datavorous/entrance-exam-dataset huggingface dataset:
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}$, then the value of $x^2+y^2+z^2+2 x y z$ is equal to

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$1$

Solution:
$\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}$
Put $\sin ^{-1} x=\alpha, \quad \sin ^{-1} y=\beta, \quad \sin ^{-1} z=\gamma$
$\begin{aligned}
& \therefore \alpha+\beta+\gamma=\frac{\pi}{2} \text {, } \\
& \alpha+\beta=\frac{\pi}{2}-\gamma \quad \text { or } \cos (\alpha+\beta)=\cos \left(\frac{\pi}{2}-\gamma\right) \\
&
\end{aligned}$
So,
$\cos \alpha \cos \beta-\sin \alpha \sin \beta=\sin \gamma$
and, we have $\sin \alpha=x \Rightarrow \cos \alpha=\sqrt{1-x^2}$
Similarly, $\quad \cos \beta=\sqrt{1-y^2}$
$\therefore$ From equation (i), we get $\sqrt{1-x^2} \cdot \sqrt{1-y^2}=x y+z$
Squaring both sides, we have $x^2+y^2+z^2+2 x y z=1$.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin \theta=-\frac{1}{\sqrt{2}}$ and $\tan \theta=1$, then $\theta$ lies in which quadrant

Options:
A First
B Second
C Third
D Fourth

The correct answer is:
Third

Solution:
$\begin{aligned}
\sin \theta=-\frac{1}{\sqrt{2}} \text { and } \tan \theta & =1 \\
\Rightarrow \sin \theta & =\sin 225^{\circ} \Rightarrow \theta=225^{\circ}
\end{aligned}$
Since $\sin \theta$ is $-v e$ and $\tan \theta$ is $+v e$ in third quadrant.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sin⁡2A=12 and sin⁡2B=-12, then which of the following is false?

Options:
A sin⁡A+B may be 0
B cos⁡A-B may be 0
C sin⁡A+B or cos⁡A-B is zero
D sin⁡A+B=0

The correct answer is:
sin⁡A+B=0

Solution:
sin⁡2A=12,sin⁡2B=-12
⇒sin⁡2A+sin⁡2B=0
⇒2sin⁡A+Bcos⁡A-B=0
⇒sin⁡A+B=0 or cos⁡A-B=0

Hence, 4th option is false.

Topic: Mathematics, Trigonometric Equations, JEE Main
If sin⁡2β is the geometric mean between sin⁡α and cos⁡α, then cos⁡4β is equal to

Options:
A 2sin2⁡π4-α
B 2cos2⁡π4-α
C 2cos2⁡π2+α
D 2sin2⁡π4+α

The correct answer is:
2sin2⁡π4-α

Solution:
sin2⁡2β=sin⁡αcos⁡α
cos⁡4β=1-2sin2⁡2β
=sin2⁡α+cos2⁡α-2sin⁡αcos⁡α
=sin⁡α-cos⁡α2
=2sinα·12-cosα122
=2sin⁡αcos⁡π4-cos⁡αsin⁡π42
=2sin⁡α-π42=2sin2⁡π4-α

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\sin 2 \theta+\sin 2 \phi=1 / 2$ and $\cos 2 \theta+\cos 2 \phi=3 / 2$, then $\cos ^2(\theta-\phi)=$

Options:
A $3 / 8$
B $5 / 8$
C $3 / 4$
D $5 / 4$

The correct answer is:
$5 / 8$

Solution:
Given,
$\sin 2 \theta+\sin 2 \phi=1 / 2$
and $\cos 2 \theta+\cos 2 \phi=3 / 2$
Square and adding ,
$\therefore\left(\sin ^2 2 \theta+\cos ^2 2 \theta\right)+\left(\sin ^2 2 \phi+\cos ^2 2 \phi\right)$
$\begin{aligned}
& \Rightarrow \cos 2 \theta \cos 2 \phi+\sin 2 \theta \sin 2 \phi=1 / 4 \\
& \Rightarrow \cos (2 \theta-2 \phi)=1 / 4 \Rightarrow \cos ^2(\theta-\phi)=5 / 8
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\sin ^2 x+2 \cos y+x y=0$, then $\frac{d y}{d x}=$

