datavorous/entrance-exam-dataset huggingface dataset:
If speed V, force F and acceleration a are chosen as the fundamental physical quantities, then the dimension of Young's modulus in terms of V, F and a are

Options:
A V-3Fa
B V-4F2a2
C V-4Fa2
D V-4F2a

The correct answer is:
V-4Fa2

Solution:
Let Y=fV, F, a
∴Y=KVxFyaz, K→Unitless
∴Y=VxFyaz
∴ML-1T-2=LT-1xMLT-2yLT-2z
∴ML-1=T-2= MyLx+y+zT-x-2y-2z
∴y=1; x+y+z=-1; -x-2y-2z=2
∴x+z=-2 ∴x+2y+2z=2
∴x+2z=0 ∴z=2, x=-4
∴Y=V-4Fa2

Topic: Physics, Units and Dimensions, JEE Main
If stopping potentials corresponding to wavelength 4000 Aâ and 4500 Aâ are 1.3 V and 0.9 V, respectively, then the work function of the metal is

Options:
A 0.3 eV

B 1.3 eV
C 2.3 eV
D 5 eV

The correct answer is:
2.3 eV

Solution:
eV s = hc Î» - Ï 0 or eV s + Ï 0 = hc Î»
or Î» = hc eV s + Ï 0
â Î» 2 Î» 1 = eV s 1 + Ï 0 eV s 2 + Ï 0
4 5 0 0 4 0 0 0 = 1.3 + Ï 0 0.9 + Ï 0
or Ï 0 = 851.3 - 9 Ã 0.9
Solving Ï 0 = 10.4 - 8.1 = 2.3 eV

Topic: Physics, Dual Nature of Matter, JEE Main
If sucrolose is n×100 times of more sweet than Aspartame. What is the value of 'n' here?

Options:

The correct answer is:
6

Solution:
Aspartame is 100 times more sweet than Sugar while sucrolose is 600 times more sweet than sugar. It means the value of 'n' is 6.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
If suddenly the gravitational force of attraction between Earth and a satellite revolving around it
becomes zero, then the satellite will

Options:
A
continue to move in its orbit with same velocity

B
move tangentially to the originally orbit in the same velocity

C
become stationary in its orbit

D
move towards the earth.

The correct answer is:
become stationary in its orbit

Solution:
When gravitational force becomes zero so centripetal force on satellite becomes zero so satellite will escape its round orbit and becomes stationary.

Topic: Physics, Gravitation, JEE Main, JEE Main 2002
If sum of all the solutions of the equation 8cosx·cosπ6+x·cosπ6-x-12=1 in 0, π is kπ, then k is equal to:

Options:
A 209
B 23
C 139
D 89

The correct answer is:
139

Solution:
Given equation 8cosx.cosπ6+x.cosπ6-x-12=1

We would use the formula

cosA+B.cosA-B=cos2A-sin2B

Now, cosπ6+x.cosπ6-x=cos2π6-sin2x

=34-sin2x

⇒8cosx 34-sin2x-12=1

⇒8cosx 14-sin2x=1

⇒8cosx14-1-cos2x=1

⇒8cosxcos2x-34=1

⇒8cos3x-6cosx=1

⇒24cos3x-3cosx=1

⇒2cos3x=1

⇒cos3x=12

General Solution 3x=2nπ±π3

⇒x=2nπ3±π9

x∈0, π

∴ x=π9, 2π3-π9, 2π3+π9

Sum=13π9, k=139

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2018 (08 Apr)
If sum of an infinite geometric series is $\frac{4}{5}$ and its 1 st term is $\frac{3}{4}$, then its common ratio is

Options:
A $\frac{7}{16}$
B $\frac{9}{16}$
C $\frac{1}{9}$
D $\frac{7}{9}$

The correct answer is:
$\frac{7}{16}$

Solution:
Hints: $\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{4}{3} \quad$ Then $\frac{\frac{3}{4}}{1-\mathrm{r}}=\frac{4}{3} \Rightarrow \mathrm{r}=1-\frac{9}{16}=\frac{7}{16}$

