datavorous/entrance-exam-dataset huggingface dataset:
Let $a, b, c, d$ be real numbers between $-5$ and 5 such that $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}}$,
$$
|c|=\sqrt{4-\sqrt{5+c}},|d|=\sqrt{4+\sqrt{5+d}}
$$
Then the product abcd is

Options:
A 11
B $-11$
C 121 ,
D $-121$

The correct answer is:
11

Solution:
Given $|\mathrm{a}|=\sqrt{4-\sqrt{5-\mathrm{a}}}$
squaring
$\begin{array}{c}
a^{2}=4-\sqrt{5-a} \\
\Rightarrow a^{4}+16-8 a^{2}=5-a \\
\Rightarrow a^{4}-8 a^{2}+a+11=0
\end{array}$
Sirnilarly squaring other given equations
\& solving we can say that $a, b,-c,-d$ are roots of $x^{4}-8 x^{2}+x+11=0$
$\therefore$ product of roots
$a b(-c)(-d)=11$
abcd $=11$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $a, b, c, d \in \mathbf{R}$ be such that $a d-b c \neq 0$ and $e$ be a positive number other than 1 . If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$, then the values of $x$ and $y$ are respectively.

Options:
A $e^{\frac{\Delta_1}{\Delta_3}}, e^{\frac{\Delta_2}{\Delta_3}}$
B $e^{\frac{\Delta_3}{\Delta_2}}, e^{\frac{\Delta_1}{\Delta_2}}$
C $e^{\frac{-\Delta_1}{\Delta_3}}, e^{\frac{-\Delta_2}{\Delta_3}}$
D $e^{\frac{\Delta_2}{\Delta_1}}, e^{\frac{\Delta_3}{\Delta_1}}$

The correct answer is:
$e^{\frac{\Delta_1}{\Delta_3}}, e^{\frac{\Delta_2}{\Delta_3}}$

Solution:
We have,
$\begin{aligned}
\Delta_1 & =\left|\begin{array}{ll}
m & b \\
n & d
\end{array}\right|=m d-b n \\
\Delta_2 & =\left|\begin{array}{ll}
a & m \\
c & n
\end{array}\right|=a n-m c \\
\Delta_3 & =\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c \\
x^a y^b & =e^m, x^c y^d=e^n \\
y & =\left(\frac{e^m}{x^a}\right)^{1 / b} \text { and } y=\left(\frac{e^n}{x^c}\right)^{1 / d}
\end{aligned}$
$\begin{array}{ll}
\therefore \quad & \frac{e^{m / b}}{x^{a / b}}=\frac{e^{n / d}}{x^{c / d}} \Rightarrow x^{c / d-a / b}=e^{n / d-m / b} \\
\Rightarrow \quad x & =\left(e^{n / d-m / b}\right)^{\frac{1}{c}-\frac{a}{b}} \Rightarrow x=e^{\left(\frac{b n-m d}{b d} \times \frac{b d}{b c-a d}\right)} \\
\Rightarrow & x=e^{\frac{b n-m d}{b c-a d}} \Rightarrow x=e^{\frac{m d-b n}{a d-b c}}=e^{\frac{\Delta_1}{\Delta_3}}
\end{array}$
Putting the value of $x$ in $y=\frac{e^{m / b}}{x^{a / b}}$ we get
$y=e^{\frac{a n-m c}{a d-b c}}=e^{\frac{\Delta_2}{\Delta_3}}$

Topic: Mathematics, Determinants, JEE Main
Let ∆ABC is an isosceles triangle with AB=AC. If B=0,a, C=2a,0 and the equation of AB is 3x-4y+4a=0, then the equation of side AC is

Options:
A y=8x-16a
B 3y=4x-8a
C x=2a
D y+8x=16a

The correct answer is:
x=2a

Solution:
Let, â ABC=â ACB=Î¸ and slope of AC=m
Slope of BC =-12
Slope of AB=34
Now, tan â ABC=tanâ¡â ACB
â 34--121+34-12=-12-m1+-12m
â 2=-12-m1-m2â2-m=-12-mâm=not defined
â equation of AC is x=2a

