datavorous/entrance-exam-dataset huggingface dataset:
If α≠β but, α2=4α-2 and β2=4β-2, then the quadratic equation with roots αβ and βα is

Options:
A x2-4x+2=0
B x2-6x+1=0
C x2+6x-1=0
D x2+4x-2=0

The correct answer is:
x2-6x+1=0

Solution:
Since, α2=4α-2,β2=4β-2
⇒ α,β are the roots of x2=4x-2
⇒ α+β=4,αβ=2
The quadratic equation with roots αβ,βα is,
x2-αβ+βαx+αβ×βα=0
⇒ x2-(α+β)2-2αβαβx+1=0
⇒ x2-(16-4)2x+1=0
⇒ x2-6x+1=0

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha \neq \beta$ but $\alpha^2=5 \alpha-3$ and $\beta^2=5 \beta-3$ then the equation having $\alpha / \beta$ and $\beta / \alpha$ as its roots is

Options:
A
$3 x^2-19 x+3=0$

B
$3 x^2+19 x-3=0$

C
$3 x^2-19 x-3=0$

D
$x^2-5 x+3=0$

The correct answer is:
$3 x^2-19 x+3=0$

Solution:
We have $\alpha^2=5 \alpha-3$
$$
\begin{aligned}
& \Rightarrow \alpha^2-5 \alpha+3=0 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2} \text {. Similarly, } \beta^2=5 \beta-3 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2} \\
& \therefore \alpha=\frac{5+\sqrt{13}}{2} \text { and } \beta=\frac{5-\sqrt{13}}{2} \text { or vice - versa } \\
& \alpha^2+\beta^2=\frac{50+26}{4}=19 \& \alpha \beta=\frac{1}{4}(25-13)=3
\end{aligned}
$$
Thus, the equation having $\frac{\alpha}{\beta} \& \frac{\beta}{\alpha}$ as its roots is
$$
x^2-x\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+\frac{\alpha \beta}{\alpha \beta}=0 \Rightarrow x^2-x\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right)+1=0 \text { or } 3 x^2-19 x+1=0
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2002
If C0, C1, C2, ....., C20 are the binomial coefficients in the expansion of 1+x20, then the value of C1C0+2C2C1+3C3C2+.....+19C19C18+20C20C19 is equal to (where Cr represents Crn)

Options:
A 120
B 210
C 180
D 240

The correct answer is:
210

Solution:
CrCr-1=n-r+1r=20-r+1r=21-rr
⇒r⋅CrCr-1=21-r
⇒∑r=120r⋅CrCr-1=∑r=12021-r
=21∑r=1201-∑r=120r
=21×20-20×212
=210

Topic: Mathematics, Binomial Theorem, JEE Main
If $c_{0}, c_{1}, c_{2}, \ldots, c_{n}$ are binomial coefficients of order $n$, then the value of $\frac{c_{1}}{2}+\frac{c_{2}}{4}+\frac{c_{3}}{6}+\ldots$

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

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

Solution:
We have,
$c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots=\frac{1}{2}\left\{(1+x)^{n}-(1-x)^{n}\right\}$
$\Rightarrow \int_{0}^{1}\left(c_{1} x+c_{3} x^{3}+c_{5} x^{5}+\ldots\right) d x$
$=\frac{1}{2} \int_{0}^{1}(1+x)^{n}-(1-x)^{n} d x$
$\Rightarrow \frac{c_{1}}{2}+\frac{c_{3}}{4}+\frac{c_{5}}{6}+\ldots=\frac{2^{n}-1}{n+1}$
$\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{4.2}+\frac{1}{4}\left(\frac{1}{2}\right)^{2}+\ldots \infty\right)$

Topic: Mathematics, Binomial Theorem, JEE Main
If C0,C1,C2,â¦..,Cn are binomial coefficients, (where Cr= nCr), then the value of C0-C1+C2-C3+...+-1nCn is equal to

Options:
A 2n-1
B 2n
C 0
D 1

The correct answer is:
0

Solution:
1+xn=C0+C1x+C2x2+..Cnxn
Put x=-1
C0-C1+C2-C3+...+-1nCn=0

Topic: Mathematics, Binomial Theorem, JEE Main
If $\mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots \ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}}$ denote the binomial

coefficientsin the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, then the value of $\mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots$

