datavorous/entrance-exam-dataset huggingface dataset:
If $\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0,$ then the value of

$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}$ is

Options:
A 0
B 1
C 2
D $4 \mathrm{pqr}$

The correct answer is:
2

Solution:
$\left|\begin{array}{ccc}\mathrm{p} & \mathrm{q}-\mathrm{y} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q} & \mathrm{r}-\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$

Apply $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},$ we get

$\left|\begin{array}{ccc}\mathrm{x} & 0 & -\mathrm{z} \\ 0 & \mathrm{y} & -\mathrm{z} \\ \mathrm{p}-\mathrm{x} & \mathrm{q}-\mathrm{y} & \mathrm{r}\end{array}\right|=0$

$\Rightarrow \mathrm{x}[\mathrm{yr}+\mathrm{z}(\mathrm{q}-\mathrm{y})]-\mathrm{z}[0-\mathrm{y}(\mathrm{p}-\mathrm{x})]=0$

[Expansion along first row $]$

$\Rightarrow x y r+z x q+y z p=2 x y z \Rightarrow \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$

Topic: Mathematics, Determinants, JEE Main
If p, q and r are 3 statements, then the truth value of ~p∨q∧~r⇒p is

Options:
A True if truth values of p,q,r are T,F,T respectively
B False if truth values of p,q,r are T,F,T respectively
C False if truth values of p,q,r are T,F,F respectively
D False if truth values of p,q,r are T,T,T respectively

The correct answer is:
True if truth values of p,q,r are T,F,T respectively

Solution:
pqr~p~r~p∨q~p∨q∧~r~p∨q∧~r⇒pTFTFFFFTTFFFTFFTTTTFFTFT

Topic: Mathematics, Mathematical Reasoning, JEE Main
If P, Q and R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

Options:
A PQ−Q2R
B PQ-R
C PQR
D PR-Q2R

The correct answer is:
PQ−Q2R

Solution:
P-QQR
Since P and Q have different dimensions, therefore, their subtraction is not possible

Topic: Physics, Units and Dimensions, JEE Main
If p,q and r are three logical statements then the truth value of the statement pâ§~qâ¨qâr, where p is true, is

Options:
A True if q is true
B False if q is true
C True if q is false
D False if q is false

The correct answer is:
True if q is false

Solution:
Clearly, if q is False, p is True, then pâ§~q is True
Hence, pâ§~qâ¨qâr is True irrespective of r

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p, q and r are three logical statements, then the truth value of the statement p∧q∨∼q→r, where p is true, is

Options:
A True if q is false
B False if q is false
C True if q is true
D False if q is true

The correct answer is:
True if q is true

Solution:
Clearly, if q is true, p is true then p∧q is true
Hence, p∧q∨~q→r is true irrespective of r

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p,q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression {(p∨q)∧((~p)∨r)}→((~q)∨r) false ?

Options:
A p=T,q=F,r=T
B p=T,q=T,r=F
C p=F,q=T,r=F
D p=T,q=F,r=F

The correct answer is:
p=F,q=T,r=F

Solution:
Given,

{(p∨q)∧((~p)∨r)}→((~q)∨r)

Now solving the above expression using truth table from given options we get,

p	q	r	p∨q	~p∨r	A≡p∨q∧~p∨r	B≡~q∨r	A→B

A	T	F	T	T	T	T	T	T

B	T	T	F	T	F	F	F	T

C	F	T	F	T	T	T	F	F

D	T	F	F	T	F	F	T	T

So, only Option C will be false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (29 Jan Shift 1)
If $p, q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively, then the area of the square (in sq. units) formed by the points of intersection of the ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ and the pair of lines $x^2-y^2=0$ is

Options:
A 4
B $\sqrt{2}$
C $\frac{\sqrt{3}}{2}$
D 16

The correct answer is:
4

Solution:
$p$ and $q$ are eccentricity of hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate of hyperbola,
$\therefore \quad p^2=\frac{a^2+b^2}{a^2} \text { and } q^2=\frac{a^2+b^2}{b^2}$
Solving Eqs. (i) and (ii), we get
$(1,1),(-1,-1),(-1,1)(1,-1)$
Area of square $A B C D,=A B^2=(2)^2=4$

Topic: Mathematics, Hyperbola, JEE Main
If \(p, q\) are true statements and \(r, s\) are false statements, then the truth value of \(\sim[(p \wedge \sim r) \vee(\sim q \vee s)]\) i

Options:
A true.
B false.
C false for any \(p\).
D false for any \(q\).