Topic: Mathematics, Sequences and Series, JEE Main
If sum of the first 21 terms of the series log91/2x+log91/3x+log91/4x+….. where x>0 is 504, then x is equal to

Options:
A 243
B 9
C 7
D 81

The correct answer is:
81

Solution:
Let

S=log91/2x+log91/3x+log91/4x+…..+log91/22x

⇒S=2log9x+3log9x+……+22log9x

⇒S=log9x(2+3+…..+22)

⇒S=log9x212(2+22)

⇒S=252log9x

Given,

S=504

⇒252log9x=504

⇒log9x=2

⇒x=81

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

Options:
A S1/2I1/2h0
B S1/2I3/2h-1
C S3/2I1/2h0
D S1/2I1/2h-1

The correct answer is:
S1/2I1/2h0

Solution:
P∝SaIbhc
[MLT-1]=MT-2aML2bML2T-1C
MLT-1=[Ma+b+cL2b+2cT-2a-c]
From comparing the power on both side
a+b+c=1
2b+2c=1
-2a-c=-1
From the above three equations we will get
a=12, b=12 & c=0
P∝S12I12h0

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula units, the dimensional formula for linear momentum would be:

Options:
A S1/2I1/2h0
B S1/2I3/2h-1
C S3/2I1/2h0
D S1/2I1/2h-1

The correct answer is:
S1/2I1/2h0

Solution:
P∝SaIbhc
[MLT-1]=MT-2aML2bML2T-1C
MLT-1=[Ma+b+cL2b+2cT-2a-c]
From comparing the power on both side
a+b+c=1
2b+2c=1
-2a-c=-1
From the above three equations we will get
a=12, b=12 & c=0
P∝S12I12h0

Topic: Physics, Units and Dimensions, JEE Main
If system of equation ax+y+z=a,x+by+z=b and x+y+cz=c is inconsistent, then which of the following is correct?

Options:
A abc-a-b-c+2=0
B abc-a-b-c+3=0, a=1
C abc-a-b-c+3=0
D abc-a-b-c+2=0, a≠1,b≠1,c≠1

The correct answer is:
abc-a-b-c+2=0, a≠1,b≠1,c≠1

Solution:
We have,

a x+y+z=a ...(i)

x+by+z=b ...(ii)

x+y+cz=c ...(iii)

are inconsistent.

∴a111b111c=0 and a11bb1c1c≠0

⇒ a111b111c=0

⇒ a(bc-1)-1(c-1)+1(1-b)=0

abc-a-c+1+1-b=0

abc-a-b-c+2=0

and a11bb1c1c≠0

It is possible only when a≠1,b≠1 and c≠1

Topic: Mathematics, Determinants, JEE Main
If $t_1$ and $t_2$ are the times of flight of two particles having the same initial velocity $u$ and range $\mathrm{R}$ on the horizontal, then $\mathrm{t}_1^2+\mathrm{t}_2^2$ is equal to

Options:
A
$\frac{u^2}{g}$

B
$\frac{4 u^2}{g^2}$

C
$\frac{u^2}{2 g}$

D
1

The correct answer is:
$\frac{4 u^2}{g^2}$

Solution:
$\mathrm{t}_1=\frac{2 \mathrm{u} \sin \alpha}{\mathrm{g}}, \mathrm{t}_2=\frac{2 \mathrm{u} \sin \beta}{\mathrm{g}}$ where $\alpha+\beta=90^{\circ}$
$\therefore \mathrm{t}_1^2+\mathrm{t}_2^2=\frac{4 \mathrm{u}^2}{\mathrm{~g}^2}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2004
If $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}^{3}}{3 \mathrm{Yq}}}$ then find the dimensions of q. Where $\mathrm{T}$ is the time period of bar of mass $\mathrm{M}$ length $\mathrm{L}$ and Young modulus $\mathrm{Y}$.