$+\left(C_{0}+C_{1}+\ldots . .+C_{n-1}\right)$

Options:
A $\mathrm{n} \cdot 2^{\mathrm{n}-1}$
B $\mathrm{n} \cdot 2^{\mathrm{n}}$
C $(n-1) \cdot 2^{n-1}$
D $(n-1) \cdot 2^{n}$

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

Solution:
$\begin{aligned} & \mathrm{C}_{0}+\left(\mathrm{C}_{0}+\mathrm{C}_{1}\right)+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\mathrm{C}_{2}\right)+\ldots \ldots \ldots \ldots \ldots \\ &+\left(\mathrm{C}_{0}+\mathrm{C}_{1}+\ldots \ldots \ldots . \mathrm{C}_{\mathrm{n}-1}\right) \\ &=\mathrm{n} \mathrm{C}_{0}+(\mathrm{n}-1) \mathrm{C}_{1}+(\mathrm{n}-2) \mathrm{C}_{2}+\ldots \ldots \mathrm{C}_{\mathrm{n}-1} \\ &=\mathrm{C}_{1}+2 \mathrm{C}_{2}+3 \mathrm{C}_{3}+4 \mathrm{C}_{4} \ldots \ldots \mathrm{n} \mathrm{C}_{\mathrm{n}}=\mathrm{n} .2^{\mathrm{n}-1} \end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
If C0, C1, C2, ……Cn denotes the binomial coefficients in the expansion of (1+x)n, then the value of

C0+C12+C23+...+Cnn+1 is equal to

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

The correct answer is:
2n+1-1n+1

Solution:
We know that ,
1+xn=C0+C1x+C2x2+…+Cnxn
On integrating both sides, from 0 to 1, we get
1+xn+1n+101=C0x+C1x22+C2x33+…+Cnxn+1n+1 01
⟹2n+1-1n+1=C0+C12+C23+...+Cnn+1

Topic: Mathematics, Binomial Theorem, JEE Main
If C1112+C2113+…..+C91110=nm with gcd(n,m)=1, then n+m is equal to

Options:

The correct answer is:
2041

Solution:
Given,

C1112+C2113+C3114.........+C91110=mn

Now, we know that,

1+x11=C011+C111x+C211x2+..........C1111x11

Now integrating the above expression both side we get,

∫011+x11dx=∫01C011+C111x+C211x2+..........C1111x11dx

⇒1+x121201=C011x+C111x22+C211x33+..........C1111x121201

⇒212-112=1+C1112+C2113+C3114.........+C91110+1111+112​​​​​​

⇒212-112-1-1-112=C1112+C2113+C3114.........+C91110

⇒C1112+C2113+C3114.........+C91110=212-2-2412

⇒C1112+C2113+C3114.........+C91110=4096-2612

⇒C1112+C2113+C3114.........+C91110=407012

⇒C1112+C2113+C3114.........+C91110=20356

Hence, on comparing we get, m+n=2041

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If \( c < 1 \) and the system of equations \( x+y-1=0,2 x-y-c=0 \) and \( b x+3 b y-c=0 \) is consistent, then the possible real values of \( b \) are

Options:
A \( b \in\left(-3, \frac{3}{4}\right) \)
B \( b \in\left(-\frac{3}{2}, 4\right) \)
C \( b \in\left(-\frac{3}{4}, 3\right) \)
D none of these

The correct answer is:
\( b \in\left(-\frac{3}{4}, 3\right) \)

Solution:
The given system is consistent.

So, value of determinant of coefficient matrix is zero.

∴Δ=11-12-1-c-b3b-c=0

⇒c+bc-6b+b+2c+3bc=0

⇒3c+4bc-5b=0

⇒c=5b4b+3

Now, c<1⇒5b4b+3<1

⇒5b4b+3-1<0

⇒b-34b+3<0

⇒b∈-34, 3

Topic: Mathematics, Determinants, JEE Main
If $|\vec{c}|^2=60$ and $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$, then a value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is:

