datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{R}$ is the Rydberg Constant in $\mathrm{cm}^{-1}$, then hydrogen atom does not emit any radiation of wave-length in the range of

Options:
A $\frac{1}{\mathrm{R}}$ to $\frac{4}{3 \mathrm{R}} \mathrm{cm}$
B $\frac{7}{5 \mathrm{R}}$ to $\frac{19}{5 \mathrm{R}} \mathrm{cm}$
C $\frac{4}{\mathrm{R}}$ to $\frac{36}{5 \mathrm{R}} \mathrm{cm}$
D $\frac{9}{\mathrm{R}}$ to $\frac{144}{7 \mathrm{R}} \mathrm{cm}$

The correct answer is:
$\frac{7}{5 \mathrm{R}}$ to $\frac{19}{5 \mathrm{R}} \mathrm{cm}$

Solution:
Hint:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}^{2}}}-\frac{1}{\mathrm{n}_{\mathrm{i}^{2}}}\right]$
For range of wavelengths:$\mathrm{n}_{\mathrm{i}}=1,2,3, \ldots . .$ for Lyman, Balmer, Paschen, $\ldots \ldots$
$\mathrm{n}_{\mathrm{f}}=\mathrm{n}_{\mathrm{i}}+1$ and $\mathrm{n}_{\mathrm{f}}=\infty$ for upper and lower range
Thus, Lyman : $\left[\frac{1}{\mathrm{R}}\right.$ to $\left.\frac{4}{3 \mathrm{R}}\right]$, Balmer : $\left[\frac{4}{\mathrm{R}}\right.$ to $\left.\frac{36}{5 \mathrm{R}}\right]$, Paschen : $\left[\frac{9}{\mathrm{R}} \mathrm{to} \frac{144}{36 \mathrm{R}}\right]$, Bracket : $\left[\frac{16}{\mathrm{R}}\right.$ to $\left.\frac{400}{9 \mathrm{R}}\right]$, Pfund : $\left[\frac{25}{\mathrm{R}}\right.$ to $\left.\frac{900}{11 \mathrm{R}}\right]$

Topic: Physics, Atomic Physics, JEE Main
If $R$ is the set of all real numbers and $f: R-\{2\} \rightarrow R$ is defined by $f(x)=\frac{2+x}{2-x}$ for $x \in R-\{2\}$

Options:
A $R-\{-2\}$
B $R$
C $R-\{1\}$
D $R-\{-1\}$

The correct answer is:
$R-\{-1\}$

Solution:
Let $y=f(x)=\frac{2+x}{2-x}$
$$
\begin{aligned}
& \Rightarrow \quad 2 y-x y=2+x \\
& \Rightarrow \quad x(1+y)=2 y-2 \\
& \Rightarrow \quad x=\frac{2(y-1)}{1+y}, y \neq-1
\end{aligned}
$$
$\therefore$ Domain of $y$ is $R-\{-1\}$.
Hence, range of $f(x)$ is $R-\{-1\}$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2014
If $R$ is the set of all real numbers and $f: R-\{2\} \rightarrow R$ is defined by $f(x)=\frac{2+x}{2-x}$ for $x \in R-\{2\}$

Options:
A $R-\{-2\}$
B $R$
C $R-\{1\}$
D $R-\{-1\}$

The correct answer is:
$R-\{-1\}$

Solution:
Let $y=f(x)=\frac{2+x}{2-x}$
$$
\begin{aligned}
& \Rightarrow \quad 2 y-x y=2+x \\
& \Rightarrow \quad x(1+y)=2 y-2 \\
& \Rightarrow \quad x=\frac{2(y-1)}{1+y}, y \neq-1
\end{aligned}
$$
$\therefore$ Domain of $y$ is $R-\{-1\}$.
Hence, range of $f(x)$ is $R-\{-1\}$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2014
If $\alpha \in R, n \in N$ and $n+2(n-1)+3(n-2)+\ldots+$ $(n-1) 2+n .1=\alpha n(n+1)(n+2)$, then $\alpha=$

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{5}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{6}$

