datavorous/entrance-exam-dataset huggingface dataset:
The lanthanide ion that would show colour is:

Options:
A Lu3+
B La3+
C Gd3+
D Sm3+

The correct answer is:
Sm3+

Solution:
The elements or ions having unpaired electrons show colour. The valence shell electronic configuration of La3+ and Lu3+ are 4f0 and 4f14 respectively. Hence, they are colourless. Gd3+ is also colourless, this is because, here 4f orbitals are half filled,hence stable.As such f-f transitions are not possible because they will destroy the symmetry of 4f and make them unstable. Sm3+ has the valence shell electronic configuration 4f5, hence, it is coloured due to the presence of unpaired electrons.

La+3= Colourless
Gd+3= Colourless
Sm+3= Pale yellow colour
Lu+3= Colourless

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The lanthanoid that does NOT show + 4 oxidation state is:

Options:
A Dy
B Ce
C Eu
D Tb

The correct answer is:
Eu

Solution:
Eu(z=63)=4F76s2
Dy(z=66)=4F106s2
Tb(z=65)=4F96s2
Ce(58)=4F15d16s2
Dy, Tb and Ce show +4 oxidation state while Eu do not show +4 oxidation state.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The larger of $\cos (\log \theta)$ and $\log (\cos \theta)$, if $e^{-\pi / 2} < \theta < \pi / 2$ is

Options:
A $\cos (\log \theta)$
B $\log (\cos \theta)$
C none of function is larger
D one of the two function is undefined on domain even to compare

The correct answer is:
$\cos (\log \theta)$

Solution:
$$
\begin{aligned}
& \text { } \cos (\log \theta), \theta \in\left(e^{\frac{-\pi}{2}}, \frac{\pi}{2}\right) \\
& e^{-\frac{\pi}{2}} < \theta < \frac{\pi}{2} \\
& \Rightarrow \quad-\frac{\pi}{2} < \log \theta < \log \frac{\pi}{2} \leq \frac{\pi}{2}
\end{aligned}
$$
So, $\cos (\log \theta) \geq 0$
$$
\log (\cos \theta) \leq 0
$$
$$
0 \leq \cos \theta \leq 1 \text { in } \theta \in\left[e^{-\frac{\pi}{2}}, \frac{\pi}{2}\right]
$$
$$
\therefore \quad \log (\cos \theta) \leq 0
$$
Hence, $\cos (\log \theta) \geq \log (\cos \theta)$.

Topic: Mathematics, Trigonometric Ratios & Identities, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 1)
The larger of two angles made with the X-axis of a straight line drawn through (1,2) so that it intersects the line x+y=4 at a point distant 6/3 from the point (1,2) is

Options:
A 60°
B 75°
C 105°
D None of these

The correct answer is:
75°

Solution:
Let A=(1,2) equation of line is

x-1cosθ=y-2sinθ=r=63

∴ x=63cosθ+1,y=63sinθ+2

This point lies on line x+y=4

∴ 63cosθ+63sinθ=1

⇒ sinθ+cosθ=32

⇒ sinπ4+θ=32

⇒ π4+θ=60° or 120°

⇒ θ=15° or 75°

∴ Larger angle =75°

Topic: Mathematics, Straight Lines, JEE Main
The largest interval lying in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for which the function $\left[f(x)=4^{-x^2}+\cos ^{-1}\left(\frac{x}{2}-1\right)+\log (\cos x)\right]$ is defined, is

Options:
A
$[0, \pi]$

B
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

C
$\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$

D
$\left[0, \frac{\pi}{2}\right)$

The correct answer is:
$\left[0, \frac{\pi}{2}\right)$

Solution:
$f(x)$ is defined if $-1 \leq \frac{x}{2}-1 \leq 1$ and $\cos x>0$
or $0 \leq x \leq 4$ and $-\frac{\pi}{2} < x < \frac{\pi}{2}$
$\therefore \mathrm{x} \in\left[0, \frac{\pi}{2}\right) \text {. }$

Topic: Mathematics, Functions, JEE Main, JEE Main 2007
The largest magnitude the electric field on the axis of a uniformly charged ring of radius 3 m is at a distance h from its centre. What is the value of h? Take 12=0.7

Options:

The correct answer is:
2.1

Solution:
Eaxial =Qh4ÏÎµ0h2+a232

where,

a=radius of the ring,

h=distance of point from the centre.