Options:
A
$4 \sqrt{2}$

B
12

C
24

D
$12 \sqrt{2}$

The correct answer is:
$12 \sqrt{2}$

Solution:
Let, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
Given, $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$
$$
\begin{aligned}
&\Rightarrow\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a & b & c \\
1 & 2 & 5
\end{array}\right|=\overrightarrow{0} \\
&\Rightarrow(5 b-2 c) \hat{i}-(5 a-c) \hat{j}+(2 a-b) \hat{k} \\
&=0 \hat{i}+0 \hat{j}+0 \hat{k}
\end{aligned}
$$
Comparing both sides, we get $5 b-2 c=0 ; 5 a-c=0 ; 2 a-b=0$ or $5 b=2 c ; 5 a=c ; 2 a=b$
Also given $|\vec{c}|^2=60$
$$
\Rightarrow a^2+b^2+c^2=60
$$
Putting the value of $b$ and $c$ in above eqn., we get
$$
\begin{aligned}
&a^2+(2 a)^2+(5 a)^2=60 \\
&\Rightarrow a^2+4 a^2+25 a^2=60 \Rightarrow 30 a^2=60 \\
&a^2=2 \\
&a=\pm \sqrt{2} ; b=2 \sqrt{2} ; c=5 \sqrt{2}
\end{aligned}
$$
Now, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
$$
\therefore \vec{c}=\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}
$$
Value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is
$$
\begin{aligned}
&(\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\
&=-7 \sqrt{2}+4 \sqrt{2}+15 \sqrt{2}=12 \sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If $|\vec{c}|^2=60$ and $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$, then a value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is:

Options:
A
$4 \sqrt{2}$

B
12

C
24

D
$12 \sqrt{2}$

The correct answer is:
$12 \sqrt{2}$

Solution:
Let, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
Given, $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$
$$
\begin{aligned}
&\Rightarrow\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a & b & c \\
1 & 2 & 5
\end{array}\right|=\overrightarrow{0} \\
&\Rightarrow(5 b-2 c) \hat{i}-(5 a-c) \hat{j}+(2 a-b) \hat{k} \\
&=0 \hat{i}+0 \hat{j}+0 \hat{k}
\end{aligned}
$$
Comparing both sides, we get $5 b-2 c=0 ; 5 a-c=0 ; 2 a-b=0$ or $5 b=2 c ; 5 a=c ; 2 a=b$
Also given $|\vec{c}|^2=60$
$$
\Rightarrow a^2+b^2+c^2=60
$$
Putting the value of $b$ and $c$ in above eqn., we get
$$
\begin{aligned}
&a^2+(2 a)^2+(5 a)^2=60 \\
&\Rightarrow a^2+4 a^2+25 a^2=60 \Rightarrow 30 a^2=60 \\
&a^2=2 \\
&a=\pm \sqrt{2} ; b=2 \sqrt{2} ; c=5 \sqrt{2}
\end{aligned}
$$
Now, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
$$
\therefore \vec{c}=\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}
$$
Value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is
$$
\begin{aligned}
&(\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\
&=-7 \sqrt{2}+4 \sqrt{2}+15 \sqrt{2}=12 \sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2014 (11 Apr Online)
If λ=c2n2n2-22 for Balmer series, what is the value of c2 ?

Options:
A 4RH
B 2RH
C 2RH
D 4RH

The correct answer is:
4RH

Solution:
λ=c2n2n2-22
or 1λ=1c2n2-22n2=1c21-22n2
=22c2122-1n2
So, RH=22c2 or c2=4RH

Topic: Chemistry, Structure of Atom, JEE Main
If C32n:C3n=10:1, then the ratio n2+3n:n2-3n+4 is

Options:
A 35:16
B 27:11
C 65:37
D 2:1

The correct answer is:
2:1

Solution:
We have been given that C32n:C3n=10

We know that Crn=n!r!n-r!

⇒2n!3!2n-3!n!3!n-3!=10

⇒2n!3!2n-3!×3!n-3!n!=10

⇒2n2n-12n-22n-3!2n-3!×n-3!nn-1n-2n-3!=10

⇒2n2n-12n-2nn-1n-2=10

⇒42n-1n-2=10

⇒8n-4=10n-20

⇒n=8

Hence, the value of n2+3nn2-3n+4 is =82+3882-38+4=8844=2

Therefore, the value of n2+3nn2-3n+4 is 2:1.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If $C(28,2 r)=C(28,2 r-4)$, then what is $r$ equal to ?