The correct answer is:
false.

Solution:
A sentence which is either true or false but never both is called a statement.
Hence x is good actor is not a statement and y is bad actress is also not a statement.
Hence given sentence is not a statement

Topic: Mathematics, Mathematical Reasoning, JEE Main
If PQ be a double ordinate of the parabola, y2=-4x, where P lies in the second quadrant. If R divides PQ in the ratio 2:1, then the locus of R is:

Options:
A 3y2=-2x
B 9y2=4x
C 9y2=-4x
D 3y2=2x

The correct answer is:
9y2=-4x

Solution:
Let P-t2,2t & Q-t2,-2t

R divides in the ratio 2:1

-2t2-t22+1,-4t+2t2+1

-t2,23t

x=-t2

y=-23t

∴ 9y2=4t2

⇒9y2=-4x

Topic: Mathematics, Parabola, JEE Main, JEE Main 2015 (11 Apr Online)
If \( P Q \) be a vertical tower subtending angles \( \alpha, \beta \) and \( \gamma \) at the points \( A, B \) and \( C \), respectively on the line in the horizontal plane through the foot \( Q \) of tower and on the same side of it, then \( B C \cot \alpha-C A \cot \beta+A B \cot \gamma \) is equal to

Options:
A \( 0 \)
B \( 1 \)
C \( 2 \)
D None of these

The correct answer is:
\( 0 \)

Solution:
Let h be the height of the tower, then h=AQ tan α=BQ tan β=CQ tan γ
⇒ BC=BQ-CQ=hcot β-cot γ,
CA=hcot α-cot γ
and AB=hcot α-cot β
Now, BC cot α-CA cot β+AB cot γ

=hcot α cot β-cot γ-cot β cot α-cot γ+cot γ cot α-cot β
=0

Topic: Mathematics, Heights and Distances, JEE Main
If PQ is the focal chord of the parabola y2=-x and P is -4,2, then the ordinate of the point of intersection of the tangents at P and Q is

Options:
A 1516
B 78
C 4
D 1718

The correct answer is:
1516

Solution:
The tangent at the end points of the focal chord meets at the directrix.

The equation of the tangent at P is y⋅2=− x−4 2
⇒4y=-x+4

This tangent intersects the directrix at 14,1516

Topic: Mathematics, Parabola, JEE Main
If p∧~q∧p∧r→~p∨q is false, then the truth values of p, q and r are respectively

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

The correct answer is:
T, F, T

Solution:
Recall p⇒q≡F if and only if T⇒F≡F

So, p∧~q∧p∧r→~p∨q

Taking Option A : T∧F∧T∧T→F∨T

≡F∧T→T

≡F→T≡T

Hence, option A is false.

Taking Option B : F∧F∧F∧F→T∨T

≡F→T≡T

Hence, option B is false.

Taking Option C : T∧T∧T∧T→F∨F

≡T→F≡F

Hence, option C is correct.

Taking Option D : F∧T∧F∧F→T∨F

≡F∧F→T

≡F→T≡T

Hence, option D is false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2018 (15 Apr)
If $(\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\mathrm{p} \wedge \mathrm{r}) \rightarrow \sim p \vee q$ is false, then the truth values of $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ are respectively

Options:
A
$\mathrm{F}, \mathrm{T}, \mathrm{F}$

B
$\mathrm{T}, \mathrm{F}, \mathrm{T}$

C
F, F, F

D
T, T, T

The correct answer is:
$\mathrm{T}, \mathrm{F}, \mathrm{T}$

Solution:
As the truth table for the $(p \wedge \sim q) \wedge(p \wedge r)$ $\rightarrow \sim p \vee q$ is false, then only possible values of $(p, q, r)$ is $(T, F, T)$
$\begin{array}{llllll}p & q & r & \sim q & p \wedge \sim q & p \wedge r \\ \sim p & \sim p \vee q & (p \wedge \sim q) \wedge(p \wedge r)(p \wedge \sim q) \wedge(p \wedge r) \\ \rightarrow \sim p \vee q\end{array}$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If P→×Q→=Q→×P→, the angle between P→ and Q→ is θ0°<θ<360°. The value of θ will be ___°.