Topic: Mathematics, Straight Lines, JEE Main
Let $a, b, c, p, q$ be real numbers.
Suppose, $\alpha, \beta$ are the roots of the equation $x^2+2 p x+q=0$ and $\alpha, \frac{1}{\beta}$ are the roots of the equation $a x^2+2 b x+c=0$, where $\beta^2 \notin\{-1,0,1\}$.
Statement $1\left(p^2-q\right)\left(b^2-a c\right) \geq 0$.
Statement $2 b \neq p a$ or $c \neq q a$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
Given, $\alpha$ and $\beta$ are the roots of $x^2+2 p x+q=0$.
$$
\begin{array}{rlrl}
& \therefore & \alpha+\beta & =-2 p \\
\text { and } & \alpha \beta & =q
\end{array}
$$
$\alpha$ and $\frac{1}{\beta}$ are the roots of $a x^2+2 b x+c=0$
and
$$
\alpha+\frac{1}{\beta}=\frac{-2 b}{a}
$$
and
$$
\frac{\alpha}{\beta}=\frac{c}{a}
$$
$$
\text { Now, } \begin{aligned}
\left(p^2-q\right)\left(b^2-a c\right) & =\left[\left(\frac{\alpha+\beta}{-2}\right)^2-\alpha \beta\right]\left[\left(\frac{\alpha+\frac{1}{\beta}}{2}\right)^2-\frac{\alpha}{\beta}\right] a^2 \\
& =\frac{(\alpha-\beta)^2}{16}\left(\alpha-\frac{1}{\beta}\right)^2 \cdot a^2 \geq 0
\end{aligned}
$$
Statement 1 is true.
$$
\begin{array}{ll}
\text { Again now, } & p a=-\left(\frac{\alpha+\beta}{2}\right) a=-\frac{a}{2}(\alpha+\beta) \\
\text { and } & b=-\frac{a}{2}\left(\alpha+\frac{1}{\beta}\right) \\
& p a \neq b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta \\
\Rightarrow & \beta \neq 1 \\
\because & \beta^2 \neq\{-1,0,1\}, \text { correct. } \\
\text { Similarly, if } & c \neq q x \\
\Rightarrow & a \frac{\alpha}{\beta} \neq a \alpha \beta \Rightarrow \alpha\left(\beta-\frac{1}{\beta}\right) \neq 0
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \\
\Rightarrow & \beta \neq\{-1,0,1\}
\end{array}
$$
Statement 2 is true.
Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.

Topic: Mathematics, Quadratic Equation, JEE Main
Let a,b,c∈R+ and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0 and x+y+(1-c)z=0 has infinitely many solutions, the minimum value of abc is

Options:
A 33
B 9
C 27
D 3

The correct answer is:
27

Solution:
We have,

1-ax+y+z=0

x+1-by+z=0

x+y+1-cz=0

has many solutions

∴ 1-a1111-b1111-c=0

⇒ ab+ac+bc=abc

∴ ab+bc+ac3≥a2b2c21/3⇒abc1/3≥3

∴ Minimum of abc=33=27

Topic: Mathematics, Determinants, JEE Main
Let a, b, c∈R be all non-zero and satisfies a3+b3+c3=2. If the matrix A=abcbcacab satisfies ATA=I, then a value of abc can be

Options:
A -13
B 13
C 3
D 23

The correct answer is:
13

Solution:
∵ATA=I

⇒abcbcacababcbcacab=100010001

⇒a2+b2+c2ab+bc+acab+bc+acab+bc+aca2+b2+c2ab+bc+acab+bc+acab+bc+aca2+b2+c2=100010001

On comparing each element both sides, we get

a2+b2+c2=1 & ab+bc+ca=0 .........i

We know that a3+b3+c3-3abc=a+b+ca2+b2+c2-ab-bc-ac.

⇒2-3abc=a+b+c1-0 (from equation i)

⇒2-3abc=a+b+c .........ii

Now, we also know that a+b+c2=a2+b2+c2+2ab+bc+ca.

⇒a+b+c2=1+20 (from equation i)

⇒a+b+c=±1

Putting it in equation ii, we get

2-3abc=±1

⇒3abc=2∓1

⇒abc=13 or abc=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let a, b, c∈R be such that a2+b2+c2=1. If acosθ=bcosθ+2π3=ccosθ+4π3,where θ=π9, then the angle between the vectors ai^+bj^+ck^ and bi^+cj^+ak^ is:

Options:
A 0
B 2π3
C π2
D π9

The correct answer is:
π2

Solution:
Let acosθ=bcos θ+2π3=ccos θ+4π3=k

⇒a=kcosθ, b=kcosθ+2π3, c=kcosθ+4π3

⇒ab+bc+ca=k2cosθ+4π3+cosθ+cosθ+2π3cosθ+4π3cosθcosθ+2π3

=k2cosθ+2cosθ+π.cosπ3cosθ.cosθ+2π3.cosθ+4π3

=k2cosθ-2cosθ.12cosθ.cosθ+2π3.cosθ+4π3=0

Let ϕ be the angle between two given vectors.

