datavorous/entrance-exam-dataset huggingface dataset:
If x=sin2tan-13 and y=sin12tan-143, then

Options:
A 2x=1-y
B x2=1-2y
C x2=1+y
D y2=2x-1

The correct answer is:
y2=2x-1

Solution:
Let, x=sin⁡2θ (where tan⁡θ=3)
⇒x=2tan⁡θ1+tan2θ=61+9=35
If α=tan-143

⇒y=sinα2=121-cosα
=121-35=15
∴y2=15⇒y2=2x-1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x$ satisfies the inequality $\log _{25} x^2+\left(\log _5 x\right)^2 < 2$, then $x$ belongs to

Options:
A $\left(\frac{1}{5}, 5\right)$
B $\left(\frac{1}{25}, 5\right)$
C $\left(\frac{1}{5}, 25\right)$
D $\left(\frac{1}{25}, 25\right)$

The correct answer is:
$\left(\frac{1}{25}, 5\right)$

Solution:
Domain $=(0, \infty)$
$\begin{aligned}
&\log _5 x+\left(\log _5 x\right)^2-2 < 0 \\
&\Rightarrow\left(\log _5 x-1\right)\left(\log _5 x+2\right) < 0 \\
&\therefore-2 < \log _5 x < 1 \\
&\therefore \frac{1}{25} < x < 5
\end{aligned}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2022
If x satisfies the inequality tan-1x2+3tan-1x-4>0, then the complete set of values of x is

Options:
A -tan⁡4,π4
B -∞,tan⁡4∪π4,∞
C tan⁡1,∞
D tan⁡4,tan⁡1

The correct answer is:
tan⁡1,∞

Solution:
Let, tan-1x=t

the given inequality becomes t2+3t-4>0ât+4t-1>0
â´t<-4 or t>1
tan-1x<-4 (not possible) or tan-1x>1
âx>tanâ¡1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If x=secθ-cosθ and y=secnθ-cosnθ, then x2+4dydx2=

Options:
A n(y+4)
B n2y2+4
C n(y+2)
D n2y2+2

The correct answer is:
n2y2+4

Solution:
x=secθ-cosθ

⇒dxdθ=secθtanθ+sinθ

y=secnθ-cosnθ

⇒dydθ=nsecnθtanθ+ncosn-1θsinθ

dydx=dydθdxdθ=nsecnθtanθ+cosn-1θsinθsecθtanθ+sinθ

dydx=n1cosn+1θ+cosn-1θ1cos2θ+1

x2+4dydx2=1-cos2θcosθ2+4n21+cos2nθ21+cos2θ2cos2n-2θ

=n21+cos2nθ2cos2nθ

=n21+cos4nθ+2cos2nθcos2nθ

=n21-cos2nθ2+4(cosnθ)2(cosnθ)2

=n2secnθ-cosnθ2+4

=n2y2+4

Topic: Mathematics, Differentiation, JEE Main
If x=sec⁡t+tan⁡t and y=sec⁡t-tan⁡t, where t is a parameter, then the value of dydx when x=13 is

Options:
A 0
B -3
C 3
D 13

The correct answer is:
-3

Solution:
âµxy=1⇒y=1x
dydx=-1x2
at x=13,dydx=-3

Topic: Mathematics, Differentiation, JEE Main
If x=sin⁡2tan-1⁡23 and y=sin⁡12tan-1⁡125, then

Options:
A x=1-y
B x2=1-2y
C x2=1+y
D y2=x3

The correct answer is:
y2=x3

Solution:
Let x=sin⁡2θ, where tan⁡θ=23
⇒x=2tan⁡θ1+tan2⁡θ=431+12=4313
If α=tan-1⁡125,y=sin⁡α2=121-cos⁡α
=121-513=213
∴y2=413⇒y2=x3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$,then

