datavorous/entrance-exam-dataset huggingface dataset:
The number of stereoisomers possible for a compound of the molecular formula $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{Me}$ is :

Options:
A
3

B
2

C
4

D
6

The correct answer is:
4

Solution:
About the double bond, two geometrical isomers are possible and the compound is having one chiral carbon.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2009
The number of structural isomers for C6H14 is:

Options:
A 4
B 5
C 6
D 3

The correct answer is:
5

Solution:
Firstly we calculate DU (Degree of unsaturation) of given molecular formula as :-

DU=C+1-H+X-N2

for C6H14

DU=6+1-142=0

∴ possibility is for alkanes only.

(PCC stands for parent carbon chain)

∴ Total structural isomers of C6H14 are 5.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)
The number of subshells associated with n=4 and m=-2 quantum numbers is:

Options:
A 8
B 2
C 16
D 4

The correct answer is:
2

Solution:
For n=4 possible values of ℓ=0,1,2,3 because values ℓ lie between 0 to n-1

We have to check values of m which lie between -ℓ to+ℓ.

ℓ=2 & ℓ=3 can have m=-2 So possible subshells are 2.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9} is

Options:
A 610
B 106
C 910
D 109

The correct answer is:
106

Solution:
Given that A={0,1,2,3,…8,9}

We need to find the number of symmetric matrices.

We know that a matrix is symmetric if Aτ=A.

A=a11a12a13a21a22a23a31a32a33

Now a11,a22,a33 can have any digit from 0 to 9.

Number of ways this can be done is 10×10×10=103

Now a12=a21. So number of ways this can be done is 10×1.

Similarly a13=a31, So number of ways this can be done is 10×1.

Also, a23=a32. So number of ways this can be done is 10×1.

∴ Total symmetric matrix=103×10×10×10=106

Hence this is the correct option.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The number of symmetric relations defined on the set {1,2,3,4} which are not reflexive is _______.

Options:

The correct answer is:
960

Solution:
We know that,

Total number of relation both symmetric and reflexive =2n2-n2, where n is number of elements in given set,

Also, total number of symmetric relation =2n2+n2

Then number of symmetric relation which are not reflexive will be,

=2n(n+1)2-2n(n-1)2

=210-26

=1024-64

=960

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The number of tangents that can be drawn from $(0,0)$ to the circle $x^2+y^2+2 x+6 y-15=0$ is

Options:
A None
B one
C Two
D Infinite

The correct answer is:
None

Solution:
Since $S_1 \lt 0$. Point lies inside the circle, hence no tangent can be drawn.

Topic: Mathematics, Circle, JEE Main
The number of tangents that can be drawn from 2,0 to the curve y=x6 is/are

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
Let A be α,α6 & B be 2,0
So, slope of AB=6α5=α6α-2
⇒α=0 or 6α-12=α
⇒α=0 or α=125
⇒2 tangents are possible

Topic: Mathematics, Application of Derivatives, JEE Main
The number of tangents that can be drawn to y=ex from π,0 is

Options:
A 0
B 1
C 4
D 5

The correct answer is:
1

Solution:
Slope of AB=ex-0x-π=dydxB
⇒exx-π=ex
⇒x-π=1
⇒x=π+1
∴B is π+1,eπ+1
Hence, only one tangent can be drawn.

Topic: Mathematics, Application of Derivatives, JEE Main
The number of tangents with positive slope that can be drawn from the origin to the curve y=sin⁡x is

Options:
A 0
B 2
C 4
D infinitely many

The correct answer is:
infinitely many

Solution:
Slope of OP=cosα=sin⁡αα
⇒tan⁡α=α
(which is true for infinitely many positive values of ‘α’ )

Topic: Mathematics, Application of Derivatives, JEE Main
The number of terminal oxygen atoms present in the product B obtained from the following reaction is

FeCr2O4+Na2CO3+O2→A+Fe2O3+CO2

A+H+→B+H2O+Na+

Options:

The correct answer is:
6

Solution:
4FeCr2O4+8Na2CO3+7O2→8Na2CrO4+2Fe2O3+8CO2

2Na2CrO4+2H+→Na2Cr2O7⏟B+2Na++H2O

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The number of terms common to the two A.P.’s 3,7,11,…,407 and 2,9,16,…,709 is ____________.

