datavorous/entrance-exam-dataset huggingface dataset:
The maximum number of points on the parabola y2=16x which are equidistant from a variable point P (which lie inside the parabola) are

Options:

The correct answer is:
4

Solution:
The maximum number of points on the parabola will be obtained when they are intersection point with a circle.
A circle can intersect a parabola at maximum 4 points

Topic: Mathematics, Parabola, JEE Main
The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$, where $\lambda$ is the wavelength of light used, in a Young's double slit experiment is

Options:
A
zero

B
3

C
infinite

D
5

The correct answer is:
3

Solution:
As $\sin \theta=\frac{n \lambda}{d}$ and $\sin \theta$ cannot be $\ngtr 1$
$$
\therefore 1=\frac{n \lambda}{1.8 \lambda}
$$
or $n=1.8$
Hence maximum number of possible interference maximas, $0, \pm 1$ i.e. 3

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (12 May Online)
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

Options:
A
infinite

B
five

C
three

D
zero

The correct answer is:
five

Solution:
For interference maxima, $d \sin \theta=n \lambda$
Here $d=2 \lambda$
$\therefore \sin \theta=\mathrm{n} / 2$ and is satisfied by 5 integral values of $\mathrm{n}(-2,-1,0,1,2)$, as the maximum value of $\sin \theta$ can only be 1 .

Topic: Physics, Wave Optics, JEE Main, JEE Main 2004
The maximum number of possible oxidation states of actinoids are shown by:

Options:
A nobelium (No) and lawrencium (Lr)
B berkelium (Bk) and californium (Cf)
C actinium (Ac) and thorium (Th)
D neptunium (Np) and plutonium (Pu)

The correct answer is:
neptunium (Np) and plutonium (Pu)

Solution:
Np and Pu show maximum number of oxidation states starting from +3 to +7 all oxidation states.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The maximum number of structural isomers of hexane is

Options:

The correct answer is:
5.00

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The maximum number of unpaired electrons is present in

Options:
A Fe
B Cu
C Co
D Ni

The correct answer is:
Fe

Solution:
Thus, maximum number of unpaired electrons is present in Fe.

Topic: Chemistry, d and f Block Elements, JEE Main
The maximum possible area bounded by the parabola y=x2+x+10 and a chord of the parabola of length 1 is

Options:
A 112
B 16
C 13
D 12

The correct answer is:
16

Solution:
We have,

y=x2+x+10

⇒x2+x+14=y-10+14

⇒x+122=y-394

Latusrectum of parabola is 1.

∴ Length of chord is also 1.

Area is maximum when chord is latusrectum.
Area of shaded region=Area of rectangle OABC-2∫-1/20parabola

=10-2∫-120x2+x+10dx

=10-2x33+x22+10-120

=10-2-124+18-102=16

Topic: Mathematics, Area Under Curves, JEE Main
The maximum possible denticities of a ligand given below towards a common transition and inner-transition metal ion, respectively, are:

Options:
A 8 and 6
B 6 and 6
C 8 and 8
D 6 and 8

The correct answer is:
6 and 8

Solution:
Towards common transition element and inner transition metal ion given ligand can have maximum denticity of 6 and 8 respectively.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The maximum possible number of real roots of equation $x^5-6 x^2-4 x+5=0$ is

Options:
A $\mathrm{O}$
B $3$
C $4$
D $5$

The correct answer is:
$3$

Solution:
Let $f(x)=x^5-6 x^2-4 x+5=0$
Then the number of change of sign in $f(x)$ is 2 therefore $f(x)$ can have at most two positive real roots.
Now, $f(-x)=-x^5-6 x^4+4 x+5=0$
Then the number of change of sign is 1 .
Hence $f(x)$ can have at most one negative real root. So that total possible number of real roots is 3 .

