datavorous/entrance-exam-dataset huggingface dataset:
If 3+cot⁡80ocot⁡20ocot⁡80o+cot⁡20o=tanπk, then the value of k is (where, πk is an acute angle)

Options:

The correct answer is:
3.6

Solution:
3sin⁡80osin⁡20o+cos⁡80ocos⁡20osin⁡20ocos⁡80o+cos⁡20osin⁡80o
=2sin⁡20osin⁡80o+cos⁡80ocos⁡20o+sin⁡80osin⁡20sin20o+80o
=cos⁡60o-cos⁡100o+cos80o-20osin⁡100o
=2cos⁡60o-cos⁡100osin⁡100o=1-cos⁡100osin⁡100o
=2sin250∘2sin⁡50ocos⁡50o=tan⁡50o
=tan5π18=tanπ185
⇒k=185=3.6

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If (3+i)100=299(p+iq), then p and q are roots of the equation :

Options:
A x2-(3+1)x+3=0
B x2+(3+1)x+3=0
C x2+(3-1)x-3=0
D x2-(3-1)x-3=0

The correct answer is:
x2-(3-1)x-3=0

Solution:
(3+i)100=299(p+iq)

2100cosπ6+isinπ6100=299p+iq

2100ei50π3=299p+iq

2cos50π3+isin50π3=p+iq

2cos17π-π3+isin17π-π3=p+iq

2-cosπ3+isinπ3=p+iq

2-12+i32=p+iq

p=-1 & q=3

p and q Roots of the equation
x2-(3-1)x-3=0

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (26 Aug Shift 2)
If 3+isinθ4-icosθ,θ∈0,2π, is a real number, then an argument of sinθ+icosθ is

Options:
A π-tan-143
B π-tan-134
C -tan-134
D tan-143

The correct answer is:
π-tan-143

Solution:
z=3+isinθ4-icosθ×4+icosθ4+icosθ

Im(z)=3cosθ+4sinθ16+cos2θ
As z is purely real ⇒3cosθ+4sinθ=0⇒tanθ=-34
Argsinθ+icosθ=π+tan-1cosθsinθ=π+tan-1-43=π-tan-143

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If \( 3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3} \), then the value of \( x \) lying in \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is

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

The correct answer is:
\( \frac{1}{\sqrt{3}} \)

Solution:
3sin-1⁡2x1+x2-4 cos-1⁡ 1-x21+x2+2tan-1⁡ 2x1-x2=π3
On putting x=tan ⁡θ, we get
3sin-1⁡2tan⁡θ1+tan2 ⁡θ-4cos-1⁡1-tan2 ⁡θ1+tan2 ⁡θ+2tan-1⁡2tan⁡θ1-tan2 ⁡θ=π3
⇒3sin-1⁡(sin⁡2θ)-4cos-1⁡cos⁡2θ+2tan-1⁡(tan⁡2θ)=π3
⇒32θ-42θ+22θ=π3
⇒6θ-8θ+4θ=π3
⇒θ=π6⇒tan-1⁡x=π6
⇒x=tan⁡π6⇒x=13

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $3 \sin x+4 \cos x=5$, then $6 \tan \frac{x}{2}-9 \tan ^2 \frac{x}{2}$ is equal to

Options:
A $0$
B $1$
C $3$
D $4$

The correct answer is:
$1$

Solution:
Given, $3 \sin x+4 \cos x=5$
$\begin{aligned} & \Rightarrow 3\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+4\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)=5 \\ & \Rightarrow 6 \tan \frac{x}{2}+4-4 \tan ^2 \frac{x}{2}=5\left(1+\tan ^2 \frac{x}{2}\right) \\ & \Rightarrow 6 \tan \frac{x}{2}-9 \tan ^2 \frac{x}{2}=1\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If 3sin⁡x+cos⁡x-2=y-12 for 0≤x≤8π, then the number of values of the pair x,y is equal to

Options:

The correct answer is:
4

Solution:
max.â¡3sinâ¡x+cosâ¡x=32+12=2
and y-12â¥0
So, LHS â¤0 and RHS â¥0
So, the possibility of equality is only when
LHS = RHS =0
RHS =0ây=1
LHS =0â3sinâ¡x+cosâ¡x=2
â32sinâ¡x+12cosâ¡x=1
âcosâ¡x-Ï3=1
âx-Ï3=2nÏâx=2nÏ+Ï3
0â¤2nÏ+Ï3â¤8Ïân=0,1,2,3
â There are four values of x and hence four pairs of x,y

Topic: Mathematics, Trigonometric Equations, JEE Main
If 3 tan-112+3-tan-113=tan-11x, then the value of x is equal to

Options:

The correct answer is:
2

Solution:
The given equation can be written as
3tan-12-3=tan-11x+tan-113
⇒315o=tan-11x+131-1x13⇒1=3+x3x-1⇒x=2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If 3x-1,3,x-3 are the first three terms of an arithmetic progression, then the sum of the first five terms can be

Options:
A 5
B 10
C 20
D 30

The correct answer is:
5

Solution:
Case-I : x<13â-3x+1,3,-x+3 are in A.P.
â6=-3x+1-x+3
â4x=-2âx=-12
â terms are 52, 3, 72, 4, 92
â sum =352
Case-II: 13â¤x<3
3x-1,3,-x+3 are in A.P.
â6=3x-1-x+3=2x+2âx=2
â terms are 5, 3, 1, â1, â3
â sum =5
Case-III: xâ¥3
â terms are 3x-1,3,x-3
â6=3x-1+x-3=4x-4
â4x=10âx=52 not possible

Topic: Mathematics, Sequences and Series, JEE Main
If -3<x2-Î»x-2x2+x+1<2 for all xâR, then the value of Î» belongs to

Options:
A -1,7
B -6,2
C -1,2
D -6,7

The correct answer is:
-1,2

Solution:
-3<x2-λx-2x2+x+1<2
⇒-3x2-3x-3<x2-λx-2<2x2+2x+2 since x2+x+1>0, ∀x∈R
⇒4x2+x3-λ+1>0,x2+x2+λ+4>0
i 4x2-xλ-3+1>0
⇒D<0⇒λ-32-4×4×1<0
⇒λ-3+4λ-3-4<0
⇒λ+1λ-7<0⇒λ∈-1,7
ii x2+xλ+2+4>0
⇒D<0⇒λ+22-4×4<0
λ+2-4λ+2+4<0λ-2λ+6<0⇒λ∈-6,2
Taking the intersection of the solutions of (i) and (ii), we get,

λ∈-1,2

Topic: Mathematics, Quadratic Equation, JEE Main
If 3x+4y=122 is a tangent o the ellipse x2a2+y29=1 for some a∈R, then the distance between the foci of the ellipse is

Options:
A 27
B 4
C 25
D 22

The correct answer is:
27

Solution:
3x+4y=122
⇒4y=-3x+122
⇒y=-34x+32
Condition of tangency c2=a2m2+b2
18=a2∙916+9
a2∙916=9
a2=16
a=4
e=1-b2a2=1-916=74
∴ae=74∙4=7
∴ Focus are ±7,0
∴ Distance between foci =27

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If $\frac{2 z_1}{3 z_2}$ is a purely imaginary number, then $\left|\frac{z_1-z_2}{z_1+z_2}\right|=$

Options:
A $3/2$
B $1$
C $2 / 3$
D $4 / 9$

The correct answer is:
$1$

Solution:
As given, let $\frac{2 z_1}{3 z_2}=$ iy $\frac{z_1}{z_2}=\frac{3}{2}$ iy , so that
$\left|\frac{z_1-z_2}{z_1+z_2}\right|=\left|\frac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right|=\left|\frac{\frac{3}{2} i y-1}{\frac{3}{2} i y+1}\right|=\left|\frac{1-\frac{3}{2} i y}{1+\frac{3}{2} i y}\right|=1$ {$|z|=|\bar{z}|\}$

Topic: Mathematics, Complex Number, JEE Main
If $(4,0)$ and $(-4,0)$ be the vertices and $(6,0)$ and $(-6,0)$ be the foci of a hyperbola, then its eccentricity is

