datavorous/entrance-exam-dataset huggingface dataset:
Let PRâ=3i^+j^-2k^ and SQâ=i^-3j^-4k^ represent the diagonals of the parallelogram PQRS. If PTâ=2i^-j^+k^ is another vector, then the volume (in cubic units) of the parallelepiped formed by the vectors PTâ, PQâ and PSâ is

Options:

The correct answer is:
20

Solution:
Here,

PQ→+PS→=PR→PQ→-PS→=SQ→

⇒PQ→=12PR→+SQ→
⇒PQ→=2i^-j^-3k^
Also, PS→=12PR→-SQ→
=i^+2j^+k^
Volume of parallelepiped =PT→ PQ→ PS→
=2-112-1-3121
=25+15+15=20 cubic units

Topic: Mathematics, Vector Algebra, JEE Main
Let P(r)=QπR4r be the charge density distribution for a solid sphere of radius R and total charge Q. For a point p inside the sphere at distance r1 from the centre of the sphere, the magnitude of the electric field is

Options:
A Zero
B Q4π ε0 r12
C Qâr124Ï Îµ0 R4
D Q r123π ε0 R4

The correct answer is:
Qâr124Ï Îµ0 R4

Solution:
P(r)=QπR4r
From Gauss law
∮E.ds=qenε0=∫ρVdVε0
=∫Or1Qr4πr2drπR4/ε0
E.4πr12=QπR44πr144ε0
E=Qr124πε0R4

Topic: Physics, Electrostatics, JEE Main
Let $P(r)=\frac{Q}{\pi R^4} r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. for a point ' $p$ ' inside the sphere at distance $r_1$ from the centre of the sphere, the magnitude of electric field is

Options:
A
0

B
$\frac{Q}{4 \pi \varepsilon_0 r_1^2}$

C
$\frac{Q r_1^2}{4 \pi \varepsilon_0 R^4}$

D
$\frac{Q_1^2}{3 \pi \varepsilon_0 R^4}$

The correct answer is:
$\frac{Q r_1^2}{4 \pi \varepsilon_0 R^4}$

Solution:
$$
E 4 \pi r_1^2=\frac{\int_0^{r_1} \frac{Q}{\pi R^4} r 4 \pi r^2 d r}{\varepsilon_0} \Rightarrow E=\frac{Q r_1^2}{4 \pi \varepsilon_0 R^4} .
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2009
Let
p: Ramesh listens to music.
q: Ramesh is out of his village
r: It is Sunday
s: It is Saturday

Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as

Options:
A ~q∧r∨s⇒p
B q∧r∨s⇒p
C p⇒q∧r∨s
D p⇒(~q)∧r∨s

The correct answer is:
p⇒(~q)∧r∨s

Solution:
Given,
p: Ramesh listens to music.
q: Ramesh is out of his village
r: It is Sunday
s: It is Saturday

Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as

Now on breaking each statement and combining them we get,

p≡ Ramesh listens to music

~q≡ He is in village.

r∨s≡ Saturday or Sunday

So, on combining all statement we get,

p⇒~q∧r∨s

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let PS be the median of the triangle with vertices P(2,2), Q(6,-1) and R(7,3). The equation of the line passing through (1,-1) and parallel to PS is

Options:
A 4x+7y+3=0
B 2x-9y-11=0
C 4x-7y-11=0
D 2x+9y+7=0

The correct answer is:
2x+9y+7=0

Solution:
P(2,2); Q(6,-1); R(7,3)
S is the midpoint of Q and R.
∴ S≡132, 1
∴ Slope of PS=2-12-132=1-92=-29
Therefore, the equation of the line passing through (1,-1) and parallel to PS is
y+1=-29x-1
∴ 9y+9=-2x+2
∴ 2x+9y+7=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (06 Apr)
Let P=θ:sinθ-cosθ=2cosθ and Q=θ:sinθ+cosθ=2sinθ, be two sets. Then

