
$I_{n}=\int_{0}^{1 / 2} \frac{x^{n}}{n !} d x+\int_{1 / 2}^{1} \frac{(1-x)^{n}}{n !} d x=\frac{1}{(n+1) !}\left(\left(\frac{1}{2}\right)^{n+1}+\left(\frac{1}{2}\right)^{n+1}\right)=\frac{\left(\frac{1}{2}\right)^{n}}{(n+1) !}$
$\sum_{n=1}^{\infty} I_{n}=\left(\frac{1 / 2}{2 !}+\frac{(1 / 2)^{2}}{3 !}+\ldots \ldots\right)=2 \sqrt{e}-3$

Topic: Mathematics, Definite Integration, JEE Main
For each t∈R, let t be the greatest integer less than or equal to t. Then, limx→1+1-x+sin1-xsin1-xπ21-x1-x

Options:
A equals 0
B equals -1
C does not exist
D equal 1

The correct answer is:
equals 0

Solution:
limx→1+1−x+sin1−xsin1−xπ21−x1−x

=limh→01−1+h+sin1−1+hsinπ21−1+h1−1+h1−1+h

=limh→01−h−1+sin1−h−1sinπ21−1−h1−1−h1−1−h

=limh→0−h+sinhsinπ2−h-h−h

=limh→0−h+sinhsin−π2h−1=limh→0−h+sinh−1−h

=1−1=0.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (10 Jan Shift 1)
For each t∈R, let t be the greatest integer less than or equal to t. Thenlimx→0+x1x+2x+…+15x

Options:
A does not exist in R
B is equal to 0
C is equal to 15
D is equal to 120

The correct answer is:
is equal to 120

Solution:
We know that

1x−1<1x≤1x

Hence,

1+2+3. . .+15x−15<1x+2x+....+15x≤1+2+...+15x

⇒limx→0+x1x+2x+....+15x=15162

=120

Topic: Mathematics, Limits, JEE Main, JEE Main 2018 (08 Apr)
For each x∈R, let x be the greatest integer less than or equal to x. Then limx→0-xx+xsin⁡xx is equal to

Options:
A 1
B 0
C -sin⁡1
D sinâ¡1

The correct answer is:
-sin⁡1

Solution:
=limx→0-⁡x x+xsin⁡xx
=limx→0-x-1-xsin⁡-1-x x=-1,∀ x∈[-1,0)
=limx→0--1+xsin⁡1 sin-θ=-sinθ
=-sin⁡1.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If 1, log331-x+2, log34.3x-1 are in arithmetic progression, then x equals

Options:
A log34
B 1-log34
C 1-log43
D log43

The correct answer is:
1-log34

Solution:
Since the given numbers are in AP, we get,
2log3(31 - x+2)1/2 =1+log3(4.3x-1)
⇒ log 3 3 1 - x + 2 = log 3 3 + log 3 4.3 x - 1
⇒ 31-x+2=3(4.3x-1)
⇒ 3.3-x+ 2=12.3x-3
Let, 3x=t
∴ 3 t + 2 = 1 2 t - 3
⇒ 12t2-5t-3=0
⇒ t = - 1 3 , 3 4
∴ 3 x = 3 4 ∵ 3 x cannot be negative
⇒ x = log 3 3 4 ⇒ x = 1 - log 3 4

Topic: Mathematics, Sequences and Series, JEE Main
If 1,log104x-2 and log104x+185 are in arithmetic progression for a real number x then the value of the determinant 2x-12x-1x210xx10 is equal to:

Options:

The correct answer is:
2

Solution:
If three numbers a, b, c are in arithmetic progression, then 2b=a+c.