Options:
A $[\mathrm{L}]$
B $\left[\mathrm{L}^{2}\right]$
C $\left[\mathrm{L}^{4}\right]$
D $\left[\mathrm{L}^{3}\right]$

The correct answer is:
$\left[\mathrm{L}^{4}\right]$

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}^{3}}{3 \mathrm{Yq}}},$ writing dimensions of both
the sides, we get $[\mathrm{T}]=\left[\frac{\mathrm{ML}^{3}}{\mathrm{ML}^{-1} \mathrm{~T}^{-2} \mathrm{q}}\right]^{1 / 2}$
or $\quad q=\left[L^{4}\right]$

Topic: Physics, Units and Dimensions, JEE Main
If tan-112x+1+tan-114x+1=cot-1x22, then the number of all possible values of x is/are

Options:
A 1
B 2
C 3
D 0

The correct answer is:
2

Solution:
tan-112x+1+14x+11-12x+1⋅14x+1=tan-12x2
tan-16x+28x2+6x=tan-12x2
Therefore, 3x+14x2+3x=2x2
x23x+1=24x2+3x
3x3-7x2-6x=0
x3x2-7x-6=0
xx-33x+2=0
We take x=3,0 because when x=-23, LHS of the given equation will be negative whereas tan-12x2 is positive.
Hence, the number of values =2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If Î=â«tan-1exex+e-xdx=tan-1fx22+C (where C is the constant of integration), then the range of y=fx âxâR is

Options:
A -∞,∞
B 0,∞
C 0,∞
D -∞,0

The correct answer is:
0,∞

Solution:
Given integral =â«tan-1ex1+e2xexdx
Let, tan-1ex=t
â11+e2xâexdx=dt
âÎ=â«tdt=t22+C
=tan-1ex22+C
Hence, fx=exâ0,â

Topic: Mathematics, Indefinite Integration, JEE Main
If t is a real number and k=t2-t+1t2+t+1, then the system of equations
3x-y+4z=3
x+2y-3z=-2
6x+5y+kz=-3 for any allowable value of k, has

Options:
A a unique solution
B infinite solutions
C no solution
D 2 solutions

The correct answer is:
a unique solution

Solution:
k=t2-t+1t2+t+1ât2k-1+tk+1+k-1=0
as tâR
Dâ¥0âk+12-4k-12â¥0
âkâ13,3
for the given equation,
Î=3-1412-365k=7k+5>0 âkâ13,3
hence, unique solution

Topic: Mathematics, Determinants, JEE Main
If $\mathrm{T}$ is the half-life of a radioactive substance then its instantaneous rate of change of activity is proportional to

Options:
A $\sqrt{\mathrm{T}}$
B $\mathrm{T}$
C $\mathrm{T}^{2}$
D $\mathrm{~T}^{-2}$

The correct answer is:
$\mathrm{~T}^{-2}$

Solution:
No Solution

Topic: Physics, Nuclear Physics, JEE Main
If $T$ is the reverberation time of an auditorium of volume $V$ then

Options:
A $T \propto \frac{1}{V}$
B $T \propto \frac{1}{V^2}$
C $T \propto V^2$
D $T \propto V$

The correct answer is:
$T \propto V$

Solution:
Reverberation time ${ }^{T=\frac{k V}{\alpha S}} \Rightarrow T \propto V$.

Topic: Physics, Waves and Sound, JEE Main
If $t_n$ denotes the number of triangles formed with $n$ points in a plane, no three of which are collinear and if $t_{n+1}-t_n=36$, then $n$ is equal to