Topic: Mathematics, Quadratic Equation, JEE Main
The maximum possible value of $x^{2}+y^{2}-4 x-6 y, x, y$ real, subject to the condition $|x+y|+|x-y|=4$

Options:
A is 12
B is 28
C is 72
D does not exist

The correct answer is:
is 28

Solution:
$|x+y|+|x-y|=4$ represent a square
$x^{2}+y^{2}-4 x-6 y=(x-2)^{2}+(y-3)^{2}-13$
$=$ (distance point on square from $(2,3))^{2}-13$
Maximum $=(-2-2)^{2}+(-2-3)^{2}-13=28$

Topic: Mathematics, Straight Lines, JEE Main
The maximum potential energy of a block executing simple harmonic motion is 25 J. A is amplitude of oscillation. At A2, the kinetic energy of the block is

Options:
A 37.5 J
B 9.75 J
C 18.75 J
D 12.5 J

The correct answer is:
18.75 J

Solution:
The maximum potential energy of a block executing simple harmonic motion is given by Umax=12mω2A2=25 J

As, kinetic energy of particle executing SHM is KE=12mω2A2-x2.

So, kinetic energy of block KE at A2=12mv12=12mω2A2-A24

⇒KE=12mω23A24=3412mω2A2

⇒KE=34×25=18.75 J

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The maximum power drawn out of the cell from a source is given by (where $r$ is internal resistance)

Options:
A $E^2 / 2 r$
B $E^2 / 4 r$
C $E^2 / r$
D $E^2 / 3 r$

The correct answer is:
$E^2 / 4 r$

Solution:
$i=\frac{E}{r+R} \Rightarrow P=i^2 R \Rightarrow P=\frac{E^2 R}{(r+R)^2}$
Power is maximum when $r=R \Rightarrow P_{\text {max }}=E^2 / 4 r$

Topic: Physics, Current Electricity, JEE Main
The maximum prescribed concentration of copper in drinking water is:

Options:
A 3 ppm
B 5 ppm
C 0.5 ppm
D 0.05 ppm

The correct answer is:
3 ppm

Solution:
The maximum prescribed concentration of Cu in drinking water is 3 ppm. If it is the prescribed conc. of Cu is above, it is not good for health.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The maximum probability of finding electron in the dxy orbital is:

Options:
A along the x- axis
B along the y- axis
C at the angle of 45ο from the x-and y-axis
D at the angle of 90ο from the x-and y-axis

The correct answer is:
at the angle of 45ο from the x-and y-axis

Solution:
The maximum probability of finding electron in the dxy orbital lies at the angle of 45ο from the x-and y-axis.

Topic: Chemistry, Structure of Atom, JEE Main
The maximum range of a bullet fired from a toy pistol mounted on a car at rest is $R_0=40 \mathrm{~m}$. What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the
direction of firing with uniform velocity $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$ on a horizontal surface? $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A
$30^{\circ}$

B
$60^{\circ}$

C
$75^{\circ}$

D
$45^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
No Solution is given.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2013 (25 Apr Online)
The maximum reabsorption of water without the influence of ADH takes place at

Options:
A proximal convoluted tubule
B Henle's loop
C distal convoluted tubule
D collecting duct

The correct answer is:
proximal convoluted tubule

Solution:
PCT is lined by simple cuboidal brush border epithelium. Brush border increases the surface area for
reabsorption.
Functions of PCT:
1. PCT absorbs almost all essential nutrients.
2. PCT absorbs about 70-80 % electrolytes and water.
3. As PCT selectively secretes H+ ions, ammonia and potassium ions into the filtrate and also absorbs HCO3- from it, it maintains the pH and ionic balance of the body fluids. The maximum reabsorption (65 to 70%) takes place in proximal convoluted tubule of the nephron.

Topic: Biology, Excretory Products and their Elimination, NEET
The maximum rectification efficiency of a half-wave rectifier is

Options:
A $100 \%$
B $55.6 \%$
C $81.2\%$
D $40.64\%$

The correct answer is:
$40.64\%$

Solution:
Concept:
The efficiency of a rectifier is defined as the ratio of dc output power to input power.
The efficiency of a half-wave rectifier will be:
$\begin{gathered}
\eta=\frac{P_{d x}}{P_{a c}} \\
\eta=\frac{\frac{v_2}{R_L}}{\frac{v_{\text {m }}}{R_L}}
\end{gathered}$
VDC $=$ DC or average output voltage
RL = Load Resistance
For a half-wave rectifier, the output DC voltage or the average voltage is given by:
$V_{D C}=\frac{V_m}{\pi}$
Also, the RMS voltage for a half-wave rectifier is given by:
$V_{r m s}=\frac{V_m}{2}$
Calculation:
The efficiency for a half-wave rectifier will be:
$\eta=\frac{\left(\frac{v_m}{\tau}\right)^2}{\left(\frac{v_m}{2}\right)^2}=40.6 \%$
For Half wave rectifier maximum efficiency $=40.6 \%$
Note: For Full wave rectifier maximum efficiency $=81.2 \%$