Options:
A $x>y$
B $x=y$
C $x=0=y$
D $x < y$

The correct answer is:
$x>y$

Solution:
We have, $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$ $\Rightarrow x=\sin \left(2 \tan ^{-1} 2\right)$
Let $\tan ^{-1} 2-\alpha \Rightarrow \tan \alpha=2$
$\therefore \quad x=\sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan ^2 \alpha}=\frac{2(2)}{1+(2)^2}=\frac{4}{5}$
Now, $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$
Let $\tan ^{-1} \frac{4}{3}=\beta$
$\tan \beta=\frac{4}{3}, \cos \beta=\frac{1}{\sqrt{1+\tan ^2 \beta}}=\frac{3}{5}$
Then, $y=\sin \left(\frac{\beta}{2}\right)$
$\begin{aligned}
& =\sqrt{\frac{1-\cos \beta}{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{1}{5}} \\
& \therefore \quad x>y
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$ and

$x \sin \theta=y \cos \theta,$ then $x^{2}+y^{2}=$

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

The correct answer is:
1

Solution:
$x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

and $x \sin \theta=y \cos \theta$

Equation (i) may be written as $x \sin \theta \cdot \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta=\sin \theta \cos \theta \therefore y=\sin \theta \quad \ldots$

Putting the value of y from

(iii) in (ii), we get $\mathrm{x} \sin \theta=\sin \theta \cdot \cos \theta \Rightarrow \mathrm{x}=\cos \theta \quad \ldots$ (iv)

Squaring (iii) and (iv) and adding, we get $\mathrm{x}^{2}+\mathrm{y}^{2}=\cos ^{2} \theta+\sin ^{2} \theta=1$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\int x \sin x d x=-x \cos x+A$, then $A=$

Options:
A $\sin x+$ constant
B $\cos x+$ constant
C Constant
D None of these

The correct answer is:
$\sin x+$ constant

Solution:
Since, $\int x \sin x d x=-x \cos x+A$$
\Rightarrow-x \cos x+\sin x+\text { constant }=-x \cos x+A$
Equating it, we get $A=\sin x+$ constant.

Topic: Mathematics, Indefinite Integration, JEE Main
If x takes all permissible negative values, then sin-1â¡x is equal to

Options:
A -cos-1⁡1-x2
B cos-1⁡x2-1
C π-cos-1⁡1-x2
D cos-1⁡1-x2

The correct answer is:
-cos-1⁡1-x2

Solution:
Let x=-y, y>0
∴sin-1⁡x=sin-1⁡-y
=-sin-1⁡y
=-cos-1⁡1-y2
=-cos-1⁡1-x2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x$ takes non-positive permissible value, then $\sin ^{-1} x=$

Options:
A $\cos ^{-1} \sqrt{1-x^2}$
B $-\cos ^{-1} \sqrt{1-x^2}$
C $\cos ^{-1} \sqrt{x^2-1}$
D $\pi-\cos ^{-1} \sqrt{1-x^2}$

The correct answer is:
$-\cos ^{-1} \sqrt{1-x^2}$

Solution:
Let $\sin ^{-1} x=y$. Then $x=\sin y$
Since $-1 \leq x \leq 0$, therefore $\frac{-\pi}{2} \leq \sin ^{-1} x \leq 0$ and so $\frac{-\pi}{2} \leq y \leq 0$
We have $\cos y=\sqrt{1-\sin ^2 y}$
$\Rightarrow \cos y=\sqrt{1-x^2}$, for $0 \leq y \leq \pi$ ....(i)
Now $-\frac{\pi}{2} \leq y \leq 0 \Rightarrow \frac{\pi}{2} \geq-y \geq 0$
$\Rightarrow \cos (-y)=\sqrt{1-x^2}$
{from (i)}
$\Rightarrow-y=\cos ^{-1} \sqrt{1-x^2} \Rightarrow y=-\cos ^{-1} \sqrt{1-x^2}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x=\tan 15^{\circ}, y=\operatorname{cosec} 75^{\circ}$ and $z=4 \sin 18^{\circ}$, then :

Options:
A $x < y < z$
B $y < z < x$
C $z < x < y$
D $x < z < y$

The correct answer is:
$x < y < z$

Solution:
$\because \quad x=\tan 15^{\circ}=\tan\left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}$
$=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
and $y=\operatorname{cosec} 75^{\circ}$
$=\frac{1}{\sin \left(45^{\circ}+30^{\circ}\right)}$
$=\frac{1}{\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}}$
$=\frac{1}{\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{2 \sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2}$
and $\quad z=4 \sin 18^{\circ}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1$
It is clear from above that
$(2-\sqrt{3}) < (\sqrt{6}-\sqrt{2}) < (\sqrt{5}-1)$
$\Rightarrow \quad x < y < z$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $x, v$ and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $\mathrm{T}$, then, which of the following does not change with time?