Options:
A $7$
B $8$
C $9$
D $10$

The correct answer is:
$9$

Solution:
$t_n=$ The number of triangles formed with $n$ points in a plane, no three of which are collinear.
i.e.,
$$
\begin{aligned}
& t_n={ }^n C_3 \\
& t_{n+1}={ }^{n+1} C_3
\end{aligned}
$$
$$
\Rightarrow \quad t_{n+1}={ }^{n+1} C_3
$$
Now, $t_{n+1}-t_n=36$
$$
\begin{array}{crrl}
\Rightarrow & { }^{n+1} C_3-{ }^n C_3=36 \\
\Rightarrow & \frac{(n+1) !}{(n-2) ! 3 !}-\frac{n !}{(n-3) ! 3 !}=36 \\
\Rightarrow & \frac{(n+1) n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=36 \\
\Rightarrow & & n(n-1)(n+1-n+2)=36 \times 6 \\
\Rightarrow & & 3 n(n-1)=36 \times 6 \\
\Rightarrow & & n^2-n-72=0 \\
\Rightarrow & & n^2-9 n+8 n-72=0 \\
\Rightarrow & & n(n-9)+8(n-9)=0 \\
\Rightarrow & & (n-9)(n+8)=0 \\
\therefore & & n=9 \quad(\because n \neq-3)
\end{array}
$$

Topic: Mathematics, Permutation Combination, JEE Main
If t=x+4, then dxdtt=4 is:

Options:
A 4
B Zero
C 8
D 16

The correct answer is:
Zero

Solution:
Given that t=x+4

⇒x=t-42=t2-8t+16

Differentiating the above equation with respect to time,

⇒dxdt=2t-8

Putting the value of t=4

⇒dxdt=2×4−8=0

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$, where $x>0$, then $x=$

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

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

Solution:
$$
\begin{aligned}
& \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4} \\
& \therefore \tan ^{-1}\left[\frac{2 x+3 x}{1-(2 x)(3 x)}\right]=\frac{\pi}{4} \Rightarrow \tan \frac{\pi}{4}=\frac{5 x}{1-6 x^2}=1 \\
& \therefore 6 x^2+5 x-1=0 \Rightarrow(6 x-1)(x+1)=0 \Rightarrow x=-1, \frac{1}{6}
\end{aligned}
$$
Since $x>0$, we get $x=\frac{1}{6}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1} 3 x$, then $x=$

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

The correct answer is:
$0, \pm \frac{1}{2}$

Solution:
$\begin{aligned}
& \tan ^{-1}(x-1)+\tan ^{-1}(x)+\tan ^{-1}(x+1)=\tan ^{-1} 3 x \\
& \Rightarrow \tan ^{-1}(x-1)+\tan ^{-1}(x)=\tan ^{-1} 3 x-\tan ^{-1}(x+1) \\
& \Rightarrow \tan ^{-1}\left[\frac{(x-1)+x}{1-(x-1)(x)}\right]=\tan ^{-1}\left[\frac{3 x-(x+1)}{1+3 x(x+1)}\right] \\
& \Rightarrow \frac{2 x-1}{1-x^2+x}=\frac{2 x-1}{1+3 x^2+3 x} \\
& \Rightarrow\left(1-x^2+x\right)(2 x-1)=\left(1+3 x^2+3 x\right)(2 x-1) \\
& \text { On simplification } x=0, \pm \frac{1}{2} .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If tan-1⁡xπ≤π6, then the maximum value of 3x is (Use π=3.14)

Options:

The correct answer is:
3.14

Solution:
We have tan-1⁡xπ≤π6
⇒tan⁡tan-1⁡xπ≤tan⁡π6
⇒xπ≤13⇒x≤π3
∴ the maximum value of 3x is π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\text { If } \tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$
then $1-x y-y z-z x$ is equal to

Options:
A 1
B 0
C -1
D 2

The correct answer is:
0

Solution:
$\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow\left(\tan ^{-1} x+\tan ^{-1} y\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)+\tan ^{-1} z=\frac{\pi}{2}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{(x+y)}{1-x y}+z}{1-z\left(\frac{x+y}{1-x y}\right)}\right]=\frac{\pi}{2}$
$\Rightarrow \frac{(x+y)+z(1-x y)}{(1-x y)-z(x+y)}=\tan \frac{\pi}{2}=\infty$
ie, $\quad(1-x y)-z(x+y)=0$
$\Rightarrow \quad 1-x y-z x-y z=0$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\tan \left(\frac{B-C}{2}\right)=x \cot \left(\frac{A}{2}\right)$, find the value of $x$