Solution:
We have,
$$
\begin{aligned}
T_r & =r[n-(r-1)] \\
& =r(n-r+1) \\
& =n r-r^2+r \\
\therefore \quad S_n & =\sum_{r=1}^n T_r=\sum_{r=1}^n\left(n r-r^2+r\right) \\
& =\sum_{r=1}^n\left[(n+1) r-r^2\right]=(n+1) \sum_{r=1}^n r-\sum_{r=1}^n r^2
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{(n+1) n(n+1)}{2}-\frac{n(n+1)(2 n+1)}{6} \\
& =\frac{n(n+1)}{2}\left[(n+1)-\left(\frac{2 n+1}{3}\right)\right] \\
& =\frac{n(n+1)}{2}\left[\frac{3 n+3-2 n-1}{3}\right] \\
& =\frac{n(n+1)}{2}\left[\frac{n+2}{3}\right] \\
& =\frac{n(n+1)(n+2)}{6} \\
\therefore \quad \alpha & =\frac{1}{6}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main
If Δr=r2r-13r-2n2n-1a12nn-1 n-1212n-13n+4, then the value of ∑r=1n−1Δr

Options:
A Is independent of both a and n
B Depends only on a
C Depends only on n
D Depends both on a and n

The correct answer is:
Is independent of both a and n

Solution:
Δr=r2r-13r-2n2n-1a12nn-1 n-1212n-13n+4

Since, the second and the third rows are independent of r, hence the sum is applied to the first row only.

⇒∑r=1n−1Δr=∑r=1n−1r2∑r=1n−1r−∑r=1n−113∑r=1n−1r−2∑r=1n−11n2n−1a12nn−1n−1212n−13n+4

Using ∑r=1nr=nn+12, we get

∑r=1n−1Δr=n-1n2 2n-1n2-n-1 3n-1n2-2n-1n2n-1a12nn-1n-1212n-13n+4

⇒∑r=1n−1Δr=12nn-1 n-121 2n-13n+4n2 n-1a12nn-1 n-12 12n-13n+4

⇒∑r=1n−1Δr=0, (∵ R1 and R3 are identical)

Hence, the sum is independent of both n and a.

Topic: Mathematics, Determinants, JEE Main
If Δr=r2r-13r-2n2n-1a12nn-1 n-1212n-13n+4, then the value of ∑r=1n−1Δr

Options:
A Is independent of both a and n
B Depends only on a
C Depends only on n
D Depends both on a and n

The correct answer is:
Is independent of both a and n

Solution:
Δr=r2r-13r-2n2n-1a12nn-1 n-1212n-13n+4

Since, the second and the third rows are independent of r, hence the sum is applied to the first row only.

⇒∑r=1n−1Δr=∑r=1n−1r2∑r=1n−1r−∑r=1n−113∑r=1n−1r−2∑r=1n−11n2n−1a12nn−1n−1212n−13n+4

Using ∑r=1nr=nn+12, we get

∑r=1n−1Δr=n-1n2 2n-1n2-n-1 3n-1n2-2n-1n2n-1a12nn-1n-1212n-13n+4

⇒∑r=1n−1Δr=12nn-1 n-121 2n-13n+4n2 n-1a12nn-1 n-12 12n-13n+4

⇒∑r=1n−1Δr=0, (∵ R1 and R3 are identical)

Hence, the sum is independent of both n and a.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2014 (19 Apr Online)
If $r$ represents the radius of the orbit of a satellite of mass $m$ moving around a planet of mass $M$, the velocity of the satellite is given by

Options:
A $v^2=g \frac{M}{r}$
B $v^2=\frac{G M m}{r}$
C $v=\frac{G M}{r}$
D $v^2=\frac{G M}{r}$

The correct answer is:
$v^2=\frac{G M}{r}$

Solution:
Given, $\mathrm{M}$ is mass of planet, $\mathrm{m}$ is mass of satellite, $\mathrm{r}$ is radius of orbit.
Let, $\mathrm{v}$ be velocity of the satellite
Now, for the satellite to revolve in the orbit the centripetal force must be balanced by gravitational force i.e.,
$\mathrm{F}_{\mathrm{c}}=\mathrm{F}_{\mathrm{g}} \rightarrow(1)$
we know that,
$\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^2}{\mathrm{r}} \text { and } \mathrm{F}_{\mathrm{g}}=\mathrm{GMmr}^2$
Substituting these values in (1),
$\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{GMm}}{\mathrm{r}^2}$
After solving this equation we get
$v^2=\frac{G M}{r}$