∴cosϕ=ai^+bj^+ck^.bi^+cj^+ak^a2+b2+c2 ·b2+c2+a2 =ab+bc+caa2+b2+c2=0

⇒ϕ=π2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Let a, b, c∈R . If fx=ax2+bx+c is such that a+b+c=3 and fx+y=fx+fy+xy, ∀ x, y∈R , then ∑ n=1 10 f(n) is equal to:

Options:
A 330
B 165
C 190
D 255

The correct answer is:
330

Solution:
Given, fx+y=fx+fy+xy ∀x, y∈R
Put y=1, we get
⇒fx+1=fx+f1+x
⇒fx+1-fx=3+x

On taking Sigma on both sides

⇒ ∑x=1n-1fx+1-fx=∑x=1n-13+x

⇒f2-f1+f3-f2+f4-f3+f5-f4+......+fn-fn-1

=3n-1+n-1n2

⇒fn-f1=3n-1+n-1n2

⇒fn-3=3n-1+n-1n2

⇒fn=n22+5n2

Again taking Sigma on both the sides

⇒∑n=110fn=∑n=110n22+5n2

⇒∑n=110fn=12.10.11.216+52.10.112

⇒∑n=110fn=330.

Topic: Mathematics, Functions, JEE Main, JEE Main 2017 (02 Apr)
Let \( a, b \in R \). If the mirror image of the point \( P(a, 6,9) \) with respect to the line \( \frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9} \) is \( (20, b,-a-9) \), then \( |a+b| \) is equal to:

Options:
A \( 86 \)
B \( 90 \)
C \( 84 \)
D \( 88 \)

The correct answer is:
\( 88 \)

Solution:
P9,6,9

x-37=y-25=z-1-9

Q=20,b,-a-9

20+a2-37=b+62-25=-92-1-9

14+914=b+210=a+218

⇒a=-56 and b=-32

⇒a+b=88

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let AB is the focal chord of a parabola and D and C be the foot of the perpendiculars from A and B on its directrix respectively. If CD=6 units and area of trapezium ABCD is 36 square units, then the length (in units) of the chord AB is

Options:

The correct answer is:
12

Solution:
Let S be focus, AS=AD and BS=BC
Area of trapezium
=12AD+BC.6
=3(AS+BC)
=3AB
Hence, AB=12 units

Topic: Mathematics, Parabola, JEE Main
Let a,b∈R,a≠0 be such that the equation, ax2-2bx+5=0 has a repeated root α, which is also a root of the equation, x2-2bx-10=0. If β is the other root of this equation, then α2+β2 is equal to:

Options:
A 25
B 26
C 28
D 24

The correct answer is:
25

Solution:
Given, ax2-2bx+5=0 has repeated root α.

∴2α=2ba⇒α=baand α2=5a⇒b2a2=5a

⇒b2=5a ...i a≠0

α+β=2b ...ii

and αβ=-10 ...iii

α=ba is also root of x2-2bx-10=0

⇒b2-2ab2-10a2=0

by i⇒5a-10a2-10a2=0

⇒20a2=5a

⇒a=14 and b2=54

Now α2+β2=α+β2-2αβ

=5+20

=25

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let a,b∈R,a≠0. If the function f, defined as
fx=2x2a, 0≤x<1a, 1≤x<22b2-4bx3, 2≤x<8,is continuous in the interval 0, ∞, then an ordered pair a,b can be

Options:
A -2,1-3
B 2,-1+3
C 2,1-3
D -2, 1+ 3

The correct answer is:
2,1-3

Solution:
Here, fx is continuous at x=1 ⇒limx→1-fx=limx→1+fx
⇒2a=a ⇒a= ± 2

Since, f(x) is continuous at x=2

⇒limx→2-fx=limx→2+fx

⇒a=2b2-4b22

Put a= 2

⇒2=b2-2b ⇒b2-2b-2=0

⇒b=2± 4+4.22=1± 3

Hence, a,b≡2,1±3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2016 (10 Apr Online)
Let a,b∈R,b≠0. Defined a function, fx=asinπ2(x−1),     for x≤0tan2x−sin2xbx3,     for x>0

If f is continuous at x=0, then 10-ab is equal to

Options:

The correct answer is:
14

Solution:
We have,

fx=asinπ2x−1,    for x≤0tan2x−sin2xbx3,    for x>0

If function is continuous at x=0, then

LHL=RHL=f0

Now,

LHL=limx→0-asinπ2x−1

⇒LHL=limh→0asinπ20-h−1

⇒LHL=limh→0asinπ2−1

⇒LHL=-a

Now,

RHL=limx→0+ tan2x−sin2xbx3

∵sinx=x−x33!+x55!+……

∵tanx=x+x33+2x515+….