Options:
A 7
B 8
C 12
D 16

The correct answer is:
8

Solution:
$\mathrm{C}(28,2 \mathrm{r})=\mathrm{C}(28,2 \mathrm{r}-4)$
${ }^{28} \mathrm{C}_{2 \mathrm{r}}={ }^{28} \mathrm{C}_{2 \mathrm{r}-4}$
$\Rightarrow 2 \mathrm{r}+2 \mathrm{r}-4=28$
$\Rightarrow 4 \mathrm{r}=32$
$\Rightarrow \mathrm{r}=8$

Topic: Mathematics, Binomial Theorem, JEE Main
If Ca3PO42 and H3PO3 contain same number of 'P' atom then the ratio of oxygen atoms in these compounds respectively is

Options:
A 8/3
B 2/3
C 3
D 4/3

The correct answer is:
4/3

Solution:
2x=y (Same 'P' atoms given)
x=y2
Ratio of 'O' atom =8x3y=8×y23y=43

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
If $C$ and $D$ are two events such that $C \subset D$ and $P(D) \neq 0$, then the correct statement among the following is

Options:
A
$P(C \mid D) \geq P(C)$

B
$P(C \mid D) < P(C)$
C
$P(C \mid D)=\frac{P(D)}{P(C)}$

D
$P(C \mid D)=P(C)$

The correct answer is:
$P(C \mid D) \geq P(C)$

Solution:
$\mathrm{C} \cap \mathrm{D}=\mathrm{C} \Rightarrow \mathrm{P}(\mathrm{C} \cap \mathrm{D})=\mathrm{P}(\mathrm{C}) \Rightarrow \mathrm{P}\left(\frac{\mathrm{C}}{\mathrm{D}}\right)=\frac{\mathrm{P}(\mathrm{C} \cap \mathrm{D})}{\mathrm{P}(\mathrm{D})} \geq \mathrm{P}(\mathrm{C})$

Topic: Mathematics, Probability, JEE Main, JEE Main 2011
If C and V represent capacity and voltage respectively then what are the dimensions of λ where C/V=λ ?

Options:
A M-2 L-4 I3 T7
B M-2 L-3I2 T6
C M-1 L-3I-2 T-7
D M-3 L-4I3 T7

The correct answer is:
M-2 L-4 I3 T7

Solution:
λ=CV=Q/VV=QV2

V= work Q

λ=Q3work2=It3F.s2

=I3T3ML2T-22=M-2L-4I3T7

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If Î±, Î²âC are the distinct roots of the equation x2-x+1=0, then Î±101+Î²107 is equal to

Options:
A 2
B -1
C 0
D 1

The correct answer is:
1

Solution:
x2-x+1=0
Roots are -ω, -ω2.
α101+β107=-ω101+-ω2107
=-ω2-ω ∵ω3=1

=-ω2+ω

=1 ∵1+ω+ω2=0

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2018 (08 Apr)
If Î±, Î²âC are the distinct roots of the equation x2-x+1=0, then Î±101+Î²107 is equal to

Options:
A 2
B -1
C 0
D 1

The correct answer is:
1

Solution:
x2-x+1=0
Roots are -ω, -ω2
α101+β107=-ω101+-ω2107
=-ω2-ω
=1

Topic: Mathematics, Complex Number, JEE Main
If c is a point at which Rolle’s theorem holds for the function, fx=loge⁡x2+α7x in the interval 3,4, where α∈R, then f''c is equal to

Options:
A -112
B 112
C -124
D 37

The correct answer is:
112

Solution:
f3=f4⇒Î±=12
f'x=x2-12xx2+12
â´f'c=0
â´c=12
â´f''c=112

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (08 Jan Shift 1)
If $c$ is any arbitrary constant, then the general solution of the differential equation $y d x-x d y=x y d x$ is given by

Options:
A $y=c x e^{-x}$
B $x=c y e^{-x}$
C $y+e^x=c x$
D None of these

The correct answer is:
$y=c x e^{-x}$

Solution:
Given $y d x-x d y=x y d x$
$\Rightarrow \frac{y d x-x d y}{x y}=d x \Rightarrow d\left[\ln \left(\frac{x}{y}\right)\right]=d x$
Integrating both sides, we get $\ln \frac{x}{y}+\ln c=x$
$y e^x=c x$.