Options:

The correct answer is:
180

Solution:
-PQsinθ

=PQsinθ

⇒θ=180°

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If $p+q+r=0=a+b+c$, then the value of the determinant $\left|\begin{array}{lll}p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a\end{array}\right|$ is

Options:
A 0
B $p a+q b+r c$
C 1
D None of these

The correct answer is:
0

Solution:
We have $\left|\begin{array}{lll}p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a\end{array}\right|$
$\begin{aligned} & \quad=p q r\left(a^3+b^3+c^3\right)-a b c\left(p^3+q^3+r^3\right) \\ & =p q r(3 a b c)-a b c(3 p q r)=0,\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
If $p, q, r$ are 3 real numbers satisfying the matrix equation, $[p q r]\left[\begin{array}{lll}3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}3 & 0 & 1\end{array}\right]$ then $2 p+q-r$ equals :

Options:
A
$-3$

B
$-1$

C
4

D
2

The correct answer is:
$-3$

Solution:
Given
$$
\begin{aligned}
& {\left[\begin{array}{lll}
p & q & r
\end{array}\right]\left[\begin{array}{lll}
3 & 4 & 1 \\
3 & 2 & 3 \\
2 & 0 & 2
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 1
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{lll}
3 p+3 q+2 r & 4 p+2 q & p+3 q+2 r
\end{array}\right]=\left[\begin{array}{lll}
3 & 0 & 1
\end{array}\right] \\
& \Rightarrow 3 p+3 q+2 r=3 \\
& 4 p+2 q=0 \Rightarrow q=-2 p \\
& p+3 q+2 r=1 \\
&
\end{aligned}
$$
$p=1, q=-2, r=3$
$\therefore 2 p+q-r=2(1)+(-2)-(3)=-3$.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2013 (22 Apr Online)
If $p, q, r$ are simple propositions, then $(p \wedge q) \wedge(q \wedge r)$ is true then

Options:
A $p, q, r$ are all false
B $p, q, r$ are all true
C $p, q$ are true and $r$ is false
D $p$ is true and $q$ and $r$ are false

The correct answer is:
$p, q, r$ are all true

Solution:
$(p \wedge q) \wedge(q \wedge r)$ is true means $p \wedge q, q \wedge r$ are both true. $\Rightarrow p, q, r$ are all true.

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p,q,r are simple propositions, then the truth value of ∼p∨q∧∼r⇒p is

Options:
A True if the truth values of p,q,r are T,F,T respectively
B False if the truth values of p,q,r are T,F,T respectively
C False if the truth values of p,q,r are T,F,F respectively
D True if the truth values of p,q,r are F,T,F respectively

The correct answer is:
True if the truth values of p,q,r are T,F,T respectively

Solution:
~T∨F∧~T⇒T
F∨F∧F⇒T
F∨F→T
F→T which is T

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $p, q, r$ are simple propositions with truth alues $\mathrm{T}, \mathrm{F}, \mathrm{T}$, then the truth value of $(\sim p \vee q) \wedge \sim r \Rightarrow p$ is

Options:
A True
B False
C True if $r$ is false
D True if $q$ is true

The correct answer is:
True

Solution:
$\sim p \vee q$ means $F \vee F=F, \sim r$ means $F$
$(\sim p \vee q) \wedge \sim r$ means $F$
$[(\sim p \vee q) \wedge \sim r] \Rightarrow p$ means $\mathrm{T}$

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p⇒q∨r is False, then the truth values of p, q, r are respectively, (where T is True and F is False)

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

The correct answer is:
T, F, F

Solution:
We know that the conditional statement a⇒b is False only when a is False and b is True.

