datavorous/entrance-exam-dataset huggingface dataset:
If ex+efx=e, then the domain of the function f is

Options:
A -∞,1
B -∞,1
C -1,∞
D 1,∞

The correct answer is:
-∞,1

Solution:
We have,
ex+efx=e⇒efx=e-ex
⇒fx=log⁡e-ex
For fx to be defined,

e-ex>0
⇒e1>ex⇒x<1
∴ Domain of f=-∞,1

Topic: Mathematics, Functions, JEE Main
If $\int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$, for all $x \in R-\{0\}$, then the value of $f\left(\frac{\pi}{6}\right)$ is

Options:
A
1/2

B
1

C
0

D
$-1 / 2$

The correct answer is:
$-1 / 2$

Solution:
Let $\int_e^x t f(t) d t=\sin x-x \cos x-\frac{x^2}{2}$
By using Leibnitz rule, we get
$$
\begin{aligned}
& \frac{d}{d x}\left[\int_e^x t f(t) d t\right]=\frac{d}{d x}\left[\sin x-x \cos x-\frac{x^2}{2}\right] \\
& \Rightarrow x f(x)-e f(e) \cdot 0=x \sin x-x
\end{aligned}
$$
Now, put $x=\frac{\pi}{6}$, we get
$$
\begin{aligned}
& \frac{\pi}{6} \cdot f\left(\frac{\pi}{6}\right)=\frac{\pi}{6} \cdot \sin \frac{\pi}{6}-\frac{\pi}{6} \\
& \Rightarrow f\left(\frac{\pi}{6}\right)=\sin \frac{\pi}{6}-1=\frac{1}{2}-1=-\frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2012 (07 May Online)
If ey+xy=e, the ordered pair dydx,d2ydx2 at x=0 is equal to

Options:
A -1e,-1e2
B -1e,1e2
C 1e,-1e2
D 1e,1e2

The correct answer is:
-1e,1e2

Solution:
ey+xy=e

differentiate w. r. t. x

eydydx+xdydx+y=0

⇒dydxx+ey=-y,  dydx(0,1)=-1e

Again differentiate w. r. t. x

ey.d2ydx2+dydx.ey.dydx+x.d2ydx2+dydx+dydx=0

⇒x+eyd2ydx2+dydx2.ey+2dydx=0

Now, at 0,1

ed2ydx2+1e2e+2-1e=0

∴ d2ydx2=1e2

Hence, ordered pair dydx,d2ydx2=-1e,1e2

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If each capacitor has capacitance C, then equivalent capacitance between A and B is

Options:
A 4C3
B 8C3
C 12C
D 5C12

The correct answer is:
8C3

Solution:
Mark equipotential points and redraw the circuit in order to identify series and parallel combinations.
Ceq.=4 Ã 2CÃ C2C+C=4Ã 2C3=8C3

Topic: Physics, Electrostatics, JEE Main
If each diode has a forward bias resistance of 25Ω in the below circuit,

Which of the following options is correct?

Options:
A I1I2=1
B I2I3=1
C I1I2=2
D I3I4=1

The correct answer is:
I1I2=2

Solution:
As the diode D2 is connected in reverse order, there is no current through it.

The effective resistances of the circuit can, then, be calculated as follows:

Reff=25+25+125225+125+25+125 Ω= 100 Ω

Each current through the diodes can be calculated as follows:

I1=5 V100 Ω= 0.05 A

As the diode D2 is non-conducting, the current I1 will be divided equally among the other two diodes D1 and D3.

Hence, the currents in the diodes can be written as

I2=0.05 A2= 0.025 AI4=0.025 A

Thus,

I1I2=0.05 A0.025 A= 2

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If each element in a row of a determinant is multiplied by the same factor $\mathrm{r}$, then the value of the determinant:

Options:
A is multiplied by $\mathrm{r}^{3}$.
B is increased by $3 \mathrm{r}$
C remains unchanged
D is multiplied by r

The correct answer is:
is multiplied by r

Solution:
As we know that if each element of any row (or column) of a determinant is multiplied by the same number, then the value of determinant is multiplied by that number.