Topic: Physics, Gravitation, JEE Main
If Î»âR such that the origin and the non-real roots of the equation 2z2+2z+Î»=0 form the vertices of an equilateral triangle in the argand plane, then 1Î» is equal to _____. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Let, z1 and z2 are roots of the equation 2z2+2z+λ=0
⇒z1+z2=-1 and z1z2=λ2
If 0, z1 & z2 form an equilateral triangle, then z12+z22=z1z2
⇒z1+z22=3z1z2
⇒-12=3λ2⇒1λ=32=1.5

Topic: Mathematics, Complex Number, JEE Main
If Î»âR such that the origin and the non-real roots of the equation 2z2+2z+Î»=0 form the vertices of an equilateral triangle in the argand plane, then 1Î» is equal to

Options:

The correct answer is:
1.5

Solution:
Let, z1 and z2 are roots of the equation 2z2+2z+λ=0
⇒z1+z2=-1 and z1z2=λ2
If 0, z1 & z2 form an equilateral triangle, then z12+z22=z1z2
⇒z1+z22=3z1z2
⇒-12=3λ2⇒1λ=32=1.5

Topic: Mathematics, Complex Number, JEE Main
If R, XL and XC represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless:

Options:
A R XL XC
B RXLXc
C RXLXc
D RXLXc

The correct answer is:
RXLXc

Solution:
As resistance, inductive reactance and capacitive reactance have same unit ohm. So, all three have same dimension.

Thus, RXLXC is dimensionless.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If $R=\left\{(x, y): x, y \in W, x^2+y^2=25\right\}$, then find the domain and range of $R$.

Options:

None

Solution:
We have, $R=\left\{(x, y): x, y \in W, x^2+y^2=25\right\}$
$=\{(0,5),(3,4),(4,3),(5,0)\}$
Domain of $R=\{0,3,4,5\}$
Range of $R=\{5,4,3,0\}$.

Topic: Mathematics, Relations and Functions (Class 11)
If R=x, y:x, y∈Z, x2+3y2≤8 is a relation on the set of integers Z, then the domain of R-1 is

Options:
A -2, -1, 1, 2
B 0, 1
C -2, -1, 0, 1, 2
D −1, 0, 1

The correct answer is:
−1, 0, 1

Solution:
Given x28+y283≤1; x, y∈Z

Domain of R-1= Range of R is possible integral y-coordinates

=-1,0,1

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If R=x, y|x, yâZ, x2+y2â¤4 is a relation in Z , then domain of R is

Options:
A 0,Â 1,Â 2
B 0,Â -Â 1,Â -Â 2
C - 2, - 1, 0, 1, 2
D None of these

The correct answer is:
- 2, - 1, 0, 1, 2

Solution:
âµ R=x, yx, yâZ, x2+y2â¤4
R=-2, 0,-1, 0,(0, -1),-1, 1,(1, -1),0, -10, 1,0, 2,0,-2,1, 0,(0, 1),1, 1,(-1, -1),2, 0,(0, 0)
Hence, Domain of R=-2, -1, 0, 1, 2

Topic: Mathematics, Sets and Relations, JEE Main
If radius of ${ }_{13}^{27} \mathrm{Al}$ nucleus is estimated to be $3.6$ Fermi then the radius ${ }_{52}^{125} \mathrm{Te}$ nucleus be nearly

Options:
A
6 fermi

B
8 fermi

C
4 fermi

D
5 fermi

The correct answer is:
6 fermi

Solution:
$\frac{\mathrm{R}}{3.6}=\left(\frac{125}{27}\right)^{\frac{1}{3}} \Rightarrow \mathrm{R}=6$ fermi

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2005
If radius of the director circle of the ellipse $\frac{(3 x+4 y-2)^2}{100}+\frac{(4 x-3 y+5)^2}{625}=1$ is