Options:
A $\frac{y+2 \sin x}{2 \sin y+x}$
B $\frac{y+\sin 2 x}{2 \sin y-x}$
C $\frac{y+2 \sin x}{\sin y+x}$
D None of these

The correct answer is:
$\frac{y+\sin 2 x}{2 \sin y-x}$

Solution:
$\begin{aligned} & \sin ^2 x+2 \cos y+x y=0 \\ & \Rightarrow 2 \sin x \cos x-2 \sin y \frac{d y}{d x}+y+x \frac{d y}{d x}=0\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{y+\sin 2 x}{2 \sin y-x}$.

Topic: Mathematics, Differentiation, JEE Main
If sin3â¡Î¸-cos3â¡Î¸sinâ¡Î¸-cosâ¡Î¸-cosâ¡Î¸1+cot2â¡Î¸-2tanâ¡Î¸cotâ¡Î¸=-1 âÎ¸â0,2Ï, then

Options:
A θ∈0,π2-π4
B θ∈π2,π-3π4
C θ∈π,3π2-5π4
D θ∈0,π-π4,π2

The correct answer is:
θ∈0,π-π4,π2

Solution:
Since, sin⁡θ-cos⁡θ≠0⇒tan⁡θ≠1

To define cot⁡θ, θ≠0,π,2π and

to define tan⁡θ,θ≠π2,3π2
⇒θ≠π4,5π4,π2,3π2,0,π,2π
Now, sin2⁡θ+cos2⁡θ+sin⁡θcos⁡θ-sin⁡θcos⁡θ-2=-1
⇒1+sin⁡θcos⁡θ-sin⁡θcos⁡θ=1
⇒cos⁡θsin⁡θ-sin⁡θ=0
⇒cos⁡θ=0 or sin⁡θ=sin⁡θ
⇒θ=π2,3π2 or θ=0,π
⇒θ∈0,π-π4,π2

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\sin \theta=\frac{-4}{5}$ and $\theta$ lies in the third quadrant, then $\cos \frac{\theta}{2}=$

Options:
A $\frac{1}{\sqrt{5}}$
B $-\frac{1}{\sqrt{5}}$
C $\sqrt{\frac{2}{5}}$
D $-\sqrt{\frac{2}{5}}$

The correct answer is:
$-\frac{1}{\sqrt{5}}$

Solution:
Given that $\sin \theta=-\frac{4}{5}$ and $\theta$ lies in the III quadrant.
$\begin{aligned}
& \Rightarrow \cos \theta=\sqrt{1-\frac{16}{25}}= \pm \frac{3}{5} \\
& \cos \frac{\theta}{2}= \pm \sqrt{\frac{1+\cos \theta}{2}}=\sqrt{\frac{1-3 / 5}{2}}= \pm \sqrt{\frac{1}{5}} \\
& \text { But } \cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}} \text {. since } \frac{\theta}{2} \text { will be in II quadrant. } \\
& Hence \cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}} \\
&
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sin6⁡θ+cos6⁡θ+kcos2⁡2θ=1, then k is equal to

Options:
A 12tan2⁡2θ
B 14tan2⁡2θ
C 4cot2⁡2θ
D 34tan2⁡2θ

The correct answer is:
34tan2⁡2θ

Solution:
sin2⁡θ+cos2⁡θsin4⁡θ-sin2⁡θcos2⁡θ+cos4⁡θ+kcos2⁡2θ=1
or sin2⁡θ+cos2⁡θ2-3sin2⁡θcos2⁡θ+kcos2⁡2θ=1
or 1-3sin2⁡θcos2⁡θ+kcos2⁡2θ=1
or kcos2⁡2θ=3sin2⁡θcos2⁡θ
or kcos2⁡2θ=342sin⁡θcos⁡θ2
or kcos2⁡2θ=34sin2⁡2θ
or k=34tan2⁡2θ

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\sin \theta=\frac{24}{25}$ and $\theta$ lies in the second quadrant, then $\operatorname{sec} \theta+\tan \theta=$