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 2)
If each of the lines $5 x+8 y=13$ and $4 x-y=3$ contains a diameter of the circle $x^2+y^2-2\left(a^2-7 a+11\right)$
$x-2\left(a^2-6 a+6\right) y+b^3+1=0$, then :

Options:
A
$a=5$ and $b \notin(-1,1)$

B
$a=1$ and $b \notin(-1,1)$

C
$\quad a=2$ and $b \notin(-\infty, 1)$

D
$a=5$ and $b \in(-\infty, 1)$

The correct answer is:
$a=5$ and $b \in(-\infty, 1)$

Solution:
Point of intersection of two given lines is $(1,1)$. Since each of the two given lines contains a diameter of the given circle, therefore the point of intersection of the two given lines is the centre of the given circle.
Hence centre $=(1,1)$
$$
\therefore a^2-7 a+11=1 \Rightarrow a=2,5
$$
and $a^2-6 a+6=1 \Rightarrow a=1,5$
From both (i) and (ii), $a=5$
Now on replacing each of $\left(a^2-7 a+11\right)$ and $\left(a^2-6 a+6\right)$ by 1 , the equation of the given circle is $x^2+y^2-2 x-2 y+b^3+1=0$
$$
\begin{aligned}
& \Rightarrow(x-1)^2+(y-1)^2+b^3=1 \\
& \Rightarrow b^3=1-\left[(x-1)^2+(y-1)^2\right] \\
& \therefore b \in(-\infty, 1)
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (09 Apr Online)
If each of the observations $x_1, x_2, \ldots, x_n$ is increased or decreased by $k$, where $k$ is a positive number, then the variance of the da

Options:
A increases by $\mathrm{k}$
B do not change
C is equal to $\mathrm{k}^2$
D is equal to $2 \mathrm{k}$

The correct answer is:
do not change

Solution:
If each of the observations $x_1, x_2, \ldots x_n$ is increased or decreased by $k$ then mean and variance of the obtained data remains unchanged.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2023 (17 May Shift 1)
If earth has uniform density, and radius 'R'. The value of acceleration due to gravity at distance d above the surface is same as acceleration due to gravity at distance d below the surface. If d=x-12R, then find x.

Options:

The correct answer is:
5

Solution:
gh=g.R2R+d2 h=d
gd=g0 1-dR
R2R+d2=1-dR
R2.R=R+d2R-d
R3=R2+d2+2dR R-d
R3=R3+Rd2+2dR2-dR2-d3-2d2R
0=-d2R+dR2-d3
d2+d.R-R2=0
d=5-12R
∴x=5

Topic: Physics, Gravitation, JEE Main
If earth is supposed to be a sphere of radius $R$, if $g_{30}$ is value of acceleration due to gravity at latitude of $30^{\circ}$ and $g$ at the equator, the value of $g-g_{30^{\circ}}$ is

Options:
A $\frac{1}{4} \omega^2 R$
B $\frac{3}{4} \omega^2 R$
C $\omega^2 R$
D $\frac{1}{2} \omega^2 R$

The correct answer is:
$\frac{3}{4} \omega^2 R$

Solution:
Acceleration due to gravity at latitude $\lambda$ is given by
$\begin{aligned} & g^{\prime}=g-R \omega^2 \cos ^2 \lambda \\ & \text { At } 30^{\circ}, g_{30^{\circ}}=g-R \omega^2 \cos ^2 30^{\circ}=g-\frac{3}{4} R \omega^2 \\ & \therefore \quad g-g_{30}=\frac{3}{4} \omega^2 R .\end{aligned}$

Topic: Physics, Gravitation, JEE Main
If eccentricity of the ellipse x2a2+1+y2a2+2=1 is 16, then the ratio of the length of the latus rectum to the length of the major axis is