Topic: Physics, Semiconductors, JEE Main
The maximum refractive index of a prism which permits the passage of light through it is n. If the refracting angle of the prism is 90°, then what is the value of n?

Options:

The correct answer is:
2

Solution:
For passing the ray from the prism,
μ<cosecA2
μ<cosec90°2
μ<2
μmax⁡=2

Topic: Physics, Ray Optics, JEE Main
The maximum slope of the curve y=12x4-5x3+18x2-19x occurs at the point

Options:
A

3,212

B

2,2

C

2,9

D 0,0

The correct answer is:

2,2

Solution:
dydx=2x3-15x2+36x-19

Since, slope is maximum so,

d2ydx2=6x2-30x+36=0

⇒x2-5x+6=0

x=2,3

at x=2

d3ydx3=12x-30

at x=2,d3ydx3<0

So, maxima

y=12×16-5×8+18×4-19×2

=8-40+72-38=80-78=2

point 2,2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The maximum slope of the curve y=-x3+3x2-4x+9 is

Options:
A 1
B –1
C 2
D 3

The correct answer is:
–1

Solution:
dydx=-3x2+6x-4=Zlet
For maximum value of Z,
dZdx=0
-6x+6=0
x=1
d2Zdx2=-6<0 so, maxima occurs at x=1
Zmax=-3+6-4=-1

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum three digit integer in the decimal system will be represented in the binary system by which one of the following?

Options:
A 1111110001
B 111111110
C 1111100111
D 1111000111

The correct answer is:
1111100111

Solution:
The maximum three digit integer in decimal system $=999$. We go on dividing till we get a dividend $ < 2$ and write remainders from last to firstas shown below:
\begin{array}{l|l|l}
\hline 2 & 999 & \\
\hline 2 & 499 & 1 \\
\hline 2 & 249 & 1 \\
\hline 2 & 124 & 1 \\
\hline 2 & 62 & 0 \\
\hline 2 & 31 & 0 \\
\hline 2 & 15 & 1 \\
\hline 2 & 7 & 1 \\
\hline 2 & 3 & 1 \\
\hline & 1 & 1 \\
\hline
\end{array}
Hence, $(999)_{10}=(1111100111)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
The maximum valency of an element with atomic number 7 is

Options:
A 2
B 5
C 4
D 3

The correct answer is:
5

Solution:
Valency is according to valence shall configuration which here is $1 s^2, 2 s^2, 2 p^3$, ie. 5

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The maximum value of $2 x^3-24 x+107$ in the interval $[-3,3]$ is

Options:
A $75$
B $89$
C $125$
D $139$

The correct answer is:
$139$

Solution:
Let $f(x)=2 x^3-24 x+107$
At $x=-3, f(-3)=2(-3)^3-24(-3)+107=125$
At $x=3, f(3)=2(3)^3-24(3)+107=89$
For maxima or minima, $f^{\prime}(x)=6 x^2-24=0$ $\Rightarrow x=2,-2$
So at $x=2, f(2)=2(2)^3-24(2)+107=75$
at $x=-2, f(-2)=2(-2)^3-24(-2)+107=139$
Thus the maximum value of the given function in $[-3,3]$ is 139.

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum value of $3 \cos \theta-4 \sin \theta$ is

Options:
A 3
B 4
C 5
D None of these

The correct answer is:
5

Solution:
Let $3=r \cos \alpha, 4=r \sin \alpha$, so $r=5$
$f(\theta)=r \cdot(\cos \alpha \cos \theta+\sin \alpha \sin \theta)=5 \cdot \cos (\theta-\alpha)$
$\therefore$ The maximum value of $f(\theta)=5.1=5$.
$\{$ Since the maximum value of $\cos (\theta-\alpha)=1\}$.
Aliter : As we know that, the maximum value of $a \sin \theta+b \cos \theta$ is $+\sqrt{a^2+b^2}$ and the minimum value is $-\sqrt{a^2+b^2}$. Therefore, the maximum value is $(3 \cos \theta+4 \sin \theta)=+\sqrt{3^2+(-4)^2}=5$ and the minimum value is -5 .

