datavorous/entrance-exam-dataset huggingface dataset:
The products obtained when benzyl phenyl ether is heated with HI in the mole ratio \( 1: 1 \) are
1. Phenol
2. Benzyl alcohol
3. Benzyl iodide
4. Iodobenzene

Options:
A \( 1 \) and \( 3 \) only
B \( 3 \) and \( 4 \) only
C \( 1 \) and \( 4 \) only
D \( 2 \) and \( 4 \) only

The correct answer is:
\( 1 \) and \( 3 \) only

Solution:
Benzyl phenyl ether is an unsymmetrical ether so halide ion of HI attached to the simple alkyl group and reaction takes place by following mechanism.
1. Protonation of ether
Benzyl phenyl ether
2. Nucleophilic attack

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

Options:
A ClO2- and ClO3-
B Cl- and ClO-
C Cl- and ClO2-
D ClO- and ClO3-

The correct answer is:
Cl- and ClO-

Solution:
Cl2 undergoes disproportionation in cold NaOH.
Cl2+2 NaOH →NaCl+NaOCl+H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2017 (02 Apr)
The products of the following reactions are
(i) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{Cl}+\mathrm{Cu}_{2} \mathrm{Cl}_{2} \rightarrow \)
(ii) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{Br}_{2} \) (aq) \( \rightarrow \)

Options:
A \( { }_{6} \mathrm{H}_{5} \mathrm{Cl} \) and

B \( \mathrm{H}_{5} \mathrm{Cl} \) and

C \( \mathrm{H}_{5} \mathrm{Cl} \) and

D ne of these

The correct answer is:
\( { }_{6} \mathrm{H}_{5} \mathrm{Cl} \) and

Solution:
The Cl, Br and CN nucleophiles can easily be introduced in the benzene ring of benzene diazonium salt in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction.

(i) C6 H5 N2 Cl+Cu2Cl2→ C6 H5Cl+N2

When aniline treated with bromine water, the bromine water gets decolourized and white precipitate is formed. This reaction results in the formation of 2,4,6-tribromo phenylamine.

(ii)

Topic: Chemistry, Hydrocarbons, JEE Main
The projectile motion of a particle of mass 5 g is shown in the figure.

The initial velocity of the particle is 52 ms-1 and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x×10-2 kgms-1. The value of x, to the nearest integer, is ___ .

Options:

The correct answer is:
5

Solution:
|u→|=|v→|

u→=ucos45i^+usin45j^

v→=vcos45i^-vsin45j^

|ΔP→|=mv→-u→

ΔP=2musin45°=2×5×10-3×52×12

ΔP=50×10-3=5×10-2.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The projection of 2i^-3j^+4k^ on the line whose equation is r→=3+λi^+3-2λj^+5+6λk^, where λ is a scalar parameter, is

Options:
A 641
B 3241
C 1641
D 75

The correct answer is:
3241

Solution:
Given line is parallel to i^-2j^+6k^=bâ (let)
Let aâ=2i^-3j^+4k^
Projection of aâ on bâ is a → ⋅ b → b → = 2+6+24 41 = 32 41

Topic: Mathematics, Vector Algebra, JEE Main
The projection of the line joining the points \( (3,4,5) \) and \( (4,6,3) \) on the line joining the points \( (-1,24) \) and \( (1,0,5) \) is

Options:
A \( 4 / 3 \)
B \( 2 / 3 \)
C \( -4 / 3 \)
D \( 1 / 2 \)

The correct answer is:
\( -4 / 3 \)

Solution:
Let A=3, 4 , 5, B=4, 6, 3, C=-1, 2, 4, D≡(1, 0, 5)
For AB, x2-x1=4-3=1, y2-y1=6-4=2
z2-z1=3-5=-2
Let l, m, n for CD are 23, -23,13.
∴ Projection of AB on CD=∑l(x2-x1)
=213+-232+13(-2)
=-43

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The projection of the line segment joining the point 1,-1,3 and 2,-4,11 on the line joining the points -1,2,3 and 3,-2,10 is _______