Options:
A P⊂Q and Q-P ≠ ϕ
B Q⊄P
C P=Q
D P⊄Q

The correct answer is:
P=Q

Solution:
P: sinθ-cosθ= 2cosθ

⇒sinθ=cosθ+ 2cosθ

tanθ= 2+1

Q:sinθ+cosθ= 2sinθ⇒2-1sinθ=cosθ

=tanθ= 12-1= 2+1

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2016 (10 Apr Online)
Let Px0,y0 be the point on the hyperbola 3x2-4y2=36, which is nearest to the line 3x+2y=1. Then 2y0-x0 is equal to :

Options:
A -3
B 9
C -9
D 3

The correct answer is:
-9

Solution:
Given,

Equation of hyperbola, 3x2-4y2=36

And nearest line 3x+2y=1

Now slope of the given line will be, m=-32

Now slope of tangent which will be parallel to given line of hyperbola 3x2-4y2=36 is given by,

m=3secθ12·tanθ

Now putting the value of m=-32 we get,

⇒312×1sinθ=-32

⇒sinθ=-13

So, point will be 12secθ,3tanθ

=12·32,-3×12

=62,-32,

Hence, 2y0-x0=2-32-62=-9

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let px4+qx3+rx2+sx+t=x2+3xx-1x+3x+1-2xx-4x-3x+43x be an identity, where p, q, r, s and t are constants, then the value of s is equal to

Options:

The correct answer is:
71

Solution:
Differentiating both sides, we get,
4px3+3qx2+2xr+s=2x+311x+1-2xx-4x-3x+43x+x2+3xx-1x+31-21x-3x+43x+x2+3xx-1x+3x+1-2xx-4113
Putting x=0, we get, s=31110-4-340+0-131-21-340+0-1310-4113
s=64-3+10=71

Topic: Mathematics, Determinants, JEE Main
Let $p(x)$ be a function defined on $R$ such that $p^{\prime}(x)=p^{\prime}(1-x)$, for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then $\int_0^1 p(x) d x$ equals

Options:
A
21

B
41

C
42

D
$\sqrt{41}$

The correct answer is:
21

Solution:
$\mathrm{p}^{\prime}(\mathrm{x})=\mathrm{p}^{\prime}(1-\mathrm{x})$
$\Rightarrow \mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+\mathrm{c}$
at $x=0$
$\mathrm{p}(0)=-\mathrm{p}(1)+\mathrm{c} \quad \Rightarrow 42=\mathrm{c}$
now $\mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+42$
$\Rightarrow \mathrm{p}(\mathrm{x})+\mathrm{p}(1-\mathrm{x})=42$
$\mathrm{I}=\int_0^1 \mathrm{p}(\mathrm{x}) \mathrm{dx}=\int_0^1 \mathrm{p}(1-\mathrm{x}) \mathrm{dx}$
$2 \mathrm{I}=\int_0^1(42) \mathrm{dx} \quad \Rightarrow \mathrm{I}=21$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2010
Let $\mathrm{p}(\mathrm{x})$ be a polynomial such that $p(x)-p^{\prime}(x)=x^{n}$, where $\mathrm{n}$ is a positive integer. Then $p(0)$ equals.

Options:
A $\mathrm n !$
B $\mathrm(n-1) !$
C $\frac{1}{n !}$
D $\frac{1}{(n-1) !}$

The correct answer is:
$\mathrm n !$

Solution:
Let $\mathrm{P}(\mathrm{x})=\mathrm{a}_{0} \mathrm{x}^{\mathrm{n}}+\mathrm{a}_{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{2} \mathrm{x}^{\mathrm{n}-2} \ldots \ldots . \mathrm{a}_{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}$
$$
\begin{array}{l}
P^{1}(x)=\sum_{r=0}^{n-1} a_{r}(n-r) x^{n-r-1} \\
P(x)-P^{1}(x)=a_{0} x^{n}+\sum_{r=1}^{n}\left\{a_{r}-a_{r+}(n-r+1)\right\} x^{n-r}=x^{n}
\end{array}
$$
So, $a_{0}=1$ and $a_{r}=a_{r-1}(n-r+1)$
$$
\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{r}-1}}=\mathrm{n}-\mathrm{r}+1
$$
Now, $P(0)=a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \ldots \ldots \frac{a_{2}}{a_{1}} \frac{a_{1}}{a_{0}} \cdot a_{0}$ $=(1 \times 2 \ldots . ., \mathrm{n})=\mathrm{n} !$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $P(x)$ be a polynomial, which when $\begin{array}{llll}\text { divided by } & (x-3) & \text { and } & (x-5) & \text { leaves }\end{array}$ remainders 10 and 6, respectively. If the polynomial is divided by $(x-3)(x-5),$ thes the remainder is

