datavorous/entrance-exam-dataset huggingface dataset:
The value of $\lim _{x \rightarrow \infty}\left(\frac{x^2+b x+4}{x^2+a x+5}\right)$ is

Options:
A $b / a$
B $1$
C $0$
D $4 / 5$

The correct answer is:
$1$

Solution:
$\lim _{x \rightarrow \infty}\left[\frac{1+\frac{b}{x}+\frac{4}{x^2}}{1+\frac{a}{x}+\frac{5}{x^2}}\right]=1$

Topic: Mathematics, Limits, JEE Main
The value of limx→-∞⁡x2tan⁡2x16x2-x+1 is equal to

Options:
A 1
B 12
C -1
D -12

The correct answer is:
-12

Solution:
limx→-∞⁡2xtan⁡2x2x-x16-1x+1x2=2-16=-12

Topic: Mathematics, Limits, JEE Main
The value of ∫ln⁡cotxsin⁡2xdx is equal to (where, C is the constant of integration)

Options:
A ln⁡cot⁡x22+C
B ln⁡cot⁡x24+C
C ln⁡cot⁡x26+C
D -14ln⁡cot⁡x2+C

The correct answer is:
-14ln⁡cot⁡x2+C

Solution:
∫ln⁡cot⁡x2sin⁡xcos⁡xdx=Ι (let)
Put ln⁡cot⁡x=t
⇒1cot⁡x×cosec2⁡xdx=-dt
⇒dxcos⁡xsin⁡x=-dt
So, Ι=12∫-tdt=-12t22+C
=-14ln⁡cot⁡x2+C

Topic: Mathematics, Indefinite Integration, JEE Main
The value of log 10 âK for a reaction AâB is
(Given âHr 298 Ko=-54.07 kJ mol-1,
âSr 298 Ko=10 JK-1mol-1 and
R=8.314 JK-1mol-1, 2.303Ã8.314Ã298=5705 )

Options:
A 5
B 10
C 95
D 100

The correct answer is:
10

Solution:
For the equilibrium, AâB
ÎG o â=â ÎH o â TÎS o
ÎG o =â2.303 RTâ log 10 âK ( K is equilibrium constant)
â2.303 RTâ log 10 âK=â ÎH o â TÎS o
2.303 RTâ log 10 âKâ=â TÎS o â ÎH o
log10â¡K=TâSo-âHo2.303 RT=298Ã10+54.07Ã10002.303Ã8.314Ã298=10
Hence (B) is correct option.

Topic: Chemistry, Thermodynamics (C), JEE Main
The value of log 10 K for reaction, A â B ; is

(Given : Î r H 298âK 0 =â54.07âkJâmo l â1 ,âââ Î r S 288âK 0 =10âJâ K â1 mo l â1 and R=8.31âJâ K â1 âmo l â1 ,ââ2.303Ã8.314Ã298=5705 )

Options:

The correct answer is:
10

Solution:
log K =  ΔS o 2.303 R − ΔH o 2.303 RT
= 10 2.303×8.314 − (−54.07× 10 3 ) 2.303×8.314×298
= 10×298 5705 + 54070 5705
= 57050 5705 =10

Topic: Chemistry, Thermodynamics (C), JEE Main
The value of loge2ddxlogcosxcosecx at x=π4 is

Options:
A -22
B 22
C -4
D 4

The correct answer is:
4

Solution:
For loge2ddxlogcosxcosecx

let, y=logcosxcosecx

i.e. y=-lnsinxlncosx

⇒dydx=-cotx·lncosx+tanx·lnsinxlncosx2

Now dydxx=π4=-cotπ4·lncosπ4+tanπ4·lnsinπ4lncosπ42

=4ln2

⇒loge2·4ln2=4

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The value of $\operatorname{Lt}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}$ is

Options:
A $\mathrm{e}^2$
B $\mathrm{e}$
C $\frac{1}{\mathrm{e}}$
D $\frac{1}{\mathrm{e}^2}$