RHL=limx→0+ (2x)33+(2x)36bx3=83+86b=4b

And,

f(0)=asinπ2(0−1)=asin−π2=−a

Then,

RHL=f0

⇒4b=−a⇒−ab=4⇒10−ab=14

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Let a,b∈R be such that the equation ax2-2bx+15=0 has repeated root α and if α and β are the roots of the equation x2-2bx+21=0, then α2+β2 is equal to:

Options:
A 37
B 58
C 68
D 92

The correct answer is:
58

Solution:
Given ax2-2bx+15=0 ...i

Has repeated roots So D=0

4b2-4×15×a=0

⇒ b2=15a ...ii

Also given

Now α will satisfy both quadratic

ax2-2bx+15=0 & x2-2bx+21=0

Putting the value we get

aα2-2bα+15=0 α2-2bα+21=0- + - ----------a-1α2=6

α2=6a-1

Now in equation (1) product of Root α2=15a

So 15a=6a-1⇒2a=5a−5⇒a=53

Now b2=15a⇒b2=15×53 ⇒b2=25

So b=±5

Now in quadratic x2-2bx+21=0

Putting the value of b we get

x2-10x+21=0 ⇒x-7x-3=0

So x=3 or 7.

Or

x2+10x+21=0⇒x=-3 or x=-7

So α=±3 & β=±7

So α2+β2=32+72=9+49=58

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Let $a, b \in R$ be such that the function $f$ given by $f(x)=\ln |x|+b x^2+a x, x \neq 0$ has extreme values at $x=-1$ and $x=2$.
Statement 1: $f$ has local maximum at $x=-1$ and at $x=2$.
Statement 2: $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{-1}{4}$

Options:
A
Statement $1$ is false, statement $2$ is true

B
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

C
Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$

D
Statement $1$ is true, statement $2$ is false

The correct answer is:
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

Solution:
$f^{\prime}(x)=\frac{1}{x}+2 b x+a$
$f$ has extremevalues and differentiable
$\Rightarrow f^{\prime}(-1)=0 \quad \Rightarrow a-2 b=1$
$f^{\prime}(2)=0 \quad \Rightarrow a+4 b=-\frac{1}{2} \quad \Rightarrow a=\frac{1}{2}, b=-\frac{1}{4}$
$f^{\prime \prime}(-1), f^{\prime \prime}(2)$ are negative. $f$ has local maxima at $-1,2$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (Offline)
Let a,b∈R. If the mirror image of the point Pa,6,9 with respect to the line x-37=y-25=z-1-9 is 20, b,-a-9, then |a+b| is equal to:

Options:
A 86
B 90
C 84
D 88

The correct answer is:
88

Solution:
Pa,6,9

x-37=y-25=z-1-9

Q=20,b,-a-9

Mid point of PQ=a+202, b+62, -a2

lies on line

20+a2-37=b+62-25=-a2-1-9

a+20-614=b+6-410=-a-2-18

14+a14=b+210=a+218

a+1414=a+218

18a+252=14a+28

4a=-224

a=-56

b+210=a+218

b+210=-5418

b+210=-3⇒b=-32

|a+b|=|-56-32|=88

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let a, b, $x, y$ be real numbers such that $a^{2}+b^{2}=81, x^{2}+y^{2}=121$ and $a x+b y=99$. Then the set of all possible values of ay $-\mathrm{bx}$ is -

Options:
A $\left(0, \frac{9}{11}\right]$
B $\left(0, \frac{9}{11}\right)$
C $\{0\}$
D $\left[\frac{9}{11}, \infty\right)$

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

Solution:
$a^{2}+b^{2}=81$
$x^{2}+y^{2}=121$
$a x+b y=99$
$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)=(81)(121)$ ...(1)
and $(a x+b y)^{2}=(99)^{2}$ ...(2)
$(1)-(2)$
(ay $-b x)^{2}=0$
ay $-b x=0$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $A B C D$ be a parallelogram such that $\overrightarrow{A B}=\overrightarrow{\mathrm{q}}, \overrightarrow{A D}=\vec{p}$ and $\angle B A D$ be an acute angle. If $\vec{r}$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $A D$, then $\vec{r}$ is given by