Hence, p⇒(q∨r): False when p:True;  q∨r : False

Again, we know that a∨b is False when both a & b are false.

Hence, p:True; q:False; r: False.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If p⇒q∨r is False, then the truth values of p, q, r are respectively (where T is True and F is False)

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

The correct answer is:
T, F, F

Solution:
p⇒(q∨r): False
p:True;  q∨r : False
p:True;  q:False;  r : False

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p⇒q∨r is false, then the truth values of p, q and r respectively are (here, T is true and F is false)

Options:
A TFF
B FFF
C FTT
D TTF

The correct answer is:
TFF

Solution:
p is true and q∨r is false
⇒p is true, q is false and r is false

Topic: Mathematics, Mathematical Reasoning, JEE Main
If p,q,râN, then the number of points having position vector pi^+qj^+rk^ such that 8â¤p+q+râ¤12 are

Options:
A 110
B 185
C 144
D 108

The correct answer is:
185

Solution:
8â¤p+q+râ¤12
âp+q+r=8,9,10,11,12
Here, p,q,râ¥1
Hence, required number {using number of positive integral solutions = n-1Cr-1 }
=7C2+8C2+9C2+10C2+11C2
=7Ã62+8Ã72+9Ã82+10Ã92+11Ã102
=21+28+36+45+55
=185

Topic: Mathematics, Permutation Combination, JEE Main
If p,âq,âr,âs>0, then limxââp1x+q1x+r1x+s1x43x is equal to

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

The correct answer is:
pqrs 3 4

Solution:
Consider the given expression,
lim x â â p 1 x + q 1 x + r 1 x + s 1 x 4 3 x [where, p,âq,âr,âs > 0]
Put, 1 x = z , so that x â â â z â 0
= lim z â 0 p z + q z + r z + s z 4 3 z
which is in the form 1 â .
=elimzâ0pz+qz+rz+sz4â13z
= e lim z â 0 p z - 1 z + q z - 1 z + r z - 1 z + s z - 1 z 3 4
= e 3 4 log e p + log e q + log e r + log e s
=e34logepqrs=pqrs34

Topic: Mathematics, Limits, JEE Main
If p,q,r,s∈R, then the equation x2+px+3q-x2+rx+q-x2+sx-2q=0 has

Options:
A 6 real roots
B at least two real roots
C 2 real and 4 imaginary roots
D 4 real and 2 imaginary roots

The correct answer is:
at least two real roots

Solution:
For real roots,

p2-12qâ¥0 ...(i)
r2+4qâ¥0 ...(ii)
s2-8qâ¥0 ...(iii)

If q>0, then (ii) is always true.
If q<0, then (i) and (iii) are always true.
So, the equation has atleast two real roots.

Topic: Mathematics, Quadratic Equation, JEE Main
If $(p+q)^{\text {th }}$ term of a G.P. be $m$ and $(p-q)^{t h}$ term be $n$, then the $p^{\text {th }}$ term will be

Options:
A $m / n$
B $\sqrt{m n}$
C $m n$
D $0$

The correct answer is:
$\sqrt{m n}$

Solution:
Given that $m=a r^{p+q-1}$ and $n=a r^{p-q-1}$
$r^{p+q-1-p+q+1}=\frac{m}{n} \Rightarrow$ $r=\left(\frac{m}{n}\right)^{1 /(2 q)}$
and $a=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}$
$=a r^{\rho-1}=\frac{m}{\left(\frac{m}{n}\right)^{(p+q-1) / 2 q}}\left(\frac{m}{n}\right)^{(p-1) / 2 q}$
Now $p^{\text {th }}$ term $=m\left(\frac{m}{n}\right)^{(p-1) / 2 q-(p+q-1) /(2 q)}$ $=m\left(\frac{m}{n}\right)^{-1 / 2}=m^{1-1 / 2} n^{1 / 2}$
Aliter : As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $(p+q)^{\text {th }}$ and
$(p-q)^{\text {th }}$ terms are equidistant from $p^{\text {th }}$ term i.e. at a distance of $q$. Therefore, $p^{\text {th }}$ term will be G.M. of $(p+q)^{\text {th }}$ and $(p-q)^{t h}$ i.e. $\sqrt{m n}$

Topic: Mathematics, Sequences and Series, JEE Main
If Prn=Pr+1n and Crn=Cr-1n, then the value of r is equal to:

Options:
A 1
B 4
C 2
D 3

The correct answer is:
2

Solution:
We have, Prn=Pr+1n

⇒n!n-r!=n!n-r-1!