The correct answer is:
$\mathrm{e}^2$

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{1+5 x^2}{1+3 x^2}-1\right)}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{2 x^2}{x^2\left(1+3 x^2\right)}}=\mathrm{e}^2$

Topic: Mathematics, Limits, JEE Main
The value of $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}$ is

Options:
A 1
B $\frac{1}{2}$
C $-\frac{1}{2}$
D 0

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

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}=\operatorname{Lim}_{x \rightarrow 0} \frac{\cos x-\cos ^2 x}{x^2}=\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right) \cos x$
$$
=\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\left(\frac{x}{2}\right)^2 \times 4}=\frac{1}{2}
$$

Topic: Mathematics, Limits, JEE Main
The value of $m$ for which the function $f(x)=\left\{\begin{array}{r}m x^2, x \leq 1 \\ 2 x, x\gt1\end{array}\right.$ is differentiable at $x=1$,is

Options:
A 0
B 1
C 2
D Does not exist

The correct answer is:
Does not exist

Solution:
$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{m(1-h)^2-m}{-h}=\lim _{h \rightarrow 0} \frac{m\left[1+h^2-2 h-1\right]}{-h} \\ & =\lim _{h \rightarrow 0} m(2-h)=2 m \text { and } R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2(1+h)-m}{h} . \text { For differentiability, } L f^{\prime}(1)=R f^{\prime}(1)\end{aligned}$
But for any value of $m, R f^{\prime}(1)=L f^{\prime}(1)$ not possible.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of $m$, for which the line ${ }^{y=m x+\frac{25 \sqrt{3}}{3}}$, is a normal to the conic $\frac{x^2}{16}-\frac{y^2}{9}=1$, is

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

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

Solution:
We know that the equation of the normal of the conic $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at point $(a \sec \theta, b \tan \theta)$ is $a x \cos \theta+b y \cot \theta=a^2+b^2$
or $y=\frac{-a}{b} \sin \theta x+\frac{a^2+b^2}{b \cot \theta}$
Comparing above equation with equation $y=m x+\frac{25 \sqrt{3}}{3}$ and taking $a=4, b=3$ we get, $\frac{a^2+b^2}{b \cot \theta}=\frac{25 \sqrt{3}}{3} \Rightarrow \tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}$ and $m=-\frac{a}{b} \sin \theta=\frac{-4}{3} \sin 60^{\circ}=\frac{-4}{3} \times \frac{\sqrt{3}}{2}=\frac{-2}{\sqrt{3}}$.

Topic: Mathematics, Hyperbola, JEE Main
The value of $m$ for which $y=m x+6$ is a tangent to the hyperbola $\frac{x^2}{100}-\frac{y^2}{49}=1$, is

Options:
A $\sqrt{\frac{17}{20}}$
B $\sqrt{\frac{20}{17}}$
C $\sqrt{\frac{3}{20}}$
D $\sqrt{\frac{20}{3}}$

The correct answer is:
$\sqrt{\frac{17}{20}}$

Solution:
If $y=m x+c$ touches $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $c^2=a^2 m^2-b^2$. Here $c=6, a^2=100, b^2=49$ $\therefore 36=100 m^2-49 \Rightarrow 100 m^2=85 \Rightarrow m=\sqrt{\frac{17}{20}}$.

Topic: Mathematics, Hyperbola, JEE Main
The value of magnetic quantum number of the outermost electron of Zn+ ion is _________ .
(Integer answer)

Options:

The correct answer is:
0

Solution:
Zn+=|Ar|183 d104s1
For last e- :nℓms400±12

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The value of \( \mathrm{k}>0 \) such that the angle between the lines \( 4 x-y+7=0 \) and \( k x-5 y-9=0 \) is \( 45^{\circ} \) is

Options:
A \( \frac{25}{3} \)
B \( \frac{5}{3} \)
C \( 3 \)
D \( 5 \)

The correct answer is:
\( 3 \)