Options:

The correct answer is:
14

Solution:
Common Difference of first series =4

Common difference of second series =7

First common term =23

Common difference =L.C.M7×4=28

To find number of common terms we put

Last term ≤407

⇒23+n-1×28≤407

⇒n-1×28≤384

⇒n≤13.71+1

⇒n≤14.71

⇒n=14

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The number of terms in an A.P. is even, the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by 1012, then the number of terms in the A.P. is

Options:
A 4
B 8
C 16
D 12

The correct answer is:
8

Solution:
We know that the nth term and sum of nterms of an A.P. with first term a and common difference d are respectively a+n-1d and n22a+n-1d.

Let, the number of terms in the given A.P. be 2n then there are 2n2=n even terms and n odd terms.

Then, T1=a and T2n=a+(2n-1)d

Given T2n-T1=212

⇒2n-1d=212

⇒2nd-d=212 ...i

Also, the sum of odd terms i.e. a+a+2d+a+4d+... is

n22a+n-12d=24

⇒2a+n-12d=48n ...ii

And, the sum of even terms i.e. a+d+a+3d+a+5d+... is

n22a+d+n-12d=30

⇒n22a+n-12d+2d=30

Put the value from equation ii to get

n248n+2d=30

⇒24+dn=30

⇒nd=6 ...iii

Put this value in the equation i, to get

2×6-d=212

⇒d=24-212=32

Now, put d in the equation iii, to get

⇒n×32=6

⇒n=4

⇒2n=8

Thus, the number of terms in the given A.P. is 8.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (19 Apr Online)
The number of terms in the expansion of 1+x1011-x+x2100 in powers of x is

Options:
A 301
B 302
C 101
D 202

The correct answer is:
202

Solution:
1 + x 1 0 1 1 + x 2 - x 1 0 0
=1+x1+x1001-x+x2100
=1+x1+x3100 ∵a3+b3=a+ba2-ab+b2

1 + x 3 1 0 0 has 101 terms
1 + x 1 + x 3 1 0 0 = 1 1 + x 3 1 0 0 + x 1 + x 3 1 0 0
⇒ No. of terms = 101+101=202

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2014 (09 Apr Online)
The number of terms in the expansion of 516+7191824 which are integers is

Options:
A 100
B 101
C 102
D 103

The correct answer is:
102

Solution:
TK+1=1824CK 51824-K6 7K9
Where, K=0,1,2.....,1824
Now, 1824-K6 will be an integer if
K=0,6,12,18,....,1824
Also, K9 will be an integer if K=0,9,18,.....,1818.
â Both are integers if K=0,18,36,.....,1818
If the number of terms =n,

â1818=0+nâ118ân=102

Topic: Mathematics, Binomial Theorem, JEE Main
The number of terms in the expansion of $(a+b+c)^n$ will be

Options:
A $n+1$
B $n+3$
C $\frac{(n+1)(n+2)}{2}$
D None of these

The correct answer is:
$\frac{(n+1)(n+2)}{2}$

Solution:
By converse rule, put $n=1,2$, then number of terms are 3,6 .
Hence number of terms of $(a+b+c)^n$ $=\frac{(n+1)(n+2)}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main
The number of terms in the expansion of $\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$, in which powers of $x$ and $y$ are free from radical signs are

