datavorous/entrance-exam-dataset huggingface dataset:
If $a$ and $c$ are positive real numbers and the ellipse $\frac{x^2}{4 c^2}+\frac{y^2}{c^2}=1$ has four distinct points ir common with the circle $x^2+y^2=9 a^2$, then

Options:
A
$9 a c-9 a^2-2 c^2 < 0$
B
$6 a c+9 a^2-2 c^2 < 0$
C
$9 a c-9 a^2-2 c^2>0$

D
$6 a c+9 a^2-2 c^2>0$

The correct answer is:
$9 a c-9 a^2-2 c^2>0$

Solution:
Radius $=3 a$
Length of major axis $=4 c$
Now, (Radius $) < $ (Half of the length of major axis)
$$
\begin{aligned}
& 3 a < 2 c \\
& 9 a^2 < 4 c^2 \\
& 9 a c-9 a^2>9 a c-4 c^2
\end{aligned}
$$
$9 a c-9 a^2-2 c^2>9 a c-6 c^2$
Again $3 a < 2 c$
$\Rightarrow 9 a c < 6 c^2$
$\Rightarrow 9 a c-6 c^2 < 0$
From (i) and (ii),
$9 a c-9 a^2-2 c^2>0$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2013 (09 Apr Online)
If ' $a$ ' and ' $c$ ' are the segments of a focal chord of a parabola and $b$ the semi-latus rectum, then

Options:
A $a, b, c$ are in A.P.
B $a, b, c$ are in G.P.
C $a, b, c$ are in H.P.
D None of these

The correct answer is:
$a, b, c$ are in H.P.

Solution:
Semi latus rectum is harmonic mean between segments of focal chords of a parabola.
$\therefore \quad b=\frac{2 a c}{a+c} \Rightarrow a, b, c$ are in H.P.

Topic: Mathematics, Parabola, JEE Main
If $|\mathbf{a}|=|\mathbf{b}|=1$ and $|\mathbf{a}+\mathbf{b}|=\sqrt{3}$, then the value of $(3 \mathbf{a}-4 \mathbf{b}) \cdot(2 \mathbf{a}+5 \mathbf{b})$ is

Options:
A $-21$
B $-21 / 2$
C 21
D $21 / 2$

The correct answer is:
$-21 / 2$

Solution:
$(3 a-4 b)(2 a+5 b)=6|\mathbf{a}|^2-20|\mathbf{b}|^2+7 \mathbf{a} \cdot \mathbf{b}=6-20+7 \mathbf{a} \cdot \mathbf{b}$
Given, $|\mathbf{a}+\mathbf{b}|^2=(\sqrt{3})^2 \Rightarrow$
$\Rightarrow$ $|\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=3$ $\Rightarrow 2 \mathbf{a} \cdot \mathbf{b}=1 \Rightarrow \mathbf{a} \cdot \mathbf{b}=\frac{1}{2}$
Therefore, $(3 a-4 b) \cdot(2 a+5 b)$
$=6-20+7 \times \frac{1}{2}=-14+\frac{7}{2}=\frac{-21}{2}$

Topic: Mathematics, Vector Algebra, JEE Main
If $(\mathbf{a} \times \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{b})^2=144$ and $|\mathbf{a}|=4$, then $|\mathbf{b}|=$

Options:
A 16
B 8
C 3
D 12

The correct answer is:
3

Solution:
We know that $(\mathbf{a} \times \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{b})^2=|\mathbf{a}|^2|\mathbf{b}|^2$ $\therefore 144=16|\mathbf{b}|^2 \Rightarrow|\mathbf{b}|=3$.