Options:
A -3
B -5
C -7
D -9

The correct answer is:
-7

Solution:
$\begin{aligned}
\sin \theta=\frac{24}{25} \Rightarrow \cos \theta & =\frac{-7}{25}, \tan \theta=\frac{-24}{7} \\
\therefore \quad \sec \theta+\tan \theta & =\frac{-25}{7}+\frac{-24}{7}=-7
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sin⁡A and cos⁡A are the roots of the equation 4x2-3x+a=0, sin⁡A+cos⁡A+tan⁡A+cot⁡A+sec⁡A+cosecA=7 and 0<A<π2, then the value of a must be

Options:
A 725
B 257
C 2825
D 2528

The correct answer is:
2825

Solution:
sinâ¡A and cosâ¡A are the roots of the equation 4x2-3x+a=0
⇒sinâ¡A+cosâ¡A=34 and sinâ¡Acosâ¡A=a4
Also, sinâ¡A+cosâ¡A+tanâ¡A+cotâ¡A+secâ¡A+cosecA=7
⇒sinâ¡A+cosâ¡A+1sinâ¡Acosâ¡A+sinâ¡A+cosâ¡Asinâ¡Acosâ¡A=7
⇒34+1a4+34a4=7
⇒34+4a+3a=7
⇒7a=7-34=254
⇒a=2825

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\sin A+\sin B=a$ and $\cos A+\cos B=b$, then $\cos (A+B)$ equals?

Options:
A $\frac{a^2+b^2}{b^2-a^2}$
B $\frac{2 a b}{a^2+b^2}$
C $\frac{b^2-a^2}{a^2+b^2}$
D $\frac{a^2-b^2}{a^2+b^2}$

The correct answer is:
$\frac{b^2-a^2}{a^2+b^2}$

Solution:
Given, $\sin A+\sin B=a$
$$
\cos A+\cos B=b
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{array}{ll}
& \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{a}{b} \\
\Rightarrow \quad & \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{a}{b} \\
\Rightarrow \quad & \tan \frac{A+B}{2}=\frac{a}{b} \\
\text { Now, } \quad & \cos (A+B)=\frac{1-\tan ^2\left(\frac{A+B}{2}\right)}{1+\tan ^2\left(\frac{A-B}{2}\right)}
\end{array}
$$
$$
\left(\because \text { using } \cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)
$$
$$
\begin{aligned}
& \Rightarrow \quad \cos (A+B)=\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}} \\
& \therefore \quad \cos (A+B)=\frac{b^2-a^2}{b^2+a^2}
\end{aligned}
$$
[Using Eq. (iii)]

Topic: Mathematics, Trigonometric Equations, JEE Main
If sinθ+cosθ=12, then 16sin2θ+cos4θ+sin6θ is equal to:

Options:
A 23
B -27
C -23
D 27

The correct answer is:
-23

Solution:
sinθ+cosθ=12

sin2θ+cos2θ+2sinθcosθ=14⇒sin2θ=-34

Now:

cos4θ=1-2sin22θ

=1-2-342

=1-2×916=-18

And sin6θ=3sin2θ-4sin32θ

=3-4sin22θ·sin2θ

=3-4916·-34

=34×-34=-916

So, 16sin2θ+cos4θ+sin6θ

=16-34-18-916=-23

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2021 (27 Jul Shift 1)
If $\sin \theta+\cos \theta=1$, then $\sin \theta \cos \theta=$

Options:
A 0
B 1
C 2
D $1 / 2$

The correct answer is:
0

Solution:
$\sin \theta+\cos \theta=1$
Squaring on both sides, we get
$\begin{aligned}
& \Rightarrow \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=1 \\
& \therefore \sin \theta \cos \theta=0 .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sinâ¡Î¸=cos2â¡Î¸, then the value of cos12â¡Î¸+3cos10â¡Î¸+3cos8â¡Î¸+cos6â¡Î¸-1 is equal to

Options:

The correct answer is:
0

Solution:
sin⁡θ=cos2⁡θ⇒sin⁡θ+sin2⁡θ=1
Now, cos6⁡θcos6⁡θ+3cos4⁡θ+3cos2⁡θ+1-1
=cos6⁡θcos2⁡θ+13-1
=cos2⁡θ3sin⁡θ+13-1
=sin3⁡θsin⁡θ+13-1
=sin2⁡θ+sin⁡θ3-1=1-1=0

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\sin \theta+\cos \theta=p$ and $\sin ^3 \theta+\cos ^3 \theta=q$, then $p\left(p^2-3\right)$ is equal to

Options:
A $q$
B $2 q$
C $-q$
D $-2 q$

The correct answer is:
$-2 q$

Solution:
$\begin{aligned} & \text { Given, } \quad \sin \theta+\cos \theta=p \\ & \text { and } \quad \sin ^3 \theta+\cos ^3 \theta=q \\ & \Rightarrow \quad(\sin \theta+\cos \theta) \\ & \quad\left(\sin ^2 \theta-\sin \theta \cdot \cos \theta+\cos ^2 \theta\right)=q\end{aligned}$
$$
\Rightarrow \quad p(1-\sin \theta \cdot \cos \theta)=q
$$
[From Eq. (i) and $\sin ^2 \theta+\cos ^2 \theta=1$ ]
$$
\begin{aligned}
\Rightarrow & 1-\sin \theta \cdot \cos \theta=\frac{q}{p} \\
\Rightarrow \quad & \sin \theta \cdot \cos \theta=1-\frac{q}{p}
\end{aligned}
$$
On squaring both sides of Eq. (i), we get
$$
\begin{aligned}
& \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cdot \cos \theta=p^2 \\
& \Rightarrow \quad 1+2\left(1-\frac{q}{p}\right)=p^2 \quad \text { [from Eq. (iii)] } \\
& \Rightarrow \quad p+2(p-q)=p^3 \\
& \Rightarrow \quad 3 p-2 q=p^3 \\
& \Rightarrow \quad p^3-3 p=-2 q \\
& \Rightarrow \quad p\left(p^2-3\right)=-2 q \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\sin \theta+\operatorname{cosec} \theta=2$, then $\sin ^2 \theta+\operatorname{cosec}^2 \theta=$

Options:
A 1
B 4
C 2
D None of these

The correct answer is:
2

Solution:
$\begin{aligned}
\sin ^2 \theta+\operatorname{cosec}^2 \theta & =(\sin \theta+\operatorname{cosec} \theta)^2-2 \sin \theta \operatorname{cosec} \theta \\
& =(2)^2-2=4-2=2, \text { since }(\sin \theta+\operatorname{cosec} \theta)=2 .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sin⁡cot-1⁡1-x=cos⁡tan-1⁡-x, then x is

Options:
A 1
B 12
C 0
D -12

The correct answer is:
12

Solution:
sin⁡cot-1⁡1-x=cos⁡tan-1⁡-x
Let, cot-1⁡1-x=θ
⇒cot⁡θ=1-x
⇒sin⁡θ=12-2x+x2
Let, tan-1⁡-x=ϕ
⇒tan⁡ϕ=-x⇒cos⁡ϕ=11+x2
∴sin⁡θ=cos⁡ϕ
⇒12-2x+x2=11+x2
⇒1+x2=2-2x+x2
⇒x=12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin \left(\cot ^{-1}(x+1)=\cos \left(\tan ^{-1} x\right)\right.$, then $x=$

Options:
A $-\frac{1}{2}$
B $\frac{1}{2}$
C $0$
D $\frac{9}{4}$

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

Solution:
$\sin \left[\cot ^{-1}(x+1)\right]=\sin \left(\sin ^{-1} \frac{1}{\sqrt{x^2+2 x+2}}\right)$
$=\frac{1}{\sqrt{x^2+2 x+2}}$
$\begin{aligned} & \cos \left(\tan ^{-1} x\right)=\cos \left(\cos ^{-1} \frac{1}{\sqrt{1+x^2}}\right)=\frac{1}{\sqrt{1+x^2}} \\ & \text { Thus, } \frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}} \\ & \Rightarrow x^2+2 x+2=1+x^2 \Rightarrow x=-\frac{1}{2} .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin x=-\frac{3}{5}$, where $\pi < x < \frac{3 \pi}{2}$, then $80\left(\tan ^2 x-\cos x\right)$ is equal to