Options:
A $5/2$
B $2$
C $3 / 2$
D $\sqrt{2}$

The correct answer is:
$3 / 2$

Solution:
$\begin{aligned} & \text { Vertices }( \pm 4,0) \equiv( \pm a, 0) \Rightarrow a=4 \\ & \text { Foci }( \pm 6,0) \equiv( \pm a e, 0) \Rightarrow e=\frac{6}{4}=\frac{3}{2}\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
If $\left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|^2=\left|\begin{array}{ll}3 & 2 \\ 1 & x\end{array}\right|-\left|\begin{array}{cc}x & 3 \\ -2 & 1\end{array}\right|$, then $x=$

Options:
A -14
B 2
C 6
D 7

The correct answer is:
6

Solution:
$\begin{aligned} & \left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|\left|\begin{array}{ll}4 & 1 \\ 2 & 1\end{array}\right|=\left|\begin{array}{cc}3 & 2 \\ 1 & x\end{array}\right|-\left|\begin{array}{cc}x & 3 \\ -2 & 1\end{array}\right| \\ & \quad=\left|\begin{array}{cc}17 & 9 \\ 9 & 5\end{array}\right|=(3 x-2)-(x+6) \\ & \Rightarrow 85-81=2 x-8 \Rightarrow 4+8=2 x \Rightarrow x=6\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
If $(-4,5)$ is one vertex and $7 x-y+8=0$ is one diagonal of a square, then the equation of second diagonal is

Options:
A $x+3 y=21$
B $2 x-3 y=7$
C $x+7 y=31$
D $2 x+3 y=21$

The correct answer is:
$x+7 y=31$

Solution:
One vertex of square is $(-4,5)$ and equation of one diagonal is $7 x-y+8=0$
Diagonal of a square are perpendicular and bisect each other
Let the equation of the other diagonal be $y=m x$ $+\mathrm{c}$ where $\mathrm{m}$ is the slope of the line and $\mathrm{c}$ is the $y-$ intercept.
Since this line passes through $(-4,5)$
$$
\therefore \quad 5=-4 \mathrm{~m}+\mathrm{c}
$$
Since this line is at right angle to the line
$$
7 x-y+8=0 \text { or } y=7 x+8 \text {, having slope }=7 \text {, }
$$
$$
\therefore \quad 7 \times \mathrm{m}=-1 \text { or } \mathrm{m}=\frac{-1}{7}
$$
Putting this value of $m$ in equation (i) we get
$$
5=-4 \times\left(\frac{-1}{7}\right)+\mathrm{c}
$$
or $5=\frac{4}{7}+\mathrm{c} \quad$ or $\quad \mathrm{c}=5-\frac{4}{7}=\frac{31}{7}$
Hence equation of the other diagonal is
$$
y=-\frac{1}{7} \mathrm{x}+\frac{31}{7} \text { or } 7 \mathrm{y}=-\mathrm{x}+31
$$
or $x+7 y-31=0$ or $x+7 y=31$.

Topic: Mathematics, Straight Lines, JEE Main
$\quad$ If $4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0,$ the family of lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is concurrent at one or the other of the two points-

Options:
A $\left(-1,-\frac{1}{2}\right),(-2,-1)$
B $(-1,-1),\left(-2,-\frac{1}{2}\right)$
C $(-1,2),\left(\frac{1}{2},-1\right)$
D $(1,2),\left(\frac{1}{2},-1\right)$

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

Solution:
$$

\text { } \begin{array}{l}

4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c \\

\quad \equiv(2 a+b)^{2}-3(2 a+b) c+2 c^{2}=0 \\

\quad \Rightarrow(2 a+b-2 c)(2 a+b-c)=0 \Rightarrow c=2 a+b \\

\quad \text { or } c=a+\frac{1}{2} b

\end{array}

$$

The equation of the family of lines is

$$

a(x+2)+b(y+1)=0 \text { or } a(x+1)+b\left(y+\frac{1}{2}\right)=0

$$

giving the point of concurrence (-2,-1) or

$$

\left(-1,-\frac{1}{2}\right)

$$

Topic: Mathematics, Straight Lines, JEE Main
If 4cos⁡36o+cot712∘=n1+n2+n3+n4+n5+n6, then the value of ∑i=16ni210 is equal to _______. Roundoff answer to nearest integer.