Options:
A $-2 x+16$
B 16
C $2 x-16$
D 60

The correct answer is:
$-2 x+16$

Solution:
$\because P(x)=(x-3)(x-5)$
$=O(x)+(a x+b)$
Given, $\quad P(3)=10$ and $P(5)=6$
$\Rightarrow \quad 3 a+b=10$
and $\quad 5 a+b=6$
On solving Bqs.
(i) and (ii), we get $a=-2$ and $b=16$
$\therefore$ Remainder $=-2 x+16$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2015
Let px be a quadratic polynomial such that p0=1. If px leaves remainder 4 when divided by x-1 and it leaves remainder 6 when divided by x+1 then:

Options:
A p-2=19
B p2=19
C p-2=11
D p2=11

The correct answer is:
p-2=19

Solution:
Assume , px=ax2+bx+c
Given , p0=1⇒c=1
Now,p1=4p-1=6
a+b+c =4a-b+ c=6⇒a+b+1 =4a-b+1=6 a+b =3a-b=5 ⇒ a=4 b=-1
px=4x2-x+1
p-2=16+2+1=19

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2017 (08 Apr Online)
Let $p(x)$ be a quadratic polynomial with constant term 1. Suppose $p(x),$ when divided by $x-1$ leaves remainder 2 and when divided by $x+1$ leaves remainder $4 .$ Then, the sum of the roots of $p(x)=0$ is

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

The correct answer is:
$\frac{1}{2}$

Solution:
Let $\quad p(x)=a x^{2}+b x+c$
Given, constant term 'c' $=1$
$\therefore \quad p(x)=a x^{2}+b x+1$
Now, by given condition, $p(1)=2$ (remainder)
$\begin{array}{lr}\Rightarrow & a+b+1=2 \\ \Rightarrow & a+b=1\end{array}$
$\begin{array}{ll}\text { and } & p(-1)=4 \quad \text { (remainder) }\end{array}$
$\begin{array}{lr}\Rightarrow & a-b+1=4 \\ \Rightarrow & a-b=3\end{array}$
On adding Eqs. (iii) and (iv), we get
$2 a=4 \Rightarrow a=2$ form eys
(iii) $b=-1$
On putting the values of a and b in Eq. (ii), we get $\begin{aligned} p(x)=2 x^{2}-x+1=0 \\ \text { Sum of the roots } &=-\frac{\text { (Coefficient of } x)}{\text { (Coefficient of } x^{2} \text { ) }}=\frac{-(-1)}{2} \\ &=\frac{1}{2} \end{aligned}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2013
Let P(x) be a real polynomial of degree 3 which vanishes at x=-3. Let P(x) have local minima at x=1, local maxima at x=-1 and ∫-11Pxdx=18, then the sum of all the coefficients of the polynomial P(x) is equal to ___ .

Options:

The correct answer is:
8

Solution:
Let P'(x)=a(x-1)(x+1)=ax2-1

Px=a∫x2-1dx+c

=ax33-x+c

Now P(-3)=0

⇒a-273+3+c=0

⇒-6a+c=0 …(1)

Now ∫-11ax33-x+cdx=18

=2c=18⇒c=9 ...2

From (1) & (2) ⇒-6a+9=0⇒a=32

⇒Px=32x33-x+9

Then, the Sum of coefficient=12-32+9

=8

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let $p(x)=x^{2}-5 x+a$ and $q(x)=x^{2}-3 x+b$, where $a$ and $b$ are positive integers. Suppose $\operatorname{hof}(p(x), q(x)$ $=x-1$ and $k(x)=1 c m(p(x), q(x))$. If the coefficient of the highest degree term of $k(x)$ is 1, the sum of the roots of $(x-1)+k(x)$ is.