For the maximum value,

dEdh=0

â´ ddhQx4ÏÎµ0h2+a232=0

â´h2+a232-h32h2+a212Â·(2h)h2+a23=0

â´ h2+a2-3h2=0

â´ h=Â±a2

=32 m â¦.(âµa=3 m)

=3Ã0.7=2.1 m

Topic: Physics, Electrostatics, JEE Main
The largest natural number n such that 3n divides 66! is _______

Options:

The correct answer is:
31

Solution:
We know that,

Maximum value of n for pn where p is prime divides the number a! is given by,

ap+ap2+ap3+ap4+...

Since, 3 is a prime number, so we have

=663+6632+6633+6634+...

=22+7+2+0

=31

So, maximum value of n is 31.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The largest non-negative integer $k$ such that $24^{\mathrm{k}}$ divides $13 !$ is.

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
$24^{k} \rightarrow\left(2^{3} \times 3\right)^{k}$
Exponent of 2 in 13! $\left[\frac{13}{2}\right]+\left[\frac{13}{2^{2}}\right]+\left[\frac{13}{2^{3}}\right]=10$
Exponent of 3 in $13 !$ $\left[\frac{13}{3}\right]+\left[\frac{13}{3^{2}}\right]=5$
So $\left(2^{3} \times 3\right)^{3}$ So $\mathrm{K}=3$

Topic: Mathematics, Permutation Combination, JEE Main
The largest perfect square that divides $2014^{3}-2013^{3}+2012^{3}-2011^{3}+\ldots .+2^{3}-1^{3}$ is

Options:
A $\mathrm1^{2}$
B $2^{2}$
C $1007^{2}$
D $2014^{2}$

The correct answer is:
$1007^{2}$

Solution:
$S=2014^{3}-2013^{3}+2012^{3}-2011^{3}+\ldots .+2^{3}-1^{3}$
$\sum_{t=1}^{2014} r^{2}+\sum_{t=1}^{107}(2 r-1)(2 r)$
$\mathrm{S}=\sum_{r=1}^{2014} r^{2}+\sum_{r=1}^{1007}\left(4 r^{2}-2 r\right)$
$\mathrm{S}=\frac{(2014)(2015)(4029)}{6}+\frac{4 \times 1007 \times 1008 \times 2015}{6}-\frac{2 \times 1007 \times 1008}{2}=(1007)^{2} \times 4031$

Topic: Mathematics, Sequences and Series, JEE Main
The largest value of a, for which the perpendicular distance of the plane containing the lines r→=i^+j^+λi^+aj^-k^and r→=i^+j^+μ-i^+j^-ak from the point 2,1,4 is 3, is ______.

Options:

The correct answer is:
2

Solution:
Given,

r→=i^+j^+λi^+aj^-k^

r→=i^+j^+μ-i^+j^-ak

Direction ratio of plane containing these lines is given by,

i^j^k^1a-1-11-a=i^1-a2-j^-a-1+k^1+a

n→=1-ai^+j^+k^

Also plane is passing through the point 1,1,0

So, equation of plane will be,

1-ax-1+y-1+z-0=0

1-a x+y+z+a-2=0

Now distance of point 2,1,4 from this plane is 3

∴ D=1-a2+1+4+a-21-a2+1+1=3

⇒5-a=3·a2-2a+3

⇒a2+2a-8=0

⇒ a=2,-4

∴ largest value of a=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The largest value of r, for which the region represented by the set ω∈Cω-4-i≤r is contained in the region represented by the set z∈Cz-1≤z+i, is equal to :

Options:
A 22
B 322
C 17
D 522

The correct answer is:
522

Solution:
z-1≤z+i
Putting z=x+iy and simplifying, we get x+y≥0
The region in shown shaded right side of the line x+y=0
The largest value of r would be the length of perpendicular from A4,1, on the line x+y=0
4+12= 52= 522

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2015 (10 Apr Online)
The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122 \mathrm{~nm}$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