Options:
A
$\operatorname{six}$

B
twelve

C
seven

D
five

The correct answer is:
$\operatorname{six}$

Solution:
Given expansion is $\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$ The general term is
$$
T_{r+1}={ }^{55} \mathrm{C}_r\left(y^{1 / 5}\right)^{55-r} \cdot\left(x^{\frac{1}{10}}\right)^r
$$
$T_{r+1}$ would free from radical sign if powers of $y$ and $x$ are integers.
i.e. $\frac{55-r}{5}$ and $\frac{r}{10}$ are integer.
$\Rightarrow r$ is multiple of 10 .
Hence, $r=0,10,20,30,40,50$
It is an A.P.
Thus, $50=0+(k-1) 10$ $50=10 k-10 \Rightarrow k=6$
Thus, the six terms of the given expansion in which $x$ and $y$ are free from radical signs.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2012 (12 May Online)
The number of terms in the series $101+99+97+\ldots . .+47$ is

Options:
A $25$
B $28$
C $30$
D $20$

The correct answer is:
$28$

Solution:
Given series $101+99+97+\ldots \ldots \ldots .+47$
So, first term $a=101$, common difference $d=-2$ and last term $I=47$
We know that last term of a series
$T_1=a+(n-1) d \Rightarrow 47=101+(n-1)(-2)$
$\Rightarrow-54=(n-1)(-2) \Rightarrow n=28$

Topic: Mathematics, Sequences and Series, JEE Main
The number of the correct reaction(s) among the following is ______

Options:

The correct answer is:
1

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The number of the given compounds, which can form iodoform with alkali and iodine
Diethyl ketone, Acetone, Ethyl chloride, Catechol, Ethanol, Resorcinol, Isopropyl alcohol, Acetophenone, Butanone, Lactic acid () and salicylic acid.

Options:

The correct answer is:
5

Solution:
Catechol, Mandelic acid and CH3CH2-CO-CH2-CH3âNo yellow ppt

aldehydes & ketones having âCO||âCH3 group & 1Â° or 2Â° alcohols having CH3âC||âOH give yellow ppt CH3 with I2/OHâ

Her acetone, isopropyl alcohol, ethyl alcohol, Butanone & Acetophenone can from CHI3 and rest are not

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of the positive integral solutions x,y,z of the equation xyz=24 is t, then the number of all possible factors of t is

Options:

The correct answer is:
8

Solution:
24=1.1.24=1.2.12=1.3.8=1.4.6=2.2.6=2.3.4
Each of 1,2,12, 1,3,8, 1,4,6, 2,3,4
can be permuted in 3! ways
Each of 1,1,24,2,2,6 can be permuted in 3 ways
The desired number of ways is
4×6+2×3=24+6=30
⇒t=30=2×3×5
Number of factors of 30
=2×2×2=8

Topic: Mathematics, Permutation Combination, JEE Main
The number of the real roots of the equation x+12+|x-5|=274 is ________.

Options:

The correct answer is:
2

Solution:
Case-I

x≤5

x+12-x-5=274

x+12-x+1-34=0

x+1=32,-12

x=12,-32

Case-II

x>5

x+1+x-5=274

x+12+x+1-514=0

x=-1±522 (rejected as x>5)

So, the equation have two real root.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The number of those tangents to the curve y2-2x3-4y+8=0 which pass through the point 1,2 is

Options:
A 0
B 2
C 1
D 3

The correct answer is:
2

Solution:
Let Pα,β be any point on the curve

y2-2x3-4y+8=0

Now, 2ydydx-6x2-4dydx=0

⇒ 2y-4dydx-6x2=0

⇒ dydx=6x22y-4

Thus, slope =m=dydxP=6α22β-4
Now, the equation of the tangent at P is
y-β=6α22β-4x-α
which is passing through 1,2
⇒ 2-β=6α22β-41-α

⇒ 2-β=3α2β-21-α

⇒-β-22=3α21-α

⇒ 3α3-3α2-β2+4β-4=0 ...i

Also, Pα,β lies on y2-2x3-4y+8=0

So, β2-2α3-4β+8=0 ...ii

From (i) and (ii), we get,

⇒-α3+3α2-4=0

⇒α3-3α2+4=0

⇒α3+α2-4α2-4α+4α+4=0

⇒α2α+1-4αα+1+4α+1=0

⇒ α+1α2-4α+4=0

⇒ α+1α-22=0

when α=-1, then β2+2-4β+8=0

Thus, β is imaginary.