Topic: Mathematics, Differential Equations, JEE Main
If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$, then

Options:
A
$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}=2 \overrightarrow{\mathrm{PC}}$

B
$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+2 \overrightarrow{\mathrm{PC}}=0$

C
$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+ \overrightarrow{\mathrm{PC}}=0$

D
None of these

The correct answer is:
$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}=2 \overrightarrow{\mathrm{PC}}$

Solution:
$$
\begin{aligned}
& \overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CP}}=0 \\
& \overline{\mathrm{PB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CP}}=0
\end{aligned}
$$
Adding, we get
$$
\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BC}}+2 \overrightarrow{\mathrm{CP}}=0
$$
Since $\overrightarrow{A C}=-\overrightarrow{B C}$
$$
\begin{aligned}
& \& \overrightarrow{\mathrm{CP}}=-\overrightarrow{\mathrm{PC}} \\
& \Rightarrow \overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}-2 \overrightarrow{\mathrm{PC}}=0 .
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2005
If $\mathrm{C}$ is the reflecton of $\mathrm{A}(2,4)$ in $\mathrm{x}$-axis and $\mathrm{B}$ is the reflection of $\mathrm{C}$ in $\mathrm{y}$-axis, then $|\mathrm{AB}|$ is

Options:
A 20
B $2 \sqrt{5}$
C $4 \sqrt{5}$
D 4

The correct answer is:
$4 \sqrt{5}$

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{A} \equiv(2,4) ; \mathrm{C} \equiv(2,-4) ; \mathrm{B} \equiv(-2,-4) \\
& |A B|=\sqrt{(2-(-2))^2+(4-(-4))^2}=\sqrt{4^2+8^2} \\
& =\sqrt{16+64}=\sqrt{80}=\sqrt{16 \times 5}=4 \sqrt{5}
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
If cos Î¸-11cos 2Î¸ 432 77=0, then the number of values of Î¸ in 0,2Ï is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
cos⁡θ7+17cos⁡2θ-6+17cos⁡2θ-8=0
⇒2cos⁡2θ+cos⁡θ-2=0⇒4cos2θ-2+cos⁡θ-2=0
⇒4cos2θ+cos⁡θ-4=0⇒cos⁡θ=-1±658
cos⁡θ=-1-658rejected,cos⁡θ=-1+658
Hence, 2 solutions are possible

Topic: Mathematics, Determinants, JEE Main
If α=cos-135 , β=tan-113 , where 0<α,β<π2, then α-β is equal to

Options:
A tan-1914
B cos-19510
C sin-1⁡9510
D tan-19510

The correct answer is:
sin-1⁡9510

Solution:
α=cos-135=tan-143
∴a-β=tan-143-tan-113
=tan-143-131+43×13
=tan-11139
=tan-1913

=sin-1⁡9510

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If cos-1sin⁡x≥sin-1sin⁡x, then the number of integral values of x in the interval x∈0,3π are

Options:
A 7
B 6
C 4
D 5

The correct answer is:
5

Solution:
If cos-1x≥sin-1x, then x∈-12,12 {using graph}
For given inequality, we must have sin⁡x∈-12,12

⇒x∈nπ-π4,nπ+π4,n∈Ι
Possible integral values of x are 0,3,6,7,9

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cosα+β=35 ,sin⁡(α-β)=513 and 0<α,β<π4, then tan⁡2α is equal to:

Options:
A 2116
B 6352
C 3352
D 6316

The correct answer is:
6316

Solution:
cosα+β=35⇒tanα+β=43

sin(α-β)=513⇒tan(α-β)=512

Now,

tan2α=tanα+β+α-β

=tan(α+β)+tan(α-β)1-tan(α+β)tan(α-β)

=43+5121-43×512=6316

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If â«cos4xdxsin3xsin5x+cos5x35=-1K1+cotPxKP+C, then the value of K+P is equal to (where C is the constant of integration)

Options:

The correct answer is:
7

Solution:
Let, I=∫cos4xdxsin6x1+cot5x35=∫sec2xdxtan6x1+1tanx535
Substituting, 1+1tan⁡x5=t and -5tan6⁡xsec2⁡xdx=dt
I=-15∫dtt3/5=-15t2525+C
=-12∫1+cot5x25+C
Hence, K=2, P=5

Topic: Mathematics, Indefinite Integration, JEE Main
If cosxdydx-ysinx=6x, 0<x<π2 and yπ3=0, then yπ6 is equal to

Options:
A -π243
B π223
C -π22
D -π223

The correct answer is:
-π223

Solution:
The given D.E. can be written as

dydx-ytanx=6xsecx ...i, which is a liner differential equation of the form dydx+Py=Q, where P & Q are the function of x or constants.