Options:
A
$\vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$

B
$\vec{r}=-\overrightarrow{\mathrm{q}}+\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}}}\right) \overrightarrow{\mathrm{p}}$

C
$\vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$

D
$\vec{r}=-3 \vec{q}+\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$

The correct answer is:
$\vec{r}=-\overrightarrow{\mathrm{q}}+\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}}}\right) \overrightarrow{\mathrm{p}}$

Solution:
$\overrightarrow{\mathrm{AE}}=$ vector component of $\overrightarrow{\mathrm{q}}$ on $\overrightarrow{\mathrm{p}}$
$\overrightarrow{\mathrm{AE}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \overrightarrow{\mathrm{p}} \quad \therefore$ From $\triangle \mathrm{ABE} ; \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \quad \Rightarrow \overrightarrow{\mathrm{r}}=-\overrightarrow{\mathrm{q}}+\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}})} \overrightarrow{\mathrm{p}}$

Topic: Mathematics, Vector Algebra, JEE Main
Let $A B C D$ be a parallelogram such that $\overrightarrow{A B}=\overrightarrow{\mathrm{q}}, \overrightarrow{A D}=\vec{p}$ and $\angle B A D$ be an acute angle. If $\vec{r}$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $A D$, then $\vec{r}$ is given by

Options:
A
$\vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$

B
$\vec{r}=-\overrightarrow{\mathrm{q}}+\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}}}\right) \overrightarrow{\mathrm{p}}$

C
$\vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p}$

D
$\vec{r}=-3 \vec{q}+\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p}$

The correct answer is:
$\vec{r}=-\overrightarrow{\mathrm{q}}+\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}}}\right) \overrightarrow{\mathrm{p}}$

Solution:
$\overrightarrow{\mathrm{AE}}=$ vector component of $\overrightarrow{\mathrm{q}}$ on $\overrightarrow{\mathrm{p}}$
$\overrightarrow{\mathrm{AE}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \overrightarrow{\mathrm{p}} \quad \therefore$ From $\triangle \mathrm{ABE} ; \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}=\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})} \quad \Rightarrow \overrightarrow{\mathrm{r}}=-\overrightarrow{\mathrm{q}}+\frac{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})}{(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{p}})} \overrightarrow{\mathrm{p}}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (Offline)
Let A be a $2 \times 2$ matrix
Statement-1 $: \operatorname{adj}(\operatorname{adj} A)=A$
Statement-2 : $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|$

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Solution:
$$
\begin{aligned}
& |\operatorname{adj} A|=|A|^{n-1}=|A|^{2-1}=|A| \\
& \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A=|A|^0 A=A
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2009
Let $A$ be a $2 \times 2$ matrix with $\operatorname{det}(A)=-1$ and $\operatorname{det}((A+I)(\operatorname{Adj}(A)+I))=4$. Then the sum of the diagonal elements of $A$ can be:

Options:
A -1
B 2
C 1
D -2

The correct answer is:
2

Solution:
Let A=abcd, so adjA=a-b-cd

Now given A=-1

So, ad-bc=-1

Now A+IadjA+I=4

⇒AadjA+adjA+A+I=4

⇒AI+adjA+A+I=4

⇒-I+adjA+A+I=4

⇒adjA+A=4

⇒a+d00a+d=4

⇒a+d2=4

So, a+d=2 or -2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2=1$, where 1 is $2 \times 2$ identity matrix. Define $\operatorname{Tr}(\mathrm{A})=$ sum of diagonal elements of $A$ and $|A|=$ determinant of matrix $A$.
Statement-1: $\operatorname{Tr}(\mathrm{A})=0$
Statement-2: $|\mathrm{A}|=1$

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is false

C
Statement-1 is false, Statement-2 is true

D
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:
Let $A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right), a b c d \neq 0$
$$
\begin{aligned}
& A^2=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \cdot\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \\
& \Rightarrow A^2=\left(\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right) \\
& \Rightarrow a^2+b c=1, b c+d^2=1 \\
& a b+b d=a c+c d=0 \\
& c \neq 0 \text { and } b \neq 0 \\
& \text { Trace } A=a+d=0 \\
& |A|=a d-b c=-a^2-b c=-1 .
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2010
Let $\mathrm{A}$ be a $2 \times 2$ matrix with real entries. Let I be the $2 \times 2$ identity matrix. Denote by $\operatorname{tr}(\mathrm{A})$, the sum of diagonal entries of $A$. Assume that $A^2=1$.
Statement -1: If $A \neq 1$ and $A \neq-1$, then $\operatorname{det} A=-1$.
Statement $-2$ : If $A \neq 1$ and $A \neq-1$, then $\operatorname{tr}(A) \neq 0$.