Options:
A
$a^2 T^2+4 \pi^2 v^2$

B
$\frac{a T}{x}$

C
$a T+2 \pi v$

D
$\frac{a T}{v}$

The correct answer is:
$\frac{a T}{x}$

Solution:
$$
\frac{\text { aT }}{\mathrm{X}}=\frac{\omega^2 \mathrm{XT}}{\mathrm{X}}=\frac{4 \pi^2}{\mathrm{~T}^2} \times \mathrm{T}=\frac{4 \pi^2}{\mathrm{~T}}=\text { constant. }
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2009
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T , then which of the following do not change with time?

Options:
A aTv
B aT+2πv
C a2T2+4π2v2
D aT

The correct answer is:
aTv

Solution:
∵ In option ,aTv=ms2×sms=m×s2m×s2=M0L0T0
So, according to the result above relation does not depend on time.

Topic: Physics, Units and Dimensions, JEE Main
If $x+\frac{1}{x}=2 \cos \theta$, then $x$ is equal to

Options:
A $\cos \theta+i \sin \theta$
B $\cos \theta-i \sin \theta$
C $\cos \theta \pm i \sin \theta$
D $\sin \theta \pm i \cos \theta$

The correct answer is:
$\cos \theta \pm i \sin \theta$

Solution:
$\begin{aligned} & x+\frac{1}{x}=2 \cos \theta \Rightarrow x^2-2 x \cos \theta+1=0 \\ & \Rightarrow x=\frac{2 \cos \theta \pm \sqrt{4 \cos ^2 \theta-4}}{2} \Rightarrow x=\cos \theta \pm i \sin \theta\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $x+\frac{1}{x}=2 \cos \alpha$, then $x^{n}+\frac{1}{x^{n}}$ is equal to

Options:
A $2^{n} \cos \alpha$
B $2^{n} \cos n \alpha$
C $2 i \sin n \alpha$
D $2 \cos n \alpha$

The correct answer is:
$2 \cos n \alpha$

Solution:
We have,
$x+\frac{1}{x}=2 \cos \alpha$
$\Rightarrow \quad x^{2}+1=2 x \cos \alpha$
$\Rightarrow \quad x^{2}-(2 \cos \alpha) x+1=0$
$\Rightarrow \quad x=\frac{2 \cos \alpha \pm \sqrt{4 \cos ^{2} \alpha-4}}{2}$
$=\frac{2 \cos \alpha \pm \sqrt{-4 \sin ^{2} \alpha}}{2}$
$=\frac{2 \cos \alpha \pm 2(\sin \alpha) i}{2}=\cos \alpha \pm(\sin \alpha) i$
Now, $\quad x^{n}=(\cos \alpha \pm(\sin \alpha) i)^{n}=\cos n \alpha \pm i \sin n \alpha$
and $\frac{1}{x^{n}}=(\cos \alpha \pm(\sin \alpha) i)^{-n}=\cos n \alpha \mp i \sin n \alpha$
$\therefore x^{n}+\frac{1}{x^{n}}=(\cos n \alpha \pm i \sin n \alpha)+(\cos n \alpha \mp i \sin n \alpha)$
$=2 \cos n \alpha$

Topic: Mathematics, Quadratic Equation, JEE Main
If x, x+1 and x-1 are the first three terms of an arithmetic progression, then its sum upto 20 terms is