Options:
A 56
B 36
C 23
D 26

The correct answer is:
56

Solution:
a2+2>a2+1
a2+1=a2+21-e2
⇒a2=4⇒a=±2
Hence, the required ratio is a2+1a2+2=56

Topic: Mathematics, Ellipse, JEE Main
If Electric field intensity of a uniform plane electro magnetic wave is given as E=-301.6sinkz-ωta^x+452.4sinkz-ωta^y V m-1. Then, magnetic intensity H of this wave in A m-1 will be [Given : Speed of light in vacuum c=3×108 m s-1, Permeability of vacuum μ0=4π×10-7 N A-2]

Options:
A +0.8sinkz-ωta^y+0.8sinkz-ωta^x
B +1.0×10-6sinkz-ωta^y+1.5×10-6kz-ωta^x
C -0.8sinkz-ωta^y-1.2sinkz-ωta^x
D -1.0×10-6sinkz-ωta^y-1.5×10-6sinkz-ωta^x

The correct answer is:
-0.8sinkz-ωta^y-1.2sinkz-ωta^x

Solution:
The relation between magnetic field and electric field is given by,

c=E0B0⇒B0=E0c

And magnetic intensity is given by,

H0=B0μ0⇒H0=E0cμ0

Also, E→ and B→ are perpendicular to each other.

Therefore,

Hy0=301.63×108×4π×10-7=0.8 A m-1Hx0=452.43×108×4π×10-7=1.2 A m-1

Hence, C is the answer.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (26 Jun Shift 1)
If electron falls from $n=3$ to $n=2$, then emitted energy is

Options:
A $10.2 \mathrm{eV}$
B $12.09 \mathrm{eV}$
C $1.9 \mathrm{eV}$
D $0.65 \mathrm{eV}$

The correct answer is:
$1.9 \mathrm{eV}$

Solution:
$$
\Delta E=E_3-E_2=13.6\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=1.9 \mathrm{eV}
$$

Topic: Chemistry, Structure of Atom, JEE Main
If electronic charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental quantities, the permeability of vacuum μ0 can be expressed in units of:

Options:
A mc2he2
B hme2
C hcme2
D hce2

The correct answer is:
hce2

Solution:
Magnetic potential energy inside a solenoid is
U=12LI2 Where L=self-inductor⇒U=12μ0n2A lI2
∵n= number of turns per unit length
=Nl
U=12μ0Nl2Al I2
U=12μ0N2l2Al I2
μ0=2UlN2AI2
Dimensionally, U=hcλ
I=eT
A=l2
l=l′
μ0=hcλ [l]l2t2e2
=hλte2=hce2
μ0=hce2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2015 (11 Apr Online)
If electronic charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental quantities, the permeability of vacuum μ0 can be expressed in units of:

Options:
A mc2he2
B hme2
C hcme2
D hce2

The correct answer is:
hce2

Solution:
Magnetic potential energy inside a solenoid is
U=12LI2 Where L = self-inductor
U=12Î¼0n2A lI2
âµn= no. of turns per unit length
=Nl
U=12Î¼0Nl2Al I2
U=12Î¼0N2l2Al I2
Î¼0=2UlN2AI2
Dimensionally, U=hcÎ»
I=eT
A=l2
l=lâ²
Î¼0=hcÎ» [l]l2t2e2
=hÎ»te2=hce2
Î¼0=hce2

Topic: Physics, Units and Dimensions, JEE Main
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is

Options:
A $10!\times 2$
B $10!$
C $9!\times 2$
D None of these

The correct answer is:
$9!\times 2$

Solution:
Required number of ways $9!^* 2$.
\{By fundamental property of circular permutation\}

Topic: Mathematics, Permutation Combination, JEE Main
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be___(For surface tension, Force=Surface tension×length)