Options:
A 4
B 5
C 6
D 7

The correct answer is:
7

Solution:
$\quad \therefore$ HCF $=x-1$
$\quad \Rightarrow p(x)=x^{2}-5 x+a$
$=x^{2}-5 x+4$
$=(x-1)(x-4) \quad ......\text {(1)}$
and $\quad q(x) x^{2}-3 x+b=x^{2}-3 x+2$
$=(x-1)(x-2)$
$\Rightarrow k(x)=(x-1)(x-2)(x-4) \quad ......\text {(2)}$
Hence
$(x-1)+R(x)=(x-1)+(x-1)(x-2)(x-2)(x-4)$
$=(x-1)(x-3)^{2}$
Hence sum of roots $=7$

Topic: Mathematics, Quadratic Equation, JEE Main
Let px=x2+ax+b have two distinct real roots, where a,b are real number. Define gx=px3 for all real number x

Then, which of the following statements are true?
I. g has exactly two distinct real roots.
II. g can have more than two distinct real roots.
III. There exists a real number α such that gx≥α for all real x

Options:
A Only I
B Both I and III
C Only II
D Both II and III

The correct answer is:
Both I and III

Solution:
Let the given quadratic polynomial px=x2+ax+b has two distinct real roots α and β, then px=x2+ax+b=x-αx-β
and since gx=px3=x3-αx3-β

let α=α13 and β=β13
then gx=x3-α13x3-β13

=x-α1x-β1x2+α1x+α12x2+β1x+β12

∵ the discriminants of quadratic equations x2+α1x+α12 and x2+β1x+β12 are
negative.

∴ gx has exactly two distinct real roots and since gx=x6+ax3+b is an even degree polynomial, so there exists a real number α ' such that gx≥α for all real x

Topic: Mathematics, Quadratic Equation, JEE Main
Let Px=x2+bx+c be a quadratic polynomial with real coefficients such that ∫01Pxdx=1 and Px leaves remainder 5 when it is divided by x-2 Then the value of 9b+c is equal to:

Options:
A 9
B 15
C 7
D 11

The correct answer is:
7

Solution:
∫01x2+bx+cdx=1

13+b2+c=1 ⇒ b2+c=23

3b+6c=4 ...1

P2=5

4+2b+c=5

2b+c=1 ...2

From 1 & 2

b=29 & c=59

9b+c=7

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Let P x = x 2 + bx + c, where b and c are integers. If P(x) is a factor of both x 4 + 6 x 2 + 2 5 and 3 x 4 + 4 x 2 + 2 8 x + 5, then the value of P(1) is

Options:

The correct answer is:
4

Solution:
If x = Î± is the common root of both the equations, then
Î± 4 + 6 Î± 2 + 2 5 = 0 â Î± 4 = - 2 5 - 6 Î± 2

Using this value of Î±4 in the second equation

â3-25-6Î±2+4Î±2+28Î±+5=0
â - 7 5 - 1 8 Î± 2 + 4 Î± 2 + 2 8 Î± + 5 = 0
â2Î±2-4Î±+10=0
âÎ±2-2Î±+5=0
on comparing with x 2 + bx + c , we get
b=-2, c=5
So, P x = x 2 â2x+5
âP1=1-2+5=4

Topic: Mathematics, Quadratic Equation, JEE Main
Let P:y2=4ax, a>0 be a parabola with focus S.Let the tangents to the parabola P make an angle of π4 with the line y=3x+5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is:

Options:
A 8 only
B 2 only
C 14only
D any a>0

The correct answer is:
any a>0

Solution:
Lines making angle π4 with y=3x+5

So using formula tanα=m2-m11+m1m2, we have slope -2 & 12.