Options:
A
$802 \mathrm{~nm}$

B
$823 \mathrm{~nm}$

C
$1882 \mathrm{~nm}$

D
$1648 \mathrm{~nm}$

The correct answer is:
$823 \mathrm{~nm}$

Solution:
The series in $U-V$ region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from $n=2$ to $n=1$.
$$
\therefore \quad 122=\frac{\frac{1}{R}}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}
$$
The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series.
$$
\lambda=\frac{\frac{1}{R}}{\left(\frac{1}{3^2}-\frac{1}{\infty}\right)}
$$
Solving Eqs. (i) and (ii), we get
$\therefore$ Correct option is (b).
$$
\lambda=823.5 \mathrm{~nm}
$$

Topic: Physics, Atomic Physics, JEE Main
The Lassiagne's extract is boiled with dil HNO3 before testing for halogens because,

Options:
A AgCN is soluble in HNO3
B Silver halides are soluble in HNO3
C Ag2 S is soluble in HNO3
D Na2 S and NaCN are decomposed by HNO3

The correct answer is:
Na2 S and NaCN are decomposed by HNO3

Solution:
The Lassaigne’s extract is first boiled with dil. HNO3 to decompose cyanide or sulphide of sodium (NaCN or Na2S) formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogen.

NaCN ( aq ) + HNO3 ( aq ) → NaNO3 ( aq ) + HCN (↑)

Na2S ( aq ) + 2 HNO3 ( aq ) → 2 NaNO3 ( aq ) + H2S ( ↑ )

Hence option D is the answer.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The last digit of \( (2137)^{754} \) is

Options:
A \( 2 \)
B \( 3 \)
C \( 7 \)
D \( 9 \)

The correct answer is:
\( 9 \)

Solution:
Units digit of 74n+1 is 7, that of 74n+2 is 9, that of 74n+3 is 1 and that of 74n is 7, where n∈N.

Now, 754=4·188+2.

Thus, units digit of (2137)754 will be same as that of 74n+2 i.e. 9.

Topic: Mathematics, Binomial Theorem, JEE Main
The last term in the expansion of 213+2-12n is 3-53log38. Then, the value of the fifth term is

Options:
A 110
B 210
C 310
D 220

The correct answer is:
210

Solution:
We are given that last terms is 3-53log38
=3-533log32
3log32-5=2-5=132
Also, last term is
nCn2-12n=2-n2⇒2-5=2-n2⇒n=10
The fifth term is
10C421362-124=10×9×8×74×3×2×22×2-2
=210

Topic: Mathematics, Binomial Theorem, JEE Main
The last two digits of the number 2314 are

Options:
A 01
B 03
C 09
D 17

The correct answer is:
09

Solution:
2314=5297=530-17
=C705307-C715306+.…….-C755302+C76530-1
=100K+7×530-1
=100K+3709
⇒ last two digits are 09

Topic: Mathematics, Binomial Theorem, JEE Main
The lattice energy of solid NaCl is 180 kcal/mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1 kcal/mol. If the hydration energies of Na+ and Cl- are in the ratio 6 : 5, what is the enthalpy of hydration of Na+ ion ?

Options:
A - 8.5 kcal mol-1
B - 97.64 kcal mol-1
C + 82.6 kcal mol-1
D + 100 kcal mol-1

The correct answer is:
- 97.64 kcal mol-1

Solution:
Î H solu = Î H hydration + Î H Lattice
1 = Î H hydration + 1 8 0
Î H hydration = - 1 7 9 kcal m - 1
Î H hyd Na + + Î H hyd cl - = -1 7 9 kcal m - 1
Î H hyd Na + = - 1 7 9 Ã 6 1 1
= - 9 7 Â· 6 4 kcal m - 1

Topic: Chemistry, Thermodynamics (C), JEE Main
The latusrectum of the parabola \( y^{2}=4 a x \), whose focal chord is \( P S Q \), such that \( S P=3 \) and \( S Q=2 \) is given by

Options:
A \( \frac{24}{5} \)
B \( \frac{12}{5} \)
C \( \frac{6}{5} \)
D \( \frac{1}{5} \)

The correct answer is:
\( \frac{24}{5} \)

Solution:
Since, the semi latus-rectum of a parabola is the HM of segments of a focal chord.

