datavorous/entrance-exam-dataset huggingface dataset:
If P→=3i^+3j^+2k^ and Q→=4i^+3j^+2.5k^ then, the unit vector in the direction of P→×Q→ is 1x(3i^+j^-23k^). The value of x is

Options:

The correct answer is:
4

Solution:
Unit vector in the direction of P→×Q→ is n^=P→×Q→|P→×Q→|.

Here, P→×Q→=i^j^k^332432.5=3i^2+j^2-3k^

And P→×Q→=322+122+32=4=2

⇒P→×Q→|P→×Q→|=123i^2+j^2-3k^

=14(3i^+j^-23k^) ⇒x=4

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
If \( p: 3 \) is an odd number and \( q: 15 \) is a prime number, then the truth value of \( [\sim(p \Longleftrightarrow q)] \) is equivalent to that of
\( \begin{array}{llll}\text { (i) } p & \Longleftrightarrow(\sim q) & \text { (ii) }(\sim p) \Longleftrightarrow q & \text { (iii) } \sim(p \wedge q)\end{array} \)

Options:
A Only (i)
B Only (iii)
C Both (i) and (ii)
D (i), (ii) and (iii)

The correct answer is:
(i), (ii) and (iii)

Solution:
There are two methods.

Either, use truth table of double implication.

OR
Apply the identity, ~p⟺~q≡~p⟺q≡p⟺~q

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p: "4 is an odd number" and q: "43 is an even number" are two statements, then which of the following statements is equivalent to ∼p⇒q?

Options:
A "4 is an odd number and 43 is an even number"
B The negation of the statement "4 is not an odd number or 43 is not an even number"
C Both ("4 is an odd number and 43 is an even number") and (The negation of the statement "4 is not an odd number or 43 is not an even number")
D "4 is an odd number and 43 is not an even number"

The correct answer is:
"4 is an odd number and 43 is not an even number"

Solution:
∼(p⇒q)≡∼∼p∨q≡p∧ ∼q
Hence, the equivalent statement of ∼p⇒q is ''4 is an odd number and 43 is not an even number''

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $\mathrm{P}(6,1)$ be the orthocentre of the triangle whose vertices are $\mathrm{A}(5,-2), \mathrm{B}(8,3)$ and $\mathrm{C}(\mathrm{h}, \mathrm{k})$, then the point $C$ lies on the circle:

Options:
A $x^2+y^2-61=0$
B $x^2+y^2-52=0$
C $x^2+y^2-65=0$
D $x^2+y^2-74=0$

The correct answer is:
$x^2+y^2-65=0$

Solution:
Slope of $A D=3$
Slope of $\mathrm{BC}=-\frac{1}{3}$
equation of $\mathrm{BC}=3 \mathrm{y}+\mathrm{x}-17=0$
slope of $\mathrm{BE}=1$
Slope of $\mathrm{AC}=-1$
equation of $\mathrm{AC}$ is $\mathrm{x}+\mathrm{y}-3=0$
point $C$ is $(-4,7)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If $\mathrm{P}\left(\frac{\pi}{6}\right)$ is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \mathrm{~S}, \mathrm{~S}^{\prime}$ are its foci and $\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}-2\left|\mathrm{SP}-\mathrm{S}^{\prime} \mathrm{P}\right|$, then $\mathrm{e}=$

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

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

Solution:
$$
\begin{aligned}
& \text { } \because\left|P S-P S^{\prime}\right|=2 a \\
& \Rightarrow 2|P S-P S|=4 a \\
& \text { now } P S+P S=2 e \cdot \frac{2 a}{\sqrt{3}}
\end{aligned}
$$
$\left(\right.$ where $\frac{2 \mathrm{a}}{\sqrt{3}}$ is the $\mathrm{x}$ - coordinate of point $\left.\mathrm{P}\right)$
$$
\Rightarrow \mathrm{e}=\sqrt{3}
$$

Topic: Mathematics, Hyperbola, JEE Main
If P=1,2,3,4,5 and Q=a,b,c, then the number of onto functions from P to Q is

Options:
A 150
B 144
C 147
D 154

The correct answer is:
150

Solution:
Total number of functions =35
(since each of 1,2,3,4 or 5 can correspond to any of a,b or c)
The number of functions that corresponds to only one element of B is 3C1Ã15 and the number of functions that correspond to atmost two elements of B is 3C2Ã25.
Total number of onto functions =35-3C2Ã25+3C1Ã15=243-96+3=150
3C1Ã15 is repeated twice in 3C2Ã25