Options:
A EÂ V-2Â T-1
B EÂ V-1Â T-2
C E V-2 T-2
D E-2Â V-1Â T-3

The correct answer is:
E V-2 T-2

Solution:
Dimensional formula of surface tension, S=M1L0T-2
Dimensional formula of energy, E=M1L2T-2
Dimensional formula of velocity, V=M0L1T-1
By applying dimensional analysis,
S=EaVbTc
M1L0T-2=M1L2T-2aM0L1T-1bTc
M1L0T-2=MaL2a+bT-2a-b+c
On comparing,
a=1
2a+b=0⇒b= -2
-2a-b+c=-2⇒c= -2
∴S=E1V-2T-2

Topic: Physics, Units and Dimensions, JEE Main
If enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g}), \mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure are 52,394 and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively, the change in enthalpy for combustion of $\mathrm{C}_2 \mathrm{H}_4$ is equal to

Options:
A $-141.2 \mathrm{~kJ} / \mathrm{mol}$
B $-1412 \mathrm{~kJ} / \mathrm{mol}$
C $+14.2 \mathrm{~kJ} / \mathrm{mol}$
D $+1412 \mathrm{~kJ} / \mathrm{mol}$

The correct answer is:
$-1412 \mathrm{~kJ} / \mathrm{mol}$

Solution:
Enthalpy of formation of $\mathrm{C}_2 \mathrm{H}_4, \mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ are $52,-394$ and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively.
(Given)
The reaction is
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
change in enthalpy,
$(\Delta H)=\Delta H_{\text {products }}-\Delta H_{\text {reactants }}$
$=2 \times(-394)+2 \times(-286)-(52+0)$
$=-1412 \mathrm{~kJ} / \mathrm{mol}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
If enthalpy of atomization for Br2(l) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relation between them

Options:
A is x=y.
B does not exist.
C is x>y.
D is x<y.

The correct answer is:
is x>y.

Solution:
∆Hatomisation=∆HVap+Bond Energy
Hence x>y

Enthalpy of atomization is enthalpy change involved when one mole of gaseous atom is formed from a substance in an elemental state (Elemental state of Bromine is Br2(l)).

While, bond energy is energy required to break one mole of bonds in gaseous state to form separated atoms.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (09 Jan Shift 1)
If enthalpy of hydrogenation of C 6 H 6 l into C 6 H 12 l is â205 kJ and resonance energy of C 6 H 6 l is â152 kJ molâ1 then enthalpy of hydrogenation of l to C 6 H 12 l is?
[Assume enthalpiies of vapourisation of all liquids involved are equal].

Options:
A – 535.5 kJ/mol
B – 238 kJ/mol
C – 357 kJ/mol
D – 119 kJ/mol

The correct answer is:
– 119 kJ/mol

Solution:
C 6 H 6 lâ¡ + 3 H 2 g â¶ C 6 H 1 2 lâ¡
Î H hyd C 6 H 6 lâ¡ â = - 2 0 5 kJ
Benzene on reactant side so add resonance energy of benzene with sign. Heat evolved for hydrogenation of three double bonds
=â205â152=â357 kJâ mol â1
Î H hyd â = - 3 5 7 3 = - 1 1 9 kJ / mole

Topic: Chemistry, Thermodynamics (C), JEE Main
If equal volumes of BaCl2 and NaF solutions are mixed, which of these combinations will not give a precipitate?

Substance
Ksp

BaF2
1.7×10-7

Options:
A 0.004 M BaCl2 and 0.02 M NaF
B 0.010 M BaCl2 and 0.015 M NaF
C 0.015 M BaCl2 and 0.010 M NaF
D 0.020 M BaCl2 and 0.002 M NaF

The correct answer is:
0.020 M BaCl2 and 0.002 M NaF

Solution:
In the mixture of 0.02 M BaCl2 and 0.002 M NaF, the ionic product is less than solubility product, hence there will be no precipitate of BaF2 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
If equation of hyperbola passing through \( (-2,1) \) and having asymptotes \( 2 x+y+5=0 \) and \( x-6 y+3=0 \) is \( a x^{2}+b y^{2}+2 h x y+2 g x+2 f y+c=0 \), then value of \( c \) is

Options:

The correct answer is:
25

Solution:
Equation of hyperbola will be

2x+y+5 x-6y+3+λ=0

It passes through (-2,1), so

-4+1+5 -2-6+3+λ=0 ⇒λ=10

Hence, equation of hyperbola wiil be 2x2-6y2-11xy+6x-10y+25=0

On comparing with ax2+by2+2hxy+2gx+2fy+c=0, we get,

c=25

Topic: Mathematics, Hyperbola, JEE Main
If equation of transverse wave is $y=x_{0} \cos$ $2 \pi\left(\mathrm{nt}-\frac{\mathrm{x}}{\lambda}\right) .$ Maximum velocity of particle is twice of wave velocity, if $\lambda$ is-

Options:
A $\pi / 2 \mathrm{x}_{0}$
B $2 \pi x_{0}$
C $\pi$
D $\pi x_{0}$

The correct answer is:
$\pi x_{0}$

Solution:
$y=x_{0} \cos 2 \pi\left(\right.$ nt $\left.-\frac{x}{\lambda}\right)$

$\mathrm{y}=\mathrm{x}_{0} \cos \frac{2 \pi}{\lambda}(\mathrm{vt}-\mathrm{x}) \quad[\because \mathrm{v}=\mathrm{n} \lambda]$

$\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)_{\max }=\mathrm{x}_{0} \times \frac{2 \pi}{\lambda} \mathrm{v}=2 \mathrm{v}($ given $) \therefore \lambda=\pi \mathrm{x}_{0}$

Topic: Physics, Waves and Sound, JEE Main
If equation Z-1n=Zn=1 ânâN has solutions, then n can be

Options:
A 4
B 12
C 15
D 21

The correct answer is:
12

Solution:
Z-1n=ZnâZ-1=Zâx=12 â¦(i)
Zn=1âZ=1âx2+y2=1 â¦(ii)
From (i) & (ii),

14+y2=1ây=Â±32
â12+32i and 12-32i can be the solutions
Which would exist when n is a multiple of 6 âµarg12+32i=tan-13=Ï3âleast value of n is equal to 2ÏÏ3=6

Topic: Mathematics, Complex Number, JEE Main
If equations ax2+bx+c=0, a, b, c∈R, a≠0 and 2x2+3x+4=0 have a common root, then a:b:c equals :

Options:
A 2:3:4
B 4:3:2
C 1:2:3
D 3:2:1

The correct answer is:
2:3:4

Solution:
ax2+bx+c=0 ....(i)
2x2+3x+4=0 ....(ii)
For equation (ii), D=9-424=-23<0
Roots are complex conjugates.
⇒ Both roots are common in equation (i) and (ii).
∴ a 2 = b 3 = c 4
⇒ a : b : c : : 2 : 3 : 4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (09 Apr Online)
If equivalent conductance of $1 \mathrm{M}$ benzoic acid is $12.8 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ and if the conductance of benzoate ion and $\mathrm{H}^{+}$ion are 42 and $288.42 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ respectively. its degree of dissociation is

Options:
A $39 \%$
B $3.9 \%$
C $0.35 \%$
D $0.039\%$

The correct answer is:
$3.9 \%$

Solution:
$A_{m\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)}^o=A_{\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right)}^o+A_{\left(\mathrm{H}^{+}\right)}^o$
$=42+288.42=330.42$
$\alpha=\frac{\Lambda_m^c}{\Lambda_m^o}=\frac{12.8}{330.42}=3.9 \%$

Topic: Chemistry, Electrochemistry, JEE Main
If every pair of the equations
$x^2+p x+q r=0, x^2+q x+r p=0, x^2+r x+p q=0$
have a common root, then the sum of three common roots is

Options:
A $\frac{-(p+q+r)}{2}$
B $\frac{-p+q+r}{2}$
C $-(p+q+r)$
D $-p+q+r$

The correct answer is:
$\frac{-(p+q+r)}{2}$

Solution:
Let the roots be $\alpha, \beta ; \beta, \gamma$ and $\gamma, \alpha$ respectively.
$\therefore \quad \alpha+\beta=-p, \beta+\gamma=-q, \gamma+\alpha=-r$
Adding all, we get $\Sigma \alpha=-(p+q+r) / 2$ etc.