Hence the semi latus rectum is:-
=2SP .SQSP+SQ
=2×3×23+2=125
∴ Latus rectum of the parabola=245

Topic: Mathematics, Parabola, JEE Main
The law of thermodynamics which leads to the conclusion that it is impossible to convert 'all' the heat extracted from a hot source into work is :

Options:
A zeroth law
B first law
C second law
D third law

The correct answer is:
second law

Solution:
Second law of thermodynamics- It is impossible to construct an engine, which will produce no effect other than extracting heat from the reservoir and performing an equivalent amount of work.

This is applicable to a heat engine. It indicates that a working substance operating in a cycle cannot convert all the heat extracted from the source into mechanical work. It must reject some heat to the sink at a lower temperature.

Topic: Physics, Thermodynamics, JEE Main
The law of triad is applicable to a group of

Options:
A \( \mathrm{Cl}, \mathrm{Br}, \mathrm{I} \)
B \( \mathrm{C}, \mathrm{N}, \mathrm{O} \)
C \( \mathrm{Na}, \mathrm{K}, \mathrm{Rb} \)
D \( \mathrm{H}, \mathrm{O}, \mathrm{N} \)

The correct answer is:
\( \mathrm{Cl}, \mathrm{Br}, \mathrm{I} \)

Solution:
According to Dobereiner, the atomic weight of the middle element is nearly the same as the average atomic weights of the other two elements.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The law related to the integrated magnetic field around a closed loop to the electric current passing through the loop is called ______

Options:
A Coulomb's Law
B Lenz's Law
C Ampere's Law
D More than one of the above

The correct answer is:
Ampere's Law

Solution:
- Ampere's Law: Line integral of the magnetic field B around any closed curve is equal to $\mu_0$ times the net current I threading through the area enclosed by the curve i.e.
$\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 I$
where I - current flowing

Topic: Physics, Magnetic Properties of Matter, JEE Main
The laws of electromagnetic induction have been used in the construction of a

Options:
A galvanometer
B voltmeter
C electric motor
D generator

The correct answer is:
generator

Solution:
Hint:
The laws of Electromagnetic Induction give us idea about the various ways in which we can produce Electric Current using Electric or Magnetic field.
Correct Option is D.
Explanation for correct answer:
- Galvanometer is a device which detects current in the circuit.
- Voltmeter is a device which measures potential difference between two points in the circuit.
- Electric motor is a device which converts Electrical Energy into Mechanical Energy.
- Electric Generator is an device which produces Electric Current from Mechanical energy, Chemical energy, or by using Magnetic or Electric Fields. The laws if Electromagnetic Induction are used in Electric Generator which produces Electric Current from Magnetic or Electric Fields.

Topic: Physics, Electromagnetic Induction, JEE Main
The layer of atmosphere between 10 km to 50 km above the sea level is called as:

Options:
A stratosphere
B troposphere
C mesosphere
D thermosphere

The correct answer is:
stratosphere

Solution:
Based on fact layer of atmosphere between 10 km to 50 km above sea level is called as stratosphere.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The layers of atmosphere are heated through

Options:
A Convection
B Conduction
C Radiation
D (b) and (c) both

The correct answer is:
Convection

Solution:
The layers if the atmosphere are heated through Radiation from sun and convection currents, are responsible for heating.

Topic: Physics, Thermal Properties of Matter, JEE Main
The least count of a measuring scale is \( 1 \mathrm{~mm} \). The heights two students is measured with it. What will be the error in the difference between their heights?