Topic: Mathematics, Trigonometric Equations, JEE Main
The maximum value of 3cos⁡θ+5sin⁡θ-π6 for any real value of θ is :

Options:
A 19
B 31
C 792
D 34

The correct answer is:
19

Solution:
Given fθ=3cos⁡θ+5sin⁡θ-π6

⇒fθ=3cos⁡θ+5sin⁡θ·32-cos⁡θ·12

⇒fθ=532sin⁡θ+12cos⁡θ

Now using the concept acosx+bsinx+c ∈ c-a2+b2, c+a2+b2, we can write

Maximum value of fθ is 5322+122=754+14=19

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The maximum value of $4 \sin ^2 x+3 \cos ^2 x$ is

Options:
A 3
B 4
C 5
D 7

The correct answer is:
4

Solution:
$f(x)=4 \sin ^2 x+3 \cos ^2 x \sin ^2 x+3$ and $0 \leq|\sin x| \leq 1$
$\therefore$ Maximum value of $\sin ^2 = x+3$ is 4.

Topic: Mathematics, Trigonometric Equations, JEE Main
The maximum value of $a \cos x+b \sin x$ is

Options:
A $a+b$
B $a-b$
C $|a|+|b|$
D $\left(a^2+b^2\right)^{1 / 2}$

The correct answer is:
$\left(a^2+b^2\right)^{1 / 2}$

Solution:
Let $f(x)=a \cos x+b \sin x$
Suppose that $a=r \sin \theta$ and $b=r \cos \theta$ i.e., $r=\sqrt{a^2+b^2}$
Now, $f(x)=r \sin \theta \cos x+r \cos \theta \sin x$
$\begin{gathered}
=\sqrt{a^2+b^2}\{\sin (\theta+x)\} \\
\text {But }-1 \leq \sin (\theta+x) \leq 1 \Rightarrow-r \leq r \sin (\theta+x) \leq r \\
\Rightarrow-\sqrt{a^2+b^2} \leq \sqrt{a^2+b^2} \sin (\theta+x) \leq \sqrt{a^2+b^2}
\end{gathered}$
Thus, maximum value of $f(x)$ is $\sqrt{a^2+b^2}$.

Topic: Mathematics, Trigonometric Equations, JEE Main
The maximum value of $\cos ^2\left(\frac{\pi}{3}-x\right)-\cos ^2\left(\frac{\pi}{3}+x\right)$ is

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

The correct answer is:
$\frac{\sqrt{3}}{2}$

Solution:
$\begin{aligned} & \cos ^2\left(\frac{\pi}{3}-x\right)-\cos ^2\left(\frac{\pi}{3}+x\right) \\ & =\left\{\cos \left(\frac{\pi}{3}-x\right)+\cos \left(\frac{\pi}{3}+x\right)\right\}\left\{\cos \left(\frac{\pi}{3}-x\right)-\cos \left(\frac{\pi}{3}+x\right)\right\} \\ & =\left\{2 \cos \frac{\pi}{3} \cos x\right\}\left\{2 \sin \frac{\pi}{3} \sin x\right\} \\ & =\sin \frac{2 \pi}{3} \sin 2 x=\frac{\sqrt{3}}{2} \sin 2 x\end{aligned}$
Its maximum value is $\frac{\sqrt{3}}{2},\{-1 \leq \sin 2 x \leq 1\}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The maximum value of fx=sin2x1+cos2xcos2x1+sin2xcos2xcos2xsin2xcos2xsin2x, x∈R is