Topic: Mathematics, Functions, JEE Main
If P–3,2 is one end of the focal chord PQ of the parabola (y+2)2=-4x-1, then the slope of the normal at Q is

Options:
A -12
B 2
C 12
D -2

The correct answer is:
-12

Solution:
The equation of the tangent at –3,2 to the parabola y2+4x+4y=0 is
2y+2x-3+2y+2=0
or 2x+4y-2=0⇒x+2y-1=0
Since the tangent at one end of the focal chord is parallel to the normal at the other end, the slope of the normal at the other end of the focal chord is -12

Topic: Mathematics, Parabola, JEE Main
If $P(A)=\frac{2}{3}, P(B)=\frac{2}{5}$ and $P(A \cup B)-P(A \cap B)=\frac{2}{5}$, then
what is $P(A \cap B)$ equal to ?

Options:
A $\frac{3}{5}$
B $\frac{5}{11}$
C $\frac{1}{3}$
D None of these

The correct answer is:
$\frac{1}{3}$

Solution:
$\quad P(A \cup B)-P(A \cap B)=\frac{2}{5}$
We know that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$\frac{2}{3}+\frac{2}{5}-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5}-\frac{2}{3}-\frac{2}{5}$
$-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6-16}{15} \Rightarrow \frac{-10}{15}=-\frac{2}{3} \times \frac{-1}{2}=\frac{1}{3}$

Topic: Mathematics, Probability, JEE Main
If $P(A)=0.8, P(B)=0.9, P(A B)=p$, which one ofthe following is correct?

Options:
A $0.72 \leq p \leq 0.8$
B $\quad 0.7 \leq p \leq 0.8$
C $0.72 < p < 0.8$
D $0.7 < p < 0.8$

The correct answer is:
$\quad 0.7 \leq p \leq 0.8$

Solution:
We know,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.8+0.9-p \leq 1$
$\Rightarrow \quad 1.7-p \leq 1$
$\Rightarrow \quad 0.7 \leq p$
Now, $\quad P(A) < P(B)$
$\therefore \quad P(A \cap B) \leq P(A)$
$\Rightarrow \quad p \leq 0.8$
Hence, $0.7 \leq p \leq 0.8$

Topic: Mathematics, Probability, JEE Main
If $\mathrm{P}(\mathrm{A})=\frac{3}{10}, \mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{5}$, then $\mathrm{P}(\mathrm{A} / \mathrm{B}) \times \mathrm{P}(\mathrm{B} / \mathrm{A})=$

Options:
A $\frac{1}{3}$
B $\frac{1}{12}$
C $\frac{1}{10}$
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{12}$

Solution:
$$
\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right) \times \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \times \frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}
$$
We know that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$$
\begin{aligned}
& \frac{3}{5}=\frac{3}{10}+\frac{2}{5}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{10} \\
& \therefore \text { Given expression }=\frac{\left(\frac{1}{10}\right)}{\left(\frac{2}{5}\right)} \times \frac{\left(\frac{1}{10}\right)}{\left(\frac{3}{10}\right)}=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main
If p: A man is happy and q: A man is rich are two statements, then the statement, "If a man is not happy, then he is not rich" can be written using logical operators as

Options:
A â¼ p â â¼ q
B ∼ q → p
C ∼ q → ∼ p
D q → ∼ p

The correct answer is:
â¼ p â â¼ q

Solution:
Given, p: A man is happy

âand q: A man is rich.
'If man is not happy, then he is not rich' is written as ∼ p → ∼ q

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $p \neq a, q \neq b, r \neq c$ and the system of equations
$\begin{aligned}
& p x+a y+a z=0 \\
& b x+q y+b z=0 \\
& c x+c y+r z=0
\end{aligned}$
has a non-trivial solution, then the value of
$\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

Options:
A
$1$

B
$2$

C
$\frac{1}{2}$

D
$0$

The correct answer is:

$2$

Solution:
As the given system of equations has a non-trivial solution.
$\Delta=\left|\begin{array}{lll}p & a & a \\ b & q & b \\ c & c & r\end{array}\right|=0$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$
$\Delta=\left|\begin{array}{ccc}p & a-p & a-p \\ b & q-b & 0 \\ c & 0 & r-c\end{array}\right|=0$
Expanding along $C_3$, we get
$(a-p)\left|\begin{array}{cc}b & q-b \\ c & 0\end{array}\right|+(r-c)\left|\begin{array}{cc}p & a-p \\ b & q-b\end{array}\right|=0$
$\begin{aligned}
& \Rightarrow(a-p)(-c)(q-b)+(r-c) \\
& \{p(q-b)-b(a-p)\}=0 \\
& \Rightarrow \\
& (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0
\end{aligned}$
Dividing by $(p-a)(q-b)(r-c)$, we get
$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$

Topic: Mathematics, Determinants, JEE Main
If $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $p$ and $p^{\circ}$ is

Options:
A $p^{\circ}=\rho\left[\frac{n_{1}}{n_{1}+n_{2}}\right]$
B $D^{\circ}=p\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$
C $p=p^{\circ}\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$
D $p=p^{\circ}\left[\frac{n_{1}}{n_{1}+n_{2}}\right]$

The correct answer is:
$p=p^{\circ}\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$

Solution:
Given,
$$
\begin{array}{l}
p^{\circ}= \text {vapour pressure of pure solvent}\\
p=\text { vapour pressure of solution } \\
n_{1}=\text { moles of solute } \\
n_{2}=\text { moles of solvent }
\end{array}
$$
Thus, according to Raoult's law.
$$
\frac{p^{\circ}-p}{p^{\circ}}=x_{1}=\frac{n_{1}}{n_{1}+n_{2}}
$$
$\left(x_{1}=\right.$ mole fraction of solute)
or, $\quad 1-\frac{p}{p^{\circ}}=\frac{n_{1}}{n_{1}+n_{2}}$
$$
\frac{p}{p^{\circ}}=1-\frac{n_{1}}{n_{1}+n_{2}}
$$
$\Rightarrow$
$$
\begin{array}{c}
\frac{n_{1}+n_{2}-n_{1}}{n_{1}+n_{2}} \\
\frac{p}{p_{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}
\end{array}
$$
$\Rightarrow p=p^{\circ} \left( \frac{n_{2}}{n_{1}+n_{2}} \right)$

Topic: Chemistry, Solutions, JEE Main
If $P(\theta)$ and $Q(\pi / 2+\theta)$ are two points on the ellipse $\frac{z^2}{a^2}+\frac{y^2}{b^2}=1$. Locus of the mid-point of $P Q$ is

Options:
A $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$
B $\frac{x^2}{a^2}+\frac{y^2}{b^2}=4$
C $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
D None of these

The correct answer is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$

Solution:
The coordinates of $P$ and $Q$ are $(a \cos \theta, b \sin \theta)$ and $(-a \sin \theta, b \cos \theta)$ respectively.
Let $(h, k)$ be the co-ordinates of the mid-point of $P Q$. Then,
$\begin{aligned}
& 2 h=a(\cos \theta-\sin \theta) \text { and } 2 k=b(\sin \theta+\cos \theta) \\
& \Rightarrow \frac{4 h^2}{a^2}+\frac{4 k^2}{b^2}=2
\end{aligned}$
Hence, the locus of $(h, k)$ is
$\frac{4 x^2}{a^2}+\frac{4 y^2}{b^2}=2 \text { or, } \frac{2 x^2}{a^2}+\frac{2 y^2}{b^2}=1$

Topic: Mathematics, Ellipse, JEE Main
If p and q are logical statements, then ~p→p→q is equivalent to

Options:
A p∧q
B p→p∨q
C p∨q
D p∨q↔p∧q

The correct answer is:
p→p∨q

Solution:
∴s→t≡∼s∨t
∴~p→p→q≡p∨p→q
≡p∨∼p∨q
≡p∨∼p∨q
≡t∨q≡t
Also, p→p∨q≡∼p∨p∨q≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p and q are logical statements, then p⇒~q⇒p is equivalent to

Options:
A p⇒p⇒q
B pâpâ¨q
C p⇒pâ§q
D p⇒pâq

The correct answer is:
p⇒p∨q

Solution:
p⇒~q⇒p≡∼p∨q∨p
≡∼p∨p∨q≡t∨q≡t
Hence, the given statement p⇒~q⇒p is a tautology
Now, p⇒p⇒q is false when p is true and q is false