Options:
A 90 or 175
B 180 or 350
C 360 or 700
D 720 or 1400

The correct answer is:
180 or 350

Solution:
2x+1=x+x-1
(i) Case- Î :-

xâ¤-1
2-x-1=x+-x+1â-2x-2=1
âx=-32
Then the A.P. is â-32,12,52......
S20=202-3+19Ã2=202Ã35=350
(ii) Case- ÎÎ :-
-1â¤x<1
2x+1=x+-x+1â2x+2=1âx=-12
Then the A.P. is â-12,12,32......
S20=202-1+19Ã1=10Ã18=180
(ii) Case- ÎÎÎ :-
xâ¥1
2x+1=x+x-1â2x+2=2x-1â2=-1
(not possible)
Hence, the required sum is 350 or 180

Topic: Mathematics, Sequences and Series, JEE Main
If ∫xx+1+exdx=px+qln⁡x+1+ex+c, where c is the constant of integration, then p+q is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
∫x+1+ex-ex+1x+1+exdx=∫1-ex+1x+1+exdx=x-ln⁡x+1+ex+c

On comparing,

p=1, q=-1
Hence, p+q=0

Topic: Mathematics, Indefinite Integration, JEE Main
If xϕ(x)=∫5x3t2-2ϕ'(t)dt, x>-2, ϕ(0)=4, then ϕ(2) is

Options:

The correct answer is:
4

Solution:
Given that

xϕ(x)=∫5x3t2-2ϕ'(t)dt

ϕ(0)=4, x>-2

Differentiating on both the sides and using Newton Leibnitz rule

⇒xϕ'(x)+1ϕ(x)=3x2-2ϕ'(x)

∵ddxUV = UdVdx + VdUdx ; ddx∫fxgxhxdx = g'xhgx-f'xhfx

⇒(x+2)ϕ'(x)+ϕ(x)=3x2

It is a linear differential equation in the form of dydx+Py = Q

⇒I.F.=e∫1x+2dx=x+2

⇒ϕx·x+2=∫x+23x2x+2dx

⇒ϕ(x)·(x+2)=x3+c

Given that ϕ(0)=4

⇒c=8

⇒ϕ2=8+84=4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If $\int \frac{d x}{x+x^7}=p(x)$ then, $\int \frac{x^6}{x+x^7} d x$ is equal to:

Options:
A
$\ln |x|-p(x)+c$

B
$\ln |x|+p(x)+c$

C
$x-p(x)+c$

D
$x+p(x)+c$

The correct answer is:
$\ln |x|-p(x)+c$

Solution:
$\int \frac{x^6}{x+x^7} d x=\int \frac{x^6}{x\left(1+x^6\right)} d x$
$=\int \frac{\left(1+x^6\right)-1}{x\left(1+x^6\right)} d x$
$=\int \frac{1}{x} d x-\int \frac{1}{x+x^7} d x$
$=\ln |x|-p(x)+c$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2013 (09 Apr Online)
If x=xt is the solution of the differential equation t+1dx=2x+t+14dt,x0=2, then x1 equals ________

Options:

The correct answer is:
14

Solution:
Given: t+1dx=2x+t+14dt

⇒dxdt=2x+t+14t+1

⇒dxdt-2xt+1=t+13

⇒IF=e∫-2t+1dt

⇒IF=e-2logt+1

⇒IF=1t+12

Solution of the differential equations is given by,

⇒x×1t+12=∫t+1dt

⇒xt+12=t22+t+c

It is given that x0=2

⇒20+12=02+0+c

⇒c=2

⇒xt+12=t22+t+2

Putting t=1,

⇒x1+12=122+1+2

⇒x4=12+3

⇒x=2+12

⇒x=14

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If x=xy is the solution of the differential equation ydx dy=2x+y3y+1ey,x1=0; then xe is equal to

Options:
A eee3-1
B e3ee-1
C ee-1
D eee2-1

The correct answer is:
e3ee-1

Solution:
Given

dxdy-2xy=y2y+1ey

Finding integrating factor I.F.=e∫−2ydx=e−2lny=1y2

⇒xy2=∫y+1eydy

⇒xy2=y·ey+C

0=e+C⇒C=-e

⇒xy2=yey-e

Putting y=e in the equation we get

⇒xe2=e.ee-e

⇒x=e3ee-1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If $x^{x}=y^{y}$, then $\frac{d y}{d x}$ is