Options:
A $\frac{c-a}{c+a}$
B $\frac{a-b}{a+b}$
C $\frac{b-c}{b+c}$
D None

The correct answer is:
$\frac{b-c}{b+c}$

Solution:
By the formulae
$\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \left(\frac{A}{2}\right)$

Topic: Mathematics, Properties of Triangles, JEE Main
If tan2⁡x+y+cot2⁡x+y=1-2x-x2, then where, n∈Z

Options:
A x=1,y=nπ±π4-1
B x=-1,y=nπ±π4+1
C x=-1,y=nπ2±π4-1
D x=+1,y=nπ2±π4-1

The correct answer is:
x=-1,y=nπ±π4+1

Solution:
Left hand side is greater than or equal to 2.
⇒1-2x-x2≥2
⇒x+12≤0
⇒x=-1
Hence, the given equation is tan2⁡y-1+cot2⁡y-1=2
⇒tan⁡y-1=±1
⇒y-1=nπ±π4
⇒y=nπ±π4+1
So, x=-1 and y=nπ±π4+1

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\tan \theta=\frac{-4}{3}$, then $\sin \theta=$

Options:
A $-4 / 5$ but not $4 / 5$
B $-4 / 5$ or $4 / 5$
C $4 / 5$ but not $-4 / 5$
D None of these

The correct answer is:
$-4 / 5$ or $4 / 5$

Solution:
Since $\operatorname{cosec}^2 \theta=1+\cot ^2 \theta=1+\frac{9}{16}=\frac{25}{16}$
$\left(\tan \theta=-\frac{4}{3}\right)$
$\sin ^2 \theta=\frac{1}{\operatorname{cosec}^2 \theta}=\frac{16}{25} \Rightarrow \sin \theta= \pm \frac{4}{5},$
Both the values are acceptable, since
$\tan \theta=-\frac{4}{3}$
i.e., $\theta$ lies in $2^{\text {nd }}$ or $4^{\text {th }}$ quadrant.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tanâ¡Î¸=3tanâ¡Ï, then the maximum value of tan2â¡Î¸-Ï is (where, tanâ¡Ï>0)

Options:
A 1
B 13
C 2
D 4

The correct answer is:
13

Solution:
tan⁡θ=3tan⁡ϕ
tan⁡θ-ϕ=tan⁡θ-tan⁡ϕ1+tan⁡θtan⁡ϕ
=2tan⁡ϕ1+3tan2⁡ϕ=2cot⁡ϕ+3tan⁡ϕ
Using AM≥GM

cot⁡ϕ+3tan⁡ϕ2≥3
⇒ 2cot⁡ϕ+3tan⁡ϕ≤13
⇒tan⁡θ-ϕ≤13
⇒tan2⁡θ-ϕ≤13

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tanπ9,x,tan7π18 are in arithmetic progression and tanπ9,y,tan5π18 are also in arithmetic progression, then x-2y is equal to :

Options:
A 4
B 3
C 0
D 1

The correct answer is:
0

Solution:
x=12tanπ9+tan7π18

and 2y=tanπ9+tan5π18

so, x-2y=12tanπ9+tan7π18-tanπ9+tan5π18

⇒x-2y=cotπ9-tanπ92-tan5π18

⇒x-2y=cot2π9-12cotπ9-tan5π18

⇒x-2y=cot2π9-tan5π18

=0

∵tan5π18=cotπ2-5π18=cot2π9∵tan7π18=cotπ2-7π18=cotπ9

∵cot2θ-12cotθ=cot2θ

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (27 Jul Shift 2)
If $\tan \theta=-\frac{1}{\sqrt{10}}$ and $\theta$ lies in the fourth quadrant, then $\cos \theta=$