Options:

The correct answer is:
8

Solution:
Let A 1,-1,3 , B 2,-4,11
AB→=i^-3j^+8k^
C -1,2,3 , D 3,-2,10
CD→=4i^-4j^+7k^
Projection of AB→ on CD→ =AB→.CD→CD→=4+12+5616+16+49=729=8

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The projection of the line segment joining the points $(-1,0,3)$ and $(2,5,1)$ on the line whose direction ratios are $6,2,3$ is

Options:
A $\frac{10}{7}$
B $\frac{22}{7}$
C $\frac{18}{7}$
D None of these

The correct answer is:
$\frac{22}{7}$

Solution:
Obviously, the projection
$\begin{aligned}
& =[2-(-1)] \frac{6}{7}+[5-0] \frac{2}{7}+[1-3] \frac{6}{7} \\
& =\frac{18+10-6}{7}=\frac{22}{7} .
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The projection of the line x-12=y+11=z-23 on a plane P is x-11=y+12=z-21, then the equation of the plane P is

Options:
A 5x-8y+11z=35
B 5x+8y–21z+45=0
C 5x+8y+11z=35
D 5x–8y+21z=45

The correct answer is:
5x-8y+11z=35

Solution:
The required plane passes through 1,–1,2
The required plane is parallel to i^j^k^213121
=i^-5-j^-1+k^3
=-5i^+j^+3k^
So, normal vector to the plane is i^j^k^-513121
=i^-5-j^-8+k^-11
=-5i^+8j^-11k^
Equation of the required plane is
-5x-1+8y+1-11z-2=0
-5x+5+8y+8-11z+22=0
-5x+8y-11z+35=0
5x-8y+11z-35=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The projection of the vector $2 \mathbf{i}+\mathbf{j}-3 \mathbf{k}$ on the vector $\mathbf{i}-2 \mathbf{j}+\mathbf{k}$ is

Options:
A $-\frac{3}{\sqrt{14}}$
B $\frac{3}{\sqrt{14}}$
C $-\sqrt{\frac{3}{2}}$
D $\frac{3}{\sqrt{2}}$

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

Solution:
$\frac{(2 \mathbf{i}+\mathbf{j}-3 \mathbf{k}) \cdot(\mathbf{i}-2 \mathbf{j}+\mathbf{k})}{|\mathbf{i}-2 \mathbf{i}+\mathbf{k}|}=\frac{2-2-3}{\sqrt{1+4+1}}=\frac{-3}{\sqrt{6}}=-\sqrt{\frac{3}{2}}$

Topic: Mathematics, Vector Algebra, JEE Main
The projection of the vector $\mathbf{i}+\mathbf{j}+\mathbf{k}$ along the vector $\mathbf{j}$ is

Options:
A 1
B 0
C 2
D -1

The correct answer is:
1

Solution:
Projection of the vector $\mathbf{i}+\mathbf{j}+\mathbf{k}$ along vector
$\mathbf{j}=\frac{(\mathbf{i}+\mathbf{j}+\mathbf{k}) \cdot \mathbf{j}}{|\mathbf{j}|}=\frac{1}{1}=1 .$

Topic: Mathematics, Vector Algebra, JEE Main
The projections of a line on the cl-ordinate axes are $4,6,12$. The direction cosines of the line are

Options:
A $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$
B $2,3,6$
C $\frac{2}{11}, \frac{3}{11}, \frac{6}{11}$
D None of these

The correct answer is:
$\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$

Solution:
Direction cosines
$=\left(\frac{4}{\sqrt{16+36+144}}, \frac{6}{14}, \frac{12}{14}\right)$
or $\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The projections of a vector on the three coordinate axis are $6,-3,2$ respectively. The direction cosines of the vector are