Options:
A $6$
B $\sqrt{34}$
C $\sqrt{29}$
D $\sqrt{26}$

The correct answer is:
$\sqrt{29}$

Solution:
$\begin{aligned} & \frac{(3 x+4 y-2)^2}{100}+\frac{(4 x-3 y+5)^2}{625}=1 \\ & \Rightarrow \frac{\left(\frac{3 x+4 y-2}{5}\right)^2}{4}+\frac{\left(\frac{4 x-3 y+5}{5}\right)^2}{25}=1 \\ & \Rightarrow a^2=4 \text { and } b^2=25\end{aligned}$
Radius of director circle $=\sqrt{a^2+b^2}=\sqrt{29}$

Topic: Mathematics, Ellipse, JEE Main
If relations R1 and R2 from set A to set B are defined as R1=1,2, 3,4, 5,6 and R2=2,1, 4,3, 6,5, then nAÃB can be equal to

Options:
A 35
B 53
C 36
D 55

The correct answer is:
36

Solution:
nAÃB=nAÃnB
Now minimum nA=6 and minimum nB=6
⇒nAÃnB can be equal to 7Ã13=91

Topic: Mathematics, Sets and Relations, JEE Main
If Rolle's theorem for a function fx=exsin⁡x-cos⁡x is verified on π4,5π4 then the value of c is

Options:
A π3
B π2
C 3π4
D π

The correct answer is:
π

Solution:
Given, fx=ex(sinâ¡Â x-cosâ¡Â x)
â´Â Â f′x=excosâ¡Â x+sinâ¡Â x+sinâ¡Â x-cosâ¡Â xex
â´Â Â f′x=2exsinâ¡Â x
To verify Rolle's Theorem.
f′c=0
2ecsinâ¡Â c=0
âÂ sinâ¡Â c=0
â´Â Â c=Ï

Topic: Mathematics, Application of Derivatives, JEE Main
If Rolle's theorem holds for the function fx=2x3+bx2+cx, x∈-1,1 at the point x=12, then 2b+c is equal to

Options:
A 2
B 1
C -1
D -3

The correct answer is:
-1

Solution:
Since, Rolle's theorem is satisfied in [-1, 1] and corresponding to x=12 we have f(−1)=f(1) & f'12=0

So, −2+b−c=2+b+c

⇒ c=−2

Further f'x=6x2+2bx+c

Put x=12

So f'12=32+b+c=0

⇒b=12

Hence, 2b+c=-1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (10 Apr Online)
If Rolle's theorem holds for the function fx=x3-ax2+bx-4,x∈1,2 with f'43=0, then ordered pair a, b is equal to :

Options:
A -5,-8
B -5,8
C 5,8
D 5,-8

The correct answer is:
5,8

Solution:
f1=f2

⇒ 1-a+b-4=8-4a+2b-4

⇒ 3a-b=7 …1

Also f'43=0 (given)

⇒ 3x2-2ax+bx=43=0

⇒ 163-8a3+b=0

⇒ 8a-3b-16=0 …2

Solving 1 and 2

a=5, b=8

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If Rolle's theorem holds for the function $f(x)=x^3+b x^2+a x+5$ on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$, then the values of $a$ and $b$ respectively are

Options:
A $-11,-6$
B 11,6
C $11,-6$
D 6,11

The correct answer is:
$11,-6$

Solution:
Since $\mathrm{f}(x)$ satisfies the Rolle's theorem,
$f(1)=f(3)$
$\therefore \quad 1+b+a+5=27+9 b+3 a+5$
$\Rightarrow 2 a+8 b=-26$
$\Rightarrow a+4 b=-13$ ...(i)
$\vec{f}(x)=x^3+b x^2+a x+5$
$\therefore \quad \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{a}$
Now, $\mathrm{f}^{\prime}(\mathrm{c})=0$
$$
\begin{aligned}
& \Rightarrow \mathrm{f}^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0 \\
& \Rightarrow 3\left(2+\frac{1}{\sqrt{3}}\right)^2+2 \mathrm{~b}\left(2+\frac{1}{\sqrt{3}}\right)+\mathrm{a}=0 \\
& \Rightarrow 3\left(4+\frac{4}{\sqrt{3}}+\frac{1}{3}\right)+4 \mathrm{~b}+\frac{2 \mathrm{~b}}{\sqrt{3}}+\mathrm{a}=0 \\
& \Rightarrow \mathrm{a}+4 \mathrm{~b}+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \\
& \Rightarrow-13+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \\
& \Rightarrow \frac{2 \mathrm{~b}+12}{\sqrt{3}}=0 \\
& \Rightarrow \mathrm{b}=-6
\end{aligned}
$$
Substituting $b=-6$ in (i), we get $a=11$