Options:
A $1 / \sqrt{11}$
B $-1 / \sqrt{11}$
C $\sqrt{\frac{10}{11}}$
D $-\sqrt{\frac{10}{11}}$

The correct answer is:
$\sqrt{\frac{10}{11}}$

Solution:
We have $\tan \theta=-\frac{1}{\sqrt{10}}$, therefore $\theta$ is in IV quadrant. So $\cos \theta=+v e$
$\begin{aligned}
& \text { Now } 1+\tan ^2 \theta=\sec ^2 \theta \Rightarrow 1+\frac{1}{10}=\sec ^2 \theta \\
& \Rightarrow \sec ^2 \theta=\frac{11}{10} \Rightarrow \cos \theta=\sqrt{\left(\frac{10}{11}\right)}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tan15°+1tan75°+1tan105°+tan195°=2a, then the value of a+1a is :

Options:
A 4
B 4-23
C 2
D 5-323

The correct answer is:
4

Solution:
Given:

2a=tan15°+1tan75°+1tan105°+tan195°

⇒2a=tan15°+1tan90°-15°+1tan90°+15°+tan180°+15°

⇒2a=tan15°+1cot15°-1cot15°+tan15°

⇒2a=2tan15°

⇒a=tan15°

⇒a=tan45°-30°

⇒a=1-131+13

⇒a=3-13+1

So,

a+1a=3-13+1+3+13-1

⇒a+1a=3-12+3+122

⇒a+1a=82=4

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If $\tan \theta=\frac{20}{21}, \cos \theta$ will be

Options:
A $\pm \frac{20}{41}$
B $\pm \frac{1}{21}$
C $\pm \frac{21}{29}$
D $\pm \frac{20}{21}$

The correct answer is:
$\pm \frac{21}{29}$

Solution:
$\tan \theta=\frac{20}{21} \Rightarrow \cos \theta= \pm \frac{21}{29}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tan 25 o =a, then the value of tanâ¡205o-tanâ¡115otanâ¡245o+tanâ¡335o in terms of a is

Options:
A 1-a21+a2
B 1-a2a
C 2a1+a2
D 1+a21-a2

The correct answer is:
1+a21-a2

Solution:
tan⁡205o-tan⁡115otan⁡245o+tan⁡335o=tan⁡25o+cot⁡25ocot⁡25o-tan⁡25o
=tan2⁡25o+11-tan2⁡25o=a2+11-a2
=1+a21-a2

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112,0<A,B,C<π2, then A+B is equal to:

Options:
A C
B π−C
C 2π−C
D π2−C

The correct answer is:
C

Solution:
Given, tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112=1+x+x2xx

We know that,

tanA+B=tanA+tanB1−tanAtanB

⇒tanA+B=1xx2+x+1+xx2+x+11−1x2+x+1

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=x2+x+1xx

⇒tanA+B=tanC

⇒A+B=C

Topic: Mathematics, Trigonometric Equations, JEE Main
If tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112,0<A,B,C<π2, then A+B is equal to:

Options:
A C
B π−C
C 2π−C
D π2−C

The correct answer is:
C

Solution:
Given, tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112=1+x+x2xx

We know that,

tanA+B=tanA+tanB1−tanAtanB

⇒tanA+B=1xx2+x+1+xx2+x+11−1x2+x+1

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=x2+x+1xx

⇒tanA+B=tanC

⇒A+B=C

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If $\tan A=\frac{1}{2}$, then $\tan 3 A=$

Options:
A $\frac{9}{2}$
B $\frac{11}{2}$
C $\frac{7}{2}$
D $-\frac{1}{2}$

The correct answer is:
$\frac{11}{2}$

Solution:
We have $\tan A=\frac{1}{2}$
$\Rightarrow \tan 3 A=\frac{3 \tan A-\tan ^3 A}{1-\tan ^2 A}=\frac{3 \cdot \frac{1}{2}-\frac{1}{8}}{1-3 \cdot \frac{1}{4}}=\frac{12-1}{2}=\frac{11}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If tan⁡A and tan⁡B are the roots of the quadratic equation 3x2-10x-25=0, then the value of 3sin2⁡A+B-10sin⁡A+Bcos⁡A+B-25cos2⁡A+B is :

Options:
A -25
B 10
C -10
D 25

The correct answer is:
-25

Solution:
Since, tan⁡A and tan⁡B are roots of the equation 3x2-10x-25=0.