Options:
A
$6,-3,2$

B
$\frac{6}{5},-\frac{3}{5}, \frac{2}{5}$

C
$\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$

D
$-\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$

The correct answer is:
$\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$

Solution:
Projection of a vector on coordinate axis are $x_2-x_1, y_2-y_1, z_2-z_1$
$$
\begin{aligned}
& x_2-x_1=6, y_2-y_1=-3, z_2-z_1=2 \\
& \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}=\sqrt{36+9+4}=7
\end{aligned}
$$
The D.C's of the vector are $\frac{6}{7},-\frac{3}{7}, \frac{2}{7}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2009
The property of hydrogen which distinguishes it from alkali metals is

Options:
A Its electropositive character
B Its affinity for non metal
C Its reducing character
D Its non-metallic character

The correct answer is:
Its non-metallic character

Solution:
Hydrogen is a non - metal while all other members of group 1 (alkali metals) are metals.

Topic: Chemistry, Hydrogen, JEE Main
The property of the alkaline earth metals that increases with their atomic number is

Options:
A lonisation energy
B Electronegativity
C Solubility of their sulphates
D Solubility of their hydroxides

The correct answer is:
Solubility of their hydroxides

Solution:
Solubility of alkaline earth metal hydroxides increase with their atomic number. In other words, as we go down the group IIA, the solubilities of the hydroxides increases.

Topic: Chemistry, s Block Elements, JEE Main
The proposition p→~p∧~q is equivalent to :

Options:
A q
B ~p∨q
C ~p∧q
D ~p∨~q

The correct answer is:
~p∨q

Solution:
~p∧~q

=~p∨~~q By De Morgan's Laws.

=~p∨q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The proposition ~p∨p∧~q is equivalent to

Options:
A p→∼q
B p∧∼q
C q→p
D none

The correct answer is:
p→∼q

Solution:
~p∨p∧~q

p	q	∼p	∼q	p ∧∼q

~p∨p∧~q

p→~q

T	T	F	F	F	F	F

T	F	F	T	T	T	T

F	T	T	F	F	T	T

F	F	T	T	F	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2017 (08 Apr Online)
The proposition $\sim(p \vee \sim q) \vee \sim(p \vee q)$ is logically equivalent to:

Options:
A
$\mathrm{p}$

B
$\mathrm{q}$

C
$\sim \mathrm{p}$

D
$\sim \mathrm{q}$

The correct answer is:
$\sim \mathrm{p}$

Solution:
Given $\sim(p \vee \sim q) \vee \sim(p \vee q)$
$$
\begin{aligned}
&\equiv(\sim p \vee q) \vee(\sim p \vee \sim q) \\
&\equiv \sim p \vee(q \vee \sim q) \\
&\equiv \sim p
\end{aligned}
$$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2014 (11 Apr Online)
The proposition \( (\sim p) \vee(p \wedge \sim q) \) is equivalent to

Options:
A \( p \rightarrow \sim q \)
B \( p \wedge \sim q \)
C \( q \rightarrow p \)
D none

The correct answer is:
\( p \rightarrow \sim q \)

Solution:
~p∨p∧~q

p	q	∼p	∼q	p ∧∼q

~p∨p∧~q

p→~q

T	T	F	F	F	F	F

T	F	F	T	T	T	T

F	T	T	F	F	T	T

F	F	T	T	F	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main
The proteins which are insoluble in water are

Options:
A Fibrous proteins
B Globular proteins
C Both (a) and (b)
D None of these

The correct answer is:
Fibrous proteins

Solution:
Fibrous proteins are insoluble in water.

Topic: Chemistry, Biomolecules, JEE Main
The P-T graph as given below was observed for a process of an ideal gas, which of the following statement is true?