Options:

The correct answer is:
9

Solution:
4cos⁡36o=45+14=5+1
cot⁡712o=1+cos⁡15°sin⁡15°=1+3+1223-122=22+3+13-1
=22+3+13+13-13+1=26+22+3+3+3+12
=26+23+22+42=6+3+2+2=6+4+3+2
⇒4cos⁡36o+cot712o=5+1+6+4+3+2
=1+2+3+4+5+6
=n1+n2+n3+n4+n5+n6
⇒∑i=16ni2=12+22+33+42+52+62
=6×7×136=91
⇒∑i=16ni210=9110=9.1

Topic: Mathematics, Trigonometric Equations, JEE Main
If 4cos⁡36o+cot712∘=n1+n2+n3+n4+n5+n6, then the value of ∑i=16ni210 is equal to

Options:

The correct answer is:
9.1

Solution:
4cos⁡36o=45+14=5+1
cot⁡712o=1+cos⁡15°sin⁡15°=1+3+1223-122=22+3+13-1
=22+3+13+13-13+1=26+22+3+3+3+12
=26+23+22+42=6+3+2+2=6+4+3+2
⇒4cos⁡36o+cot712o=5+1+6+4+3+2
=1+2+3+4+5+6
=n1+n2+n3+n4+n5+n6
⇒∑i=16ni2=12+22+33+42+52+62
=6×7×136=91
⇒∑i=16ni210=9110=9.1

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If 4 dice are rolled once, the number of ways of getting the sum as 10 is K, then the value of K10 is equal to

Options:

The correct answer is:
8

Solution:
x1+x2+x3+x4=10; 1≤xi≤6
The desired number is the coefficient of x10 in
x+x2+…+x64=x41+x+…+x54
or the coefficient of x6 in 1+x+x2+…+x54
=1-x61-x4=1-x641-x-4
=1-4x6+…1-x-4
or the coefficient of x6 in 1-x-4-4× coefficient of x0 in 1-x-4
=6+4-1C4-1-4×0+4-1C4-1
=9C3-4×1=9×8×73×2-4=84-4=80
⇒K=80⇒K10=8

Topic: Mathematics, Permutation Combination, JEE Main
If 4 distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all 4 of them are either divisible by 3 or divisible by 5 is

Options:
A Â 6C4Â 100C4
B Â 33C4Â 100C4
C Â 20C4Â 100C4
D 47C4 100C4

The correct answer is:
47C4 100C4

Solution:
Total cases = 100C4
Numbers which are divisible by either 3 or 5
=n3+n5-n3â©5
=33+20-6
=47
Favourable cases = 47C4
Hence, required probability = 47C4 100C4

Topic: Mathematics, Probability, JEE Main
If $0.4 \mathrm{gm~} \mathrm{NaOH}$ is present in 1 litre solution, then its $\mathrm{pH}$ will be

Options:
A 2
B 10
C 11
D 12

The correct answer is:
12

Solution:
$[\mathrm{NaOH}]=\frac{0.4}{40}=0.01 \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=10^{-2} \mathrm{M}$
$\left[H^{+}\right]=10^{-12}, p H=-\log \left[H^{+}\right]=12$

Topic: Chemistry, Ionic Equilibrium, JEE Main
If 4i3-i2i+12=rcos⁡θ+isin⁡θ, then cos⁡θ+sin⁡θ is equal to (where, i2=-1)

Options:

The correct answer is:
1.4

Solution:
4i3-i2i+12=-4i-i2i+12
⇒-5i24i2+1+4i=25-1-3+4i
=253-4i×3+4i3+4i=2532+423+4i
=535+45i=rcos⁡θ+isin⁡θ
⇒r=5, cos⁡θ=35, sin⁡θ=45
⇒cos⁡θ+sin⁡θ=75=1.4

Topic: Mathematics, Complex Number, JEE Main
If 4 letter words are formed using letters of the word âMORADABADâ, then the probability that D comes exactly once in the 4 letter word is