Solution:
m1=4, m2=k5
tan45o= 4-k51+4k5 ⇒ 5k=15, k=3

Topic: Mathematics, Straight Lines, JEE Main
The value of $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right), i=\sqrt{-1}$ is

Options:
A $i$
B $i-1$
C 1
D 0

The correct answer is:
$i-1$

Solution:
We have.
$\begin{aligned} & \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right) \\=& \sum_{n=1}^{13} i^{n}+\sum_{n=1}^{13} i^{n+1} \\=& i \left(\frac{1-i^{13}}{1-i}\right)+i^{2}\left(\frac{1-i^{13}}{1-i}\right) \\=& i \left(\frac{1-i}{1-i}\right)-\left(\frac{1-i}{1-i}\right) \\=& i-1 \end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The value of ∑n=1100∫n-1nex-xdx, where x is the greatest integer ≤x, is:

Options:
A

100e-1

B

100e

C

1001-e

D 1001+e

The correct answer is:

100e-1

Solution:
∑n=1100∫n-1nexdx, period of x=1

∑n=1100∫01exdx=∑n=1100∫01exdx

∑n=1100e-1=100e-1

Topic: Mathematics, Definite Integration, JEE Main
The value of ∑n=1100∫n-1nex-xdx, where x is the greatest integer ≤x, is:

Options:
A

100e-1

B

100e

C

1001-e

D 1001+e

The correct answer is:

100e-1

Solution:
∑n=1100∫n-1nexdx, period of x=1

∑n=1100∫01exdx=∑n=1100∫01exdx

∑n=1100e-1=100e-1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The value of n+2·C0n⋅2n+1-n+1·C1n⋅2n+n·C2n⋅2n-1-.... to n+1 terms is equal to

Options:
A 4
B 4n
C 4n+1
D 2n+2

The correct answer is:
4n+1

Solution:
Tr+1=-1rn-r+2Crn2n-r+1
=n+22n+1-1rCnr12r-2n+1-1rrCnr12r
=n+22n+1-1rCrn12r+2n⋅n-1r-1Cr-1n-112r-1
So, sum =n+22n+1∑-1rCrn12r+2nn∑-1r-1Cr-1n-112r-1
=n+22n+11-12n+2nn1-12n-1
=n+22n+112n+2nn12n-1
=n+22+2n
=4n+4=4n+1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of $\mathrm{n}$ for which $\frac{\mathrm{x}^{\mathrm{n}+1}+\mathrm{y}^{\mathrm{n}+1}}{\mathrm{x}^{\mathrm{n}}+\mathrm{y}^{\mathrm{n}}}$ is the geometric mean of $\mathrm{x}$ and $\mathrm{y}$ is

Options:
A $\mathrm{n}=-\frac{1}{2}$
B $\mathrm{n}=\frac{1}{2}$
C $\mathrm{n}=1$
D $\mathrm{n}=-1$

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

Solution:
Hints: $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=\sqrt{x y} \Rightarrow x^{n+1}+y^{n+1}=\sqrt{x y}\left(x^n+y^n\right)$
$$
\mathrm{x}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right)=\mathrm{y}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right),\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{n}+\frac{1}{2}}=1 \quad \mathrm{n}=-\frac{1}{2}
$$

Topic: Mathematics, Sequences and Series, JEE Main
The value of n in the molecular formula BenAl2Si6O18 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
For BenAl2Si6O18
According to charge balance in a molecule,
2n + 6 + 24 - 36 = 0
This gives, n = 3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The value of net resistance of the network as shown in the given figure is :

Options:
A $6 \Omega$
B $(5 / 2) \Omega$
C $(15 / 4) \Omega$
D $(30 / 11) \Omega$

The correct answer is:
$6 \Omega$

Solution:
Diode 2 is in reverse bias
So current will not flow in branch of $2^{\text {nd }}$ diode, So we can assume it to be broken wire.
Diode 1 is in forward bias
So it will behave like conducting wire. So new circuit will be