Options:
A w=+ve,ΔH=+ve
B w=-ve,ΔH=-ve
C w=-ve,ΔH=+ve
D w=+ve,ΔH=-ve

The correct answer is:
w=-ve,ΔH=+ve

Solution:
VB>VA and TB>TA
ÎH=ncp(TB-TA)>0
w=-pextVB-VA<0

Topic: Chemistry, Thermodynamics (C), JEE Main
The pulley system, shown in the figure, has two pulleys of radii 2 m and 1 m, respectively. If the acceleration of block A is 5 m s-2 in the downward direction, then the acceleration of block B will be

Options:
A 0 m s-2
B 5 m s-2
C 10 m s-2
D 52 m s-2

The correct answer is:
52 m s-2

Solution:
Given aA=α×2 m=5 m s-2
⇒α=52 m s-2
Hence, acceleration of B, aB=α×1 m=52 m s-2

Topic: Physics, Rotational Motion, JEE Main
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle Î¸ should be

Options:
A 45°
B 60°
C 0°
D 30°

The correct answer is:
45°

Solution:
The tension in both the strings will be same.

For equilibrium in vertical direction for body B we have
2 mg=2TcosÎ¸
â´ââââââââââââââ 2 mg=2( mg )cosÎ¸
( âµââT=mg at equilibrium)
â´âââââââââââcosÎ¸= 1 2 âââââââââââââââÎ¸=45Â°

Topic: Physics, Laws of Motion, JEE Main
The pure inductors each of inductance $6 \mathrm{H}$ are connected as shown in the figure. Their equivalent inductance between the points ' $\mathrm{P}$ ' and ' $\mathrm{Q}$ ' is

Options:
A $0.5 \mathrm{H}$
B $18 \mathrm{H}$
C $6.3 \mathrm{H}$
D $2 \mathrm{H}$

The correct answer is:
$2 \mathrm{H}$

Solution:
In the given circuit, one end of each inductor is joined together. Similarly, the other end of each inductor is joined together. Hence the three inductors are connected in parallel. Their equivalent inductance is given by
$$
\begin{aligned}
& \frac{1}{L}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6} \\
& \therefore \mathrm{L}=2 \mathrm{H}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
The purest form of commercial iron is:

Options:
A pig iron
B wrought iron
C cast iron
D scrap iron and pig iron

The correct answer is:
wrought iron

Solution:
Purest form is wrought iron.

Wrought iron is an iron alloy with a very low carbon content (less than 0.08%) in contrast to that of cast iron (2.1% to 4%). It consists of around 99.4% iron by mass. The presence of slag is beneficial for blacksmithing operations, and gives the material its unique fibrous structure.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The purification of the colloidal particles from crystalloid dimensions through semipermeable membrane is known as

Options:
A Coagulation
B Dialysis
C Ultrafiltration
D Peptisation

The correct answer is:
Dialysis

Solution:
This property of animal membranes is utilized for the purification of sols. The process involved is called dialysis. It may be defined as follows. The process of separating the impurity particles of true solution dimensions (crystalloids) from an impure sol by means of diffusion through a suitable membrane such as parchment paper or cellophane membrane is called dialysis. The apparatus used in this method is called dialyzer. It consists of a bag made of parchment or cellophane. The bag is filled with the impure sol to be purified and is suspended in a tank through which pure water is circulated. The impurities of electrolytes present in the sol diffuse out of the bag leaving behind pure sol in the bag.

Topic: Chemistry, Surface Chemistry, JEE Main
The pyrimidine bases present in DNA are

Options:
A
cytosine and adenine

B
cytosine and guanine

C
cytosine and thymine

D
cytosine and uracil

The correct answer is:
cytosine and thymine

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2006
The pyrimidine bases present in DNA are

Options:
A Cytosine and Uracil
B Cytosine and Thymine
C Cytosine and Guanine
D Cytosine and Adenine

The correct answer is:
Cytosine and Thymine

Solution:
In DNA four bases are present adenine (A), guanine (G), cytosine (C) and thymine (T). Out of which pyrimidine bases are cytosine and thymine.

Topic: Chemistry, Biomolecules, JEE Main
The pyrimidne bases present in DNA are

Options:
A cytosine and adenine
B cytosine and guanine
C cytosine and thymine
D cytosine and uracil

The correct answer is:
cytosine and thymine

Solution:
DNA contains cytosine and thymine as pyrimidine bases and guanine and adenine as purine bases.