Given 1, log104x-2, log104x+185 are in arithmetic progression, then we have

2log104x-2=1+log104x+185

Using log10mn=nlog10m, we get

log104x-22=log1010+log104x+185

Using log10m+log10n=log10mn, we get

log104x-22=log10104x+185

⇒4x-22=104x+185

⇒4x2+4-4·4x=10·4x+36

⇒4x2-144x-32=0

⇒4x-164x+2=0

⇒4x=16 as 4x+2≠0

⇒x=2

Now, 2x-12x-1x210xx10

=314102210

=30-2-10-4+41-0

=-6+4+4=2.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (17 Mar Shift 2)
If \( 1, \log _{10}\left(4^{x}-2\right) \) and \( \log _{10}\left(4^{x}+\frac{18}{5}\right) \) are in arithmetic progression for a real number \( x \) then the value of the determinant
\( \left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right| \) is equal to :

Options:

The correct answer is:
2

Solution:
2log104x-2=1+log104x+185

4x-22=104x+185

4x2+4-44x-32=0

4x-16(4x+2)=0

4x=16

x=2

=-6+4+4=2

Topic: Mathematics, Determinants, JEE Main
If 1n+1 nCn+1n nCn-1+...+12 nC1+nC0=102310 then n is equal to

Options:
A 9
B 8
C 7
D 6

The correct answer is:
9

Solution:
Given that 1n+1 nCn+1n nCn-1+...+12 nC1+nC0=102310

⇒∑r=0n1r+1 nCr=102310 ∴ n+1Cr+1=n+1r+1 nCr

⇒∑r=0n1n+1 n+1Cr+1=102310

⇒1n+1n+1C1+n+1C2+...+n+1Cn+1=102310

We know that C0n+C1n+C2n+.......+Cnn=2n

⇒2n+1-1n+1=102310=210-110

⇒n+1=10

⇒n=9

Hence this is the correct option.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (12 Apr Shift 1)
If $(1-p)$ is a root of quadratic equation $x^2+p x+(1-p)=0$, then its roots are

Options:
A
0,1

B
$-1,2$

C
$0,-1$

D
$-1,1$

The correct answer is:
$0,-1$

Solution:
(3)
$$
\begin{aligned}
& (1-p)^2+p(1-p)+(1-p)=0 \quad\left(\text { since }(1-p) \text { is a root of the equation } x^2+p x+(1-p)=0\right) \\
& \Rightarrow(1-p)(1-p+p+1)=0 \\
& \Rightarrow 2(1-p)=0 \Rightarrow(1-p)=0 \Rightarrow p=1 \\
& \text { sum of root is } \alpha+\beta=-p \text { and product } \alpha \beta=1-p=0 \quad \text { (where } \beta=1-p=0) \\
& \Rightarrow \alpha+0=-1 \Rightarrow \alpha=-1 \Rightarrow \text { Roots are } 0,-1
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2004
If 1+sinâ¡Î¸+sin2â¡Î¸+sin3â¡Î¸ +...â=4+23, 0<Î¸<Ï, then

Options:
A θ=π3
B θ=π6
C θ=π3 or π6
D θ=π3 or 2π3

The correct answer is:
θ=π3 or 2π3

Solution:
Given, 1+sinâ¡Î¸+sin2â¡Î¸+sin3â¡Î¸ +â¦â=4+23
â 11-sinâ¡Î¸=4+23 [âµ0<sinâ¡Î¸<1]
â 1-sinâ¡Î¸=4-2316-12=1-32
â sinâ¡Î¸=32
â Î¸=Ï3 or 2Ï3

Topic: Mathematics, Trigonometric Equations, JEE Main
If 1âÎ±,Î² where Î±,Î² are the roots of the equation x2-ax+1+3=0, then

Options:
A a>2
B a<-6 or a>2
C -6<a<2
D a<2

The correct answer is:
a>2

Solution:
Coefficient of x2 is +ve, so parabola is opening upward.
1 lies between the roots âf1<0
â1âaâ2+3<0
â2a>4âa>2

Topic: Mathematics, Quadratic Equation, JEE Main
If ∫1x1-x1+xdx=gx+c,g1=0, then g12 is equal to