Topic: Mathematics, Application of Derivatives, JEE Main
If Rolle's theorem holds for the function \( f(x)=2 x^{3}+b x^{2}+c x, x \in[-1,1] \), at the point \( x=\frac{1}{2} \), then \( 2 \mathrm{~b}+\mathrm{c} \) equals:

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

The correct answer is:
\( -1 \)

Solution:
If Rolle's theorem is satisfied in the interval [-1, 1], then
f-1=f(1)
-2+b-c=2+b+c
c=Â -2 also f′x=6x2+2bx+c
Also if f′12=0 them
614+2b12+c=0
32+b+c=0
âµÂ Â c=Â -2,
b=12
â´Â Â 2b+c=212+(-2)
= 1-2
= -1.

Topic: Mathematics, Application of Derivatives, JEE Main
If roots of the equation $x^2-b x+c=0$ be two consectutive integers, then $b^2-4 c$ equals

Options:
A
$-2$

B
3

C
2

D
1

The correct answer is:
1

Solution:
Let $\alpha, \alpha+1$ be roots
$$
\begin{aligned}
& \alpha+\alpha+1=b \\
& \alpha(\alpha+1)=c \\
& \therefore b^2-4 c=(2 \alpha+1)^2-4 \alpha(\alpha+1)=1 .
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2005
If Rydberg's constant is R, the longest wavelength of radiation in Paschen series will be α7R, where α=______.

Options:

The correct answer is:
144

Solution:
In the Paschen radiation series, the longest wavelength corresponds to the transition between n=3 and n=4.

The formula for the wavelength in the given transition can be written as,

1λ=RZ21n12-1n22 ...1

From equation (1), it follows that

1λ=RZ2132-142=RZ219-116=7RZ29×16⇒λ=1447R

Hence, α=144.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If S=125+224+323+......+242+251, then the value of S900 is equal to

Options:

The correct answer is:
3.25

Solution:
S=∑r=125r26-r=26∑r=125r-∑r=125r2
=26×25×262-25×26×516
=25×26226-513=25×26132×9
⇒S900=134=3.25

Topic: Mathematics, Sequences and Series, JEE Main
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, fx=9x4+12x3-36x2+25,x∈R, then

Options:
A S1=-2;S2={0,1}
B S1=-1;S2=0,2
C S1=-2,0;S2={1}
D S1=-2,1;S2={0}

The correct answer is:
S1=-2,1;S2={0}

Solution:
y=9x4+12x3-36x2+25

Differentiating,

dydx=36x3+36x2-72x
=36xx2+x-2

=36xx+2x-1

Sign convention for dydx
∴S1=-2,1 & S2=0

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If S=75+952+1353+1954+…, then 160 S is equal to

.

Options:

The correct answer is:
305

Solution:
Given, if S=75+952+1353+1954+……∞ ...i

Dividing by 5 on both sides of series

S5=752+953+1354+1955+……∞ ...ii

Subtracting equations i&ii

S-S5=75+952+1353+1954+……∞-752+953+1354+1955+……∞

4S5=75+952-752+1353-953+1954-1354+……∞

4S5=75+252+453+654+855+……∞

4S5-75=k=252+453+654+855+…..∞...iii

Let 4S5-75=k

k5=253+454+655+856+…..∞...iv

Subtracting equation iii & iv

4k5=252+253+254+255……∞

Common ratio

r=253252=15, r=254253=15

4k5=22511-15=110

k=18

Now 4S5-75=18

4S5=75+18

S=6132

160S=160×6132

=305

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If $S=\{a \in \mathbf{R}:|2 a-1|=3[a]+2\{a\}\}$, where $[t]$ denotes the greatest integer less than or equal to $t$ and $\{t\}$ represents the fractional part of $t$, then $72 \sum_{a \in S} a$ is equal to ______