Options:
A \( 1 \mathrm{~mm} \)
B \( 0 \mathrm{~mm} \)
C \( 2 \mathrm{~mm} \)
D \( 0.5 \mathrm{~mm} \)

The correct answer is:
\( 2 \mathrm{~mm} \)

Solution:
Given,
The least count, ∆l=1 mm
The error in the reading of the first student, ∆l1=∆l=1 mm
The error in the reading of the second student, ∆l2=∆l=1 mm
During addition or subtraction of two measurements, the error gets added up.
Hence, the net error becomes, ∆lnet=∆l1+∆l2=2 mm

Topic: Physics, Mathematics in Physics, JEE Main
The least count of the main scale of a screw gauge is $1 \mathrm{~mm}$. The minimum number of divisions on its circular scale required to measure $5 \mu \mathrm{m}$ diameter of a wire is

Options:
A 50
B 200
C 100
D 500

The correct answer is:
200

Solution:
Least count of the main scale of screw gauge $=1 \mathrm{~mm}$
Least count of screw gauge $=$ Pitch/Number of division on the circular scale
$5 \times 10^{-6}=10^{-3} / \mathrm{N}$
$N=200$
Answer: (2) 200

Topic: Physics, Mathematics in Physics, JEE Main
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is:

Options:
A 50
B 100
C 500
D 200

The correct answer is:
200

Solution:
Least count of screw gauge =5 μm
L.C=pitchno. of div on cirxular  scale
5 μm=1 mmN
N=200

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The least count of the main scale of a vernier calipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of the vernier scale coincides with a division of the main scale and the zero of vernier scale is lying right side of the zero of the main scale. When this vernier is used to measure the length of the cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4th VSD coincides with the main scale division. The length of the cylinder is (VSD is vernier scale division)

Options:
A 3.2cm
B 3.21cm
C 3.07cm
D 2.99cm

The correct answer is:
3.07cm

Solution:
Zero Error = 0+7×0.1=0.070

Vernier reading = (3.1+4×0.01) –0.07=3.07

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The least integral value of a for which the graphs $y=2 a x+1$ and $y=(a-6) x^2-2$ do not intersect is

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

The correct answer is:
$-5$

Solution:
$(a-6) x^2-2=2 a x+1$
$(a-6) x^2-2 a x-3=0$
Since the graphs do not intersect,
$\therefore \mathrm{D} \lt 0$
$\Rightarrow 4 a^2+4 \times 3(a-6) \lt 0$
$\Rightarrow a^2+3 a-18 \lt 0$
$\Rightarrow(\mathrm{a}+6)(\mathrm{a}-3) \lt 0$
$\Rightarrow \mathrm{a} \in(-6,3)$
Hence, the least integer value of a is -5

Topic: Mathematics, Quadratic Equation, JEE Main
The least integral value of a for which the graphs y=2ax+1 and y=a-6x2-2 do not intersect is

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

The correct answer is:
-5

Solution:
a-6x2-2=2ax+1
a-6x2-2ax-3=0

Since the graphs do not intersect,
∴D<0⇒4a2+4×3a-6<0

⇒a2+3a-18<0
⇒a+6a-3<0
⇒a∈-6,3

Hence, the least integer value of a is -5

Topic: Mathematics, Quadratic Equation, JEE Main
The least integral value $\alpha$ of $x$ such that $\frac{x-5}{x^2+5 x-14}>0$, satisfies :

Options:
A
$\alpha^2+3 \alpha-4=0$

B
$\alpha^2-5 \alpha+4=0$

C
$\alpha^2-7 \alpha+6=0$

D
$\alpha^2+5 \alpha-6=0$

The correct answer is:
$\alpha^2+3 \alpha-4=0$

Solution:
$$
\text { } \begin{aligned}
& \frac{x-5}{x^2+5 x-14}>0 \Rightarrow x^2+5 x-14 < x-5 \\
& \Rightarrow x^2+4 x-9 < 0 \\
& \Rightarrow \alpha=-5,-4,-3,-2,-1,0,1
\end{aligned}
$$
$\alpha=-5$ does not satisfy any of the options
$\alpha=-4$ satisfy the option (a) $\alpha^2+3 \alpha-4=0$

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2013 (23 Apr Online)
The least number of oxyacids are formed by:

Options:
A Chlorine
B Fluorine
C Sulphur
D Nitrogen

The correct answer is:
Fluorine

Solution:
Fluorine does not form oxyacids as it is the most electronegative However, at -40oC it forms HOF. (Unstable at room temperature). Chlorine forms four types of oxyacids. Sulphur forms many oxy acids. Nitrogen mainly forms two types of oxyacids.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2015 (10 Apr Online)
The least positive integer n for which 1+i31-i3n=1 is