Options:
A 108
B 109
C 18
D 19

The correct answer is:
109

Solution:
$\begin{aligned} & \sin x=\frac{-3}{5}, \pi < x < \frac{3 \pi}{2} \\ & \tan x=\frac{3}{4} \cos x=-\frac{4}{5} \\ & 80\left(\tan ^2 x-\cos x\right) \\ & =80\left(\frac{9}{16}+\frac{4}{5}\right)=45+64=109\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2024 (08 Apr Shift 1)
If $\sin x=\frac{-24}{25}$, then the value of $\tan x$ is

Options:
A $\frac{24}{25}$
B $\frac{-24}{7}$
C $\frac{25}{24}$
D None of these

The correct answer is:
$\frac{-24}{7}$

Solution:
$\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\left(\frac{-24}{25}\right)^2}=\frac{7}{25}$
$\Rightarrow \tan x=\frac{\sin x}{\cos x}=\frac{-24}{7}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If sinâ¡x+cosâ¡x=72, where xâ0,Ï4, then the value of tanx2 is equal to

Options:
A 3-73
B 7-23
C 4-74
D 5-32

The correct answer is:
7-23

Solution:
sin⁡x+cos⁡x=72
⇒2tanx21+tan2x2+1-tan2x21+tan2x2=72
⇒4tanx2+2-2tan2x2=7+7tan2x2
⇒2+7tan2x2-4tanx2+7-2=0
⇒tanx2=4±16-42+77-222+7
=4±16-4×322+7
tanx2=2±12+7=12+7 or 32+7
tanx2=7-23 or 7-2
But, as x2∈0,π8
⇒tanx2=7-23

Topic: Mathematics, Trigonometric Equations, JEE Main
If sinâ¡x+cosâ¡y=13 and cosâ¡x+sinâ¡y=34, then the value of tanâ¡x-y2 is equal to

Options:
A 513
B 1213
C -513
D -512

The correct answer is:
-513

Solution:
sin⁡x+sin⁡π2-y=13
⇒2sin⁡π4+x-y2cos⁡x+y2-π4=13
cos⁡x+cos⁡π2-y=34
⇒2cos⁡π4+x-y2cos⁡x+y2-π4=34
On dividing we get,
tan⁡π4+x-y2=49
⇒1+tan⁡x-y21-tan⁡x-y2=49
Using componendo and dividendo, we get
22tan⁡x-y2=13-5⇒tan⁡x-y2=-513

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\sin (x+y)+\cos (x+y)=\log (x+y)$, then $\frac{d^{2} y}{d x^{2}}=$

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

The correct answer is:
0

Solution:
Given that, $\sin (x+y)+\cos (x+y)=\log (x+y)$
On differentiating w.r.t. $x$,
$\begin{aligned}
&\cos (x+y) \cdot\left(1+\frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right) \\
&=\frac{1}{(x+y)}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow\left(1+\frac{d y}{d x}\right)\left\{\cos (x+y)-\sin (x+y)-\frac{1}{(x+y)}\right\}=0 \\
&\Rightarrow \quad 1+\frac{d y}{d x}=0
\end{aligned}$
Again differentiating w.r.t. ' $x$ '
$0+\frac{d^{2} y}{d x^{2}}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=0$

Topic: Mathematics, Differentiation, JEE Main
If sinyx=loge|x|+α2 is the solution of the differential equation xcosyxdydx=ycosyx+x and y(1)=π3, then α2 is equal to

Options:
A 3
B 12
C 4
D 9

The correct answer is:
3

Solution:
Given: xcosyxdydx=ycosyx+x

⇒cosyxdydx=yxcosyx+1

Putting, y=vx

⇒dydx=v+xdvdx

⇒cosvv+xdvdx=vcosv+1

⇒v+xdvdx=v+1cosv

⇒xdvdx=1cosv

⇒cosvdv=dxx

⇒∫cosvdv=∫dxx

⇒sinv=logx+c

⇒sinyx=logx+c

Now, f1=π3

⇒sinπ3=log1+c

⇒32=c

⇒sinyx=logx+32

So, on comparing we get,

⇒α=3

⇒α2=3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (29 Jan Shift 2)
If $\sin y=x \sin (a+y)$, then $\frac{d y}{d x}$ is equal to :