Options:
A
Statement $-1$ is false, Statement $-2$ is true

B
Statement $-1$ is true, Statement $-2$ is true, Statement $-2$ is a correct explanation for Statement $-1$

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

D
Statement $-1$ is true, Statement $-2$ is false.

The correct answer is:
Statement $-1$ is true, Statement $-2$ is false.

Solution:
Let $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ so that $A^2=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & b c+d^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow a^2+b c=1=b c+d^2$ and $(a+d) c=0=(a+d) b$.
Since $A \neq I, A \neq 1, a=-d$ and hence $\operatorname{det} A=\left|\begin{array}{cc}\sqrt{1-b c} & b \\ c & -\sqrt{1-b c}\end{array}\right|=-1+b c-b c=-1$
Statement 1 is true.
But tr. $A=0$ and hence statement 2 is false.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2008
Let A be a 2×2 matrix with real entries such that A'=αA+1, where α∈ℝ--1,1., If det A2-A=4, the sum of all possible values of α is equal to

Options:
A 0
B 32
C 2
D 52

The correct answer is:
52

Solution:
Given,

A be a 2×2 matrix with real entries such that A'=αA+1,

Now let A=abcd

Now putting the value in A'=αA+I we get,

⇒acbd=αa+1αbαcαd+1

Now equating both side we get,

a=αa+1⇒a=11-α .........i

b=αc .... ii and c=αb ........iii

Now from equation ii & iii we get,

c=0 or α=±1 (not possible as mentioned in question)

∴c=0

⇒c=0, b=0
Also d=αd+1⇒d=11-α

Also given, A2-A=4

⇒|A||A-I|=4

⇒11-α211-α-12=4

Now let 11-α=t

So, t2t-12=4

⇒tt-1=±2

⇒t=2 or t=-1

⇒α=12,2

So, sum of all possible value will be 12+2=52

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let A be a 2×2 real matrix with entries from 0, 1 and A≠0. Consider the following two statements;

P If A≠l2, then A=-1

Q If A=1, then trA=2

Where l2 denotes 2×2 identity matrix and trA denotes the sum of the diagonal entries of A. Then

Options:
A P is false and Q is true
B Both P and Q are false
C P is true and Q is false
D Both P and Q are true

The correct answer is:
Both P and Q are true

Solution:
Let A=abcd a,b,c,d∈{0,1}

A=ad-bc≠0

⇒ ad=1,bc=0 or ad=0,bc=1

P

If A≠I2⇒ad≠1⇒ad=0,bc=1⇒|A|=-1

P is true.

Q

If A=1⇒ad=1,bc=0⇒ tr(A)=2

Q is true.

Topic: Mathematics, Matrices, JEE Main
Let A be a 2×2 real matrix with entries from 0, 1 and A≠0. Consider the following two statements;

P If A≠l2, then A=-1

Q If A=1, then trA=2

Where l2 denotes 2×2 identity matrix and trA denotes the sum of the diagonal entries of A. Then

Options:
A P is false and Q is true
B Both P and Q are false
C P is true and Q is false
D Both P and Q are true

The correct answer is:
Both P and Q are true

Solution:
Let A=abcd a,b,c,d∈{0,1}

A=ad-bc≠0

⇒ ad=1,bc=0 or ad=0,bc=1

P

If A≠I2⇒ad≠1⇒ad=0,bc=1⇒|A|=-1

P is true.

Q

If A=1⇒ad=1,bc=0⇒ tr(A)=2

Q is true.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let $A$ be a $2 \times 2$ symmetric matrix such that $A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]$ and the determinant of $A$ be 1 . If $A^{-1}=\alpha A+\beta I$, where $I$ is an identity matrix of order $2 \times 2$, then $\alpha+\beta$ equals _______

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \text { Let } A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right] \\ & {\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], a d-b^2=1} \\ & a+b=3, b+d=7,(3-b)(7-b)-b^2=1 \\ & 21-10 b=1 \rightarrow b=2, a=1, d=5 \\ & A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right] \\ & A^{-1}=\alpha A+\beta I \\ & {\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]} \\ & \alpha=-1, \beta=6 \rightarrow \alpha+\beta=5\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (04 Apr Shift 2)