Options:

The correct answer is:
18

Solution:
$\begin{aligned} & |2 \mathrm{a}-1|=3[\mathrm{a}]+2\{\mathrm{a}\} \\ & |2 \mathrm{a}-1|=[\mathrm{a}]+2 \mathrm{a}\end{aligned}$
$\begin{aligned} & \text { Case-1 }: \mathrm{a}>\frac{1}{2} \\ & 2 a-1=[a]+2 a \\ & {[a]=-1 \quad \therefore a \in[-1,0) \text { Reject }} \\ & \text { Case-2 }: \mathrm{a} < \frac{1}{2} \\ & -2 \mathrm{a}+1=[\mathrm{a}]+2 \mathrm{a} \\ & \mathrm{a}=\mathrm{I}+\mathrm{f} \\ & -2(\mathrm{I}+\mathrm{f})+1=\mathrm{I}+2 \mathrm{I}+2 \mathrm{f} \\ & \mathrm{I}=0, \mathrm{f}=\frac{1}{4} \quad \therefore \mathrm{a}=\frac{1}{4} \\ & \text { Hence } \mathrm{a}=\frac{1}{4} \\ & 72 \sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}=72 \times \frac{1}{4}=18\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If sin-15x+sin-112x=sin-12x+cos-12x, then the value of x is equal to

Options:

The correct answer is:
13

Solution:
sin-15x and sin-112x are both defined when x>12. Therefore
cos-11-25x2+sin-112x=π2 ∵sin-12x+cos-12x=π2
sin-112x=sin-11-25x2
On comparing, we get,
12x=1-25x2

Therefore, 144=x2-25 or x=13

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1x2-2sin-1x+1â¤0 (where, . represents the greatest integral part of x), then

Options:
A x∈sin⁡1,sin⁡2
B x∈-sin⁡1,sin⁡1
C x∈sin⁡1,1
D x∈-sin⁡1,sin⁡2

The correct answer is:
x∈sin⁡1,1

Solution:
sin-1x-12≤0
⇒sin-1x-1=0⇒sin-1x=1
⇒1≤sin-1x<2 but, sin-1x∈-π2,π2
∴1≤sin-1x≤π2
⇒sin⁡1≤x≤sinπ2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1x2+sin-1x-2â¤0 (where, [Â·] represents the greatest integral part of x), then the maximum value of x is

Options:

The correct answer is:
1

Solution:
sin-1x-1sin-1x+2≤0
⇒sin-1x∈-2,1⇒sin-1x∈-2,2
⇒-2≤sin-1x<2 but sin,-1x∈-π2≤π2
∴--π2≤sin-1x≤π2
-sinπ2≤x≤sinπ2

Topic: Mathematics, Functions, JEE Main
If sin2x+10-x2=9-x2+2sin2x+cos2x, then x lies in

Options:
A -8,8
B -3,3
C -17,17
D -21,21

The correct answer is:
-3,3

Solution:
sin2x+10-x2=9-x2+sin2x+sin2x+cos2x
⇒sin2x+1+9-x2=9-x2+sin2x+1
x+y+z=x+y+z, if x, y, z have the same sign
⇒sin2x9-x2≥0
⇒9-x2≥0
⇒x∈-3,3

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$, then $g(x)$ is equal to

Options:
A $\frac{-2}{\sqrt{\cot x}}$
B $\frac{-2}{\sqrt{\tan x}}$
C $\frac{2}{\sqrt{\cot x}}$
D $\frac{2}{\sqrt{\tan x}}$