Options:
A 146313
B 213313
C 167313
D 520621

The correct answer is:
146313

Solution:
Total cases =4 letter word is formed by using either 3 identical letter and 1 distinct letter or 2 alike of one kind and 2 alike of 2nd kind or 2 alike, 2 distinct or 4 distinct
Number of ways for 3 alike and 1 distinct =C11×C51×4!3!=20
Number of ways for 2 alike and 2 alike =C22×4!2!2!=6
Number of ways for 2 alike and 2 distinct =C21×C52×4!2!=240
Number of ways for 4 distinct =C64×4!=360
The total number of ways =20+6+240+360=626
For favorable cases, the following possibilities are there (one D is always included)
Number of ways for 3 alike =C11×4!3!=4
Number of ways for 2 alike +1 distinct =C11×C14×4!2!=48
Number of ways for 3 distinct =C35×4!=240
The total number of ways for favorable cases =240+48+4=292
Hence, the required probability =292626=146313

Topic: Mathematics, Probability, JEE Main
If $4 \sin ^{-1} x+\cos ^{-1} x=\pi$, then $x$ is equal to

Options:
A $0$
B $\frac{1}{2}$
C $-\frac{\sqrt{3}}{2}$
D $\frac{1}{\sqrt{2}}$

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

Solution:
We know that $4 \sin ^{-1} x+\cos ^{-1} x=\pi$
$\begin{aligned}
& \Rightarrow 3 \sin ^{-1} x+\sin ^{-1} x+\cos ^{-1} x=\pi \\
& \Rightarrow 3 \sin ^{-1} x=\pi-\frac{\pi}{2}=\frac{\pi}{2} \\
& \Rightarrow \sin ^{-1} x=\pi / 6 \Rightarrow x=\sin \frac{\pi}{6}=\frac{1}{2} .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If 4sin⁡27o=α-β, then the value of α+β is

Options:
A 5
B 3
C 8
D 2

The correct answer is:
8

Solution:
cos⁡27o+sin⁡27o2=1+2sin⁡27ocos⁡27o=1+sin⁡54o=1+cos⁡36o
⇒cos⁡27o+sin⁡27o=1+cos⁡36o
Similarly, cos⁡27o-sin⁡27o=1-cos⁡36o (since, cos⁡27o>sin⁡27o )
On subtracting, we get,
2sin⁡27o=1+cos⁡36o-1-cos⁡36o
=1+5+14-1-5+14
=5+52-3-52
⇒4sin⁡27o=5+5-3-5
⇒α=5+5 and β=3-5
⇒α+β=8

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If 4x-3x-12=3x+12 -22x-1 then value of x is equal to

Options:
A 52
B 2
C 32
D 1

The correct answer is:
32

Solution:
4x-3x-12=3x+12 -22x-1
4x+22x-1=3x+12+3x-12
4x+4x2=3x3+3x3
⇒ 4x1+12=3x3+13
⇒ 43x=23×43=833=23123=4332
⇒ x=32

Topic: Mathematics, Quadratic Equation, JEE Main
If 4x+3y-12=0 touches x-p2+y-p2=p2, then the sum of all the possible values of p is

Options:

The correct answer is:
7

Solution:
The distance of the centre p,p of the given circle from the line 4x+3y-12=0 is equal to the radius p
â4p+3p-125=p
â7p-12=5p or 7p-12=-5p
âp=6 or p=1
â sum of all the possible values of âpâ=6+1=7

Topic: Mathematics, Circle, JEE Main
If 4x-ay+3z=0, x+2y+az=0 and ax+2z=0 have a non-trivial solution, then the number of real value(s) of a is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
For non-trivial solution, using Cramer's rule, we get,
4-a312aa02=0
44-0+a2-a2+30-2a=0
a3+4a-16=0⇒a-2a2+2a+8=0
a=2

Hence, only one value of a

Topic: Mathematics, Determinants, JEE Main
If \( 4 x+3 y-12=0 \) touches \( (x-p)^{2}+(y-p)^{2}=p^{2} \), then the sum of all the possible values of \( p \) is

Options:

The correct answer is:
7

Solution:
The distance of the centre p,p of the given circle from the line 4x+3y-12=0 is equal to the radius p
â4p+3p-125=p
â7p-12=5p or 7p-12=-5p
âp=6 or p=1
â sum of all the possible values of âpâ=6+1=7