⇒n-r=1 …1

and Crn=Cr-1n

⇒n!r!n-r!=n!r-1!n-r+1!

⇒1rn-r!=1n-r+1n-r!

⇒n-r+1=r

⇒n+1=2r …2

From equation 1 & 2, we get

2r-1-r=1

⇒r=2.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If p represents radiation pressure, C represents the speed of light and q represents radiation energy striking a unit area per second, then non-zero integers a,b and c are such that paqbCc is dimensionless, then

Options:
A a=1,b=1,c=-1
B a=1,b=-1,c=1
C a=-1,b=1,c=1
D a=1,b=1,c=1

The correct answer is:
a=1,b=-1,c=1

Solution:
Here, M0L0T0=M L-1T-2aMT-3bLT-1c
Or M0L0T0=[Ma+bL-a+cT-2a-3b-c]
Comparing powers of M,L and T, we get
a+b=0,-a+c=0,-2a-3b=0
Solving a=1,b=-1,c=1

Topic: Physics, Units and Dimensions, JEE Main
If \(p \Rightarrow(q \vee r)\) is False, then the truth values of \(p, q, r\) are respectively, (where \(T\) is True and \(F\) is \(F a l s e\) )

Options:
A \(T, F, F\)
B \(F, T, T\)
C \(F, F, F\)
D \(T, T, F\)

The correct answer is:
\(T, F, F\)

Solution:
\(\boldsymbol{p} \Rightarrow \boldsymbol{q}\) is false only if \(\mathrm{p}\) is true and \(\mathrm{q}\) is false.
So, \(p \Rightarrow(q \vee r)\) is false only if \(\mathrm{p}\) is true and \((q \vee r)\) is false.
\((q \vee r)\) is false only if \(q\) and \(r\) are false.
Hence, the truth values of \(p, q, r\) are respectively \(T, F, F\)

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $P(S)$ denotes the set of all subsets of a given set $S$, then the number of one-to-one functions from the set $S=\{1,2,3\}$ to the $\operatorname{set} P(S)$ is

Options:
A
24

B
8

C
336

D
320

The correct answer is:
336

Solution:
Let $S=\{1,2,3\} \Rightarrow n(S)=3$
Now, $P(S)=$ set of all subsets of $S$ total no. of subsets $=2^3=8$
$$
\therefore n[P(S)]=8
$$
Now, number of one-to-one functions from
$$
S \rightarrow P(S) \text { is }{ }^8 P_3=\frac{8 !}{5 !}=8 \times 7 \times 6=336 .
$$

Topic: Mathematics, Functions, JEE Main, JEE Main 2012 (19 May Online)
If p=sin2⁡x+cos4⁡x, then

Options:
A 34â¤pâ¤1
B 316â¤pâ¤14
C 1 4 ≤p≤ 1 2
D None of these

The correct answer is:
34≤p≤1

Solution:
p=sin2⁡x+cos2⁡x(1-sin2⁡x)
⇒p=(sin2⁡x+cos2⁡x)-sin2⁡xcos2⁡x
⇒p=1−14(2sinxcosx)2⇒p=1−14sin2(2x)

Max. value = 1Min. value = 34

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If pth, 2pth and 4pth terms of an arithmetic progression are in geometric progression, then the common ratio of the geometric progression is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
tp=a+p-1d
t2p=a+2p-1d
t4p=a+4p-1d
r=t2ptp=t4pt2p=t4p-t2pt2p-tp
=2pdpd=2