Topic: Chemistry, Biomolecules, JEE Main
The quadratic equation 1-sin⁡θx2+21-sin⁡θx-3sin⁡θ=0 has both roots complex for all θ lying in the interval

Options:
A -π2,π2
B 0,3π2
C π6,7π6
D 7π6,11π6

The correct answer is:
7π6,11π6

Solution:
D=41-sinâ¡Î¸2+41-sinâ¡Î¸â3sinâ¡Î¸<0
â 41-sinâ¡Î¸1-sinâ¡Î¸+3sinâ¡Î¸<0
â 41-sinâ¡Î¸1+2sinâ¡Î¸<0
â 8sinâ¡Î¸-1sinâ¡Î¸+12>0
â sinâ¡Î¸<-12 or sinâ¡Î¸>1
sinâ¡Î¸<-12âÎ¸â7Ï6,11Ï6

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation
$2 x^{2}-\left(a^{3}+8 a-1\right) x+a^{2}-4 a=0$ has roots of opposite sign. Then,

Options:
A $a \leq 0$
B $0 < a < 4$
C $4 \leq a < 8$
D $a \geq 8$

The correct answer is:
$0 < a < 4$

Solution:
Since, roots are opposite sign.
Product of the roots $\leq 0$ $\begin{array}{ll}\Rightarrow & a^{2}-4 a < 0 \\ \Rightarrow & a(a-4) < 0 \\ \Rightarrow & 0 < a < 4\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation whose roots are $l$ and $m$, where
$$
\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text { is }
\end{aligned}
$$

Options:
A $x^2+5 x+6=0$
B $x^2-5 x+6=0$
C $x^2-5 x-6=0$
D $x^2+5 x-6=0$

The correct answer is:
$x^2-5 x+6=0$

Solution:
We have,
$$
l=\lim _{\theta \rightarrow 0}\left[\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right]
$$
Using $L$-hospital's rule
$$
=\lim _{\theta \rightarrow 0} \frac{3 \cos \theta-8 \sin \theta \cos \theta}{1}=3
$$
and
$$
\begin{aligned}
m & =\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{\theta} \\
& =\lim _{\theta \rightarrow 0} \frac{\tan 2 \theta}{2 \theta} \cdot 2=2
\end{aligned}
$$
The required quadratic equation is
$$
\begin{aligned}
& x^2-(l+m) x+l m=0 \\
& \Rightarrow \quad x^2-(3+2) x+3 \cdot 2=0 \\
& x^2-5 x+6=0 \\
&
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is :

Options:
A $16 x^2-12 x+1=0$
B $16 x^2+12 x+1=0$
C $16 x^2-12 x-1=0$
D $16 x^2+10 x+1=0$

The correct answer is:
$16 x^2-12 x+1=0$

Solution:
Since $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ are the roots of a quadratic equation.
$\therefore \text { Sum of roots }=\sin ^2 18^{\circ}+\cos ^2 36^{\circ}$
$=\left(\frac{\sqrt{5}-1}{4}\right)^2+\left(\frac{\sqrt{5}+1}{4}\right)^2$
$=\frac{5+1-2 \sqrt{5}}{16}+\frac{5+1+2 \sqrt{5}}{16}$
$=\frac{12}{16}=\frac{3}{4}$
and product of roots $=\sin ^2 18^{\circ} \cdot \cos ^2 36^{\circ}$
$=\left(\frac{\sqrt{5}-1}{4}\right)^2\left(\frac{\sqrt{5}+1}{4}\right)^2$
$=\left(\frac{5-1}{4 \times 4}\right)^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}$
Required equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$, is
$x^2-(\text { sum of roots }) x+(\text { product of roots })=0$
$\Rightarrow \quad x^2-\frac{3}{4} x+\frac{1}{16}=0$
$\Rightarrow \quad 16 x^2-12 x+1=0$

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation whose roots are the arithmetic mean and the harmonic mean of the roots of the equation x2+7x-1=0 is