Options:
A 2
B 5
C 6
D 3

The correct answer is:
3

Solution:
Given 1+i31-i3n=1
⇒-2ω2-2ωn=1
⇒ωn=1
Hence, the least positive integer value of n is 3.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2018 (16 Apr Online)
The least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{n-1}} < \frac{1}{100}$, is:

Options:
A
4

B
5

C
6

D
7

The correct answer is:
5

Solution:
$$
\begin{aligned}
&1-\frac{2}{3}-\frac{2}{3^2} \ldots \cdot \frac{2}{3^{n-1}} < \frac{1}{100} \\
&\Rightarrow 1-\frac{2}{3}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \frac{1}{3^{n-1}}\right] < \frac{1}{100} \\
&\Rightarrow \frac{1-2\left[\frac{1}{3}\left(\frac{1}{3^n}-1\right)\right]}{\frac{1}{3}-1} < \frac{1}{100} \\
&\Rightarrow 1-2\left[\frac{3^n-1}{2.3^n}\right] < \frac{1}{100} \\
&\Rightarrow 1-\left[\frac{3^n-1}{3^n}\right] < \frac{1}{100} \\
&\Rightarrow 1-1+\frac{1}{3^n} < \frac{1}{100}
\end{aligned}
$$
$$
\Rightarrow 100 < 3^n
$$
Thus, least value of $n$ is 5

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (12 Apr Online)
The least positive integer n such that (2i)n(1-i)n-2,i=-1, is a positive integer, is ______.

Options:

The correct answer is:
6

Solution:
(2i)n(1-i)n-2

=(2i)n(-2i)n-22

=(2i)n+22(-1)n-22

=2n+22in+22(-1)n-22
Clearly n must be even n=2,4 rejected.

So the least positive integer possible is n=6

Topic: Mathematics, Complex Number, JEE Main
The least positive integer n such that (2i)n(1-i)n-2,i=-1, is a positive integer, is ______.

Options:

The correct answer is:
6

Solution:
(2i)n(1-i)n-2

=(2i)n(-2i)n-22

=(2i)n+22(-1)n-22

=2n+22in+22(-1)n-22
Clearly n must be even n=2,4 rejected.

So the least positive integer possible is n=6

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The least positive integral value of α, for which the angle between the vectors αi^-2j^+2k^ and αi^+2αj^-2k^ is acute, is _____.

Options:

The correct answer is:
5

Solution:
We know that, a→.b→=a→b→cosθ

⇒cosθ=(αi^-2j^+2k^)·(αi^+2αj^-2k^)α2+4+4α2+4α2+4

⇒cosθ=α2-4α-4α2+85α2+4

Since θ is acute,

⇒α2-4α-4>0

⇒α2-4α+4>8

⇒(α-2)2>8

⇒α-2>22 or α-2<-22

α>2+22 or α<2-22

So, the least positive integral value of α is 5.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The least positive integral value of k for which cosâ¡2Ï7-sinâ¡2Ï7sinâ¡2Ï7cosâ¡2Ï7k=1001 is

Options:
A 0
B 3
C 7
D 14

The correct answer is:
7

Solution:
A=cosâ¡2Ï7-sinâ¡2Ï7sinâ¡2Ï7cosâ¡2Ï7
A2=cosâ¡2Ï7-sinâ¡2Ï7sinâ¡2Ï7cosâ¡2Ï7cosâ¡2Ï7-sinâ¡2Ï7sinâ¡2Ï7cosâ¡2Ï7=cosâ¡4Ï7-sinâ¡4Ï7sinâ¡4Ï7cosâ¡4Ï7
Similarly,
Ak=cosâ¡2kÏ7-sinâ¡2kÏ7sinâ¡2kÏ7cosâ¡2kÏ7=1001
â cosâ¡2kÏ7=1,sinâ¡2kÏ7=0
â 2kÏ7=2nÏ, 2kÏ7=mÏ
âk=7n, k=7m2
â Least positive integral value of k=7

Topic: Mathematics, Matrices, JEE Main
The least positive term of an arithmetic progression whose first two terms are 52 and 2312 is

Options:
A 6
B 5
C 16
D 377

The correct answer is:
16

Solution:
Tn=52+n-12312-52
=52+n-1-712=30-7n+712
Tn>0⇒37-7n12>0
⇒n<377⇒n=5
T5=37-3512=16

Topic: Mathematics, Sequences and Series, JEE Main
The least positive value of ‘ a ’ for which the equation, 2x2+a-10x+332=2a has real roots is ___________.