Options:
A loge3-13+1+π3
B loge3+13-1+π3
C loge3+13-1-π3
D 13loge3-13+1-π6

The correct answer is:
loge3-13+1+π3

Solution:
Given,

∫1x1-x1+xdx=gx+c

Put x=cos2θ

dx=-2sin2θ·dθ

=∫1cos2θtanθ-4sinθ·cosθdθ

=∫1cos2θ-4sin2θdθ

=-2∫1-cos2θcos2θdθ

=-22lnsec2θ-tan2θ+2θ+c

=lnsec2θ-tan2θ+2θ+c

=ln1-sin2θcos2θ+cos-1x+c

=ln1−1−x2x+cos−1x+c⏟gx

∴ g1=0⇒c=0

So, gx=ln1-1-x2x+cos-1x

g12=ln2-3+π3

g12=ln3-13+1+π3

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If 1+x-2x26=1+a1x+a2x2+...+a12x12, then the value of a2+a4+...+a12 is

Options:
A 31
B 32
C 64
D 1024

The correct answer is:
31

Solution:
Given , (1+x-2x2)6=1+a1x+a2x2+...+a12x12 ...(i)
On putting x=1 in Equation (i),
we get 1+1-26=1+a1+a2+â¦+a12
â¹(0)6=1+a1+a2+...+a12 ....(ii)
On putting x=-1 in Equation (i),
we get 1-1-26=1-a1+a2-a3+â¦+a12
â¹(-2)6=1-a1+a2-a3+...+a12 ...(iii)
On adding Equations (ii) and (iii) we get
-26=21+a2+a4+â¦+a12
â¹642-1=a2+a4+â¦+a12
â´a2+a4+â¦+a12=31

Topic: Mathematics, Binomial Theorem, JEE Main
If 1+x+2x220=a0+a1x+a2x2+....+a40x40, then the value of a0+a2+a4+.....+a38 is

Options:
A 220220-1
B 220220+1
C 239-219
D 239+219

The correct answer is:
239-219

Solution:
In the given identity, substituting x=1 and x=-1 on both the sides and adding, we get,
2a0+a2+a4+.....+a40=420+220
=240+220
⇒a0+a2+a4+....+a38+a40=239+219
But, a40 is the coefficient of x40 which is 220.
Therefore, a0+a2+a4+.....+a38+220=239+219
⇒a0+a2+a4+.....+a38=239+219-2.219=239-219

Topic: Mathematics, Binomial Theorem, JEE Main
If $\frac{1-x+6 x^2}{1-x^3}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1+x}$, then $A$ is equal to

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$1$

Solution:
We have,
$$
\begin{aligned}
\frac{1-x+6 x^2}{x-x^3} & =\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x} \\
\Rightarrow \quad 1-x+6 x^2 & =A\left(1-x^2\right)+B x(1+x)+C x(1-x)
\end{aligned}
$$
Put $x=0$, then $A=1$

Topic: Mathematics, Quadratic Equation, JEE Main
If $1+x^4+x^5=\sum_{i=0}^5 a_i\left(1+x^i\right)$, for all $x$ in $R$, then $a_2$ is:

Options:
A
$-4$

B
6

C
$-8$

D
10

The correct answer is:
$-4$

Solution:
$$
\begin{aligned}
&1+x^4+x^5=\sum_{i=0}^5 a_i(1+x)^i \\
&=a_0+a_1(1+x)^1+a_2(1+x)^2+a_3(1+x)^3 \\
&\quad+a_4(1+x)^4+a_5(1+x)^5 \\
&\Rightarrow 1+x^4+x^5
\end{aligned}
$$
$$
\begin{aligned}
=a_0+a_1(1+x) &+a_2\left(1+2 x+x^2\right)+a_3\left(1+3 x+3 x^2+x^3\right) \\
&+a_4\left(1+4 x+6 x^2+4 x^3+x^4\right) \\
&+a_5\left(1+5 x+10 x^2+10 x^3+5 x^4+x^5\right) \\
\Rightarrow & 1+x^4+x^5 \\
=a_0+a_1+a_1 x+a_2+2 a_2 x+a_2 x^2+a_3+3 a_3 x \\
+3 a_3 x^2+a_3 x^3+a_4+4 a_4 x+6 a_4 x^2+4 a_4 x^3+a_4 x^4+a_5 \\
&+5 a_5 x+10 a_5 x^2+10 a_5 x^3+5 a_5 x^4+a_5 x^5
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow & 1+x^4+x^5 \\
=&\left(a_0+a_1+a_2+a_3+a_4+a_5\right) \\
&+x\left(a_1+2 a_2+3 a_3+4 a_4+5 a_5\right) \\
+x^2\left(a_2+\right.&\left.3 a_3+6 a_4+10 a_5\right)+x^3\left(a_3+4 a_4+10 a_5\right) \\
&+x^4\left(a_4+5 a_5\right)+x^5\left(a_5\right)
\end{aligned}
$$
On comparing the like coefficients, we get
$$
\begin{aligned}
&a_5=1 \\
&a_4+5 a_5=1 \\
&a_3+4 a_4+10 a_5=0
\end{aligned}
$$
and $a_2+3 a_3+6 a_4+10 a_5=0$
from (1) \& (2), we get $a_4=-4 \ldots(5)$
From (1), (3) \& (5), we get $a_3=+6$...(6)
Now, from $(1),(5)$ and $(6)$, we get
$a_2=-4$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (12 Apr Online)
If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots \ldots .+C_n x^2$, then $C_0^2+C_1^2+C_2^2+C_3^2+\ldots \ldots+C_n^2=$

Options:
A $\frac{n!}{n!n!}$
B $\frac{(2 n)!}{n!n!}$
C $\frac{(2 n)!}{n!}$
D None of these

The correct answer is:
$\frac{(2 n)!}{n!n!}$

Solution:
$(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots . .+C_n x^n...(i)$
and $\left(1+\frac{1}{x}\right)^n=C_0+C_1 \frac{1}{x}+C_2\left(\frac{1}{x}\right)^2+\ldots . .+C_n\left(\frac{1}{x}\right)^n...(ii)$
If we multiply (i) and (ii), we get
$C_0^2+C_1^2+C_2^2+\ldots . .+C_n^2$ is the term independent of $x$ and hence it is equal to the term independent of $x$ in the product $(1+x)^n\left(1+\frac{1}{x}\right)^n$ or in $\frac{1}{x^n}(1+x)^{2 n}$ or term containing $x^n$ in $(1+x)^{2 n}$. Clearly the coefficient of $x^n$ in $(1+x)^{2 n}$ is $T_{n+1}$ and equal to ${ }^{2 n} C_n=\frac{(2 n)!}{n!n!}$
Trick : Solving conversely.
Put $n=1, n=2, \ldots$. then we get $S_1={ }^1 C_0^2+{ }^1 C_1^2=2$
$S_2={ }^2 C_0^2+{ }^2 C_1^2+{ }^2 C_2^2=1^2+2^2+1^2=6$
Now check the options
(a) Does not hold given condition,
(b) (i) Put $n=1$, then $\frac{2!}{1!1!}=2$
(ii) Put $n=2$, then $\frac{4!}{2!2!}=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}=6$
Note : Students should remember this question as an identity.