Topic: Mathematics, Sequences and Series, JEE Main
If $p$ times the $p^{\text {th }}$ term of an A.P. is equal to $q$ times the $q^{\text {th }}$ term of an A.P., then $(p+q)^{\text {th }}$ term is

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$0$

Solution:
$p\{a+(p-1) d\}-q\{a+(q-1) d\}$
$\Rightarrow \quad a(p-q)+\left(p^2-q^2\right) d+(q-p) d=0$
$\Rightarrow(p-q)\{a+(p+q-1) d\}=0$
$\Rightarrow a+(p+q-1) d=0 \Rightarrow T_{p+q}=0 \quad\{p \neq q\}$

Topic: Mathematics, Sequences and Series, JEE Main
If $P, V, T$ represent pressure, volume and temperature of the gas, the correct representation of Boyle's law is

Options:
A $V \propto \frac{1}{T}$ (at constant $P$ )
B $P V=R T$
C $V \propto 1 / P$ (at constant $T$ )
D $P V=n R T$

The correct answer is:
$V \propto 1 / P$ (at constant $T$ )

Solution:
Boyle's law is $V \propto \frac{1}{P}$ at constant $T$

Topic: Chemistry, States of Matter, JEE Main
If Px1,y1 is a point on the hyperbola x2-y2=a2, then SP.S'P= ________

Options:
A x12-y12a2
B x12+y12a2
C x12-y12
D x12+y12

The correct answer is:
x12+y12

Solution:
Given Hyperbola x2-y2=a2 is
Rectangular hyperbola, its e=2
Then S2a,0 and S'-2a,0
Also S'P=2x1+a and SP =2x1-a
Then SPS'P=2x12-a2∵x12-y12=a2
Hence, SPS'P=x12+y12

Topic: Mathematics, Hyperbola, JEE Main
If $P(x)=a x^2+b x+c$ and $Q(x)=-a x^2+d x+c$ where $a c \neq 0$, then $P(x) \cdot Q(x)=0$ has at least

Options:
A Four real roots
B Two real roots
C Four imaginary roots
D None of these

The correct answer is:
Two real roots

Solution:
Let all four roots are imaginary. Then roots of both equations $P(x)=0$ and $Q(x)=0$ are imaginary. Thus $b^2-4 a c \lt 0 ; d^2+4 a c \lt 0$, So $b^2+d^2 \lt 0$ which is impossible unless $b=0, d=0$

Topic: Mathematics, Quadratic Equation, JEE Main
If px be a polynomial of degree three that has a local maximum value 8 at x=1 and a local minimum value 4 at x=2 then p0 is equal to

Options:
A 6
B -12
C 24
D 12

The correct answer is:
-12

Solution:
Clearly p'(x)=λ(x-1)(x-2) where λ>0

Integrating both sides

px=λx33-3x22+2x+C

Given

p1=8⇒λ13-32+2+C=8 ⇒5λ6+C=8 ...1

p(2)=4⇒λ83-6+4+C=4 ⇒23λ+C=4 ...2

By 1 and 2 ⇒C=-12

⇒p(0)=-12

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If $\mathrm{P}(\mathrm{x})$ be a polynomial with real coefficients such that $\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(\cos ^{2} \mathrm{x}\right)$, for all $\mathrm{x} \in[0, \pi / 2]$. Consider the following statements :
I. $\mathrm{P}(\mathrm{x})$ is an even function.
II. $\mathrm{P}(\mathrm{x})$ can be expressed as a polynomial in $(2 \mathrm{x}-1)^{2}$
I. $\mathrm{P}(\mathrm{x})$ is a polynomial of even degree
Then,

Options:
A all are false
B only I and II are true
C only II and III are true
D all are true