Hence, not a tautology
p⇒q∨p≡~p∨p∨q≡~p∨p∨q
t∨q≡t
Hence, p⇒q∨p is a tautology.
Now, p⇒p∧q is false when p is true and q is false

Hence, not a tautology
Also, p⇒p⇔q is false when p is true and q is false

Hence, not a tautology

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p and q are logical statements, then p∧q→p→q is equivalent to

Options:
A p∧q
B p→p∨q
C p∨q
D p∨q⟷p∧q

The correct answer is:
p→p∨q

Solution:
∵s→t≡~s∨t
∴p∧q→p→q
≡∼p∨∼q∨~p∨q
≡~p∨q∨~q
≡∼p∨t≡t
Also, p→p∨q≡~p∨p∨q≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $p$ and $q$ are non-zero real numbers and $\alpha^3+\beta^3=-p, \alpha \beta=q$, then a quadratic equation whose roots are $\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha}$ is :

Options:
A
$p x^2-q x+p^2=0$

B
$q x^2+p x+q^2=0$

C
$p x^2+q x+p^2=0$

D
$q x^2-p x+q^2=0$

The correct answer is:
$q x^2+p x+q^2=0$

Solution:
Given $\alpha^3+\beta^3=-p$ and $\alpha \beta=q$
Let $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$ be the root of required quadratic equation.
So, $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta}=\frac{-p}{q}$
and $\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha}=\alpha \beta=q$
Hence, required quadratic equation is
$$
\begin{aligned}
&x^2-\left(\frac{-p}{q}\right) x+q=0 \\
&\Rightarrow x^2+\frac{p}{q} x+q=0 \Rightarrow q x^2+p x+q^2=0
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2013 (25 Apr Online)
If P and Q are points with eccentric angles Î¸ and Î¸+Ï6 on the ellipse x216+y24=1, then the area (in sq. units) of the triangle OPQ (where O is the origin) is equal to

Options:

The correct answer is:
2

Solution:
Let, P=4cosâ¡Î¸,2sinâ¡Î¸ and Q=4cosÎ¸+Ï6,2sinÎ¸+Ï6
âArea of ÎOPQ=124cosâ¡Î¸2sinâ¡Î¸14cosÎ¸+Ï62sinÎ¸+Ï61001
=4cosâ¡Î¸sinÎ¸+Ï6-cosÎ¸+Ï6sinâ¡Î¸
=4sinÎ¸+Ï6-Î¸=4sinÏ6=2 sq. units

Topic: Mathematics, Ellipse, JEE Main
If $p$ and $q$ are positive real numbers such that $p^2+q^2=1$, then the maximum value of $(p+q)$ is

Options:
A
$2$

B
$1 / 2$

C
$\frac{1}{\sqrt{2}}$

D
$\sqrt{2}$

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

Solution:
Using $A.M.$ $\geq$ $G.M.$
$\begin{aligned}
& \frac{p^2+q^2}{2} \geq p q \\
& \Rightarrow p q \leq \frac{1}{2} \\
& (p+q)^2=p^2+q^2+2 p q \\
& \Rightarrow p+q \leq \sqrt{2} .
\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2007
If p and q are real number such that p+q=3, p4+q4=369, then the value of 1p+1q-2 is equal to

Options:

The correct answer is:
4

Solution:
Given,

p4+q4=369⇒p2+q22-2pq2=369

⇒p+q2-2pq2-2pq2=369

⇒9-2pq2-2pq2=369

⇒pq2-18pq-144=0 ⇒pq=24(rejected),-6

Now, 1p+1q-2=pq2p+q2=pq29=4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If \( P \) and \( Q \) are symmetric matrices of the same order then \( P Q-Q P \) is

Options:
A symmetric matrix
B skew symmetric matrix
C identity matrix
D zero matrix

The correct answer is:
skew symmetric matrix

Solution:
(B)
Given $P=P^{\prime}$ and $Q=Q^{\prime}$
$(P Q-Q P)^{\prime}=(P Q)^{\prime}-(Q P)^{\prime}$
$=\left(Q^{\prime} P^{\prime}-P^{\prime} Q^{\prime}\right)$
$=Q P-P Q$
$=-[P Q-Q P]$
$\therefore(P Q-Q P)$ is skew symmetric.