Which are perpendicular to each-other so, A, S, Bare collinear for all a>0. (by property that perpendicular tangents will lie on directrix and chord of contact will pass through focus, so the points are collinear )

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let \( P=\left[\begin{array}{ccc}-30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14\end{array}\right] \) and \( A=\left[\begin{array}{ccc}2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1\end{array}\right] \) where \( \omega=\frac{-1+i \sqrt{3}}{2} \), and \( I_{3} \) be the identity matrix of order \( 3 \). If the determinant of the matrix \( \left(P^{-1} A P-I_{3}\right)^{2} \) is \( \alpha \omega^{2} \), then the value of \( \alpha \) is equal to

Options:

The correct answer is:
36

Solution:
Let M=P-1AP-I2

=P-1AP2-2P-1AP+I

=P-1A2P-2P-1AP+I

PM=A2P-2AP+P

=A2-2 A.I+I2P

⇒DetPM=DetA-I2×P

⇒DetPM==DetA-I2×DetP

⇒Det M==DetA-I2

Now A-I=17w2-1-w-110-w-w

DetA-I=w2+w+w+7-w+w3=-6w

DetA-I2=36w2

⇒α=36

Topic: Mathematics, Determinants, JEE Main
Let points A1, A2 and A3 lie on the parabola y2=8x. If â³A1A2A3 is an equilateral triangle and normals at points A1, A2 and A3 on this parabola meet at the point h,0, then the value of h is

Options:
A 24
B 26
C 38
D 28

The correct answer is:
28

Solution:
One of the normal from h,0 on the parabola y2=8x is y=0
hence, assume A3=0,0
Let, A1=2t12,4t1
and A2=2t12,-4t1
Then, the slope of OA1 is 4t12t12=2t1
âtanâ¡30o=2t1ât1=23
The equation of the normal at A1 is y=-t1x+4t1+2t13
ây=-23x+83+483
Putting h,0 in the above equation, we get,
0=-23h+563âh=28.

Topic: Mathematics, Parabola, JEE Main
Let PQ be a chord of the parabola $y^2=12 x$ and the midpoint of PQ be at $(4,1)$. Then, which of the following point lies on the line passing through the points $\mathrm{P}$ and Q?

Options:
A $(3,-3)$
B $(2,-9)$
C $\left(\frac{3}{2},-16\right)$
D $\left(\frac{1}{2},-20\right)$

The correct answer is:
$\left(\frac{1}{2},-20\right)$

Solution:
$\begin{aligned}
& T=S_1 \\
& y-6(x+4) \\
& =1-48 \\
& 6 x-y=23
\end{aligned}$
Option $4\left(\frac{1}{2},-20\right)$ will satisfy

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let \( P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \) and \( Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} \) be two sets. Then

Options:
A \( P \subset Q \) and \( P \neq \varnothing \)
B \( Q \not \subset P \)
C \( P \not \subset Q \)
D \( P=Q \)

The correct answer is:
\( P=Q \)

Solution:
In set P,
sin⁡θ=2+1cos⁡θ ⇒ tan⁡θ= 2+1
In set Q, 2-1sin⁡θ=cos⁡θ
⇒ tan⁡θ=12-1= 2+1 ⇒ P=Q

Topic: Mathematics, Sets and Relations, JEE Main
Let q→ and r→ be non-collinear vectors. If p→ is a vector such that p→.q→+r→=4 and p→×q→×r→=x2-2x+6q→+sin⁡yr→, then x,y lies on the line

Options:
A x+y=0
B x-y=0
C x=1
D y=π

The correct answer is:
x=1

Solution:
p→.r→q→-p→.q→r→=x2-2x+6q→+sin⁡yr→
on comparison, we get,
p→.r→=x2-2x+6 and
-p→.q→=sin⁡y
∵ p→.q→+p→.r→=4 (given)
⇒x2-2x+6-sin⁡y=4
⇒x-12+1=sin⁡y
∵ x-12+1≥1,  sin⁡y≤1
∴ Both sides are equal for x=1, sin⁡y=1

Topic: Mathematics, Vector Algebra, JEE Main
Let Q and R be two points on the line x+12=y+23=z-12 at a distance 26 from the point P4,2,7. Then the square of the area of the triangle PQR is ________.