when α=2, then β2-16-4β+8=0

⇒β2-4β-8=0

⇒β-22=12=232

⇒β-2=±23

⇒β=2±23

Hence, the point of contacts are
2,2+23 and 2,2-23
Slope of the tangents are 23,-23
Hence, the equations of tangents are
y-2+23=23x-2
and
y-2-23=-23x-2

Hence two tangent are possible.

Topic: Mathematics, Application of Derivatives, JEE Main
The number of three-digit even numbers, formed by the digits 0,1,3,4,6,7 if the repetition of digits is not allowed, is______.

Options:

The correct answer is:
52

Solution:
We have,

0,1,3,4,6,7

For three-digit even number, it must end with either 0, 4, 6

The number of three -digit even numbers ending with 0 is

=5×4=20

The number of three -digit even numbers ending with 4

=4×4=16

The number of three -digit even numbers ending with 6

=4×4=16

Total three digits even numbers =20+16+16=52

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The number of times the digit 0 is used in writing the numbers from 1 to 1000 is equal to

Options:
A 189
B 300
C 192
D 270

The correct answer is:
192

Solution:
Consider the number from 000, 001, 002,....., 998, 999.
Since, each digit has an equal likelihood of coming and digits used are 3Ã1000.
Hence, 0 appears =300 times
But, 000,001,002,....,098,099;
have some unwanted zeros, we subtract
these zeroâs i.e. 3Ã1+2Ã9+1Ã90=111.
Thus, total times 0 appears =300-111=189
Also, add 3 zeroâs in 1000.
â189+3=192.

Topic: Mathematics, Permutation Combination, JEE Main
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is

Options:

The correct answer is:
300

Solution:
3--=10×10=100

_3-=10×10=100

--3=10×10=100

Total is 300.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The number of translational degrees of freedom for a diatomic gas is

Options:
A $2$
B $3$
C $5$
D $6$

The correct answer is:
$3$

Solution:
Number of translational degrees of freedom (3) are same for all types of gases.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The number of tripeptides formed by three different amino acids using each amino acid once is _______.

Options:

The correct answer is:
6

Solution:
A tripeptide is a peptide consisting of three amino acids joined by peptide bonds. The exact nature and function of proteins is determined by the amino acids present and the order they occur.

In these, an amine group from one residue forms an amide bond with the carboxyl group of a second residue, the amino group of the second forms an amide bond with the carboxyl group of the third.

Let 3 different amino acid are A, B, C then following combination of tripeptides can be formed-

ABC, ACB, BAC, BCA, CAB, CBA

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The number of triplets a,b,c of positive integers satisfying the equation a3+1a2ba2cab2b3+1b2cac2bc2c3+1=30 is equal to

Options:
A 3
B 6
C 9
D 12

The correct answer is:
3

Solution:
abcabca3+1a2ba2cab2b3+1b2cac2bc2c3+1=30
⇒abca2+1aa2a2b2b2+1bb2c2c2c2+1c=30
⇒a3+1a3a3b3b3+1b3c3c3c3+1=30
Apply R1↔R1+R2+R3
⇒a3+b3+c3+1111b3b3+1b3c3c3c3+1=30
Apply C2↔C2-C1, C3↔C3-C1
⇒a3+b3+c3+1100b310c301=30
⇒a3+b3+c3+1=30⇒a3+b3+c3=29
a,b,c≡1,1,3 or 1,3,1 or 3,1,1

Topic: Mathematics, Determinants, JEE Main
The number of turns in a coil of Galvanometer is tripled, then

Options:
A Both voltage and current sensitivity remains constant
B Voltage sensitivity increases \( 3 \) times and current sensitivity remains constant
C Both voltage and current sensitivity decreases by \( 33 \% \)
D Voltage sensitivity remains constant and current sensitivity increases \( 3 \) times