Topic: Mathematics, Matrices, JEE Main
If p and q are the lengths of the perpendiculars from the origin on the lines, xcosecα-ysecα=kcot2α and xsinα+ycosα=ksin2α respectively, then k2 is equal to :

Options:
A 2p2+q2
B p2+2q2
C 4q2+p2
D 4p2+q2

The correct answer is:
4p2+q2

Solution:
L1:xcosecα-ysecα=kcot2α
xsinα-ycosα=kcos2αsin2α

xcosα-ysinαsinα cosα=kcos2α2 sinα cos α
⇒xcosα-ysinα=k2cos2α

We know that Perpendicular distance from a point x1, y1 to ax +by + c = 0 is ax1 + by1 + ca2+b2

Perpendicular distance from (0,0) is p=0-0-k2cos2αcos2α+sin2α

⇒p2=k24cos22α

L2:xsinα+ycosα-ksin2α=0

Perpendicular distance from (0,0) is q=0+0-ksin2αsin2α+cos2α

⇒q2=k2sin22α

4p2+q2=k2sin22α+cos22α

We know that sin2t + cos2t = 1.
4p2+q2=k2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If $P$ and $Q$ are the points of intersection of the circles $x^2+y^2+3 x+7 y+2 p-5=0$ and $x^2+y^2+2 x+2 y-p^2=0$, then there is a circle passing through $P, Q$ and $(1,1)$ for

Options:
A
all values of $p$

B
all except one value of $p$

C
all except two values of $p$

D
exactly one value of $p$

The correct answer is:
all values of $p$

Solution:
Given circles $S=x^2+y^2+3 x+7 y+2 p-5=0$
$$
S^{\prime}=x^2+y^2+2 x+2 y-p^2=0
$$
Equation of required circle is $S+\lambda S^{\prime}=0$
As it passes through $(1,1)$ the value of $\lambda=\frac{-(7+2 p)}{\left(6-p^2\right)}$
If $7+2 p=0$, it becomes the second circle
$\therefore$ it is true for all values of $p$

Topic: Mathematics, Circle, JEE Main, JEE Main 2009
If $p$ and $q$ are the roots of the equation $x^2+p x+q=0$, then

Options:
A
p =1, q = -2

B
p = 0, q = 1

C
p = -2, q = 0

D
p = -2, q = 1

The correct answer is:
p =1, q = -2

Solution:
$p+q=-p$ and $p q=q \Rightarrow q(p-1)=0 \Rightarrow q=0$ or $p=1$
If $q=0$, then $p=0$. i.e. $p=q \quad \therefore p=1$ and $q=-2$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2002
If p and q are two logical statements, then p⇒q⇒p is equivalent to

Options:
A p⇒p⇒q
B pâpâ¨q
C p⇒pâ§q
D p⇒pâq

The correct answer is:
p⇒p∨q

Solution:
p⇒q⇒p≡~p∨~q∨p
≡~p∨p∨~q ≡t∨~q ≡t
hence, p⇒q⇒p is a tautology
Now, p⇒p⇒q is false when p is true and q is false
p⇒q∨p ≡~p∨p∨q≡~p∨p∨q
t∨q ≡t
Hence, B is correct.
Also, p⇒p∧q is false when p is true and q is false
Also, p⇒p⇔q is false when p is true and q is false

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p and q are two logical statements, then ~p∨q→p→q is equivalent to

Options:
A p∧q
B p→p∨q
C p∨q
D p∨q↔p∧q

The correct answer is:
p→p∨q

Solution:
âµSâtâ¡~Sâ¨t
â´~pâ¨qâpâqâ¡pâ¨qâ¨pâq
â¡pâ¨qâ¨~pâ¨q
â¡pâ¨~pâ¨q
â¡tâ¨qâ¡t
Also, pâpâ¨qâ¡~pâ¨pâ¨qâ¡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p and q are two statements, then p∨∼p⇒ ∼q is equivalent to

Options:
A p∧q
B p
C p∧∼q
D ∼p∧q

The correct answer is:
p

Solution:
As we know that pâqâ¡â¼pâ¨q
Hence pâ¨ â¼pââ¼qâ¡ pâ¨ â¼â¼pâ¨â¼q
as we know that ~pâ¨qâ¡~pâ§~qHence pâ¨ â¼â¼pâ¨â¼qâ¡pâ¨pâ§q
Therefore pâ¨pâ§qâ¡p

Topic: Mathematics, Mathematical Reasoning, JEE Main
If P and Q are two statements, then which of the following compound statement is a tautology?