So, tan⁡A+tan⁡B=103 and tan⁡A.tan⁡B=-253

∴tan⁡A+B=tan⁡A+tan⁡B1-tan⁡A.tan⁡B=1031+253=1028=514

So, sinA+B=±5221& cosA+B=±14221

(Note that either both are negative or both positive)

∴3sin2⁡A+B-10sin⁡A+B.cos⁡A+B-25cos2⁡A+B

=3×25221-10×5×14221-25×142221

=252213-28-196=-25

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2018 (15 Apr)
If $\tan A$ and $\tan B$ are the roots of the quadratic equation, $3 x^2-10 x-25=0$ then the value of $3 \sin ^2(A+B)-10 \sin (A+B) \cdot \cos (A+B)-25 \cos ^2$ $(A+B)$ is

Options:
A
25

B
$-25$

C
$-10$

D
10

The correct answer is:
$-25$

Solution:
As $\tan A$ and $\tan B$ are the roots of $3 x^2-10 x-25=0$
So, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}}$ $=\frac{10 / 3}{28 / 3}=\frac{5}{14}$
Now, $\cos ^2(A+B)=-1+2 \cos ^2(A+B)$
$=\frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221}$
$\therefore 3 \sin ^2(A+B)-10 \sin (A+B)$
$=\cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right]$
$=\frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If \( \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha=\cot \alpha-n \cot n \alpha \) then value of \( n \) is

Options:

The correct answer is:
8

Solution:
Consider cotα-tanα=1-tan2αtanα=2 cot 2α ...1

Given relation can be written as

n cot nα=cotα-tanα-2 tan 2α-4 tan 4α

=2 cot2α-2 tan 2α-4 tan 4α

=22 cot 4α-4 tan 4α

=4cot 4α-tan 4α

=42 cot 8α

=8 cot8α

∴ n=8

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)$, then $x=$

Options:
A $\pm \frac{5}{3}$
B $\pm \frac{\sqrt{5}}{3}$
C $\pm \frac{5}{\sqrt{3}}$
D None of these

The correct answer is:
$\pm \frac{\sqrt{5}}{3}$

Solution:
Given that $\tan \left\{\cos ^{-1}(x)\right\}=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
Let $\cot ^{-1} \frac{1}{2}=\phi \Rightarrow \frac{1}{2}=\cot \phi$
$\Rightarrow \sin \phi=\frac{1}{\sqrt{1+\cot ^2 \phi}}=\frac{2}{\sqrt{5}}$
$\begin{aligned}
& \text {Let } \cos ^{-1} x= \theta \Rightarrow \sec \theta=\frac{1}{x} \Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1} \\
& \Rightarrow \tan \theta=\sqrt{\frac{1}{x^2}-1} \Rightarrow \tan \theta=\frac{\sqrt{1-x^2}}{x}
\end{aligned}$
So, $\tan \left\{\cos ^{-1}(x)\right\}=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
$\begin{aligned}
& \Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right) \\
& \Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \sqrt{\left(1-x^2\right) 5}=2 x
\end{aligned}$
$\Rightarrow \sqrt{5-5 x^2}=2 x$ which on squaring gives
$5-5 \mathrm{x}^2=4 \mathrm{x}^2$ Or, $x= \pm \frac{\sqrt{5}}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If tanâ¡k+1Î¸=tanâ¡Î¸, then the set of all the values of Î¸ is