Topic: Mathematics, Vector Algebra, JEE Main
If $|\vec{A} \times \vec{B}|=\sqrt{3}$ A. $B$, then the value of $|\vec{A}+\vec{B}|$ is:

Options:
A $\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{1 / 2}$
B $A+B$
C $\left(A^2+B^2+\sqrt{3} A B\right)^{1 / 2}$
D $\left(A^2+B^2+A B\right)^{1 / 2}$

The correct answer is:
$\left(A^2+B^2+A B\right)^{1 / 2}$

Solution:
$|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})$
$A B \sin \theta=\sqrt{3} A B \cos \theta \Rightarrow \tan \theta=\sqrt{3} \therefore \theta=60^{\circ}$
Now $|\vec{R}|=|\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2 A B \cos \theta}$
$=\sqrt{A^2+B^2+2 A B\left(\frac{1}{2}\right)}=\left(A^2+B^2+A B\right)^{1 / 2}$

Topic: Physics, Mathematics in Physics, JEE Main
If $|\mathbf{a} \cdot \mathbf{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=4$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is

Options:
A $\cos ^1 \frac{3}{4}$
B $\cos ^{-1} \frac{3}{5}$
C $\cos ^{-1} \frac{4}{5}$
D $\frac{\pi}{4}$

The correct answer is:
$\cos ^{-1} \frac{3}{5}$

Solution:
| a.b $\mid=a b \cos \theta=3$ ...(ii)
and
$|\mathbf{a} \times \mathbf{b}|=a b \sin \theta=4$ ...(i)
Dividing (ii) by (i),
we get
$\tan \theta=\frac{4}{3} \Rightarrow \cos \theta=\frac{3}{5} \Rightarrow \theta=\cos ^{-1} \frac{3}{5} \text {. }$

Topic: Mathematics, Vector Algebra, JEE Main
If a, b & 3c are in arithmetic progression and a, b & 4c are in geometric progression, then the possible values of ab are

Options:
A 23,2
B 32,12
C 23,32
D 12,2

The correct answer is:
32,12

Solution:
a+3c=2b and b2=4ac
⇒c=2b-a3=b24a
∴ 8ab–4a2=3b2⇒4a2–8ab+3b2=0
4a2b2–8ab+ 3= 0⇒ab=8±64-4×4×32×4
ab=2±4-32=2±12=32,12.

Topic: Mathematics, Sequences and Series, JEE Main
If Aâ B, AB=BA and A2=B2, then the value of the determinant of matrix A+B is (where A and B are square matrices of order 3Ã3 )

Options:
A 0
B 1
C 33
D 32

The correct answer is:
0

Solution:
A2-B2=A-BA+B=0 given AB=BA
Since, A≠B⇒A-B is not a null matrix
Hence, A+B is a null matrix or detA-B=detA+B=0
In both cases detA+B=0

Topic: Mathematics, Determinants, JEE Main
If $\mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c}, \mathbf{a} \times \mathbf{b}=\mathbf{a} \times \mathbf{c}$ and $\mathbf{a} \neq \mathbf{0}$, then

Options:
A $\mathbf{b}=\mathbf{0}$
B $\mathbf{b} \neq \mathbf{c}$
C $\mathbf{b}=\mathbf{c}$
D None of these

The correct answer is:
$\mathbf{b}=\mathbf{c}$

Solution:
$\mathbf{a} .(\mathbf{b}-\mathbf{c})=0, \mathbf{a} \times(\mathbf{b}-\mathbf{c})=\mathbf{0} \Rightarrow \mathbf{b}=\mathbf{c}$ or $\mathbf{a}=\mathbf{0}$, but $\mathbf{a} \neq \mathbf{0}$. Hence $\mathbf{b}-\mathbf{c}=\mathbf{0}$. i.e., $\mathbf{b}=\mathbf{c}$.

Topic: Mathematics, Vector Algebra, JEE Main
If a, b and c are distinct positive real numbers such that Î1=abcbcacab and Î2=bc-a2ac-b2ab-c2ac-b2ab-c2bc-a2ab-c2bc-a2ac-b2, then

Options:
A Δ1=Δ2
B Δ12+Δ2=0
C Δ12=Δ2
D Δ12=Δ22

The correct answer is:
Δ12=Δ2

Solution:
Elements of Î2 are co-factors of the elements of Î1
Hence Î12=Î2

Topic: Mathematics, Determinants, JEE Main
If a, b and c are non-zero real numbers and if the system of equations a-1x=y+z, b-1y=z+x and c-1z=x+y have a non-trivial solution, then 32a+32b+32c is equal to