Options:
A 14x2+14x-45=0
B 45x2-14x+14=0
C 14x2+45x-14=0
D 45x2+14x-45=0

The correct answer is:
14x2+45x-14=0

Solution:
Let, α,β are the roots of the equation x2+7x-1=0

⇒α+β=-7,αβ=-1

⇒A=α+β2=-72

and H=2αβα+β=2-1-7=27.
Now, equation whose roots are A and H is
x2-A+Hx+AH=0⇒x2-x-72+27+-7227=0
⇒x2+4514x-1=0⇒14x2+45x-14=0

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation whose roots are the numbers having arithmetic mean 34 and geometric mean 16 is

Options:
A $x^{2}+68 x-256=0$
B $x^{2}-68 x-256=0$
C $x^{2}-68 x+256=0$
D $x^{2}+68 x+256=0$

The correct answer is:
$x^{2}-68 x+256=0$

Solution:
Quad $E q^{n}: \rightarrow$
$A M \text { of } a, b=34$
$a+b=68$
GM of $a, b,=\pm 6$
$\begin{aligned} a, b \rightarrow \text { roots } \rightarrow \quad a b &=256 \\ x^{2}-68 x+256 &=0 \end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equation whose roots are the $\mathrm{x}$ and yintercepts of the line passing through (1,1) and making a triangle of area $\mathrm{A}$ with the co-ordinate axes is

Options:
A $x^{2}+A x+2 A=0$
B $x^{2}-2 A x+2 A=0$
C $x^{2}-A x+2 A=0$
D None of these

The correct answer is:
$x^{2}-2 A x+2 A=0$

Solution:
Equation of the line making intercepts a and

$b$ on the axes is $\frac{x}{a}+\frac{y}{b}=1$ Since, it passes through (1,1) $\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=1$

Also the area of the triangle formed by the line and the axes is $\mathrm{A}$.

$\therefore \frac{1}{2} \mathrm{ab}=\mathrm{A} \Rightarrow \mathrm{ab}=2 \mathrm{~A}$

From eqs. (i) and (ii), we get, $a+b=2 A$ Hence, a and b are the roots of the eq. $x^{2}-(a+b) x+a b=0 \Rightarrow x^{2}-2 A x+2 A=0$

Topic: Mathematics, Straight Lines, JEE Main
The quadratic equations $x^2-6 x+a=0$ and $x^2-c x+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4: 3$. Then the common root is

Options:
A
1

B
4

C
3

D
2

The correct answer is:
2

Solution:
Let $\alpha$ and $4 \beta$ be roots of $x^2-6 x+a=0$ and $\alpha, 3 \beta$ be the roots of $x^2-c x+6=0$, then $\alpha+4 \beta=6$ and $4 \alpha \beta=a$
$\alpha+3 \beta=c$ and $3 \alpha \beta=6$.
We get $\alpha \beta=2 \Rightarrow a=8$
So the first equation is $x^2-6 x+8=0 \Rightarrow x=2,4$
If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$
If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2 \quad$ (non-integer)
$\therefore$ common root is $x=2$.

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic equations $x^2-6 x+a=0$ and $x^2-c x+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4: 3$. Then the common root is

Options:
A
1

B
4

C
3

D
2

The correct answer is:
2

Solution:
Let $\alpha$ and $4 \beta$ be roots of $x^2-6 x+a=0$ and $\alpha, 3 \beta$ be the roots of $x^2-c x+6=0$, then $\alpha+4 \beta=6$ and $4 \alpha \beta=a$
$\alpha+3 \beta=c$ and $3 \alpha \beta=6$.
We get $\alpha \beta=2 \Rightarrow a=8$
So the first equation is $x^2-6 x+8=0 \Rightarrow x=2,4$
If $\alpha=2$ and $4 \beta=4$ then $3 \beta=3$
If $\alpha=4$ and $4 \beta=2$, then $3 \beta=3 / 2 \quad$ (non-integer)
$\therefore$ common root is $x=2$.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2008
The quadratic equations x2-6x+a=0 and x2-cx+6=0 have one root in common. The other roots of the first equation and the second equation are integers in the ratio 4:3. Then the common root is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
2