The correct answer is:
$\frac{-2}{\sqrt{\tan x}}$

Solution:
Given, $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$
$$
\text { Now, } \begin{aligned}
\int \frac{d x}{\sqrt{\sin ^4 x \cot x}} & =\int \frac{d x}{\sin ^2 x \sqrt{\cot x}} \\
& =\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} d x
\end{aligned}
$$
Put $\quad \cot x=t$
$$
\begin{array}{rlrl}
\Rightarrow-\operatorname{cosec}^2 x d x & =d t \\
\therefore \quad \int-\frac{1}{\sqrt{t}} d t & =-\frac{t^{1 / 2}}{1 / 2}+c \\
& =-2 \sqrt{\cot x}+c \\
& =-\frac{2}{\sqrt{\tan x}}+c \\
\therefore & g(x) & =-\frac{2}{\sqrt{\tan x}}
\end{array}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If sin23π24=2p-q-14r, then the value of p2+q2-r2 is equal to

Options:
A 6
B 12
C -1
D 9

The correct answer is:
9

Solution:
sin23π4=sinπ-π24=sinπ24
Now, cosπ12=cos⁡15o=3+122
Using 1-cos θ=2sin2θ2, we get
2sin2π24=1-cosπ12=1-3+122
⇒sin2π24=22-3-142
⇒sinπ24=sin23π24=22-3-142
⇒p=2,q=3,r=2
⇒p2+q2-r2=9

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+C$, then value of $(A, B)$ is

Options:
A
$(\sin \alpha, \cos \alpha)$

B
$(\cos \alpha, \sin \alpha)$

C
$(-\sin \alpha, \cos \alpha)$

D
$(-\cos \alpha, \sin \alpha)$

The correct answer is:
$(\cos \alpha, \sin \alpha)$

Solution:
Put $x-\alpha=t$
$\Rightarrow \int \frac{\sin (\alpha+t)}{\sin t} d t=\sin \alpha \int \cot t d t+\cos \alpha \int d t$
$=\cos \alpha(x-\alpha)+\sin \alpha \ln |\sin t|+c$
$A=\cos \alpha, B=\sin \alpha$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2004
If $S$ is stress and $Y$ is Young's modulus of material of a wire, the energy stored in the wire per unit volume is

Options:
A
$2 S^2 Y$

B
$S^2 / 2 Y$

C
$2 Y / S^2$

D
$S / 2 Y$

The correct answer is:
$S^2 / 2 Y$

Solution:
$U=\frac{1}{2}$ stress $\times$ strain $=\frac{S^2}{2 Y}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2005
If S is the set of distinct values of b for which the following system of linear equations
x+y+z=1
x+ay+z=1
ax+by+z=0
has no solution, then S is:

Options:
A An empty set
B An infinite set
C A finite set containing two or more elements
D A singleton

The correct answer is:
A singleton

Solution:
For no solution D=1111a1ab1=0

Expanding the determinant along the first row, we get D=-a-12=0

⇒ a=1

For a=1, first two planes are coincident.

Now, we have x+y+z=1 & ax+by+z=0

For no solution, these two planes are parallel.

⇒1a=1b=11

⇒a=1 & b=1.

∴ b=1 ⇒ it is a singleton set

Topic: Mathematics, Determinants, JEE Main, JEE Main 2017 (02 Apr)
If S is the sum of the first 10 terms of the series, tan-113+tan-117+tan-1113+tan-1121+…… then tan(S) is equal to :

Options:
A 56
B 511
C -56
D 1011

The correct answer is:
56

Solution:
S=tan-113+tan-117+tan-1113+…… up to10 terms.

S=tan-12-11+1.2+tan-13-21+2.3+tan-14-31+3.4+……+tan-111-101+11.10

⇒S=tan-12-tan-11+tan-13-tan-12+……+tan-111-tan-110

⇒S=tan-111-tan-11

⇒S=tan-111-11+11·1=tan-11012

∴tan(S)=56

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If S=ân=199991n+n+1n4+n+14, then the value of S is equal to

Options:
A 9
B 99
C 999
D 9999

The correct answer is:
9

Solution:
Given, S=ân=199991n+n+1n4+n+14
=ân=199991n+n+1n4+n+14Ãn+14-n4n+14-n4
=ân=19999n+14-n4n+n+1n+1-n
=ân=19999n+14-n4
=214-1+314-214+414-314+â¦+1000014-999914
=1000014-1=10-1=9