Options:
A 7
B 34
C 5
D 5

The correct answer is:
5

Solution:
fx=sin2x1+cos2xcos2x1+sin2xcos2xcos2xsin2xcos2xsin2x, x∈R

C1→C1+C2

=21+cos2xcos2x2cos2xcos2x1cos2xsin2x

R1→R1-R2

=0102cos2xcos2x1cos2xsin2x

Expanding w.r.t. R1

=-2sin2x-cos2x

So, fx=cos2x-2sin2x

fxmax=1+4=5

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The maximum value of fx=sin⁡2xsin⁡x+cos⁡x in the interval 0,π2 is

Options:
A 2
B 12
C 1
D 12

The correct answer is:
12

Solution:
fx=sin⁡x+cos⁡x2-1sin⁡x+cos⁡x
Substituting sin⁡x+cos⁡x=t, the expression becomes
gt=t2-1t=t-1t
Now, t=sin⁡x+cos⁡x, ∀x∈0,π2
t∈1,2
g't=1+1t2>0

Therefore, gt is an increasing function.
Maximum value =g2=2-12=12

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombus is

Options:
A 33
B 19
C -19
D 9

The correct answer is:
33

Solution:
If the parallelogram is a rhombus, the distance between the pair of parallel sides are equal
Hence, 12-232+42=±p-7122+52⇒p-7=±26

⇒p=33 or -19

So the maximum value of p is 33

Topic: Mathematics, Straight Lines, JEE Main
The maximum value of $\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)$ in the interval $\left(0, \frac{\pi}{2}\right)$ is attained at

Options:
A $x=\frac{\pi}{12}$
B $x=\frac{\pi}{6}$
C $x=\frac{\pi}{3}$
D $x=\frac{\pi}{2}$

The correct answer is:
$x=\frac{\pi}{12}$

Solution:
$\sqrt{2} \cos \left(x+\frac{\pi}{6}-\frac{\pi}{4}\right)=\sqrt{2} \cos \left(x-\frac{\pi}{12}\right)$
Hence maximum value will be at $x=\frac{\pi}{12}$.

Topic: Mathematics, Trigonometric Equations, JEE Main
The maximum value of the expression sinâ¡Î¸cos2Î¸ âÎ¸â0,Ï is

Options:
A 23
B 23
C 233
D 13

The correct answer is:
233

Solution:
As sin2θ+cos2θ=1,

⇒sin2θ+cos2θ2+cos2θ2=1
Applying AM≥GM, we have
sin2θ+cos2θ2+cos2θ23≥sin2θcos4θ413
⇒133≥sin2θcos4θ4

⇒sin⁡θ⋅cos2θ2≤427
⇒sin⁡θ⋅cos2θ≤233

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The maximum value of the function $f(x)=3 x^3-18 x^2+27 x-40$ on the set $\mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\}$ is

Options:
A 122
B 132
C 112
D 222

The correct answer is:
122

Solution:
$\begin{aligned}
& \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\} \\
& x^2+30 \leq 11 x \\
& x^2-11 x+30 \leq 0 \\
& (x-5)(x-6) \leq 0 \\
& x \in[5,6]
\end{aligned}$
Now, $\mathrm{f}(x)=3 x^3-18 x^2+27 x-40$
$\begin{aligned}
\mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \\
\mathrm{f}^{\prime}(x) & =9\left(x^2-4 x+3\right) \\
& =9\left[\left(x^2-4 x+4\right)-1\right] \\
& =9(x-2)^2-9
\end{aligned}$
$\therefore \quad \mathrm{f}^{\prime}(x)>0 \forall x \in[5,6]$
$\therefore \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$
$\therefore \quad$ Maximum value of $\mathrm{f}(x)$ when $x \in[5,6]$ is
$f(6)=122$

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum value of the function $x^3+x^2+x-4$ is

Options:
A $127$
B $4$
C Does not have a maximum value
D None of these

The correct answer is:
Does not have a maximum value

Solution:
$f^{\prime}(x) \neq 0$ at real values of $x$ so $f(x)$ does not have maximum value.