Options:

The correct answer is:
1.5

Solution:
System of equations can be written as
a-1x-y-z=0
x-b-1y+z=0
x+y-c-1z=0
Using Cramer's rule, for non-trivial solution ∆=0
a-1-1-11-b-1111-c-1=0
⇒a-1b-1c-1-1--1-c-1-1-11+b-1=0
⇒a-1bc-b-c-c-b=0
⇒abc-ab-ac-bc+b+c-c-b=0
⇒ab+bc+ca=abc⇒1a+1b+1c=1

Topic: Mathematics, Determinants, JEE Main
If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are perpendicular to $\mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}$ and $\mathbf{a}+\mathbf{b}$ respectively and if $|\mathbf{a}+\mathbf{b}|=6,|\mathbf{b}+\mathbf{c}|=8$ and $|\mathbf{c}+\mathbf{a}|=10$ then $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=$

Options:
A $5 \sqrt{2}$
B 50
C $10 \sqrt{2}$
D 10

The correct answer is:
10

Solution:
$\begin{aligned}
& \mathbf{a} \cdot(\mathbf{b}+\mathbf{c})=0 \Rightarrow \mathbf{a} \cdot \mathbf{b}+\mathbf{c} \cdot \mathbf{a}=0 \\
& \mathbf{b}(\mathbf{c}+\mathbf{a})=0 \Rightarrow \mathbf{b} \cdot \mathbf{c}+\mathbf{a} \cdot \mathbf{b}=0 \\
& \mathbf{c} .(\mathbf{a}+\mathbf{b})=0 \Rightarrow \mathbf{c . a}+\mathbf{b} \cdot \mathbf{c}=0
\end{aligned}$
From (i), (ii) and (iii),
$2(\mathbf{a} . \mathbf{b}+\mathbf{b . c}+\mathbf{c . a})=0$
Now, $|\mathbf{a}+\mathbf{b}|^2 +|\mathbf{b}+\mathbf{c}|^2+|\mathbf{c}+\mathbf{a}|^2=6^2+8^2+10^2$
$\begin{aligned}
& \Rightarrow 2\left[|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2\right]+2[\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}]=200 \\
& \Rightarrow 2|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=200 \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|=10
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If A, B and C are square matrices of order 3 and A=2, B=3 and C=4, then the value of 3adjABC-1 is equal to (where, adjA represents the adjoint matrix of A)

Options:
A 27
B 274
C 812
D 81

The correct answer is:
81

Solution:
3adjABC-1=33adjA⋅BC=33×A2⋅BC=33×22×34=81

Topic: Mathematics, Determinants, JEE Main
If A, B and C are square matrices of same order and Î is an identity matrix of the same order, such that C2=CB+AC and AB=Î, then C-A-1 is equal to

Options:
A C-A
B C-B
C B-C
D B+Ι

The correct answer is:
C-B

Solution:
Given,
âCC-B=AC-B+AB
âC-AC-B=AB=Î
Hence, C-A-1=C-B

Topic: Mathematics, Matrices, JEE Main
If a,b and c are the greatest values of Cp19,Cq20 and Cr21 respectively, then:

Options:
A a11=b22=c21
B a10=b11=c21
C a11=b22=c42
D a10=b11=c42

The correct answer is:
a11=b22=c42

Solution:
We know that,  nCr is maximum at middle term.

a=  19Cp= 19C10= 19C9

b=  20Cq= 20C10

c=  21C6= 21C10= 21C11

Hence,

a 19C9=a2010. 19C9=a2111.2010. 19C9

⇒a1=b2=c4211

⇒a11=b22=c42

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (08 Jan Shift 1)
If a→, b→ and c→ are three non-zero and non-coplanar vectors such that a→b→c→=4, then the value of a→+3b→-c→⋅a→-b→×a→-2b→-3c→ is equal to