Topic: Mathematics, Circle, JEE Main
If 514762135 16-7 62 4-74 35711634252020=a1a2a3b1b2b3c1c2c3, then the value of 2a2-b1+3a3-c1+4b3-c2 is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Let, X=ABAT2020
( âµ B is symmetric)
XT=ABAT2020T=ABATT2020=ABAT2020
X=XTâX is a symmetric matrix
âa2=b1,a3=c1,b3=c2
â Required value =0

Topic: Mathematics, Matrices, JEE Main
If 525456â¦.52n=0.04-28, then the value of n is equal to

Options:
A 7
B 5
C 6
D 3

The correct answer is:
7

Solution:
The given equation can be written as 52522523…52n=5228
521+2+3+…+n=5228
⇒nn+12=28
⇒n=7

Topic: Mathematics, Sequences and Series, JEE Main
If 5353-333 is divided by 10, then the remainder obtained is

Options:

The correct answer is:
6

Solution:
We know that,
xn-yn=x-yxn-1+xn-2y+....+xyn-2+yn-1
i.e. xn-yn is divisible by x-y
5353-333=5353-353+353-333+33-33
=5353-353+353-33-333-33
Clearly, 1st bracket is divisible by 50 and the last one is divisible by 30. So, both are divisible by 10.
353-33=33â350-33=33350-1
=3310-125-1=2710-125-27=2710k-1-27
=270k-54=270k-60+6=10Î¼+6.
Hence, the remainder obtained is 6.

Topic: Mathematics, Binomial Theorem, JEE Main
If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r can not be equal to:

Options:
A 34
B 32
C 54
D 74

The correct answer is:
74

Solution:
Given, 5, 5r, 5r2 are the length of sides of triangle, then we know that the sum of two sides of a triangle is more than the third side i.e.

5+5r>5r2 ...1

5+5r2>5r ...2

5r+5r2>5 ...3

From 1,

r2-r-1<0

r-1+52 r-1-52 <0

r∈1-52,1+52 ...4

from 2,

r2-r+1>0

The discriminant of the quadratic is D=-12-4×1×1=-3 and we know that if the discriminant of a quadratic is negative and its leading coefficient is positive then the quadratic is positive for all real numbers.

⇒r∈R ...5

from 3,

r2+r-1>0

r+1+52r+1-5 2>0

So, r∈-∞, -1+52∪-1-52, ∞ ...6

Now, taking intersection of 4, 5, 6, we get r∈-1+52, 1+52.

Out of the given options only 74 is not in the interval obtained.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If 5+9+13+â¦ n terms7+9+11+â¦ n+1 terms=1716, then n is equal to

Options:
A 7
B 12
C 8
D 15

The correct answer is:
7

Solution:
5+9+13+â¦n terms 7+9+11+â¦ n+1 terms
â n2 2Ã5+n-14n+122Ã7+n+1-12=1716
â n4n+6n+1 2n+14=1716 â n(2n+3)n+1n+7=1716
â 2n2+3nn2+8n+7=1716 â 32n2+48n=17n2+136n+119
â 15n2-88n-119=0
â n-7 15n+17=0 ân=7

Topic: Mathematics, Sequences and Series, JEE Main
If 597 is divided by 52 , then the remainder obtained is

Options:
A 3
B 5
C 4
D 0

The correct answer is:
5

Solution:
We know that, 54=625=52Ã12+1
â54=52Î»+1 , where Î» is a positive integer.
â5424=52Î»+124
= C02452Î»24+C152Î»23+C22452Î»2224+â¦+C2352Î»+C242424 (by binomial theorem)
â596=52C0245223Î»24+C15223Î»2224+â¦+C23Î»24+1
=a multiple of 52+1
On multiplying both sides by 5 , we get
597=596â5=5 a multiple of 52+5
Hence, the required remainder is 5 .