The correct answer is:
only II and III are true

Solution:
$$
\begin{array}{l}
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(\cos ^{2} \mathrm{x}\right) \\
\mathrm{P}\left(\sin ^{2} \mathrm{x}\right)=\mathrm{P}\left(1-\sin ^{2} \mathrm{x}\right) \\
\mathrm{P}(\mathrm{x})=\mathrm{P}(1-\mathrm{x}) \forall \mathrm{x} \in[0,1]
\end{array}
$$
Differentiable both sides w.r.t. $x$ $P^{\prime}(x)=-P^{\prime}(1-x)$
So $P^{\prime}(x)$ is symmetric about point $x=\frac{1}{2}$
So $\mathrm{P}^{\prime}(\mathrm{x})$ has highest degree odd
$\Rightarrow \mathrm{P}(\mathrm{x})$ has highest degree even

Topic: Mathematics, Functions, JEE Main
If Pz is a variable point in the complex plane such that Îm-1z=14, then the value of the perimeter of the locus of Pz is

(use Ï=3.14)

If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
25

Solution:
Îm-zÂ¯z2=14
Let, z=x+iy, then
zÂ¯=x-iy
â--yx2+y2=14âx2+y2-4y=0

Hence, locus is a circle
â radius of circle is 2
â perimeter of circle is 4Ï

Topic: Mathematics, Complex Number, JEE Main
If Pz is a variable point in the complex plane such that Îm-1z=14, then the value of the perimeter of the locus of Pz is

(use Ï=3.14)

Options:

The correct answer is:
12.56

Solution:
Îm-zÂ¯z2=14
Let, z=x+iy, then
zÂ¯=x-iy
â--yx2+y2=14âx2+y2-4y=0

Hence, locus is a circle
â radius of circle is 2
â perimeter of circle is 4Ï

Topic: Mathematics, Complex Number, JEE Main
If pH of a saturated solution of BaOH2 is 12, the value of its K(sp) is found to be n×10-7M3. find the value of ‘n’ here?

Options:

The correct answer is:
5

Solution:
As pH=12
pOH=14-12=2
[OH]=10-2M=0.01M/L
In case of Ba(OH)2s⇌Ba+2s+2OH-22s
As 2s=0.01m/L
s=0.012=5×10-3m/L
Ksp for BaOH2=4s3=45×10-33=5×10-7M3
Hence ‘n’ = 5

Topic: Chemistry, Ionic Equilibrium, JEE Main
If phytoplanktons are destroyed in the sea, then

Options:
A Algae will get more space to grow
B Primary consumers will grow luxuriantly
C It will effect the food chain
D No effect will be seen

The correct answer is:
It will effect the food chain

Solution:
In an aquatic ecosystem, phytoplankton is the main producer.

Topic: Biology, Ecosystem, NEET
If plant material is dried and burnt in a crucible, the residue would contain

Options:
A Oxides and carbonates of about ten elements
B Carbon and phosphorus
C Nitrates and sulphates only
D Oxides and carbonates of Ca and Mg only

The correct answer is:
Oxides and carbonates of about ten elements

Solution:
If dried plant parts are heated in silica crucible at 600Â°C all organic substances vaporize and the remaining plant ash contains only inorganic substances or essential mineral elements. The essential elements are C, H. O, N, P, K, S, Mg, Ca, Fe, Mo, Mn, Ni, Zn, B, Cl, Cu.

Topic: Biology, Mineral Nutrition, NEET
If $\mathrm{pOH}$ of a solution is 6.0 , then its $\mathrm{pH}$ will be

Options:
A 6
B 10
C 8
D 14

The correct answer is:
8

Solution:
$p H+p O H=14 ; p H=14-p O H ; p H=14-6=8$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
If potential energy function for the force between two atoms in a diatomic molecule is approximately given by Ux=ax8-bx4, where a and b are constants in standard SI units and x is in metres. Find the dissociation energy of the molecule (in J). [Take a=4 J m8 and b=20 J m4]

Options:

The correct answer is:
25

Solution:
U(x)=4x8-20x4
â´ dUdx=0 â -8Ã4x9+20Ã4x5
â 4x520-8x4=0
x=â and x4=820
Also d2Udx2=+ve at x4=820
â´ Umin=4x41x4-5=4Ã208208-5
â Umin=-4Ã208Ã208=-25 J
â´ Dissociation energy =Uinfinite-Umin=25 J

Topic: Physics, Work Power Energy, JEE Main