Options:
A ((P⇒Q)∧~Q)⇒Q
B ((P⇒Q)∧~Q)⇒~P
C ((P⇒Q)∧~Q)⇒P
D ((P⇒Q)∧~Q)⇒(P∧Q)

The correct answer is:
((P⇒Q)∧~Q)⇒~P

Solution:
LHS of all the options are same i.e. ((P→Q)∧~Q)

≡(~P∨Q)∧~Q

≡(~P∧~Q)∨(Q∧~Q)

≡~P∧~Q

(A) (~P∧~Q)→Q

≡~(~P∧~Q)∨Q

≡(P∨Q)∨Q≠ tautology

(B) (~P∧~Q)→~P

≡~(~P∧~Q)∨~P

≡(P∨Q)∨~P

(C) (~P∧~Q)→P

≡(P∨Q)∨P≠ Tautology

(D) (~P∧~Q)→(P∧Q)

≡(P∨Q)∨(P∧Q)≠ Tautology

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (18 Mar Shift 2)
If p and q are two statements, then which of the following statement is a tautology

Options:
A pâpâ¨~q
B pâ¨q⇒p
C p⇒pâ§q
D pâp⇒q

The correct answer is:
p⇒p∨~q

Solution:
p⇒p∨~q≡~p∨(p∨~q)≡~p∨p∨~q

We know, ~p∨p is always true

i.e. the given statement is always true, hence, a tautology

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p and q are two statements, then which of the following statements is not equivalent to pâpâq?

Options:
A pâ§q
B pâqâ§pâ¨q
C q⇒pâq
D ~qâpâ§pâ~q

The correct answer is:
~q⇒p∧p⇒~q

Solution:
p	q	pâq	pâpâq

T	T	T	T

T	F	F	F

F	T	T	F

F	F	T	F

âpâpâqâ¡pâ§qâA is true
Now, qâqâpâ¡pâ§qâ¡pâ§q
âqâpâqâ¡pâ§qâC is true
Now, common between
pâq & pâ¨q is
pâ§qâpâqâ§pâ¨qâ¡pâ§q
âB is correct
Now, ~qâpâ§pâ~qâ¡qâ¨pâ§~pâ¨~q
= Only p or only qâ¡~pâq

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $P$ is $(3,1)$ and $Q$ is a point on the curve $y^2=8 x$, then the locus of the mid-point of the line segment $P Q$ is

Options:
A $4 y^2-12 x-6 y+21=0$
B $4 y^2-16 x-4 y+25=0$
C $4 y^2+8 x-3 y-18=0$
D $4 y^2-12 x+8 y-15=0$

The correct answer is:
$4 y^2-16 x-4 y+25=0$

Solution:
$$
\text {} P(3,1) \text { and } Q \text { is a point of parabola } y^2=8 x
$$
$$
\text { Let } R_{(h, k)} \text { is the mid-point of } P Q \text {. }
$$
So, coordinate of $Q(2 h-3,2 k-1)$ point $Q$ lie on parabola, so $(2 k-1)^2=8(2 h-3)$
$$
\begin{array}{ll}
\Rightarrow & 4 k^2-4 k+1=16 h-24 \\
\Rightarrow & 4 k^2-4 k-16 h+25=0
\end{array}
$$
Hence, locus of $R(h, k)$ is $4 y^2-4 y-16 x+25=0$

Topic: Mathematics, Parabola, JEE Main
If P is a 3×3 real matrix such that PT=aP+(a-1)I, where a>1, then

Options:
A P is a singular matrix
B |AdjP|>1
C |AdjP|=12
D |AdjP|=1

The correct answer is:
|AdjP|=1

Solution:
Given,

PT=aP+(a-1)I......(1)

Taking transpose on both sides we get,

PTT=aP+(a-1)IT

⇒P=aPT+(a-1)I.......(2)

Subtract equation (1) and (2),

⇒PT-P=aP-PT

⇒P=PT as a≠-1

From equation (1),

P=aP+(a-1)I

⇒P=-I

Taking adjoint on both sides,

⇒Adj P=Adj(-I)

⇒Adj P=Adj (-I)=-I2

⇒|AdjP|=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If P is a point on the parabola y=x2+4 which is closest to the straight line y=4x-1, then the co-ordinates of P are:

Options:
A -2,8
B 1,5
C 2,8
D 3,13

The correct answer is:
2,8

Solution:
P:y=x2+4

k=h2+4

L:y=4x-1

y-4x+1=0

D=AB=k-4 h+117=h2+4-4 h+117

dDdh=2 h-417=0

h=2

d2Ddh2=217>0

∴k=4+4=8

∴ Point 2,8

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If p is the density and η is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is

Options:
A Re=πdρv
B Re=ρvηd
C Re=ρvdη
D Re=ηρνd

The correct answer is:
Re=ρvdη

Solution:
Reynolds number is given by, Re=ρvdη, where, v=flow velocity, d=hydraulic diameter of the pipe, ρ=density of the fluid & η=coefficient of viscosity of the fluid.

Reynolds number is the ratio of inertial forces to viscous forces within a fluid which is subjected to relative internal movement due to different fluid velocities.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be:

(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

Options:
A 49λ
B 34λ
C 23λ
D 12λ

The correct answer is:
49λ

Solution:
hν-ϕ=KE

hcλincident=KE+ϕ

Given ϕ<<KE

hcλincident≈KE

hcλincident=p22m ......(i)

hcλ'=1.5p22m …….(ii)

Divide (i) and (ii)

λ'λ=p22.25p2

λ'=λ×1002.25=49

λ'=49λ

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If $p$: It rains today, $q:$ I go to school, $r:$ I shall meet any friends and $s$ : I shall go for a movie, then which of the following is the proposition :
If it does not rain or if I do not go to school, then I shall meet my friend and go for a movie.

Options:
A $\sim(p \wedge q) \Rightarrow(r \wedge s)$
B $\sim(p \wedge \sim q) \Rightarrow(r \wedge s)$
C $\sim(p \wedge q) \Rightarrow(r \vee s)$
D None of these

The correct answer is:
$\sim(p \wedge q) \Rightarrow(r \wedge s)$

Solution:
Correct result is $(\sim p \vee \sim q) \Rightarrow(r \wedge s)$
So, $\sim(p \wedge q) \Rightarrow(r \wedge s)$

Topic: Mathematics, Mathematical Reasoning, JEE Main
If pKb for CN- at 25oC is 4.7, the pH of 0.5 M aqueous NaCN solution is

Options:
A 10
B 11.5
C 11
D 12

The correct answer is:
11.5

Solution:
pkb of CN-=4.7
pka of HCN=4.7
NaCN is salt of weak acid HCN and strong base NaOH So
pH=7+12pka+12logC
=7+12Ã9.3+12logâ¡0.5
â´pH=11.5

Topic: Chemistry, Ionic Equilibrium, JEE Main
If Pn-12n+1:Pn2n-1=11:21, then n2+n+15 is equal to :

Options:

The correct answer is:
45

Solution:
Given,

Pn-12n+1Pn2n-1=1121

⇒2n+1!n+2!2n-1!n-1!=1121

⇒21×2n+12n2n-1!n+2n+1nn-1!=11×2n-1!n-1!

⇒21×22n+1=11n+2n+1

⇒2n+1n+2n+1=1142

Now on comparing both side we get, n=5

So, the value of n2+n+15=52+5+15=45

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If p→~p∨~q is false, then the truth values of p and q are, respectively

Options:
A F, F
B T, T
C F, T
D T, F

The correct answer is:
T, T

Solution:
p→q has truth value false only when p is true & q is false. Hence, truth value of p is true.
Now, ~p∨~q is false only when ~p & ~q both are false.

Hence, p and q are both true.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2018 (16 Apr Online)
If p→p∧~q is false, then the truth values of p and q are respectively

Options:
A F,F
B T,F
C T,T
D F,T

The correct answer is:
T,T

Solution:
p	q	∽q	p∧∽q	p⟶p∧∽q

T	T	F	F	F

T	F	T	T	T

F	T	F	F	T

F	F	T	F	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If pâ~ pâ¨q is false, then the truth values of p and q are respectively

Options:
A F,T
B F,F
C T,T
D T,F

The correct answer is:
T,F

Solution:
pâ~ pâ¨q is false means p is true and ~ pâ¨q is false.
â p is true and both ~p and q are false.
âp is true and q is false.

Topic: Mathematics, Mathematical Reasoning, JEE Main