Topic: Mathematics, Quadratic Equation, JEE Main
If Expiratory Reserve Volume = A, Residual Volume = B, Tidal Volume = C and Inspiratory Reserve Volume = D, then Functional Residual Capacity (FRC) is equal to

Options:
A A + B
B A + B + C
C B + D
D A + B + D

The correct answer is:
A + B

Solution:
Functional Residual Capacity FRC = ERV + RV
It refers to the volume of air that remains in the lungs after a normal expiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
If $f:[0, \infty) \rightarrow[0, \infty)$ and $f(x)=\frac{x}{1+x}$, then $f$ is

Options:
A One-one and onto
B One-one but not onto
C Onto but not one-one
D Neither one-one nor onto

The correct answer is:
One-one but not onto

Solution:
$f^{\prime}(x)=\frac{1}{(1+x)^2}\gt0, \forall x \in[0, \infty)$ and range $\in[0,1)$ $\Rightarrow$ function is one-one but not onto.

Topic: Mathematics, Functions, JEE Main
If f:(0, â)â(0,â)and fx=x1+x, then f is

Options:
A one-one and onto
B one-one but not onto
C onto but not one-one
D neither one-one nor onto

The correct answer is:
one-one but not onto

Solution:
Here, f:0, ∞→0, ∞, i.e. domain is (0, ∞) and codomain is 0, ∞.
For one-one, fx=x1+x
⇒ f⁡ ′ x=11+x2>0, ∀ x∈0, ∞
∴ f(x) is increasing in its domain. Thus, f(x) is one-one in its domain.
For onto (we find range)
fx=x1+x ⇒ y=x1+x ⇒ y+yx=x
⇒ x=y1-y ⇒ y1-y≥0 as x≥0 ∴0≤y<1
i.e., Range ≠ Codomain
∴ fx is one-one but not onto.

Topic: Mathematics, Functions, JEE Main
If $\quad f(0)=0, f(1)=1, f(2)=2 \quad$ and $f(x)=f(x-2)+f(x-3)$ for $x=3,4,5, \ldots$, then $f(9)$ is equal to

Options:
A 12
B 13
C 14
D 10

The correct answer is:
10

Solution:
$f(0)=0, f(1)=1, f(2)=2$
Given,
$f(x)=f(x-2)+f(x-3), x=3,4,5, \ldots$
The given function is known as "Reccurrence function".
put $x=3, f(3)=f(1)+f(0)$
$=1+0=1$
put $x=4, f(4)=f(2)+f(1)$
$=2+1 \Rightarrow 3$
put $x=5, f(5)=f(3)+f(2)$
$=1+2 \Rightarrow 3$
put $x=6, f(6)=f(4)+f(3)$
$=3+1 \Rightarrow 3+1 \Rightarrow 4$
put $x=7, f(7)=f(5)+f(4)$
$=3+3 \Rightarrow 6$
put $x=8, f(8)=f(6)+f(5)$
$=3+4 \Rightarrow 7$
put $x=9, f(9)=f(7)+f(6)$
$=6+4 \Rightarrow 10$
Hence, $\quad f(9)=10$

Topic: Mathematics, Functions, JEE Main
If f0=0, f3=3 and f â² 3 =4, then the value of â« 0 1 x f â³ 3x dx is equal to

Options:

The correct answer is:
1

Solution:
Let, Î= â« 0 1 x f â³ 3x dx
Put 3x=tâdx=dt3
Î= 1 9 â« 0 3 t f â³ t dt
Î= 1 9 t f â² t 0 3 â â« 0 3 f â² t dt
= 1 9 3 f â² 3 â f t 0 3
=193Ã4-3=1

Topic: Mathematics, Definite Integration, JEE Main
If $f:[0, \infty) \rightarrow[0, \infty)$ is defined by $f(x)=\frac{x}{1+x}$, then $f$ is