Options:
A 40
B 44
C 48
D 52

The correct answer is:
52

Solution:
a→+3b→-c→⋅a→-b→×a→-2b→-3c→
=13-11-101-2-3a→b→c→
=13-3-3-1-2+1×4
=13×4=52

Topic: Mathematics, Vector Algebra, JEE Main
If $A, B$ and $C$ are three sets such that $A \cap B=A \cap C$ and $A \cup B=A \cup C$, then

Options:
A
$A=B$

B
$A=C$

C
$B=C$

D
$A \cap B=\phi$

The correct answer is:
$B=C$

Solution:
No Solution Available

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2009
If aâ, bâ and câ are three vectors such that 3aâ+4bâ+6câ=0â, aâ=3, bâ=3 and câ=4, then the value of -864aâ.bâ+bââcâ+câ.aâ6 is equal to

Options:

The correct answer is:
1863

Solution:
3a→+4b→=-6c→⇒3a→+4b→2=36c→2
⇒9a→2+16b→2+24a→.b→=36c→2
⇒81+144+24a→.b→=576
⇒a→.b→=35124
Similarly, 3a→+6c→=-4b→⇒3a→+6c→2=16b→2
⇒9a→2+36c→2+36a→.c→=16b→2
⇒81+576+36a→.c→=144
⇒a→.c→=-51336
And, 6c→+4b→=-3a→⇒36c→|2+16b→|2+48b→⋅c→=9a→2
⇒576+144+48b→.c→=81⇒b→.c→=-21316
Hence, -864(a→.b→+b→.c→+c→.a→)6=-12636+12312+115026=1863

Topic: Mathematics, Vector Algebra, JEE Main
If aâ, bâ and câ are three vectors, such that aâ=2, bâ=3, câ=4, aââcâ=0, aââbâ=0 and the angle between bâ and câ is Ï3, then the value of aâÃ2bâ-3câ is equal to

Options:
A 123
B 63
C 33
D 5

The correct answer is:
123

Solution:
2b→-3c→2=4b→2+9c→2-12b→⋅c→
=36+144-12×3×4×12
=108
2b→-3c→=63
As we know that,
a→×2b→-3c→2+a→⋅2b→-3c→2=a→22b→-3c→2
⇒a→×2b→-3c→2+0=4×108
⇒a→×2b→-3c→=123

Topic: Mathematics, Vector Algebra, JEE Main
If $\hat{a}, \hat{b}$ and $\hat{c}$ are unit vectors satisfying $\hat{a}-\sqrt{3} \hat{b}+\hat{c}=\overrightarrow{0}$, then the angle between the vectors $\hat{a}$ and $\hat{c}$ is :

Options:
A
$\frac{\pi}{4}$

B
$\frac{\pi}{3}$

C
$\frac{\pi}{6}$

D
$\frac{\pi}{2}$

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

Solution:
Let angle between $\hat{a}$ and $\hat{c}$ be $\theta$.
Now, $\hat{a}-\sqrt{3} \hat{b}+\hat{c}=\overrightarrow{0}$
$$
\begin{aligned}
& \Rightarrow(\hat{a}+\hat{c})=\sqrt{3} \hat{b} \\
& \Rightarrow(\hat{a}+\hat{c}) \cdot(\hat{a}+\hat{c})=3(\hat{b} \cdot \hat{b}) \\
& \Rightarrow \hat{a} \cdot \hat{a}+\hat{a} \cdot \hat{c}+\hat{c} \cdot \hat{a}+\hat{c} \cdot \hat{c}=3 \times 1 \\
& \Rightarrow 1+2 \cos \theta+1=3 \\
& \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (22 Apr Online)
If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$, then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

Options:

The correct answer is:
3

Solution:
$\begin{array}{l}

\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\

|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\

\Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\

\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}

\end{array}$

Also,

$\begin{array}{l}

|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\

=1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\

\Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\

\therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3

\end{array}$

Topic: Mathematics, Vector Algebra, JEE Main
If $\vec{a}, \vec{b}$, and $\overrightarrow{\mathrm{c}}$ are unit vectors such that $\vec{a}+2 \vec{b}+2 \overrightarrow{\mathbf{c}}=\overrightarrow{0}$, then $|\vec{a} \times \overrightarrow{\mathbf{c}}|$ is equal to