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum value of the term independent of t in the expansion of tx15+1-x110t10where x∈0,1 is:

Options:
A 10!35!2
B 10!35!2
C 2.10!335!2
D 2.10!35!2

The correct answer is:
2.10!335!2

Solution:
Tr+1=Cr10tx1/510-r1-x110tr

=Cr10t10-2rx10-r51-xr10

For the term independent of t,

⇒10-2r=0

⇒r=5

T6=fx=C510x1-x; for maximum

f'x=0⇒x=23&f''23<0

so fxmax.=C51023·13

=2.10!335!2

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The maximum value of $x^4 y^4$ when $a^2 x^4+b^2 y^4=c^6$ is

Options:
A $\frac{c^{12}}{16 a^4 b^4}$
B $\frac{c^{12}}{4 a^2 b^2}$
C $\frac{c^6}{(a+b)^{12}}$
D $\frac{c^6}{a^4+b^4}$

The correct answer is:
$\frac{c^{12}}{4 a^2 b^2}$

Solution:
$\begin{aligned} & \text {Let } f(x)=x^4 y^4 \\ & \text { And } a^2 x^4+b^2 y^4=c^6 \\ & \Rightarrow y^4=\frac{1}{b^2}\left(c^6-a^2 x^4\right) \\ & \therefore \quad f(x)=\mathrm{x}^4 \cdot \frac{1}{b^2}\left(c^6-a^2 x^4\right) \\ & =\frac{1}{b^2} x^4\left(c^6-a^2 x^4\right) \\ & f^{\prime}(x)=\frac{1}{b^2}\left(4 x^{3 c^6}-8 a^2 x^7\right) \\ & \end{aligned}$
for extrem point:
$\begin{aligned}
& f^{\prime}(\mathrm{x})=0 \Rightarrow 4 x^3\left(c^6-2 a^2 x^4\right)=0 \\
& \Rightarrow x=0 \text { or } x^4=\frac{c^6}{2 a^2}
\end{aligned}$
For maxima : $x=\frac{c^6}{2 a^2}$
$\therefore$ The maximum value is :
$\begin{aligned}
& f\left(x=\left(\frac{c^6}{2 a^2}\right)^{\frac{1}{4}}\right) \\
& =\frac{c^6}{2 a^2} \cdot \frac{1}{b^2}\left[c^6-a^2 \cdot \frac{c^6}{2 a^2}\right] \\
& =\frac{c^6}{2 a^2 b^2} \cdot \frac{c^6}{2}=\frac{c^{12}}{4 a^2 b^2}
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum value of x that satisfies the equation sin-1215x=cos-114x is

Options:

The correct answer is:
16

Solution:
âµsin-1215x=cos-11-215x2for 0<215xâ¤1âxâ¥215
â´14x2=1-215x2âx=16âx=Â±16 which satisfies xâ¥215.
Hence, the maximum value of x=16

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The maximum value of z in the following equation z=6xy+y2, where 3x+4y≤100 and 4x+3y≤75 for x≥0 and y≥0
is

Options:

The correct answer is:
904

Solution:
z=6xy+y2=y(6x+y)

3x+4y≤100 …i

4x+3y≤75 …ii

x≥0

y≥0

x≤75-3y4

Given z=y(6x+y)

⇒z≤y6·75-3y4+y

⇒z≤12225y-7y2

Now range of 12225y-7y2 is (-∞,(225)22×4×7]

So, z≤12225y-7y2≤(225)22×4×7

=5062556

≈904.0178

≈904.02

It will be attained at y=22514

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The maximum velocity (in $\mathrm{ms}^{-1}$ ) with which a car driver must traverse a flat curve of radius $150 \mathrm{~m}$ and coefficient of friction 0.6 to avoid skidding is

Options:
A 60
B 30
C 15
D 25

The correct answer is:
30

Solution:
$v=\sqrt{\mu r g}=\sqrt{0.6 \times 150 \times 10}=30 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 \mathrm{~mm}$, is $4.4 \mathrm{~m} / \mathrm{s}$. The period of oscillation is

Options:
A
$100 \mathrm{~s}$

B
$0.01 \mathrm{~s}$

C
$10 \mathrm{~s}$

D
$0.1 \mathrm{~s}$

The correct answer is:
$0.01 \mathrm{~s}$

Solution:
$A \omega=V_{\max }$
$\mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi \mathrm{A}}{\mathrm{v}_{\max }}=0.01 \mathrm{sec}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2006
The maximum velocity of a simple harmonic motion represented by
$y=3 \sin \left(100 t+\frac{\pi}{6}\right)$
is given by