Options:
A $\frac{\sin \sqrt{a}}{\sin (a+y)}$
B $\frac{\sin ^{2}(a+y)}{\sin a}$
C $\sin (a+y)$
D None of these

The correct answer is:
$\frac{\sin ^{2}(a+y)}{\sin a}$

Solution:
Given, $\sin y=x \sin (a+y)$
$$
\Rightarrow \quad x=\frac{\sin y}{\sin (a+y)}
$$
On differentiating w.r. to $y$, we get $\frac{d x}{d y}=\frac{d}{d y}\left[\frac{\sin y}{\sin (a+y)}\right]$
$$
\begin{array}{l}
=\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^{2}(a+y)} \\
=\frac{\sin a}{\sin ^{2}(a+y)} \Rightarrow \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}
\end{array}
$$

Topic: Mathematics, Differentiation, JEE Main
If $\sin y=x \sin (a+y)$, then $\frac{d y}{d x}=$

Options:
A $\frac{\sin ^2(a+y)}{\sin (a+2 y)}$
B $\frac{\sin ^2(a+y)}{\cos (a+2 y)}$
C $\frac{\sin ^2(a+y)}{\sin a}$
D $\frac{\sin ^2(a+y)}{\cos a}$

The correct answer is:
$\frac{\sin ^2(a+y)}{\sin a}$

Solution:
$\begin{aligned} \sin y= & x \sin (a+y) \Rightarrow x=\frac{\sin y}{\sin (a+y)} \\ \Rightarrow 1 & =\frac{\cos y \cdot \frac{d y}{d x} \cdot \sin (a+y)-\sin y \cos (a+y) \frac{d y}{d x}}{\sin ^2(a+y)} \\ & =\frac{\frac{d y}{d x} \cdot \sin (a+y-y)}{\sin ^2(a+y)} \Rightarrow \frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a},\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If six identical cells each having an e.m.f. of $6 \mathrm{~V}$ are connected in parallel, the e.m.f. of the combination is

Options:
A $1 \mathrm{~V}$
B $36 \mathrm{~V}$
C $\frac{1}{6} \mathrm{~V}$
D $6 \mathrm{~V}$

The correct answer is:
$6 \mathrm{~V}$

Solution:
In parallel combination $E_{e q}=E=6 \mathrm{~V}$

Topic: Physics, Current Electricity, JEE Main
If six students, including two particular students $A$ and $B$, stand in a row, then the probability that $A$ and $B$ are separated with one student in between them is

Options:
A
$\frac{8}{15}$

B
$\frac{4}{15}$

C
$\frac{2}{15}$

D
$\frac{1}{15}$

The correct answer is:
$\frac{4}{15}$

Solution:
Consider a group of three students $A, B$ and an other student in between $A$ and $B$. Choice for a student between $A$ and $B$ is 4. $A$ and $B$ can interchange their places in the group in 2 ways.
Now the group of three students (student $A$, student $B$ and a student in between $A$ and $B$ ) and the remaining 3 students can be stand in a row in 4 ! ways.
Hence total number of ways to stand in a row such that $A$ and $B$ are separated with one student in between them $=4 \times 2 \times 4$ !
Now total number of ways to stand 6 student stand in a row without any restriction $=6 !$
Hence required probability
$$
=\frac{4 \times 2 \times 4 !}{6 !}=\frac{4 \times 2}{6 \times 5}=\frac{4}{15}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2012 (19 May Online)
If sixth term of a H.P. is $\frac{1}{61}$ and its tenth term is $\frac{1}{105}$, then first term of that H.P. is