Topic: Mathematics, Sequences and Series, JEE Main
If Sn=n2a+n4n-1d is the sum of the first n terms of an arithmetic progression, then the common difference is

Options:
A a+2d
B 2a+d
C a+d2
D 2a+d2

The correct answer is:
2a+d2

Solution:
S1=a=t1
S2=4a+d2=t1+t2
⇒t2=S2-t1=4a+d2-a=3a+d2
⇒ common difference=t2-t1=3a+d2-a=2a+d2

Topic: Mathematics, Sequences and Series, JEE Main
If s⁡ n = Σ r⁡ = 0 n 1 n C⁡ r⁡ and t n = Σ r⁡ = 0 n r⁡ n C⁡ r⁡ , then t n s⁡ n is equal to

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

The correct answer is:
1 2 n

Solution:
t n = Σ r⁡ = 0 n r⁡ n C⁡ r⁡
t n = Σ r⁡ = 0 n n - n - r⁡ n C⁡ n - r⁡
t n = n Σ r⁡ = 0 n 1 n C⁡ r⁡ - Σ r⁡ = 0 n n - r⁡ n C⁡ n - r⁡
tn=nΣ1nC⁡r⁡-Σr⁡=0nr⁡nC⁡r⁡ (replacing (n-r) by r)
t n = n s⁡ n - t n
∴ t n s⁡ n = n 2

Topic: Mathematics, Binomial Theorem, JEE Main
If $S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}$ and $t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}$, then $\frac{t_n}{S_n}$ is equal to

Options:
A
$\frac{1}{2} \mathrm{n}$

B
$\frac{1}{2} n-1$

C
$n-1$

D
$\frac{2 n-1}{2}$

The correct answer is:
$\frac{1}{2} \mathrm{n}$

Solution:
$t=\sum_{r=0}^n \frac{r}{{ }^n C_r}=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}=\sum_{r=0}^n \frac{n-r}{{ }^n C_r}\left(\because{ }^n C_r={ }^n C_{n-r}\right)$
$2 t_n=\sum_{r=0}^n \frac{r+n-r}{{ }^n C_r}=\sum_{r=0}^n \frac{n}{{ }^n C_r} \Rightarrow t_n=\frac{n}{2} \sum_{r=0}^n \frac{1}{{ }^n C_r}=\frac{n}{2} S_n \Rightarrow \frac{t_n}{S_n}=\frac{n}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main
If $S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}$ and $t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}$, then $\frac{t_n}{S_n}$ is equal to

Options:
A
$\frac{1}{2} \mathrm{n}$

B
$\frac{1}{2} n-1$

C
$n-1$

D
$\frac{2 n-1}{2}$

The correct answer is:
$\frac{1}{2} \mathrm{n}$

Solution:
$t=\sum_{r=0}^n \frac{r}{{ }^n C_r}=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}=\sum_{r=0}^n \frac{n-r}{{ }^n C_r}\left(\because{ }^n C_r={ }^n C_{n-r}\right)$
$2 t_n=\sum_{r=0}^n \frac{r+n-r}{{ }^n C_r}=\sum_{r=0}^n \frac{n}{{ }^n C_r} \Rightarrow t_n=\frac{n}{2} \sum_{r=0}^n \frac{1}{{ }^n C_r}=\frac{n}{2} S_n \Rightarrow \frac{t_n}{S_n}=\frac{n}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2004
If Sn=âr=1ntr=16n2n2+9n+13, then âr=1ntr is equal to

Options:
A 12nn+1
B 12nn+3
C n+12
D n2

The correct answer is:
12nn+3

Solution:
tr=Sr-Sr-1=16r2r2+9r+13-16r-12r-12+9r-1+13
=26r3-r-13+96r2-r-12+136r-r-1
=133r2-3r+1+322r-1+136
=r2+2r+1=r+12
⇒tr=r+1⇒∑r=1ntr=∑r=1nr+1
=∑r=1nr+∑r=1n1=nn+12+n=nn+32

Topic: Mathematics, Sequences and Series, JEE Main