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The value of numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be:

Options:
A 0.48 μm
B 0.24 μm
C 0.38 μm
D 0.12 μm

The correct answer is:
0.24 μm

Solution:
Resolving limit=0.61λd=0.61×5000×10-101.25=0.24 μm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The value of objective function is maximum under linear constraints

Options:
A At the center of feasible region
B At $(0,0)$
C At any vertex of feasible region
D The vertex which is at maximum distance from $(0,0)$

The correct answer is:
The vertex which is at maximum distance from $(0,0)$

Solution:
The value of objective function is maximum under linear constraints at any vertex of feasible region.
Explanation:
In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value. Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.

Topic: Mathematics, Linear Programming, JEE Main
The value of $p$ and $q$ for which the function $f(x)=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & x < 0 \\ q & , x=0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0\end{array}\right.$ is continuous for all $\mathrm{x}$ in $\mathrm{R}$, is

Options:
A
$p=\frac{5}{2}, q=\frac{1}{2}$

B
$p=-\frac{3}{2}, q=\frac{1}{2}$

C
$p=\frac{1}{2}, q=\frac{3}{2}$

D
$p=\frac{1}{2}, q=-\frac{3}{2}$

The correct answer is:
$p=-\frac{3}{2}, q=\frac{1}{2}$

Solution:
$\lim _{x \rightarrow 0} \frac{\sin (p+1)+\sin x}{x}=q=\lim _{x \rightarrow 0} \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}}$
$\lim _{x \rightarrow 0}(p+1) \cos (p+1) x+\cos x=q=\frac{1}{2}$
$\Rightarrow p+1+1=\frac{1}{2} \Rightarrow p=-\frac{3}{2} ; q=\frac{1}{2}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2011
The value of $p$ for which both the roots of the equation $4 x^2-20 p x+\left(25 p^2+15 p-66\right)=0$ are less than 2 , lies in

Options:
A $(4 / 5,2)$
B $(2, \infty)$
C $(-1,-4 / 5)$
D $(-\infty,-1)$

The correct answer is:
$(-\infty,-1)$

Solution:
Let $f(x)=4 x^2-20 p x+\left(25 p^2+15 p-66\right)=0 \quad \ldots .$ ...(i)
The roots of (i) are real if
$b^2-4 a c=400 p^2-16\left(25 p^2+15 p-66\right)$
$=16(66-15 p) \geq 0$
$\Rightarrow p \leq 22 / 5$ ...(ii)
Both roots of (i) are less than 2 .
Therefore $f(2)\gt0$ and sum of roots $ \lt 4$.
$\Rightarrow 4.2^2-20 p .2+\left(25 p^2+15 p-66\right)\gt0$ and $\frac{20 p}{4} \lt 4$
$\Rightarrow p^2-p-2\gt0$ and $p \lt 4 / 5$
$\Rightarrow(p+1)(p-2)\gt0$ and $p \lt 4 / 5$
$\Rightarrow p \lt -1$ or $p\gt2$ and $p \lt 4 / 5 \Rightarrow p \lt -1$.....(iii)
From (ii) and (iii), we get $p \lt -1$ i.e. $p \in(-\infty,-1)$.

Topic: Mathematics, Quadratic Equation, JEE Main
The value of Planck's constant is $6.63 \times 10^{-34} \mathrm{Js}$. The speed of light is $3 \times 10^{17} \mathrm{nms}^{-1}$. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of $6 \times 10^{15} \mathrm{~s}^{-1} ?$

Options:
A 10
B 25
C 50
D 75

The correct answer is:
50

Solution:
Given, Planck's constant,
$$
h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}
$$
Speed of light, $c=3 \times 10^{17} \mathrm{~nm} \mathrm{~s}^{-1}$
Frequency of quanta
$$
v=6 \times 10^{15} \mathrm{~s}^{-1}
$$
Wavelength, $\lambda=$ ?
We know that, $v=\frac{c}{\lambda}$
or
$$
\begin{aligned}
\lambda & =\frac{c}{v} \\
& =\frac{3 \times 10^{17}}{6 \times 10^{15}} \\
& =0.5 \times 10^2 \mathrm{~nm} \\
& =50 \mathrm{~nm}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The value of Planck's constant is 6.63 $\times 10^{-34} \mathrm{Js}$. The velocity of light is $3.0 \times$ $10^8 \mathrm{~ms}^{-1}$. Which value is closest to the wavelength in nanometers of quantum of light with frequency of $8 \times 10^{15} \mathrm{~s}^{-1}$ ?