The correct answer is:
Voltage sensitivity remains constant and current sensitivity increases \( 3 \) times

Solution:
(D)
\[
\frac{\theta}{I}=\frac{N A B}{C}
\]

Topic: Physics, Current Electricity, KCET, KCET 2019
The number of turns of primary and secondary coils of a transformer is 5 and 10 respectively and mutual inductance of the transformer is 25 H. Now, the number of turns in primary and secondary are made 10 and 5 respectively. Mutual inductance of the transformer will be

Options:
A 25 H

B 12.5 H
C 50 H
D 6.25 H

The correct answer is:
25 H

Solution:
As M âN1N2, therefore, M remains the same.

Topic: Physics, Electromagnetic Induction, NEET
The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%. The percentage change in voltage sensitivity of the galvanometer will be :

Options:
A 100%
B 50%
C 75%
D 0%

The correct answer is:
0%

Solution:
Current sensitivity = Voltage sensitivity ×R

If current sensitivity is made 1.5 times, then R also increase 1.5 times.

Hence, voltage sensitivity =1.5× current sensitivity 1.5×R

= no change.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The number of unbranched isomers (including stereoisomers) of C6H12 are

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of unit cells in 58.5 g of NaCl is approximately

Options:
A 6×1020
B 1.5×1023
C 6×1023
D 0.5×1024

The correct answer is:
1.5×1023

Solution:
58.5 g NaCl=6×1023 molecule (atoms) of NaCl (as NaCl occupies FCC system)
1-unit cell have 4 molecules of NaCl
So, no. of unit cell =6×10234=1.5×1023

Topic: Chemistry, Solid State, JEE Main
The number of unit cells present in 1 g cube shaped ideal crystal of solid A+B- (with CCP close packing) (formula mass =60 ) are

Options:
A 2.5×1021
B 6.02×1023
C 4.0×1022
D 1.0×1022

The correct answer is:
2.5×1021

Solution:
Number of moles present in 1g=160
So number of formula units in 160 moles
=160×6.023×1023=1022=10×1021
Since 4 formula units constitute a unit cell
So number of unit cells present
=104×1021=2.5×1021

Topic: Chemistry, Solid State, JEE Main
The number of unit cells present in 39 g of potassium if it crystallizes as body centred cube is ( N = Avogadro number, At. Wt. of potassium = 39 g/mol )

Options:
A N 4
B N 2
C N 3
D N

The correct answer is:
N 2

Solution:
Potassium has bcc system
Number of mole =â39/39â=â1 mole
1 mole of atoms = 6.022Ã 10 23 atoms =âN
âµ In bcc system 2 atoms are present in 1 unit cell.
∴ N number of atoms are present in N 2 unit cells.

Topic: Chemistry, Solid State, JEE Main
The number of units, which are used to express concentration of solutions from the following is _________.

(Mass percent, Mole, Mole fraction, Molarity, ppm, Molality.)

Options:

The correct answer is:
5

Solution:
Both solute and solvent(solution) terms are involved in the concentration term of solution. Mass percent, mole fraction, molarity, ppm, molality are used for measuring concentration terms. The number of moles is mass to molar mass ratio, it is not a concentration term for a solution.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of unpaired electrons in [CoF6]3- are

Options:

The correct answer is:
4

Solution:
Co3+=
In the presence of weak field ligand,
[CoF6]3-=
∴ No. of unpaired electrons =4

Topic: Chemistry, Coordination Compounds, JEE Main
The number of unpaired electrons in Gadolinium $[\mathrm{Z}=64]$ is

Options:
A
3

B
8

C
6

D
2

The correct answer is:
8

Solution:
$\mathrm{Gd}(64)=[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$
$\therefore$ No. of unpaired electrons $=8$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (26 May Online)
The number of unpaired electrons in
$\quad\left[\mathrm{NiCl}_{4}\right]^{2-}, \mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2}$
respectively are

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

The correct answer is:
2,0,1

Solution:
Configuration for metal (M) in various complex.