Options:
A {nπ :n∈I}
B n π 2 : n ∈ I
C n Ï k : n â I
D n π 4 : n ∈ I

The correct answer is:
n Ï k : n â I

Solution:
As we know the general solution of the equation
tanx=tanÎ±ââisââgivenââbyââx=nÏ+Î±, nâI
Hence the solution of the equation, tan k + 1 Î¸ = tan Î¸
is given by k + 1 Î¸ = n Ï + Î¸ â k Î¸ = n Ï, n â I
â´ Î¸ â nÏk : n â I

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\tan \theta+\sin \theta=m$ and $\tan \theta-\sin \theta=n$, then

Options:
A $m^2-n^2=4 m n$
B $m^2+n^2=4 m n$
C $m^2-n^2=m^2+n^2$
D $m^2-n^2=4 \sqrt{m n}$

The correct answer is:
$m^2-n^2=4 \sqrt{m n}$

Solution:
$(m+n)=2 \tan \theta, m-n=2 \sin \theta$
$\begin{aligned}
& \therefore m^2-n^2=4 \tan \theta \cdot \sin \theta \\
& 4 \sqrt{m n}=4 \sqrt{\tan ^2 \theta-\sin ^2 \theta}=4 \sin \theta \cdot \tan \theta \\
& \text { From (i) and (ii), } m^2-n^2=4 \sqrt{m n}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3$, then which of the following is equal to 1 ?

Options:
A $\tan 2 \theta$
B $\tan 3 \theta$
C $\tan ^2 \theta$
D $\tan ^3 \theta$

The correct answer is:
$\tan 3 \theta$

Solution:
Given,
$$
\begin{aligned}
& \tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3 \\
& \Rightarrow \tan \theta+\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}+\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}=3 \\
& \Rightarrow \quad \tan \theta+\frac{8 \tan \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad \frac{9 \tan \theta-3 \tan ^3 \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad 3 \tan 3 \theta=3 \Rightarrow \tan 3 \theta=1 \\
&
\end{aligned}
$$
Hence, option (2) is correct.

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\tan \theta+\tan 4 \theta+\tan 7 \theta=\tan \theta \tan 4 \theta \tan 7 \theta$ then the general solution is

Options:
A $\theta=\frac{n \pi}{4}$
B $\theta=\frac{n \pi}{12}$
C $\theta=\frac{n \pi}{6}$
D None of these

The correct answer is:
$\theta=\frac{n \pi}{12}$

Solution:
We have,
$\tan \theta+\tan 4 \theta+\tan 7 \theta=\tan \theta \tan 4 \theta \tan 7 \theta$
$\tan \theta+\tan 4 \theta=\tan \theta \tan 4 \theta \tan 7 \theta-\tan 7 \theta$
$\tan \theta+\tan 4 \theta=-\tan 7 \theta(1-\tan \theta \tan 4 \theta)$
$\frac{\tan \theta+\tan 4 \theta}{1-\tan \theta \tan 4 \theta}=-\tan 7 \theta$
$\tan (\theta+4 \theta)=-\tan 7 \theta$
$\tan 5 \theta=\tan (-7 \theta)$
$\therefore \quad 5 \theta=-7 \theta$
$12 \theta=0$ or $12 \theta=n \pi$
So, $\quad \theta=\frac{n \pi}{12}, \forall n \in I$

Topic: Mathematics, Trigonometric Equations, JEE Main
If ∫tanx1+tanx+tan2xdx=x-KAtan-1Ktanx+1A+C, (C is a constant of integration), then the ordered pair K,A is equal to

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

The correct answer is:
2,3

Solution:
Let I=∫tanx1+tanx+tan2xdx

⇒I=∫1-sec2x1+tanx+tan2xdx

⇒I=x-∫dt1+t+t2

Where, tanx=t⇒sec2xdx=dt

∴ I=x-132tan-1t+1232+C, where C is the constant of integration.

=x-23tan-12tanx+13+C

∴ K=2 & A=3.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2018 (16 Apr Online)