Options:
A $2 \times 10^{-25}$
B $5 \times 10^{-18}$
C $4 \times 10^1$
D $3 \times 10^7$

The correct answer is:
$4 \times 10^1$

Solution:
$\begin{aligned}
& c=v \times \lambda ; \lambda=\frac{c}{v} \\
& \Rightarrow \lambda=\frac{3 \times 10^8}{8 \times 10^{15}}=0.375 \times 10^{-7} \mathrm{~m} \\
& =0.375 \times 10^{-7} \times 10^9 \mathrm{~nm} \\
& =0.375 \times 10^2 \mathrm{~nm} \\
& =37.5 \mathrm{~nm} \\
& \approx 4 \times 10 \mathrm{~nm}
\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main
The value of power dissipated across the zener diode Vz=15 V connected in the circuit as shown in the figure is x×10-1 W.

The value of x to the nearest integer is _______.

Options:

The correct answer is:
5

Solution:
Voltage across RS=22-15=7 V

Current through RS=I=735=15 A

Current through 90Ω=I2=1590=16 A

Current through zener =15-16=130 A

Power through zener diode

P=VI

P=15×130=0.5 watt

P=5×10-1 watt

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The value of ∑r=06Cr6·C6-r6 is equal to :

Options:
A 1124
B 1324
C 1024
D 924

The correct answer is:
924

Solution:
We have ∑r=06Cr6·C6-r6=C06·C66+C16·C56+...+C66·C06

Now, we know that the expansion of 1+xn=C0nx0+C1nx1+C2nx2+...Cnnxn

⇒(1+x)6(1+x)6=C06+C16x+C26x2+...+C66x6C06+C16x+C26x2+...+C66x6

⇒(1+x)12=C06+C16x+C26x2+...+C66x6C06+C16x+C26x2+...+C66x6

On comparing the coefficient of x6 both sides, we get

C612=C06·C66+C16·C56+...+C66·C06

⇒C06·C66+C16·C56+...+C66·C06=12!6!·6!

⇒C06·C66+C16·C56+...+C66·C06=12×11×10×9×8×7×6!6×5×4×3×2×1×6!

⇒C06·C66+C16·C56+...+C66·C06=924.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The value of ∑r=077Crr+2C2 is equal to

Options:
A 128
B 25118
C 1283
D 117

The correct answer is:
25118

Solution:
∑r=077Crr+2r+12
=2∑r=077Crr+1r+2
=2∑r=079Cr8×9
=28×929-9C0-9C1
=512-1-936
=50236
=25118

Topic: Mathematics, Binomial Theorem, JEE Main
The value of ∑r=020C650-r is equal to:

Options:
A C751-C730
B C750-C730
C C650-C630
D C751+C730

The correct answer is:
C751-C730

Solution:
∑r=020C650-r=

=50C6+49C6+48C6+....+30C6

=−30C7+30C7+30C6+31C6+32C6+....+50C6

=−30C7+31C7+31C6+32C6+....+50C6

=−30C7+32C7+32C6+....+50C6

=−30C7+51C7.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The value of ∑r=0nr+2r+1Crn is equal to