Here, P=-tanx & Q=6xsecx

Integrating factor =e∫Pdx=e-∫tanx dx

=e--lncosx=elncosx=cosx

And, the solution of the linear differential equation is,

yI.F.=∫QI.F.dx+c

⇒y⋅cosx=∫6xsecx⋅cosxdx

⇒y⋅cosx=∫6xdx

⇒y·cosx=3x2+c ...ii

Given yπ3=0, hence

0=3π32+c

⇒c=-π23

Thus, the curve is, y·cosx=3x2-π23

Now, put x=π6, to get

yπ6·cosπ6=3π62-π23

⇒yπ6⋅32=3⋅π236-π23

⇒yπ6⋅32=π212-π23

⇒yπ6⋅32=-π24

⇒yπ6=-π223.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If cosx-sinx=-54, where Ï2<x<3Ï4, then cotx2 is equal to

Options:
A 4-79
B 8
C -8
D 4+79

The correct answer is:
4+79

Solution:
Given, 4cos⁡x-sin⁡x=-5
41-tan2x21+tan2x2-2tan⁡x21+tan2x2=-5
⇒4-4tan2⁡x2-8tan⁡x2=-5-5tan2⁡x2
⇒tan2x2-8tan⁡x2+9=0
⇒9cot2⁡x2-8cot⁡x2+1=0
⇒cotx2=4±79
Since, π4<x2<3π8⇒2-1<cotx2<1
⇒cotx2=4+79

Topic: Mathematics, Trigonometric Equations, JEE Main
If θ=cot-1α, φ=cot-1β and αβ=-1 α>β, then the value of θ-φ is equal to

Options:
A 0
B π4
C π2
D -π2

The correct answer is:
-π2

Solution:
∵cot-1α=θ⇒α=cot⁡θ...........i
Also, cot-1β=φ⇒β=cot⁡φ...........ii
Now, given αβ=-1

So, from (i) & (ii), we get,
cot⁡θ.cot⁡φ=-1⇒tan⁡θtan⁡φ=-1   .........iii
tanθ-φ=tan⁡θ-tan⁡φ1+tan⁡θtan⁡φ............iv
θ-ϕ=±π2 (from (iii) and (iv))
But, α>β⇒θ<φ⇒θ-φ=-π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cot-1n2Ï>2Ï3, then the maximum value of the integer n is

Options:
A 3
B 4
C -4
D -3

The correct answer is:
-4

Solution:
We have, 2Ï3<cot-1n2Ï<Ï⇒cot2Ï3>n2Ï>-â
⇒-13>n2Ï>-â⇒-2Ï3>n>-â
Maximum value of n is -4

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\mathrm{C}_p$ and $\mathrm{C}_{\mathrm{v}}$ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

Options:
A
$\mathrm{C_p-C_v=R / 28}$

B
$\mathrm{C_p-C_v=R / 14}$

C
$\mathrm{C_p-C_v=R}$

D
$\mathrm{C_p-C_v=28 R}$

The correct answer is:
$\mathrm{C_p-C_v=R / 28}$

Solution:
Mayer Formula

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2007
If $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{V}}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight $M$
where $R$ is the molar gas constant.

Options:
A $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}^2}$
B $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
C $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}$
D $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{MR}$

The correct answer is:
$\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}$

Solution:
$$
\begin{array}{r}
\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \\
\therefore \quad \mathrm{Mc}_{\mathrm{p}}-\mathrm{MC}_{\mathrm{V}}=\mathrm{R} \\
\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}
\end{array}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2010 (Mains)
If Cr20 is the co-efficient of xr in the expansion of (1+x)20, then the value of ∑r=020r2Cr20 is equal to:

Options:
A 420×218
B 380×218
C 380×219
D 420×219

The correct answer is:
420×218

Solution:
∑r=020r2Cr20=∑rr·Cr20=∑r=120r·20.Cr-119

⇒20∑r=120(r-1+1)·19Cr-1=20∑r=120(r-1)·19Cr-1+∑r=120Cr-119

⇒2019∑r=220Cr-218+∑r=120Cr-119

⇒(380)218+20·219⇒220·(95+10)=(105)220=420×218

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If Cr≡Cr25 and C0+5∙C1+9∙C2+…+101∙C25=225∙k, then k is equal to ____________.