Options:

The correct answer is:
153

Solution:
Given,

L:x+12=y+23=2-12

Plotting the diagram of given value's in question we have,

Let T be any point on line whose coordinates are T2t-1,3t-2,2t+1

Now PT⊥QR by diagram,

So, 22t-5+33t-4+22t-6=0

⇒17t=34

⇒t=2, so T3,4,5

Now the value of PT=1+4+4=3

And QT=26-9=17 {by using pythagorus theorem}

∴ Area of ∆PQR=12×217×3=317

∴ Square of ar∆PQR=153.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let Q and R be two points on the line x+12=y+23=z-12 at a distance 26 from the point P4,2,7. Then the square of the area of the triangle PQR is ________.

Options:

The correct answer is:
153

Solution:
Given,

L:x+12=y+23=2-12

Plotting the diagram of given value's in question we have,

Let T be any point on line whose coordinates are T2t-1,3t-2,2t+1

Now PT⊥QR by diagram,

So, 22t-5+33t-4+22t-6=0

⇒17t=34

⇒t=2, so T3,4,5

Now the value of PT=1+4+4=3

And QT=26-9=17 {by using pythagorus theorem}

∴ Area of ∆PQR=12×217×3=317

∴ Square of ar∆PQR=153.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let $\mathrm{Q}$ be the foot of perpendicular from the origin to the plane $4 x-3 y+z+13=0$ and $R$ be a point $(-1,-6)$ on the plane. Then length $Q R$ is :

Options:
A
$\sqrt{14}$

B
$\sqrt{\frac{19}{2}}$

C
$3 \sqrt{\frac{7}{2}}$

D
$\frac{3}{\sqrt{2}}$

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

Solution:
Let $\mathrm{P}$ be the image of $\mathrm{O}$ in the given plane.
Equation of the plane, $4 x-3 y+z+13=0$ OP is normal to the plane, therefore direction ratio of OP are proportional to $4,-3,1$
Since OP passes through $(0,0,0)$ and has direction ratio proportional to $4,-3,1$. Therefore equation of OP is
$$
\begin{aligned}
& \frac{x-0}{4}=\frac{y-0}{-3}=\frac{z-0}{1}=r \text { (let) } \\
\therefore \quad & x=4 r, y=-3 r, z=r
\end{aligned}
$$
Let the coordinate of $\mathrm{P}$ be $(4 r,-3 r, r)$
Since $Q$ be the mid point of OP
$$
\therefore \quad \mathrm{Q}=\left(2 r,-\frac{3}{2} r, \frac{r}{2}\right)
$$
Since Q lies in the given plane
$$
\begin{aligned}
& 4 x-3 y+z+13=0 \\
\therefore & 8 r+\frac{9}{2} r+\frac{r}{2}+13=0 \\
\Rightarrow & r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1
\end{aligned}
$$
$\begin{aligned} \therefore \quad & \mathrm{Q}=\left(-2, \frac{3}{2},-\frac{1}{2}\right) \\ & \mathrm{QR}=\sqrt{(-1+2)^2+\left(1-\frac{3}{2}\right)^2+\left(-6+\frac{1}{2}\right)^2} \\ & =\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3 \sqrt{\frac{7}{2}}\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (22 Apr Online)
Let Q be the mirror image of the point P1,0,1 with respect to the plane S:x+y+z=5. If a line L passing through 1,-1,-1, parallel to the line PQ meets the plane S at R, then QR2 is equal to

Options:
A 2
B 5
C 7
D 11

The correct answer is:
5

Solution:
Let parallel vector of L=

Let Qa,b,c

As Q is the mirror image on given plane x+y+z=5

so a-11=b-01=c-11=-22-53

a=3,b=2,c=3

i.e. Q≡3,2,3

Let the vector parallel to L be b→

So, b→=1,1,1

Equation of line L:x-11=y+11=z+11

Let point Rλ+1,λ-1,λ-1 is lying on plane x+y+z=5, so, 3λ-1=5

⇒λ=2
So, point R is 3,1,1

Hence, QR2=5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let Q be the mirror image of the point P1, 2, 1 with respect to the plane x+2y+2z=16. Let T be a plane passing through the point Q and contains the line r→=-k^+λ(i^+j^+2k^), λ∈R. Then, which of the following points lies on T?