$\mathrm{Ni}^{2 *}$ has 2 unpaired electrons

$\mathrm{Ni}^{0}$ has zero unpaired electrons with
$s p^{3}$ -hybridisation.
$\mathrm{Cu}^{2}$ "has one unpaired electron with $d s p^{2}$ -hybridisation.

$\therefore$ (b) is the correct answer, i.e.
Number of unpaired electron (2,0,1)

Topic: Chemistry, Coordination Compounds, JEE Main
The number of values of θ∈0, π for which the system of linear equations
x+3y+7z=0
-x+4y+7z=0
sin⁡3θx+cos⁡2θy+2z=0
has a non-trivial solution, is:

Options:
A Two
B Three
C Four
D One

The correct answer is:
Two

Solution:
For non-trivial solutions
137-147sin3θcos2θ2=0
⇒4sin3θ+4sin2θ-3sinθ=0
⇒sinθ=0 or sinθ=12 or sinθ=-32
Since θ∈0, π
⇒θ=π6 and 5π6 will satisfy the equation

because sinθ=-32 is not possible as range of sinθ is -1,1
⇒ Number of θ ' s=2

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The number of values of Î¸ââ3Ï2,4Ï3 which satisfies the system of equations 2sin2Î¸+sin22Î¸=2 and sin2Î¸+cos2Î¸=tanÎ¸ is

Options:
A 2
B 4
C 6
D

8

The correct answer is:
6

Solution:
First equation is 4sin2Î¸cos2Î¸=2cos2Î¸
â cos2â¡Î¸=0 or sin2Î¸=12=122=sin2Ï4
â Î¸=2n+1Ï2, nâI or Î¸=nÏÂ±Ï4, nâI
Second equation is not satisfied by Î¸=2n+1Ï2, nâI but satisfied by Î¸=nÏÂ±Ï4, nâI

So Î¸=nÏÂ±Ï4,nâI

â´ In [â3Ï2,4Ï3], the values of Î¸ are

â5Ï4,â3Ï4,âÏ4,Ï4,3Ï4,5Ï4

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of values of a for which the curves 4x2+a2y2=4a2 and y2=16x are orthogonal is

Options:

The correct answer is:
2

Solution:
Given curves are

4x2+a2y2=4a2 ...(i)

and y2=16x ...(ii)

If the curves intersect at Pα,β, then

α2a2+β24=1 and β2=16α.

On differentiating equation (i), we get,
2xa2+2y4y'=0
⇒ y ′ =− 4x a 2 y
⇒m1=-4αa2β

On differentiating equation (ii), we get,
2yy'=16
⇒m2=8β

For curves to be orthogonal,
m1m2=-1

i.e. -4αa2β8β=-1
⇒32α=a2β2
⇒2β2=a2β2
⇒a2=2
⇒a=±2
Hence, two values of a

Topic: Mathematics, Application of Derivatives, JEE Main
The number of values of \( a \) for which the curves \( 4 x^{2}+a^{2} y^{2}=4 a^{2} \) and \( y^{2}=16 x \) are orthogonal is

Options:

The correct answer is:
2

Solution:
Given curves are

4x2+a2y2=4a2 ...(i)

and y2=16x ...(ii)

If the curves intersect at Pα,β, then

α2a2+β24=1 and β2=16α.