Topic: Mathematics, Binomial Theorem, JEE Main
If 5a+4b+20c=t, then the value of t for which the line ax+by+c-1=0 always passes through a fixed point is

Options:
A 0
B 20
C 30
D None of these

The correct answer is:
20

Solution:
Equation of the given line is ax+by+câ1=0
â20 ax+20 by+20 câ20=0 .....(i)
From the given relation, substituting value of c, we get,
20 ax+20 by+tâ5aâ4bâ20=0
â20 a( xâ 1 4 )+20 b( yâ 1 5 )+( tâ20 )=0
Clearly for t=20, the given line will pass through the point ( 1 4 , 1 5 ) for all the values of a & b

Topic: Mathematics, Straight Lines, JEE Main
If $\frac{1}{{ }^{5} C_{r}}+\frac{1}{{ }^{6} C_{r}}=\frac{1}{{ }^{4} C_{r}},$ then the value of $r$ is

Options:
A 4
B 2
C 5
D 3

The correct answer is:
2

Solution:
Given, $\frac{1}{5 C_{r}} +\frac{1}{6 C_{r}} =\frac{1}{4 C_{r}}$
$\Rightarrow \quad \frac{r !(5-r) !}{5 !}+\frac{r !(6-r) !}{6 !}=\frac{r !(4-r) !}{4 !}$
$\Rightarrow \quad \frac{(5-r) !}{5}+\frac{(6-r) !}{30}=(4-r) !$
$\Rightarrow \quad \frac{(5-r)}{5}+\frac{(6-r)(5-r)}{30}=1$
$\Rightarrow \quad 30-6 r+30-11r+r^{2}=30$
$\Rightarrow \quad r^{2}-17 r+30=0$
$\Rightarrow \quad r^{2}-15 r-2 r+30=0$
$\Rightarrow \quad r(r-15)-2(r-15)=0$
$\Rightarrow \quad(r-2)(r-15)=0$
$\therefore \quad r=15,2$
Hence, the value of $r$ is 2 .

Topic: Mathematics, Binomial Theorem, JEE Main
If 5 distinct balls are placed at random into 5 cells, then the probability that exactly one cell remains empty, is

Options:
A $48 / 125$
B $12 / 125$
C $8 / 125$
D $1 / 125$

The correct answer is:
$48 / 125$

Solution:
Clearly. 5 distinct balls can be placed into 5 olls in $5^{5}$ ways. Now, the number of ways of selecting one cell ${ }^{5} C_{1}$
Let the selected cell be empty, Now, we are left with 5 balls and 4 cells. Now, number of ways of placing 5 distinct balls into 4 cells such that each cell have atleast one $=4^{5}-{ }^{4} C_{1}(4-1)^{5}+{ }^{4} C_{2}(4-2)^{5}-{ }^{4} C_{3}(4-3)^{6}$
$=1024-4 \times 3^{5}+6 \times 2^{5}-4$
$=1024-972+192-4=1216-976$
$=240$
Thus, the number of ways of placing 5 distinct balls such that exactly one cell remains empty $={ }^{5} C_{1} \times 240$
$\therefore$ Required probability $=\frac{{ }^{5} C_{1} \times 240}{5^{5}}=\frac{240}{625}=\frac{48}{125}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
If \( 5 f(x)+3 f\left(\frac{1}{x}\right)=x+2 \) and \( y=x f(x) \) then \( \left(\frac{d y}{d x}\right)_{x=1} \) is equal to

Options:
A \( \frac{1}{4} \)
B \( \frac{7}{8} \)
C \( 1 \)
D None of these

The correct answer is:
\( \frac{7}{8} \)

Solution:
5ƒx+3ƒ1x=x+2 …1
Replacing x by 1x
∴ 5ƒ1x + 3ƒx=1x+ 2 … 2
From 1
25ƒx+15ƒ1x=5x+10  …3
and from 2
9ƒx+15ƒ1x=3x+6 … 4
Subtracting4 from 3
∴16ƒx=5x- 3x + 4
∴xƒx=5x2-3+4x16=y
∴dydx=10x+416
dydxx=1=10+416=78 .

Topic: Mathematics, Functions, JEE Main
If α+5i2 is a root of the equation 2x2-6x+k=0, then the value of k10 is α,k,∈R

Options:

The correct answer is:
1.7

Solution:
Î±+5i2 is one root
So, Î±-5i2 is the other root
âk=24Î±2+25
=129+25=17
âk10=1.7

Topic: Mathematics, Quadratic Equation, JEE Main
If 5 moles of BaCl2 is mixed with 2 moles of Na3PO4, the maximum number of moles of Ba3PO42formed is _________ (Nearest integer)

Options:

The correct answer is:
1

Solution:
The balanced chemical equation for the reaction between BaCl2 and Na3PO4 is:

3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl

5 moles of barium chloride will require 5×23=3.3 moles of sodium phosphate.