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

The correct answer is:
1, 2, 1

Solution:
We know that by image formula, image of P1, 2, 1 in x+2y+2z-16=0 is given by Q4, 8, 7

Now equation of plane will be given by T=x-x1y-y1z-z1a1b1c1a2b2c2=0

⇒T=x-0y-0z+14-08-07+1112=0

⇒2x-z=1

Now from options, only point B1, 2, 1 lies on it.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow[0,1]$ be defined by
$$
f(x)=\left\{\begin{array}{r}
x \text { for } x \in Q \\
1-x \text { for } x \notin Q
\end{array}\right.
$$
Then, the set $S=\{x \in[0,21]:(f o f)(x)\}$ is equal to

Options:
A $[0,1]$
B $-Q$
C $[0,1]-Q$
D $(0,1)$

The correct answer is:
$[0,1]$

Solution:
Given, $f(x)=\left\{\begin{array}{cc}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{array}\right.$ is defined for
$$
f:[0,1] \rightarrow[0,1]
$$
If $x$ is rational, then
$$
\begin{array}{rlrl}
f(x) & =x \\
\therefore \quad & f(f(x)) & =f(x)=x
\end{array}
$$
If $x$ is irrational, then
$$
\begin{aligned}
f(x) & =1-x \\
\therefore \quad f \circ f(x) & =f(1-x)=1-(1-x) \\
& =x
\end{aligned}
$$
$\therefore f \circ f(x)=x$ is possible for all values of domain $[0,1]$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2014
Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow[0,1]$ be defined by
$$
f(x)=\left\{\begin{array}{r}
x \text { for } x \in Q \\
1-x \text { for } x \notin Q
\end{array}\right.
$$
Then, the set $S=\{x \in[0,21]:(f o f)(x)\}$ is equal to

Options:
A $[0,1]$
B $-Q$
C $[0,1]-Q$
D $(0,1)$

The correct answer is:
$[0,1]$

Solution:
Given, $f(x)=\left\{\begin{array}{cc}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{array}\right.$ is defined for
$$
f:[0,1] \rightarrow[0,1]
$$
If $x$ is rational, then
$$
\begin{array}{rlrl}
f(x) & =x \\
\therefore \quad & f(f(x)) & =f(x)=x
\end{array}
$$
If $x$ is irrational, then
$$
\begin{aligned}
f(x) & =1-x \\
\therefore \quad f \circ f(x) & =f(1-x)=1-(1-x) \\
& =x
\end{aligned}
$$
$\therefore f \circ f(x)=x$ is possible for all values of domain $[0,1]$.

Topic: Mathematics, Functions, JEE Main
Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow[0,1]$ be defined by
$$
f(x)=\left\{\begin{array}{r}
x \text { for } x \in Q \\
1-x \text { for } x \notin Q
\end{array}\right.
$$
Then, the set $S=\{x \in[0,21]:(f o f)(x)\}$ is equal to

Options:
A $[0,1]$
B $-Q$
C $[0,1]-Q$
D $(0,1)$

The correct answer is:
$[0,1]$

Solution:
Given, $f(x)=\left\{\begin{array}{cc}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{array}\right.$ is defined for
$$
f:[0,1] \rightarrow[0,1]
$$
If $x$ is rational, then
$$
\begin{array}{rlrl}
f(x) & =x \\
\therefore \quad & f(f(x)) & =f(x)=x
\end{array}
$$
If $x$ is irrational, then
$$
\begin{aligned}
f(x) & =1-x \\
\therefore \quad f \circ f(x) & =f(1-x)=1-(1-x) \\
& =x
\end{aligned}
$$
$\therefore f \circ f(x)=x$ is possible for all values of domain $[0,1]$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2014
Let $Q(x)$ be a polynomial of degree $n$. If $Q(\mathrm{l})=1$ and $\frac{Q(2 x)}{Q(x+1)}+\frac{56}{x+7}-8=0$, then the value of ${ }^n C_0+{ }^n C_1+\ldots+{ }^n C_n$ is equal to