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

The correct answer is:
$\frac{1+\log x}{1+\log y}$

Solution:
We have,
$x^{x}=y^{y} \Rightarrow x \log x=y \log y$
$\Rightarrow \quad 1 \cdot \log x+\frac{x}{x}=y^{\prime} \log y+\frac{y}{y} y^{\prime}$
$\Rightarrow \quad \log x+1=y^{\prime}(\log y+1)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}$

Topic: Mathematics, Differentiation, JEE Main
If $x+|y|=2 y$, then $y$ as a function of $x$, at $x=0$ is

Options:
A
differentiable but not continuous

B
continuous but not differentiable

C
continuous as well as differentiable

D
neither continuous nor differentiable

The correct answer is:
continuous but not differentiable

Solution:
Given $x+|y|=2 y$
$$
\begin{aligned}
& \Rightarrow \quad x+y=2 y \text { or } x-y=2 y \\
& \Rightarrow \quad x=y \text { or } x=3 y
\end{aligned}
$$
This represent a straight line which passes through origin.
Hence, $x+|y|=2 y$ is continuous at $x=0$. Now, we check differentiability at $x=0$
$$
\begin{gathered}
x+|y|=2 y \Rightarrow x+y=2 y, y \geq 0 \\
x-y=2 y, y < 0
\end{gathered}
$$
Thus, $f(x)=\left\{\begin{array}{ll}x, & y < 0 \\ x / 3, & y \geq 0\end{array}\right\}$
$$
\begin{aligned}
& \text { Now, L.H.D. }=\lim _{h \rightarrow 0^{-}} \frac{f(x+h-) f x}{-h}(\quad) \\
& =\lim _{h \rightarrow 0^{-}} \frac{x+h-x}{-h}=-1 \\
& \text { R.H.D }=\lim _{h \rightarrow 0^{+}} \frac{f(x+h-) f x}{h}(\quad) \\
& =\lim _{h \rightarrow 0^{+}} \frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{3}=\frac{1}{3} \\
&
\end{aligned}
$$
Since, L.H.D $\neq$ R.H.D. at $x=0$
$\therefore$ given function is not differentiable at $x=0$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (07 May Online)
If x+y=3-cos⁡4θ and x-y=4sin⁡2θ, then the value of x+y is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
2

Solution:
Adding both the given equations, we get,

2x=3+4sin⁡2θ-cos⁡4θ
=3+4sin⁡2θ-1-2sin2⁡2θ
=2sin2⁡2θ+4sin⁡2θ+2
⇒x=sin2⁡2θ+2sin⁡2θ+1
⇒x=sin⁡2θ+12
Similarly, 2y=3-cos⁡4θ-4sin⁡2θ
2y=3-4sin⁡2θ-1-2sin2⁡2θ
=2sin2⁡2θ-4sin⁡2θ+2
⇒y=sin2⁡2θ-2sin⁡2θ+1
⇒y=sin⁡2θ-12
x=sin⁡2θ+1=1+sin⁡2θ
y=sin⁡2θ-1=1-sin⁡2θ
⇒x+y=2

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $x, y$ and $z$ are greater than 1 , then the value of $\left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$ is

Options:
A $\log x \cdot \log y \cdot \log z$
B $\log x+\log y+\log z$
C 0
D $1-\{(\log x) \cdot(\log y) \cdot(\log z)\}$

The correct answer is:
0

Solution:
Let $\Delta=\left|\begin{array}{ccc}1 & \log _{x}y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$
$=\left|\begin{array}{lll}\frac{\log x}{\log x} & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & \frac{\log y}{\log y} & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & \frac{\log 2}{\log z}\end{array}\right|$
On taking $\frac{1}{\log x} \frac{1}{\log y} \frac{1}{\log z}$ common from $R_{1}, R_{2}$ and $R_{3}$
we got
$$
=\frac{1}{\log x \cdot \log y \cdot \log z}\left|\begin{array}{lll}
\log x & \log y & \log z \\
\log x & \log y & \log z \\
\log x & \log y & \log z
\end{array}\right|
$$
$\therefore \Delta=0$