Solution:
Let, Î± be the common root and the other roots of the equations be 4Î² and 3Î² respectively. Then,
Î±+4Î²=6,ââÎ±4Î²=a
Î±+3Î²=c,ââÎ±3Î²=6â43=a6âââÎ±=8
The first equation is x2-6x+8=0
Whose roots are 2 and 4
If Î±=2ââÎ²=1
So, roots of first equation is 2,â4 and that of second equation is 2,â3
If Î±=4âÎ²=12ââ3Î²=32,â4Î²=2
Here the roots are not integers âÎ±=2

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic function $f(x)=2 x^2-4 x+1$ attains its minimum value at:

Options:
A $x=0$
B $x=1$
C $x=2$
D $x=-1$

The correct answer is:
$x=1$

Solution:
The vertex form of a quadratic function $a x^2+b x+c$ is at $x=-\frac{b}{2 a}$.
Here, $a=2$ and $b=-4$.
$x=-\frac{-4}{2 \times 2}=\frac{4}{4}=1$
Thus, the function attains its minimum value at $x=1$.

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic function $f(x)=2 x^2-8 x+6$ has a range of:

Options:
A $(-\infty, 6]$
B $[0, \infty)$
C $[-2, \infty)$
D $[4, \infty)$

The correct answer is:
$[-2, \infty)$

Solution:
To find the range, we need to find the vertex of the quadratic function. The vertex form of a quadratic function $f(x)=a x^2+b x+c$ is given by $x=-\frac{b}{2 a}$.
For $f(x)=2 x^2-8 x+6$ :
- $a=2$
- $b=-8$

So,
$x=-\frac{-8}{2 \cdot 2}=\frac{8}{4}=2$
Now, substitute $x=2$ back into the function to find the minimum value:
$f(2)=2(2)^2-8(2)+6=2 \cdot 4-16+6=8-16+6=-2$
Since the parabola opens upwards $(\mathrm{a}\gt0$ ), the range of the function is:
$f(x) \geq-2$

Topic: Mathematics, Quadratic Equation, JEE Main
The quadratic function $f(x)=-3 x^2+12 x-7$ has a range of:

Options:
A $(-\infty, 1]$
B $(-\infty, 2]$
C $(-\infty, 5]$
D $(-\infty, 6]$

The correct answer is:
$(-\infty, 5]$

Solution:
For $f(x)=-3 x^2+12 x-7$ :
- $a=-3$
- $b=12$
So,
$x=-\frac{12}{2(-3)}=\frac{12}{6}=2$
Now, substitute $x=2$ back into the function to find the maximum value:
$f(2)=-3(2)^2+12(2)-7=-3 \cdot 4+24-7=-12+24-7=5$
Since the parabola opens downwards $(\mathrm{a} \lt 0$ ), the range of the function is:
$f(x) \leq 5$
Therefore, the correct answer is:
3) $(-\infty, 5]$

Topic: Mathematics, Quadratic Equation, JEE Main
The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3CH211OSO3- Na+ at room temperature. The correct assignment of the sketches is -

Options:
A I :KCl, II :CH3OH, III : CH3CH211OSO3- Na+
B I :CH3CH211OSO3- Na+, II :CH3OH, III : KCl
C I :KCl, II :CH3CH211OSO3- Na+, III : CH3OH
D I :CH3OH, II :KCl, III : CH3CH211OSO3- Na+

The correct answer is:
I :CH3OH, II :KCl, III : CH3CH211OSO3- Na+

Solution:
Water has large surface tension due to very strong interaction. Generally adding organic derivatives to water decreases its surface tension due to hydrophobic interaction.
In case III, hydrophobic interaction is stronger than case II causing surface tension to decrease more rapidly.
Due to K+ Clâ (inorganic electrolyte) intermolecular forces increases, surface tension increases.

Topic: Chemistry, Surface Chemistry, JEE Main