Topic: Mathematics, Binomial Theorem, JEE Main
If 1+xn=C0+C1x+C2x2+....+Cnxn and ∑r=050Cr2r+1=m!n!2, then the value of m+n is equal to (where Cr represents Crn)

Options:
A 149
B 152
C 155
D 146

The correct answer is:
152

Solution:
1+x50=C0+C1x+C2x2+.....+C50x50=∑r=050Crxr
∫0x1+x50dx=∑r=050∫0xCrxrdx
1+x51-151=∑r=050Crxr+1r+1
and x+150=∑r=050Crx50-r
Multiplying these two equations, we get,
1+x51-151x+150=∑r=050Crxr+1r+1∑r=050Crx50-r
1+x101-1+x5051=C0x1+C1x22+C2x33+....+C50x5151C0x50+C1x49+C2x48+....+C50
Now, comparing the coefficient of x51,we get,

151101C51-0=C021+C122+C223+...+C50251=∑r=050Cr2r+1
⇒∑r=050Cr2r+1=151×101!51!50!=101!51!2=m!n!2
⇒m+n=101+51=152

Topic: Mathematics, Binomial Theorem, JEE Main
If 1+xn=C0+C1x+C2x2+.…….+Cnxn, ∀n∈N and C021+C122+C223+.……+Cn2n+1=λ2n+1!n+1!2, then the value of λ is equal to

Options:

The correct answer is:
1

Solution:
Consider 1+xn=C0+C1x+C2x2+.…….+Cnxn .
∴∫0x1+xndx=C0x+C1x22+C2x23+.…….+Cnxn+1n+1
⇒1+xn+1-1n+1=C0x+C1x22+C2x33+.…….+Cnxn+1n+1
Also, x+1n=C0xn+C1xn-1+C2xn-2+.……+Cn
Multiply and compare xn+1
C021+C122+C223+.……+Cn2n+1=
Coefficient of xn+1 in x+1n1+xn+1-1n+1
=x+12n+1-x+1nn+1
Required coefficient =Cn+12n+1n+1
=2n+1!n+1!2⇒λ=1

Topic: Mathematics, Binomial Theorem, JEE Main
If 1+xn=C0+C1x+C2x2+....+Cnxn, âr=0nr+12 Cr=2n-2fn and if the roots of the equation fx=0 are Î± and Î², then the value of Î±2+Î²2 is equal to (where Cr denotes Crn)

Options:
A 13
B 10
C 17
D 20

The correct answer is:
17

Solution:
1+xn=∑r=0nCrxr
⇒x1+xn=∑r=0nCrxr+1
Differentiating w.r.t. x we get
xn1+xn-1+1+xn=∑r=0nr+1Crxr
Again multiplying both sides by x
1+xn-1nx+1+xx=∑r=0nr+1Crxr+1
Again differentiating w.r.t. x we get,
ddx1+xn-1nx2+x2+x=∑r=0nr+12Crxr
∑r=0nr+12Crxr=1+xn-12nx+2x+1+nx2+x2+xn-11+xn-2
Putting x=1 on both sides, we get,
∑r=0nr+12Cr=2n-12n+3+n+2⋅n-12n-2=2n-2n2+5n+4
Now, fx=x2+5x+4=x+1x+4⇒α=-1, β=-4
Hence, α2+β2=17

Topic: Mathematics, Binomial Theorem, JEE Main
If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots .+C_n x^n$ then $C_0 C_2+C_1 C_3+C_2 C_4+C_{n-2} C_n$ equals

Options:
A $\frac{(2 n)!}{(n+1)!(n+2)!}$
B $\frac{(2 n)!}{(n-2)!(n+2)!}$
C $\frac{(2 n)!}{(n)!(n+2)!}$
D $\frac{(2 n)!}{(n-1)!(n+2)!}$