Topic: Mathematics, Determinants, JEE Main
If x, y and z are the roots of the equation 2t3-tanâ¡x+y+zÏt2-111t+2020=0, then xyzyzxzxy is equal to (where, x denotes the greatest integral value less than or equal to x)

Options:
A 20
B –10
C 0
D 1

The correct answer is:
0

Solution:
x+y+z is an integer

⇒tan⁡x+y+zπ=0
Hence, the sum of roots =x+y+z=0
⇒xyzyzxzxy=-x+y+zx2+y2+z2-xy-yz-zx=0

Topic: Mathematics, Determinants, JEE Main
If $\hat{x}, \hat{y}$ and $\hat{z}$ are three unit vectors in threedimensional space, then the minimum value of $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$

Options:
A
$\frac{3}{2}$

B
3

C
$3 \sqrt{3}$

D
6

The correct answer is:
3

Solution:
$(\hat{x}+\hat{y}+\hat{z})^2 \geq 0$
$$
\begin{aligned}
&\Rightarrow 3+2 \Sigma \hat{x} \cdot \hat{y} \geq 0 \\
&\Rightarrow 2 \Sigma \hat{x} \cdot \hat{y} \geq-3
\end{aligned}
$$
Now, $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ $=6+2 \sum \hat{x} . \hat{y} \geq 6+(-3)$
$$
\Rightarrow|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2 \geq 3
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2014 (12 Apr Online)
If $x, y$ are any two non-zero real numbers, $a_{i j}=x i+y j, A=\left\{a_{i j}\right\}_{n \times n}$ and $P, Q$ are two $n \times n$ matrices such that $A=x P+y Q$, then

Options:
A $P$ is singular and $Q$ is non-singular
B $P+Q$ is symmetric and $P-Q$ is skew symmetric
C Both $P+Q$ and $P-Q$ are singular
D Both $P+Q$ and $P-Q$ are non-singular

The correct answer is:
$P+Q$ is symmetric and $P-Q$ is skew symmetric

Solution:
We have, $a_{i j}=x j+y i$,
$$
\begin{aligned}
& A=\left(a_i\right)_{n \times n} \\
& A=\left[\begin{array}{cccc}
x+y & 2 x+y & 3 x+y & \ldots . . n x+y \\
x+2 y & 2 x+2 y & 3 x+2 y & \ldots . . n x+2 y \\
x+3 y & \ldots & \ldots . . & \ldots . \\
x+n y & \ldots & \ldots . & n x+n y
\end{array}\right]_{n x n n} \\
& A=x\left[\begin{array}{cccc}
1 & 2 & 3 \ldots . . & n \\
1 & 2 & 3 \ldots . & n \\
1 & \vdots & \ldots . . & \\
1 & 2 & 3 \ldots . & n
\end{array}\right]+y\left[\begin{array}{cccc}
1 & 1 & 1 . \ldots . & 1 \\
2 & \ldots & \ldots & 2 \\
3 & \ldots . & \ldots & 3 \\
\ldots & \ldots & \ldots . & n
\end{array}\right] \\
& A=x P+y Q \\
&
\end{aligned}
$$
where,
$$
P=\left[\begin{array}{cccc}
1 & 2 & 3 \ldots \ldots & n \\
1 & 2 & 3 \ldots & n \\
1 & 2 & \ldots . & \ldots \\
1 & 2 & 3 \ldots & n
\end{array}\right] \text { and } Q=\left[\begin{array}{cccc}
1 & 1 & 1 \ldots . . & 1 \\
2 & 2 & 2 \ldots . & 2 \\
3 & \ldots & \ldots & 3 \\
\ldots n & \ldots & n \ldots & n
\end{array}\right]
$$
$\therefore(P+Q)$ is symmetric and $(P-Q)$ is skew symmetric.