Options:

The correct answer is:
8

Solution:
Dâ¥0
a-102-42332-2aâ¥0
a-102-433-4aâ¥0
a2-4a-32â¥0⇒aâ-â,-4âª8,ââ

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The least positive value of $t$, so that the lines $x=t+\alpha, y+16=0 \quad$ and $\quad y=\alpha x$ are concurrent, is

Options:
A 2
B 4
C 16
D 8

The correct answer is:
8

Solution:
Consider the given equation of lines,
$$
\begin{array}{r}
x-(t+\alpha)=0 \\
y+16=0
\end{array}
$$
and
$$
-\alpha x+y=0
$$
Since, these lines are concurrent, therefore the system of equations is consistent.
Now,
$$
\left|\begin{array}{ccc}
1 & 0 & -(t+\alpha) \\
0 & 1 & 16 \\
-\alpha & 1 & 0
\end{array}\right|=0
$$
$$
\begin{array}{lr}
\Rightarrow & 1(0-16)-(t+\alpha)(0+\alpha)=0 \\
\Rightarrow & -16-\alpha(t+\alpha)=0 \\
\Rightarrow & \alpha(t+\alpha)+16=0 \\
\Rightarrow & \alpha^{2}+t \alpha+16=0
\end{array}
$$
Clearly, $\alpha$ should be real. $\therefore t^{2}-4 \times 16 \geq 0 \Rightarrow t^{2}-64 \geq 0$
Hence, least positive value of $t$ is 8.

Topic: Mathematics, Determinants, JEE Main
The least value of n n∈N, such that the function fn,x=∫ncos⁡nxdx satisfies fn,π2=-1, is (given, fn,0=0)

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
Given function is f n,x =n⋅ sin nx n +c
As fn,0=0âc=0
â´fn,x=sinâ¡nx
If fn,Ï2=-1 , we have sinâ¡nÏ2=-1
ân=3,7,11,â¦â¦
â´nmin=3

Topic: Mathematics, Indefinite Integration, JEE Main
The least value of the natural number $n$ satisfying $C(n, 5)+C(n, 6)>C(n+1,5)$ is

Options:
A $10$
B $11$
C $12$
D $13$

The correct answer is:
$11$

Solution:
$\begin{aligned} & \text { We have, }{ }^n C_5+{ }^n C_6>{ }^{n+1} C_5 \\ & { }^{n+1} C_6>{ }^{n+1} C_5 \\ & \Rightarrow \quad \frac{(n+1) !}{6 !(n+1-6) !}>\frac{(n+1) !}{5 !(n+1-5) !} \\ & \Rightarrow \quad 5 !(n-4) !>6 !(n-5) ! \\ & \Rightarrow \quad 5 !(n-4)(n-5) !>6 \cdot 5 !(n-5) ! \\ & \Rightarrow \quad n-4>6 \\ & \Rightarrow \quad n>10 \\ & \Rightarrow \quad n=11 \\ & \end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
The least value of the product xyz (such that x,y and z are positive real numbers) for which the determinant x11 1y1 11z is non-negative is

Options:
A -1
B -162
C -8
D -22

The correct answer is:
-8

Solution:
Since A.M ≥G.M

x+y+z3 ≥xyz13

x+y+z ≥3xyz13

Now, x11 1y1 11z=xyz-x+y+z+2

Now determinant is non-negative

⇒xyz-x+y+z+2≥0

⇒xyz-3xyz13+2 ≥0

Let xyz=t3

So, t3-3t+2 ≥0

⇒t+2(t2-2t+1)≥0⇒t+2(t-1)2≥0

⇒t≥-2

⇒t3≥-8

⇒ Least value of xyz= -8

Topic: Mathematics, Determinants, JEE Main, JEE Main 2015 (10 Apr Online)