Options:

The correct answer is:
51

Solution:
∑r=0254r+125Cr

=4∑r=025r.25Cr+∑r=025Cr25
=4∑r=125r×25rCr-124+225

=100∑r=125Cr-125+225
=100.224+225

=22550+1

=51.225
So, k=51

Topic: Mathematics, Binomial Theorem, JEE Main
If Cr≡Cr25 and C0+5∙C1+9∙C2+…+101∙C25=225∙k, then k is equal to ____________.

Options:

The correct answer is:
51

Solution:
∑r=0254r+125Cr

=4∑r=025r.25Cr+∑r=025Cr25
=4∑r=125r×25rCr-124+225

=100∑r=125Cr-125+225
=100.224+225

=22550+1

=51.225
So, k=51

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If $C_r$ denotes the binomial coefficient ${ }^n C_r$, then $(-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2$ is equal to

Options:
A $(3 n-2){ }^{2 n} C_n$
B $\left(\frac{3 n-2}{2}\right){ }^{2 \pi} C_n$
C $(5+3 n){ }^{2 \pi} C_n$
D $\left(\frac{3 n-5}{2}\right){ }^{2 \pi} C_{n+1}$

The correct answer is:
$\left(\frac{3 n-2}{2}\right){ }^{2 \pi} C_n$

Solution:
Consider the given expression,
$$
\begin{aligned}
& (-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2 \\
& =\sum_{r=0}^n(3 r-1) C_r^2 \\
& =\sum_{r=0}^n 3 r \cdot C_r^2-\sum_{r=0}^n C_r^2 \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=1}^n\left(r \cdot{ }^n C_r \cdot{ }^n C_r\right)-{ }^{2 n} C_n \\
& \left.=3 \sum_{r=1}^n\left(n \cdot{ }^{n-1} C_{r-1}^2 \cdot{ }^n C_r\right)-{ }^{2 n} C_n+\ldots+C_n^2={ }^{2 n} C_n\right]
\end{aligned}
$$
$\begin{aligned} & \quad=3 n \sum_{r=0}^n\left({ }^{n-1} C_{r-1}{ }^n C_{n-r}\right)-{ }^{2 n} C_n \\ & \qquad\left[\because r \cdot{ }^n C_r=n \cdot{ }^{n-1} C_{r-1}\right] \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\qquad(1+x)^{n-1}(1+x)^n\right]-{ }^{2 n} C_n \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\quad(1+x)^{2 n-1}\right]-{ }^{2 n} C_n \\ & =3 n{ }^{2 n-1} C_{n-1}-{ }^{2 n} C_n \\ & =3 n \frac{(2 n-1) !}{(n-1) !(n) !}-\frac{(2 n) !}{(n !)(n !)}\end{aligned}$
$$
\begin{aligned}
& =\frac{3 n \cdot(2 n)(2 n-1) !}{(2 n)(n-1) !(n !)}-\frac{(2 n) !}{n ! n !}=\frac{3 n}{2} \frac{(2 n) !}{n ! n !}-\frac{(2 n) !}{n ! n !} \\
& =\frac{3 n}{2} \cdot{ }^{2 n} C_n-{ }^{2 n} C_n={ }^{2 n} C_n\left(\frac{3 n}{2}-1\right) \\
& =\left(\frac{3 n-2}{2}\right) \cdot{ }^{2 n} C_n
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If C, R, L and I denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time among the following are

(1) CR

(2) LR

(3) LC

(4) LI2

Options:
A (1) and (2) only
B (1) and (3) only
C (1) and (4) only
D (1), (2) and (3) only

The correct answer is:
(1), (2) and (3) only

Solution:
C=M-1L-2T4A2,R=ML2T-3A-2
L=ML2T-2A-2 and I=M0L0T0A
1. CR=M-1L-2T4A2 ML2T-3A-2
=M0L0TA0
2. LR=ML2T-2A-2ML2T-3A-2=M0L0TA0
3. LC = ML 2 T - 2 A - 2 × M - 1 L - 2 A 2 1 / 2
=M0L0TA0
4. LI2=ML2T-2A-2 M0L0T0A12
=ML2T-2A0

Topic: Physics, Units and Dimensions, JEE Main