Options:
A $\frac{1}{28}$
B $\frac{1}{39}$
C $\frac{1}{6}$
D $\frac{1}{17}$

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

Solution:
$T_6 \quad=\frac{1}{61} \Rightarrow T_6$ of H.P. $\quad$ A.P. $=61$
and $T_{10}$ of H.P. $=\frac{1}{105} \Rightarrow T_{10}$ of A.P. = 105,
so, $a+5 d=61 \ldots(i)$ and $a+9 d=105 \ldots(ii)$
From (i) and (ii), $a=6$
Therefore first term of H.P. minimise
$=\frac{1}{a}=\frac{1}{6}$

Topic: Mathematics, Sequences and Series, JEE Main
If slope of a tangent to the curve $x y+a x+b y=0$ at the point $(1,1)$ on it is 2 , then $a-b$ is

Options:
A 3
B 1
C 2
D -1

The correct answer is:
3

Solution:
Given curve is
$\begin{aligned}
& x y+\mathrm{a} x+\mathrm{b} y=0 \\
\therefore \quad & \text { Slope }=2=\frac{\mathrm{d} y}{\mathrm{~d} x} \\
\therefore \quad & x y+\mathrm{a} x+\mathrm{b} y=0
\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& x \frac{\mathrm{d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \therefore \quad(x+\mathrm{b}) \frac{\mathrm{d} y}{\mathrm{~d} x}=-(y+\mathrm{a}) \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-(y+\mathrm{a})}{x+\mathrm{b}} \\
& \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(1,1)}=2 \\
& \therefore \quad 2=\frac{-(1+\mathrm{a})}{1+\mathrm{b}} \\
& \therefore \quad \mathrm{a}+2 \mathrm{~b}=-3 ... (i)
\end{aligned}$
Since $(1,1)$ lies on $x y+a x+b y=0$, we get $\mathbf{a}+\mathbf{b}=-1$ ... (ii)
Solving (i), (ii), we get
$\begin{aligned}
& a=1, b=-2 \\
\therefore \quad & a-b=1-(-2)=3
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta \mathrm{T}_{\mathrm{f}}\right)$, when $0.01 \mathrm{~mol}$ of sodium sulphate is dissolved in $1 \mathrm{~kg}$ of water, is $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

Options:
A
$0.0372 \mathrm{~K}$

B
$0.0558 \mathrm{~K}$

C
$0.0744 \mathrm{~K}$

D
$0.0186 \mathrm{~K}$

The correct answer is:
$0.0558 \mathrm{~K}$

Solution:
Vant Hoff's factor (i) for $\mathrm{Na}_2 \mathrm{SO}_4=3$
$$
\begin{aligned}
\therefore \Delta \mathrm{T}_{\mathrm{f}}= & \text { (i) } \mathrm{k}_{\mathrm{f}} \mathrm{m} \\
& =3 \times 1.80 \times \frac{0.01}{1}=0.0558 \mathrm{~K}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2010
If solubility product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S , then which of the following relation between S and Ksp is correct?

Options:
A S=Ksp14416
B S=Ksp691217
C S=Ksp92919
D S=Ksp21617

The correct answer is:
S=Ksp691217

Solution:
Zr3PO44s⇌3Zr+43saq+4PO43-4saq
Ksp=Zr+43PO43-4
=3s34s4
Ksp=6912 S7
s=Ksp/691217

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss's law all the charge resides on the surface. Suppose now that Colomb's force between two charges varies as $1 / \mathrm{r}^{3}$. Then, for a charged solid metallic sphere

Options:
A field inside will be zero and charge density inside will be zero.
B field inside will not be zero and charge density inside will not be zero.
C field inside will not be zero and charge density inside will be zero.
D field inside will be zero and charge density inside will not be zero.

The correct answer is:
field inside will be zero and charge density inside will not be zero.

Solution:
If coloumb's force $\propto \frac{1}{\mathrm{r}^{3}}$ gauss's law is not valid $\therefore \phi \neq \frac{\mathrm{q}_{\mathrm{en}}}{\varepsilon_{0}}$
For static condition $\mathrm{E}=0$ in both of conductor $\therefore \phi$ through a Gaussian surface just under the surface of conductor $=0$ but as $\phi=\frac{\mathrm{q}_{\mathrm{en}}}{\varepsilon_{0}}$ is not valid
So $\mathrm{q}_{\mathrm{en}}=0$ is not correct statement. Some charge will present insider bulk of conductor.