Here, sodium phosphate is the limiting reagent.
2 moles of sodium phosphate produce 1 mole of barium phosphate.

Thus,2 moles of sodium phosphate will produce 1 mole of barium phosphate.

Hence, the maximum no. of moles of Ba3(PO4)2 that can be formed is one mole.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If 5tan2⁡x-cos2⁡x=2cos⁡ 2x+9, then the value of cos⁡4x is

Options:
A -35
B 13
C 29
D -79

The correct answer is:
-79

Solution:
51-cos⁡2x1+cos⁡2x-1+cos⁡2x2=2cos⁡2x+9
Let, cos⁡2x=y

∴    5(1−y1+y−1+y2)=2y+9
⇒ 5-y2-4y+1=4y2+22y+18
⇒ 9y2+42y+13=0
⇒ y=-13 or y=-133 (Not possible)
Now, cos⁡4x=2cos2⁡2x-1
⇒ cos⁡4x=2-132-1

⇒ cos⁡4x=-79

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2017 (02 Apr)
If $5 \tan \theta=4$, then $\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta}=$

Options:
A 0
B 1
C $1 / 6$
D 6

The correct answer is:
$1 / 6$

Solution:
$\begin{aligned}
5 \tan \theta=4 \Rightarrow \tan \theta & =\frac{4}{5} \\
\therefore \sin \theta & =\frac{4}{\sqrt{41}} \text { and } \cos \theta=\frac{5}{\sqrt{41}} \\
\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta} & =\frac{5 \times \frac{4}{\sqrt{41}}-3 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}} \\
\frac{20-15}{20+10}=\frac{5}{30} & =\frac{1}{6}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $5^{\text {th }}$ term of a H.P. is $\frac{1}{45}$ and $11^{\text{th }}$ term is $\frac{1}{69}$, then its $16^{\text {th }}$ term will be

Options:
A $1 / 89$
B $1 / 85$
C $1 / 80$
D $1 / 79$

The correct answer is:
$1 / 89$

Solution:
Here $5^{\text {th }}$ term of the corresponding
A.P. $=a+4 d=45$\ldots(i)
and $11^{\text {th }}$ term of the corresponding
A.P. $=a+10 d=69$\ldots(ii)
From (i) and (ii), we get$a=29, d=4$
Therefore $16^{\text {th }}$ term of the corresponding A.P.
$=a+15 d=29+15 \times 4=89$
Hence $16^{\text {th }}$ term of the H.P. is $\frac{1}{89}$

Topic: Mathematics, Sequences and Series, JEE Main
If $\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} \mathrm{~d} x=\mathrm{A}\left(\frac{\alpha x-1}{\beta x+3}\right)^B+\mathrm{C}$, where $\mathrm{C}$ is the constant of integration, then the value of $\alpha+\beta+20 \mathrm{AB}$ is__________

Options:

The correct answer is:
7

Solution:
$\begin{aligned} & \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^B+C \\ & I=\int \frac{1}{(x-1)^{4 / 5}(x+3)^{6 / 5}} d x \\ & I=\int \frac{1}{\left(\frac{x-1}{x+3}\right)^{4 / 5}(x+3)^2} d x \\ & \left(\frac{x-1}{x+3}\right)=t \Rightarrow \frac{4}{(x+3)^2} d x=d t \quad t^{-4 / 5+1} \\ & I=\frac{1}{4} \int \frac{1}{t^{4 / 5}} d t=\frac{1}{4} \frac{t^{1 / 5}}{1 / 5}+c \\ & I=\frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1 / 5}+C \\ & A=\frac{5}{4} \quad \alpha=\beta=1 \quad B=\frac{1}{5} \\ & \alpha+\beta+20 A B=2+20 \times \frac{5}{4} \times \frac{1}{5}=7\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2024 (08 Apr Shift 2)