On differentiating equation (i), we get,
2xa2+2y4y'=0
⇒ y ′ =− 4x a 2 y
⇒m1=-4αa2β

On differentiating equation (ii), we get,
2yy'=16
⇒m2=8β

For curves to be orthogonal,
m1m2=-1

i.e. -4αa2β8β=-1
⇒32α=a2β2
⇒2β2=a2β2
⇒a2=2
⇒a=±2
Hence, two values of a

Topic: Mathematics, Application of Derivatives, JEE Main
The number of values of a∈N such that the variance of 3, 7, 12, a, 43-a is a natural number is:

Options:
A 0
B 2
C 5
D infinite

The correct answer is:
0

Solution:
Calculating Mean =3+7+12+a+43-a5=13

Variance =9+49+144+a2+43-a25-132∈N

⇒2a2-a+15=n (where n∈N)

⇒2a2-a+1-5n=0 must have solution as natural numbers

So, D=40n-7 always has 3 at unit place for any value of n

⇒D can't be perfect square

So, a can't be integer.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The number of values of α for which the system of equations
x+y+z=α

αx+2αy+3z=-1

x+3αy+5z=4

is inconsistent, is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
Given linear equations are

x+y+z=α

αx+2αy+3z=-1

x+3αy+5z=4

For this system of linear equations to have inconsistent solution

D=111α2α313α5=0

⇒α-12=0

⇒α=1

Now for α=1

D1=111-123435

=10-9--5-12+-3-8

=1+17-11≠0

Hence, for α=1 the system of equation has inconsistent solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The number of values of α in 0, 2π for which 2sin3α-7sin2α+7sinα=2, is :

Options:
A 3
B 1
C 6
D 4

The correct answer is:
3

Solution:
Given 2sin3 α-7sin2 α+7sin α=2
⇒2sin3 α-1-7 sin α sin α-1=0
⇒sin α-12sin2 α+sin α+1-7 sin α=0 ∵a3-b3=a-ba2+ab+b2
⇒sin α-12 sin2 α-5 sin α+2=0
⇒sin α-12 sin α-1sin α-2=0
∴ sin α=1, sin α=12, sin α≠2, ∵sinx∈-1,1
Now, sin α=1 ⇒ α=π2 ∵ α∈0, 2π
sin α=12 ⇒ α=π6, 5π6 ∵ α ∈0, 2π

Hence, 3 values of α in 0 , 2 π .

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2014 (09 Apr Online)
The number of values of Î± in -10Ï,10Ï for which the equations sinâ¡Î±x-cosâ¡Î±y+3z=0, cosâ¡Î±x+sinâ¡Î±y-2z=0 and 2x+3y+cosâ¡Î±z=0 have nontrivial solution is

Options:
A 10
B 20
C 40
D 15

The correct answer is:
20

Solution:
For non-trivial solution ∆=0
sin⁡α-cos⁡α3cos⁡αsin⁡α-223cos⁡α=0
⇒sin⁡αsin⁡αcos⁡α+6+cos⁡αcos2⁡α+4+33cos⁡α-2sin⁡α=0
⇒sin2⁡αcos⁡α+6sin⁡α+cos3⁡α+4cos⁡α+9cos⁡α-6sin⁡α=0
⇒14cos⁡α=0⇒cos⁡α=0⇒α=2n+1π2,n∈I
⇒ Number of values of α in -10π,10π is 20

Topic: Mathematics, Determinants, JEE Main
The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta$ is

Options:

The correct answer is:
3

Solution:
Given, $\tan \theta=\cot 5 \theta$
$$
\begin{aligned}
& \Rightarrow \quad \tan \theta=\tan \left(\frac{\pi}{2}-5 \theta\right) \\
& \Rightarrow \quad \frac{\pi}{2}-5 \theta=n \pi+\theta \\
& \Rightarrow \quad 6 \theta=\frac{\pi}{2}-n \pi \\
& \Rightarrow \quad \theta=\frac{\pi}{12}-\frac{n \pi}{6} \\
& \text { Also, } \cos 4 \theta=\sin 2 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right) \\
& \Rightarrow \quad 4 \theta=2 n \pi \pm\left(\frac{\pi}{2}-2 \theta\right)
\end{aligned}
$$
Taking positive
$$
6 \theta=2 n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{12}
$$
Taking negative
$$
2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4}
$$
Above values of $\theta$ suggests that there are only 3 common solutions.

Topic: Mathematics, Trigonometric Equations, JEE Main