Options:
A 32
B 64
C 8
D 16

The correct answer is:
8

Solution:
$Q(x)$ be a polynomial of degree $n$.
$$
Q(1)=1 \text { and } \frac{Q(2 x)}{Q(x+1)}+\frac{56}{x+7}-8=0...(i)
$$
Put $x=0$ in Eq. (i), we get
$$
\begin{aligned}
\frac{Q(0)}{Q(1)}+\frac{56}{7}-8 & =0 \\
\Rightarrow \quad \frac{Q(0)}{1}+8-8 & =0 \\
Q(0) & =0
\end{aligned}
$$
Solving Eq. (i), we get
$$
\begin{aligned}
& \frac{Q(2 x)}{Q(x+1)}+\frac{56-8 x-56}{x+7}=0 \\
& \Rightarrow \quad \frac{Q(2 x)}{Q(x+1)}-\frac{8 x}{x+7}=0 \\
& \Rightarrow \quad \frac{Q(2 x)}{Q(x+1)}=\frac{8 x}{x+7}...(ii) \\
& \because \quad Q(0)=0 \\
&
\end{aligned}
$$
$\therefore \quad x$ is a factor of $Q(x)$.
Let $Q(x)=x P(x)$
$$
\therefore \quad \begin{aligned}
\frac{Q(2 x)}{Q(x+1)} & =\frac{8 x}{x+7} \\
\frac{2 x \cdot P(2 x)}{(x+1) P(x+1)} & =\frac{8 x}{x+7} \\
\frac{P(2 x)}{P(x+1} & =\frac{4(x+1)}{x+7}...(ii)
\end{aligned}
$$
Put $x=-1$
$$
\frac{P(-2)}{P(0)}=0 \Rightarrow P(-2)=0
$$
$\Rightarrow \quad x+2$ is factor of $P(x)$
$$
P(x)=(x+2) R(x)
$$
Now, in Eq, (ii), we get
$$
\begin{aligned}
\frac{4(x+1)}{x+7} & =\frac{P(2 x)}{P(x+1)}=\frac{(2 x+2) R(2 x)}{(x+1+2) R(x+1)} \\
\Rightarrow \quad \frac{4(x+1)}{x+7} & =\frac{2 x+2}{x+3} \frac{R(2 x)}{R(x+1)} \\
\frac{2(x+3)}{x+7} & =\frac{R(2 x)}{R(x+1)} \\
\because \quad x & =-3 \Rightarrow 0=\frac{R(-6)}{R(-2)} \Rightarrow R(-6)=0
\end{aligned}
$$
$=x+6$ is a factor of $R(x)$.
So, $R(x)=(x+6) S(x)$
$$
\begin{aligned}
\frac{2(x+3)}{x+7} & =\frac{R(2 x)}{R(x+1)}=\frac{(2 x+6) S(2 x)}{(x+6+1) S(x+1)} \\
\Rightarrow & \frac{S(2 x)}{S(x+1)}=1 \Rightarrow S(2 x)=S(x+1) \text { for all } x
\end{aligned}
$$
$\Rightarrow S(x)$ is constant function.
$$
\therefore Q(x)=x P(x)=x(x+2) R(x)=x(x+2)(x+3) S(x)
$$
$\therefore Q(x)$ is cubic polynomial.
$$
\begin{aligned}
& \therefore n=3 \\
& \begin{aligned}
\text { Required value } & ={ }^3 C_0+{ }^3 C_1+{ }^3 C_2+{ }^3 C_3 \\
& =1+3+3+1=8
\end{aligned}
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main