Options:
A 300
B $\frac{3 \pi}{6}$
C 100
D $\frac{\pi}{6}$

The correct answer is:
300

Solution:
$v_{\max }=2 \omega=3 \times 100=300$

Topic: Physics, Oscillations, JEE Main
The maximum velocity of the photoelectron emitted by the metal surface is 'v'. Charge and mass of the photoelectron is denoted by 'e' and 'm' respectively. The
stopping potential in volt is

Options:
A $\frac{v^{2}}{\left(\frac{m}{e}\right)}$
B $\frac{v^{2}}{\left(\frac{e}{m}\right)}$
C $\frac{v^{2}}{2\left(\frac{m}{e}\right)}$
D $\frac{v^{2}}{2\left(\frac{e}{m}\right)}$

The correct answer is:
$\frac{v^{2}}{2\left(\frac{e}{m}\right)}$

Solution:
$\frac{1}{2} m v^{2}=e v$
$v=\frac{m v^{2}}{2 e}=\frac{v^{2}}{2\left(\frac{e}{m}\right)}$

Topic: Physics, Dual Nature of Matter, JEE Main
The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be:

Options:
A more than 3v
B equal to 3v
C v
D less than 3v

The correct answer is:
more than 3v

Solution:
According to Einstein's photoelectric effect,

KEmax=hcλ-ϕ

The incident energy of the photon is E1=12mv2=hn-ϕ ...1
If incident frequency increased to 3n
E2=12mv22=3hn- ϕ
=3hn-ϕ+2ϕ
E2=3 E1+2ϕ
⇒12mv22=32mv2+2ϕ
⇒v22=3v2+4ϕm
⇒v2>v3

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2017 (08 Apr Online)
The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is

Options:
A 192 m
B 136 m
C 272 m
D 68 m

The correct answer is:
272 m

Solution:
Maximum height is achieved when the projection is along the vertical direction,

Hmax=v22 g=136 m

Now, for maximum range, angle of projection should be 45°. Therefore,

Rmax=v2 g=2Hmax

=2136

=272 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The maximum volume in cu.m of the right circular cone having slant height 3 m is:

Options:
A 23 π
B 33 π
C 6 π
D 43π

The correct answer is:
23 π

Solution:
Given l=3

⇒r2+h2=l2=9
Volume, V=13πr2h
⇒V=13π9-h2h=13π9h-h3
dvdh=13π9-3h2
dvdh=0⇒h=3
d2Vdh2=13π-6h<0
∴ at h=3 , cone has maximum volume
∴ Vmax=13π93-33=23πcm3

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The maximum volume of a right circular cylinder if the sum of its radius and height is
$6 \mathrm{~m}$ is

Options:
A $16 \pi m^{3}$
B $32 \pi m^{3}$
C $4 \pi m^{3}$
D $64 \pi m^{3}$

The correct answer is:
$32 \pi m^{3}$

Solution:
(A)
Let height of cylinder be $\mathrm{h}$ and radius be $\mathrm{r} \Rightarrow \mathrm{r}+\mathrm{h}=6 \Rightarrow \mathrm{h}=6-\mathrm{r}$
Volume of cylinder $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}$
$=\pi \cdot r^{2} \cdot(6-r)=\pi\left(6 r^{2}-r^{3}\right)$
For the maximum volume, $\frac{\mathrm{dV}}{\mathrm{dr}}=0$
$\pi\left(12 r-3 r^{2}\right)=0 \Rightarrow 12 r=3 r^{2} \Rightarrow r=4 \Rightarrow h=6-4=2$.
$\therefore$ Volume of cylinder $=\pi \times 4^{2} \times 2=32 \pi \cdot \mathrm{m}^{3}$

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum wavelength of radiation that can produce photoelectric effect in certain metal is 200 nm. What is the maximum kinetic energy (in electron-volt) acquired by the electron due to radiation of wavelength 100 nm?

Options:

The correct answer is:
6.2

Solution:
Here, Î»0=200nm; Î»=100nm;
maximum KE =hcÎ»e-hcÎ»0e
=hce1Î»-1Î»0
=12401100-1200
=6.2 eV

Topic: Physics, Dual Nature of Matter, JEE Main