Options:
A neither one-one not onto
B one-one but not onto
C onto but not one-one
D both one-one and onto

The correct answer is:
one-one but not onto

Solution:
We have, $f(x)=\frac{x}{1+x},[0, \infty) \rightarrow[0, \infty)$
One-one
Let $f\left(x_1\right)=f\left(x_2\right) \Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}$
$\Rightarrow \quad x_1+x_1 x_2=x_2+x_1 x_2 \Rightarrow x_1=x_2$
$\therefore f(x)$ is one-one.
Onto
Let
$$
\begin{aligned}
\text { Let } & & f(x) & =y \Rightarrow \frac{x}{1+x}=y \\
\Rightarrow & & x & =y+x y \Rightarrow x(1-y)=y
\end{aligned}
$$
$$
\Rightarrow \quad x=\frac{y}{1-y}
$$
Again, $\quad x \geq 0 \Rightarrow \frac{y}{1-y} \geq 0$
$\therefore \quad y \in[0,1)$
$\therefore$ Range of $f(x)=[0,1)$
Hence, Range $\neq$ Codomain.
So, $f(x)$ is not onto.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2018 (07 May Shift 1)
If $f:[0, \pi / 2) \rightarrow R$ is defined as
$f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$ Then, the range of $f$ is

Options:
A $(2, \infty)$
B $(-\infty,-2]$
C $[2, \infty)$
D $(-\infty, 2]$

The correct answer is:
$(2, \infty)$

Solution:
We have, $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$
$=1\left(1+\tan ^{2} \theta\right)-\tan \theta$
$(-\tan \theta+\tan \theta)+1\left(\tan ^{2} \theta+1\right)$
$=2\left(1+\tan ^{2} \theta\right)-0=2 \sec ^{2} \theta$
$\therefore$ Range of $f=(2, \infty)$

Topic: Mathematics, Determinants, JEE Main
If $f(1)=1, f^{\prime}(1)=2$, then $\operatorname{Lim}_{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1}$ is

Options:
A
2

B
4

C
1

D
1/2

The correct answer is:
2

Solution:

Topic: Mathematics, Limits, JEE Main, JEE Main 2002
If f1=1,f'1=3, then the derivative of fffx+fx2 at x=1 is:

Options:
A 9
B 12
C 15
D 33

The correct answer is:
33

Solution:
y=fffx+fx2

⇒dydx=f'ffx×f'fx×f'x+2fx×f'x

⇒dydxx=1=f'ff1×f'f1×f'1+2f1×f'1

=f'f1×f'1×3+2×1×3

=f'1×3×3+6

=3×3×3+6

=27+6

=33

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If f:[1,∞)→1,∞ is defined by fx=1+1+4log2x2, then f-13=

Options:
A

0

B 1
C 64
D 1+52

The correct answer is:
64

Solution:
fx=1+1+4log2x2

If fx=y, then x=f-1y.

1+1+4log2x2=y

⇒1+4log2x=2y-1

⇒1+4log2x=4y2+1-4y

⇒log2x=y2-y

⇒x=2y2-y

f-1y=2y2-y

⇒f-13=232-3

=29-3

=26

=64

Topic: Mathematics, Functions, JEE Main
If fα=∫1αlog10t1+tdt,α>0, then fe3+fe-3 is equal to

Options:
A 9
B 92
C 9loge10
D 92loge10

The correct answer is:
92loge10

Solution:
Given,

fα=∫1αlog10t1+tdt

So, fe3=∫1e3lntln101+tdt ⋯1

Again fα=∫1αlntln101+tdt

Now let t=1x⇒x=1t⇒dt=-1x2dx

So, fα=∫11a-lnnxln101+1x-1x2dx

⇒fα=1ln10∫11αlnxxx+1dx

So, fe-3=1ln10∫1e3lnttt+1dt …2

Now adding equation 1 and 2 we get,

fe3+fe-3=1lnn10∫1c3lnt1+t1+1tdt

⇒fe3+fe-3=1ln10∫13lnttdt

Now let lnt=r⇒dtt=dr

So, fe3+fe-3=1ln10∫03rdr

⇒fe3+fe-3=1ln10r2203

⇒fe3+fe-3=1log1092

⇒fe3+fe-3=92loge10

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If $\mathrm{f}(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|$ and
$\mathrm{A}$ and $\mathrm{B}$ are respectively the maximum and the minimum values of $f(\theta)$, then $(A, B)$ is equal to:

Options:
A
$(3,-1)$

B
$(4,2-\sqrt{2})$

C
$(2+\sqrt{2}, 2-\sqrt{2})$

D
$(2+\sqrt{2},-1)$

The correct answer is:
$(2+\sqrt{2}, 2-\sqrt{2})$

Solution:
$$
\begin{aligned}
&\text { Let } f(\theta)=\left|\begin{array}{ccc}
1 & \cos \theta & 1 \\
-\sin \theta & 1 & -\cos \theta \\
-1 & \sin \theta & 1
\end{array}\right| \\
&=(1+\sin \theta \cos \theta)-\cos \theta(-\sin \theta-\cos \theta) \\
&+1\left(-\sin ^2 \theta+1\right) \\
&=1+\sin \theta \cos \theta+\sin \theta \cos \theta+ \\
&=2+2 \sin \theta \cos \theta+\cos 2 \theta \\
&=2+\sin 2 \theta+\cos 2 \theta
\end{aligned}
$$
Now, maximum value of (1)
is $2+\sqrt{1^2+1^2}=2+\sqrt{2}$ and minimum value of (1) is
$$
2-\sqrt{1^2+1^2}=2-\sqrt{2} \text {. }
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2014 (12 Apr Online)
If f1+x=f1-x ∀x∈R, then the value of the integral Ι=∫-79fxfx+f2-xdx is

Options:
A 0
B 2
C 8
D 10

The correct answer is:
8

Solution:
As f1+x=f1-xâxâx-1fx=f2-x
â´ Using a+b-x;2Î=â«-79fx+f2-xf2-x+fxdx ⇒Î=8

Topic: Mathematics, Definite Integration, JEE Main
If $\quad f:[-2,2] \rightarrow R \quad$ is defined by $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+c x}-\sqrt{1-c x}}{x} & \text { for }-2 \leq x < 0 \\ \frac{x+3}{x+1} & \text { for } 0 \leq x \leq 2\end{array}\right.$ continuous on $[-2,2]$, then $c$ is equal to

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

The correct answer is:
$3$

Solution:
Given, $f:[-2,2] \rightarrow R$
$$
f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{1+c x}-\sqrt{1-c x}}{x}, & -2 \leq x < 0 \\
\frac{x+3}{x+1}, & 0 \leq x \leq 2
\end{array}\right.
$$
Now, $\quad$ LHL $=\lim _{x \rightarrow 0^{-}} f(x)$
$$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sqrt{1-c h}-\sqrt{1+c h}}{-h} \\
& \times \frac{\sqrt{1-c h}+\sqrt{1+c h}}{\sqrt{1-c h}+\sqrt{1+c h}} \\
& =\lim _{h \rightarrow 0} \frac{(1-c h)-(1+c h)}{-h(\sqrt{1-0}+\sqrt{1+0})} \\
& =\lim _{h \rightarrow 0} \frac{-2 c h}{-h(1+1)}=c
\end{aligned}
$$
and
$$
\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)
$$
$$
\begin{aligned}
& =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{0+h+3}{0+h+1} \\
& =\lim _{h \rightarrow 0} \frac{h+3}{h+1}=\frac{0+3}{0+1}=3
\end{aligned}
$$
Since, $f$ is continuous at $x=0$.
$$
\therefore \mathrm{LHL}=\mathrm{RHL} \Rightarrow \mathrm{c}=3
$$

Topic: Mathematics, Limits, JEE Main