Topic: Mathematics, Matrices, JEE Main
If x,y are positive real numbers and 3x+4y=5, then the largest possible value of 16x2y3 is

Options:

The correct answer is:
3

Solution:
23x2+34y3=5
3x2+3x2+4y3+4y3+4y3=5
Using A.M≥G.M for 3x2,3x2,4y3,4y3,4y3
3x2+3x2+4y3+4y3+4y35≥3x224y3315
3x+4y5≥32432233x2y315
1≥243x2y315
16x2y3≤3
Maximum value of 16x2y3=3

Topic: Mathematics, Sequences and Series, JEE Main
If $x+y=k$ is a normal to the parabola $y^2=12 x$, then $k$ is

Options:
A $3$
B $9$
C $-9$
D $-3$

The correct answer is:
$9$

Solution:
Any normal is $y+t x=6 t+3 t^3$. It is identical with $x+y=k$ if $\frac{t}{1}=\frac{1}{1}=\frac{6 t+3 t^3}{k}$ $\therefore t=1$ and $1=\frac{6+3}{k} \Rightarrow k=9$.

Topic: Mathematics, Parabola, JEE Main
If $\rho=\left\{(x, y) \mid x^{2}+y^{2}=1 ; x, y \in \mathbb{R}\right\}$. Then, $\rho$ is

Options:
A reflexive
B symmetric
C transitive
D anti-symmetric

The correct answer is:
symmetric

Solution:
Obviously, the relation is not reflexive and transitive, but it is symmetric, because $x^{2}+x^{2}=2 x^{2} \neq 1$

$$

\begin{array}{l}

\text { and } x^{2}+y^{2}=1, y^{2}+z^{2}=1 \\

\Rightarrow x^{2}+z^{2}=1 \\

\text { But } x^{2}+y^{2}=1 \Rightarrow y^{2}+x^{2}=1

\end{array}

$$

Topic: Mathematics, Sets and Relations, JEE Main
If $x y+y^{2}=\tan x+y,$ then find $\frac{d y}{d x}$ is

Options:
A $\frac{\sec ^{2} x}{x+2 y}$
B $\frac{\sec ^{2} x-y}{(x+2 y-1)}$
C $(x+2 y-1) \sec ^{2} x$
D $\sec ^{2} x \cdot y$

The correct answer is:
$\frac{\sec ^{2} x-y}{(x+2 y-1)}$

Solution:
The given relation is $x y+y^{2}=\tan x+y$. Differentiating both sides with respect to $x$, we get

$\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$

or $\left[y \cdot 1+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$

or $(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$

$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$

Topic: Mathematics, Differentiation, JEE Main
If x+y+y-x=2, then the value of d2ydx2 is equal to

Options:

The correct answer is:
0.5

Solution:
Given: x+y+y-x=2 ... (i)
⇒y+x-y-xx+y-y-x=2
⇒y+x-y-x=x ... (ii)
By adding Eq. (i) and (ii), we get
2y+x=2+x⇒4y+x=4+x2+4x
∴4dydx=2x
∴d2ydx2=12

Topic: Mathematics, Application of Derivatives, JEE Main
If $\sqrt{x+y}+\sqrt{y-x}=5$, then $\left(\frac{d^{2} y}{d x^{2}}\right)=$

Options:
A $\frac{2}{25}$
B $\frac{2}{5}$
C $\frac{-2}{5}$
D $\frac{-2}{25}$

The correct answer is:
$\frac{2}{25}$

Solution:
(D)
Given $\sqrt{x+y}+\sqrt{y-x}=5 \Rightarrow \sqrt{y-x}=5-\sqrt{x+y}$
On squaring both side, we get
$$
y-x=25+x+y-10 \sqrt{x+y} \Rightarrow 10 \sqrt{x+y}-2 x=25
$$
Differentiating w.r.t. $x$,
$10 \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)-2 \times 1=0 \Rightarrow \frac{5}{\sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=2$
$$
\therefore 1+\frac{d y}{d x}=\frac{2 \sqrt{x+y}}{5}....(1)
$$
Differentiating w.r.t. $x$,
$$
\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{2}{5} \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=\frac{1}{5 \sqrt{x+y}} \times \frac{2 \sqrt{x+y}}{5} \quad \ldots[\because \text { From (1) }] \\
&=\frac{2}{25}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If xy.yx=16, then the value of dydx at (2, 2) is