Options:
A 2nn+2-1n+1
B 2nn+1-1n+1
C 2nn+4-1n+1
D 2nn+3-1n+1

The correct answer is:
2nn+3-1n+1

Solution:
∑r=0nr+2r+1Crn=∑r=0nr+1+1r+1Crn
=∑r=0nCrn+∑r=0n1r+1Crn
=2n+∑r=0n1n+1 n+1Cr+1
=2n+1n+1∑r=0n n+1Cr+1=2n+1n+12n+1-1
=2nn+1+2⋅2n-1n+1
=2nn+3-1n+1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of ∑r=110r·CrnCr-1n is equal to

Options:
A 52n-9
B 10n
C 9(n-4)
D None of these

The correct answer is:
52n-9

Solution:
We have,

∑r=110r·CrnCr-1n=∑r=110n-r+1

⇒ ∑r=110r·CrnCr-1n=∑r=110n+1-r

⇒ ∑r=110r·CrnCr-1n=10n+1-∑r=110r

⇒ ∑r=110r·CrnCr-1n=10n+1-55

=10n-45=52n-9

Topic: Mathematics, Permutation Combination, JEE Main
The value of ∑r=115r215Cr15Cr−1 is equal to:

Options:
A 1240
B 560
C 1085
D 680

The correct answer is:
680

Solution:
Given ∑ r=1 15 r 2 ( 15 C r 15 C r−1 )
Here 15Cr15Cr-1=16-rr
∑ r=1 15 r 2 ( 15 C r 15 C r−1 ) = ∑ r=1 15 r 2 16-rr
= ∑ r=1 15 r 16-r
=16 ∑ r=1 15 r - ∑ r=1 15 r 2
=16 ×15 ×162-15 ×31 ×166
=8 ×15 ×16-5 ×8 ×31
=1920-1240=680

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2016 (09 Apr Online)
The value of ∑r=1n-1r+1Crnr+1 is equal to

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

The correct answer is:
nn+1

Solution:
∑r=1n-1r+1Crnr+1
Using Cn+1r+1=n+1r+1Cnr, we get,
∑r=1n-1r+11n+1Cr+1n+1
=1n+1∑r=1n-1r+1Cr+1n+1
=1n+10-Cn+10-Cn+11
=1n+10-1+n+1=nn+1

Topic: Mathematics, Binomial Theorem, JEE Main
The value of ∑r=1630r+2(r-3) is equal to:

Options:
A 7775
B 7785
C 7780
D 7770

The correct answer is:
7780

Solution:
Given value =∑r=1630r+2(r-3)

=∑r=130r+2(r-3)-∑r=115r+2(r-3)
=nn+12n+16-nn+12-6n-mm+12m+16-mm+12-6m
where n=30, m=15
=8810-1030
=7780

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (10 Apr Online)
The value of λ∈R such that x,y,z≠0,0,0 and 2i^+3j^-4k^x+3i^-j^+2k^y+i-2j^z=λxi^+yj^+zk^ lies in

Options:
A 1,2
B 2,3
C 3,4
D 0,1

The correct answer is:
3,4

Solution:
2x+3y+zi^+3x-y-2zj^+-4x+2yk^=λxi^+yj^+zk^
Comparing both sides, we get,
2x+3y+z=λx⇒λ-2x-3y-z=0
3x-y-2z=λy⇒-3x+λ+1y+2z=0
-4x+2y=λz⇒4x-2y+λz=0
So the system of equation has non-zero solution
⇒λ-2-3-1-3λ+124-2λ=0
⇒λ-2λ2+λ+4+3-3λ-8-16-4λ-4=0
⇒λ3+λ2+4λ-2λ2-2λ-8-9λ-24-2+4λ=0
⇒λ3-λ2-3λ-34=0
This equation has only one real root which lies in (3,4) as f3f4<0

Topic: Mathematics, Vector Algebra, JEE Main
The value of Rydberg constant \(\left(R_H\right)\) is \(2.18 \times 10^{-18} \mathrm{~J}\). The velocity of electron having mass \(9.1 \times 10^{-31} \mathrm{~kg}\) in Bohr's first orbit of hydrogen atom \(=\) _______ \(\times 10^5 \mathrm{~ms}^{-1}\) (nearest integer).