Topic: Physics, Electrostatics, KVPY, KVPY 2017 (5 Nov SB/SX)
If some three consecutive coefficients in the binomial expansion of x+1n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is:

Options:
A 227
B 964
C 625
D 232

The correct answer is:
232

Solution:
From the given condition

nCr:nCr+1:nCr+2=2:15:70

⇒ nCr nCr+1=215 and nCr+1 nCr+2=1570

⇒ n!n-r!·r!n!n-r-1!·r+1!=215 and n!n-r-1!·r+1!n!n-r-2!·r+2!=1570

⇒n-r-1!·r+1!n-r!·r!=215 and n-r-2!·r+2!n-r-1!·r+1!=314

⇒n-r-1!·r+1·r!n-r·n-r-1!·r!=215 and n-r-2!·r+2·r+1!n-r-1·n-r-2!·r+1!=314

⇒ r+1n-r=215 and r+2n-r-1=314

⇒ 17r=2n-15 and 17r=3n-31

⇒3n-31=2n-15, ⇒n=16 and r=1

Hence, average = nCr+nCr+1+nCr+23

= 16C1+16C2+16C33

=232.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If sound travels in air with the speed of 340 m s-1, then the number of tones present in an open organ pipe of length 2 m for a maximum frequency of 1200 Hz, are

Options:
A 17
B 11
C 9
D 14

The correct answer is:
14

Solution:
Given, speed of sound, v=340 m s-1
Length of open organ pipe, l=2 m

Frequency, f=1200 Hz
For open organ pipe,

Fundamental frequency, f0=v2l=3402×2=85

∴ Number of tones present in the open organ pipe
=ff0=120085=14.11≃14

Topic: Physics, Waves and Sound, JEE Main
If speed (V), acceleration (A) and force (F) are considered

as fundamental units, the dimension of Young's modulus

will be :

Options:
A $\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{-2}$
B $\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{2}$
C $\mathrm{V}^{-4} \mathrm{~A}^{-2} \mathrm{~F}$
D $\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}$

The correct answer is:
$\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}$

Solution:
Let [Y] = [V] $^{a}[\mathrm{~F}]^{\mathrm{b}}[\mathrm{A}]^{\mathrm{c}}$

$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{LT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{MLT}^{-2}\right]^{\mathrm{b}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{c}}$

$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{b} \mathrm{~L}^{\mathrm{a}+b+\mathrm{c}} \mathrm{T}^{-\mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}}\right]$

Comparing power both side of similar terms we get,

$b=1, a+b+c=-1,-a-2 b-2 c=-2$

solving above equations we get

$a=-4, b=1, c=2$

so $[\mathrm{Y}]=\left[\mathrm{V}^{-4} \mathrm{FA}^{2}\right]=\left[\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}\right]$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If speed V, acceleration A and force F are considered as fundamental units, the dimension of Young's modulus will be:

Options:
A V-2A2F2
B V-2A2F-2
C V-4A-2F
D V-4A2F

The correct answer is:
V-4A2F

Solution:
Let Y=fV, F, A
Y=KVxFyAz, K→ Unit less
Y=VxFyAz
ML-1T-2=LT-1xMLT-2yLT-2z
ML-1T-2=MyLx+y+zT-x-2y-2z
⇒ y=1, x+y+z=-1; -x-2y-2z=-2
x+z=-2 x+2y+2z=2
z=2, x=-4 x+2z=0
Y=V-4A2F

Topic: Physics, Units and Dimensions, JEE Main
If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be :

Options:
A FA2V-1
B FA2V-3
C FA2V-2
D FA-1V0

The correct answer is:
FA-1V0

Solution:
Y∝FaVbAc Y=FA

MLT-2L2∝M1L1T-2aL1T-1bL2c

M1L-1T-2∝MaLa+b+2cT-2a-b

a + b + 2c = –1

–2a + b = –2

a = 1, b = 0, c = –1

Y=F1V0A-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2020 (02 Sep Shift 1)