The correct answer is:
$\frac{(2 n)!}{(n-2)!(n+2)!}$

Solution:
We have, $(1+x)^n=C_0+C_1 x+C_2 x^2 \ldots .+C_n x^n$
$\left(1+\frac{1}{x}\right)^n=C_0+C_1 \cdot \frac{1}{x}+C_2 \cdot \frac{1}{x^2}+\ldots+C_n\left(\frac{1}{x^n}\right)$
on multiplying both expansions, we get
$\frac{(1+x)^{2 n}}{x^n}=\sum C_0^2+x \sum C_0 C_1+x^2 \sum C_0 C_2+\ldots$ $+x^r \sum C_0 C_r+\ldots \ldots$
The various sigma are the coefficient of $x^0, x, x^2, \ldots ., x^r$ in L.H.S. $\frac{(1+x)^{2 n}}{x^n}$ or coefficient of $x^n, x^{n+1}, x^{n+2}, \ldots ., x^{n+r}$ in the expansion of $(1+x)^{2 n}$ which occur in $T_{n+1}, T_{n+2}, \cdots$. and are ${ }^{2 n} C_n,{ }^{2 n} C_{n+1},{ }^{2 n} C_{n+2} \cdots \cdot{ }^{2 n} C_{n+r}$ etc.
${ }^{2 n} C_{n+2}=\frac{(2 n)!}{(n-2)!(n+2)!}$

Topic: Mathematics, Binomial Theorem, JEE Main
If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$, then the value of $C_0+C_2+C_4+C_6+\ldots .$. is

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

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

Solution:
$(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots .+C_n x^n$
Putting $x=1$, we get
$\Rightarrow 2^n=C_0+C_1+C_2+\ldots . .+C_n...(i)$
or
$C_1+C_2+C_3+\ldots .+C_n=2^n-1 \quad\left[\boxtimes C_0={ }^n C_0=1\right]$
Again, putting $x=-1$, we get
$0=C_0-C_1+C_2-C_3+\ldots$
or
$C_0+C_2+C_4+\ldots=C_1+C_3+C_5+\ldots \text { i.e. } A=B$
Also from (i), $A+B=2^n$ or $A=2^{n-1}=B$
Hence, $C_0+C_2+C_4+\ldots=2^{n-1}$

Topic: Mathematics, Binomial Theorem, JEE Main
If Δ1=xsinθcosθ-sinθ-x1cosθ1x and Δ2=xsin2θcos2θ-sin2θ-x1cos2θ1x, x≠0; then for all θ∈0, π2 :

Options:
A Δ1+Δ2=-2(x3+x-1)
B Δ1-Δ2=x(cos2θ-cos4θ)
C Δ1+Δ2=-2x3
D Δ1-Δ2=-2x3

The correct answer is:
Δ1+Δ2=-2x3

Solution:
Δ1=xsinθ-sinθ-xcosθ1 cosθ1x
=x-x2-1-sinθ-xsinθ-cosθ+cosθ(-sinθ+xcosθ)
=-x3-x+xsin2θ+sinθcos⁡θ-sinθcosθ+xcos2θ
=-x3-x+xsin2⁡θ+cos2⁡θ
=-x3
Similarly: Δ2=-x3
∴Δ1+Δ2=-2x3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If 1+x+x28=a0+a1x+a2x2+…+a16x16 for all values of x, then a5100 is equal to

Options:

The correct answer is:
5.04

Solution:
Coefficient of x5= choosing x from 5 brackets out of 8 or choosing one x2 and three x or choosing two x2 and one x
=C85+C81C73+C82C61
=8×7×63×2+8×7×6×53×2+8×7×62
=56+56×5+56×3
=56×9=504
⇒a5=504

Topic: Mathematics, Binomial Theorem, JEE Main
If 1+x+x225=a0+a1x+a2x2+.…….+a50x50, then a0+a2+a4+.…….+a50 is

Options:
A even
B odd and of the form 3n
C odd and of the form 3n-1
D odd and of the form 3n+1

The correct answer is:
even

Solution:
Putting x=1 and x=-1 and adding, we get,

2a0+a2+a4+.……..+a50=325+1=1+225+1
=C250+C251⋅2+C252⋅22+.……+C2525⋅225+1
=21+C251+C252⋅2+.……+C2525⋅224
⇒a0+a2+a4+.………+a50=1+25+C225⋅2+.……+C2525⋅224
=213+C252+C253⋅2+.………+C2525⋅223 which is an even number.

Topic: Mathematics, Binomial Theorem, JEE Main