Options:

The correct answer is:
22

Solution:
$\begin{aligned} & \mathrm{V}=2.18 \times 10^6 \times \frac{\mathrm{Z}}{\mathrm{n}} \\ & =21.8 \times 10^5 \times \frac{1}{1} \approx 22 \times 10^5 \text { (nearest) }\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The value of ∫sin3xcos⁡xdx is equal to (where, c is the constant of integration)

Options:
A 23cos⁡x32+27cos⁡x72+c
B -23cos⁡x32+27cosx72+c
C -23cos⁡x3+25cos⁡x52+c
D 32cos⁡x32+52cos⁡x72+c

The correct answer is:
-23cos⁡x32+27cosx72+c

Solution:
∫sin2xcos⁡x sin⁡xdx
Substituting, cos⁡x=t and sin⁡xdx=-dt
=-∫1-t2tdt
=-∫t12-t52dt
=-t3232+t7272+c
=-23cos⁡x32+27cos⁡x72+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of sincot-1coscot-11x is equal to x>0

Options:
A 1+x22+x2
B 1-x22+x2
C 1+x22-x2
D 2+x21+x2

The correct answer is:
1+x22+x2

Solution:
Let cot-11x=α⇒ tan⁡α= x
So, cos⁡α=11+x2
∴ sin⁡cot-1⁡cos⁡tan-1⁡x
= sincot-111+x2
Let cot-111+x2=β⇒ cot⁡β=11+x2
∴sin⁡β=(1+x2)(2+x2)
∴sin⁡β=sin⁡cot-1cos⁡cot-1⁡1x=1+x22+x2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\int \frac{\sin x-\cos x}{\sin x+\cos x} d x$ is

Options:
A $\frac{1}{\sin x+\cos x}+c$
B $\frac{1}{\sin x-\cos x}+c$
C $\log (\sin x+\cos x)+c$
D $\log \left(\frac{1}{\sin x+\cos x}\right)+c$

The correct answer is:
$\log \left(\frac{1}{\sin x+\cos x}\right)+c$

Solution:
Let \(\sin x+\cos x=t\)
So, \((\cos x-\sin x) \cdot d x=d t\)
\(\begin{aligned}
& \text {So, } \int \frac{\sin x-\cos x}{\sin x+\cos x} \cdot d x=\int \frac{-d t}{t} \\
& \Rightarrow-\log (t)+c \Rightarrow \log \left(\frac{1}{t}\right)+c \\
& \Rightarrow \log \left(\frac{1}{\sin x+\cos x}\right)+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, JEE Main
The value of ∫-ππsiny1+tan2ydy (where [x] is greatest integer function) is

Options:

The correct answer is:
2

Solution:
I=∫-ππsinx1+tan2xdx=4∫0π/2sinxcosxdx

⇒I=83sin32π2-0=83
[I]=2

Topic: Mathematics, Definite Integration, JEE Main
The value of $\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is

Options:
A $\cos ^{-1}\left(\frac{1}{2}\right)$
B $\sin ^{-1}\left(-\frac{1}{2}\right)$
C $\cos ^{-1}\left(-\frac{1}{2}\right)$
D $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

The correct answer is:
$\cos ^{-1}\left(-\frac{1}{2}\right)$

Solution:
(D)
Let $\sin ^{-1}\left(\frac{-1}{2}\right)=\alpha$, where $\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}$
$\sin \alpha=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right) \Rightarrow \alpha=-\pi / 6$
Let $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\beta$, where $0 \leq \beta \leq \pi$
$\cos \beta=\frac{-\sqrt{3}}{2}=\frac{-\cos \pi}{6} \Rightarrow \cos \left(\pi-\frac{\pi}{6}\right)=\frac{\cos 5 \pi}{6} \Rightarrow \beta=\frac{5 \pi}{6}$
$\therefore \sin ^{-1}\left(\frac{-1}{2}\right)+\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-\pi}{6}+\frac{5 \pi}{6} \quad=\frac{4 \pi}{6}=\frac{2 \pi}{3}=\cos ^{-1}\left(\frac{-1}{2}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of sin-1⁡1213-sin-1⁡35 is equal to:

Options:
A π-cos-1⁡3365
B π2-cos-1⁡965
C π-sin-1⁡6365
D π2-sin-1⁡5665

The correct answer is:
π2-sin-1⁡5665

Solution:
sin-1⁡1213-sin-1⁡35
=tan-1⁡125-tan-1⁡34 sin-1⁡x=tan-1⁡x1-x2
=tan-1⁡125-341+12×35×4
=tan-1⁡3356
=sin-1⁡3365
=cos-1⁡5665=π2-sin-1⁡5665

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The value of sin-1⁡cos⁡sin-1⁡x+cos-1⁡sin⁡cos-1⁡x is equal to

Options:
A 0
B π4
C π2
D 3π4

The correct answer is:
π2

Solution:
sin-1⁡cos⁡sin-1⁡x=π2-cos-1⁡cos⁡sin-1⁡x
=π2-sin-1⁡x
Similarly, cos-1⁡sin⁡cos-1⁡x=π2-cos-1⁡x
Adding both, we get,
sin-1⁡cos⁡sin-1⁡x+cos-1⁡sin⁡cos-1⁡x
=π-sin-1⁡x+cos-1⁡x
=π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of $\sin ^{-1}(\sin 10)$ is

Options:
A $10$
B $10-3 \pi$
C $3 \pi-10$
D None of these

The correct answer is:
$3 \pi-10$

Solution:
$\begin{aligned} & \text{Since } 3 \pi \lt 10 \lt 3 \pi+\frac{\pi}{2} \Rightarrow 0 \lt 10-3 \pi \lt \frac{\pi}{2} \\ \Rightarrow & \frac{-\pi}{2} \lt 3 \pi-10 \lt 0 \Rightarrow \sin ^{-1}\{\sin (3 \pi-10)\}=3 \pi-10\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of sin-1⁡sin⁡17+cos-1⁡cos⁡10 is equal to

Options:
A 27
B -27
C 17-5π
D 9π-27

The correct answer is:
9π-27

Solution:
sin-1⁡sin⁡17=sin-1⁡sin⁡17-5π+5π
=5π-17
cos-1⁡cos⁡10=cos-1⁡cos⁡10-3π+3π
=cos-1⁡cos⁡3π+10-3π
=cos-1⁡-cos⁡10-3π
=π-cos-1⁡cos⁡10-3π
=π-10-3π=4π-10
Hence, sin-1⁡sin⁡17+cos-1⁡cos⁡10=9π-27

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of \( \sin ^{-1} \sin 17+\cos ^{-1} \cos 10 \) is equal to

Options:
A \( 27 \)
B \( -27 \)
C \( 17-5 \pi \)
D \( 9 \pi-27 \)

The correct answer is:
\( 9 \pi-27 \)

Solution:
sin-1⁡sin⁡17=sin-1⁡sin⁡17-5π+5π
=5π-17
cos-1⁡cos⁡10=cos-1⁡cos⁡10-3π+3π
=cos-1⁡cos⁡3π+10-3π
=cos-1⁡-cos⁡10-3π
=π-cos-1⁡cos⁡10-3π
=π-10-3π=4π-10
Hence, sin-1⁡sin⁡17+cos-1⁡cos⁡10=9π-27

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of \( \sin ^{-1}\left[\cot \left(\sin ^{-1} \sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right] \) is

Options:
A \( 0 \)
B \( \pi / 4 \)
C \( \pi / 6 \)
D \( \pi / 2 \)

The correct answer is:
\( 0 \)

Solution:
Given

sin-1cotsin-12-34+cos-1124+sec-12

=sin-1cotsin-13-122+cos-132+cos-112

=sin-1cot15°+30°+45°

=sin-1cot90°=sin-10=0.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main