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

The correct answer is:
-1

Solution:
Since, xy.yx=16
∴loge⁡xy+loge⁡yx=loge⁡16
⇒yloge⁡x+xloge⁡y=4loge⁡2
Now, on differentiating both sides w.r.t. x, we get
yx+loge⁡xdydx+xydydx+loge⁡y.1=0
⇒dydx=-loge⁡y+yxloge⁡x+xy
∴dydx(2,2)=-(loge⁡2+1)(loge⁡2+1)=-1

Topic: Mathematics, Differentiation, JEE Main
If x+yy+zz+xy+zz+xx+yz+xx+yy+z=kxzyyxzzyx, then k is equal to

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

The correct answer is:
-2

Solution:
x+yy+zz+xy+zz+xx+yz+xx+yy+z
Using property of determinant we can write this determinant as a sum of
x+yy+zzy+zz+xxz+xx+yy+x+yy+zxy+zz+xyz+xx+yz
Similarly,
xyzyzxzxy+xyxyzyzxz+xzzyxxzyy+xzxyxyzyz+yyz+xzzx+yxxy+z+yzzzxxxyy+yzxzxyxyz
xzxyxyzyz+0+0+0+0+0+xyzyzxzxy
=2xyzyzxzxy=-2xzyyxzzyx
k=-2

Topic: Mathematics, Determinants, JEE Main
If $x+y+z=180^{\circ}$, then $\cos 2 x+\cos 2 y-\cos 2 z$ is equal to

Options:
A $4 \sin x \cdot \sin y \cdot \sin z$
B $1-4 \sin x \cdot \sin y \cdot \cos z$
C $4 \sin x \cdot \sin y \cdot \sin z-1$
D $\cos A \cdot \cos B \cdot \cos C$

The correct answer is:
$1-4 \sin x \cdot \sin y \cdot \cos z$

Solution:
$\begin{aligned} & \cos 2 x+\cos 2 y-\cos 2 z \\ & =2 \cos (x+y) \cos (x-y)-2 \cos ^2 z+1 \\ & =2 \cos (\pi-z) \cos (x-y)-2 \cos ^2 z+1 \\ & =1-2 \cos z\{\cos (x-y)-\cos (x+y)\} \\ & =1-2 \cos z 2 \sin x \sin y=1-4 \sin x \sin y \cos z\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If x, y, z are in arithmetic progression and tan-1â¡x,tan-1â¡y and tan-1â¡z are also in arithmetic progression, then

Options:
A x=y=z
B x=y=-z
C x=1, y=2, z=3
D x=2, y=4, z=6

The correct answer is:
x=y=z

Solution:
Since, x, y, z are in AP
â´ y=x+z2 ....(i)
And tan-1â¡x,tan-1â¡y and tan-1â¡z are also in AP.
â´ 2tan-1â¡y=tan-1â¡x+tan-1â¡z
â tan-1â¡2y1-y2=tan-1â¡x+z1-xz
â 2y1-y2=2y1-xz [from equation. (i)]
â y2=xz
â x, y, z are in GP.
â´ x=y=z

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If x, y, z are in arithmetic progression with common difference d, x≠3d, and the determinant of the matrix 342x452y5kz is zero, then the value of k2 is

Options:
A 72
B 12
C 36
D 6

The correct answer is:
72

Solution:
Given x, y, z are in arithmetic progression, hence 2y=x+z, also 342x452y5kz=0

Applying R2→R1+R3-2R2

⇒342x0k-6205kz=0

⇒(k-62)(3z-5x)=0

⇒(k-62)=0 or (3z-5x)=0

If 3z-5x=0

⇒3(x+2d)-5x=0

⇒x=3d (Not possible)

⇒k=62

⇒k2=72.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (17 Mar Shift 2)