Options:
A
$\frac{1}{4}$

B
$\frac{\sqrt{15}}{4}$

C
$\frac{15}{16}$

D
$\frac{\sqrt{15}}{16}$

The correct answer is:
$\frac{\sqrt{15}}{4}$

Solution:
$\begin{aligned} & \because \vec{a}+2 \vec{b}+2 \overrightarrow{\mathrm{c}}=\overrightarrow{0} \text { [Given] } \\ \Rightarrow & \vec{a}+2 \overrightarrow{\mathrm{c}}=-2 \vec{b} \\ \Rightarrow &(\vec{a}+2 \mathrm{c} \overrightarrow{)} \cdot(\vec{a}+2 \overrightarrow{\mathrm{c}})=(-2 \vec{b})(-2 \overrightarrow{\mathrm{b}})\\ \Rightarrow \vec{a} \cdot \vec{a}+4 \vec{c} \cdot \vec{c}+4 \vec{a} \cdot \vec{c}=4 \vec{b} \cdot \vec{b} \\ \Rightarrow & 1+4+4 \vec{a} \cdot \overrightarrow{\mathrm{c}}=4 \\ \Rightarrow & \vec{a} \cdot \overrightarrow{\mathrm{c}}=\frac{-1}{4} \\ \because &|\vec{a} \cdot \overrightarrow{\mathrm{c}}|^2+|\vec{a} \times \vec{c}|^2=1(\vec{a} \text { is unit vector }) \\ & \frac{1}{16}+|\vec{a} \times \vec{c}|^2=1 \\ &|\vec{a} \times \vec{c}|^2=\frac{15}{16} \\ &|\vec{a} \times \vec{c}|^2=\frac{\sqrt{15}}{4} \end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If a, b and c be three distinct real numbers in G.P. and a+b+c=xb, then x cannot be:

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

The correct answer is:
2

Solution:
Given a, b, c are in G.P.
Let common ratio of G.P. be r , then b=ar, c=ar2
Given equation can be written as
a+ar+ar2=x. a. r
⇒x=r+1r+1
As we know r>0, then r+1r≥2⇒x≥3
r<0, then r+1r≤-2⇒x≤-1
⇒x∈-∞,-1∪3, ∞
Hence, x ≠2.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If $a, b$ and $c$ form a geometric progression with common ratio $r$, then the sum of the ordinates of the points of intersection of the line $a x+b y+c=0$ and the curve $x+2 y^2=0$ is

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

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

Solution:
Since, $a, b$ and $c$ form a geometric progression
$\therefore \quad a=a, b=a r, c=a r^2$
Therefore, given line becomes
$\begin{aligned} & a x+a r y+a r^2 & =0 \\ \Rightarrow & x+r y+r^2 & =0 \\ \Rightarrow & x & =-r y-r^2\end{aligned}$
On putting $x=-r y-r^2$ in given curve $x+2 y^2=0$, we get
$\begin{aligned}-r y-r^2+2 y^2 & =0 \\ \Rightarrow \quad 2 y^2-r y-r^2 & =0\end{aligned}$
$\therefore$ Sum of ordinates $=\frac{r}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b$ are positive real numbers such that the lines ax $+9 y=5$ and $4 x+b y=3$ are parallel, then the least possible value of $a+b$ is

Options:
A 13
B 12
C 8
D 6

The correct answer is:
12

Solution:
$\frac{a}{4}=\frac{9}{b} \Rightarrow a b=36$
using $A M \geq G \cdot M$
$\frac{a+b}{2} \geq \sqrt{a b} \Rightarrow a+b \geq 12$

Topic: Mathematics, Straight Lines, JEE Main
If A, B are two non-singular matrices of order 3 and I is an identity matrix of order 3 such that AAT=5Î and 3A-1=2AT-Aadj4B, then B2 is equal to (where AT and adjA denote transpose and adjoint matrices of the matrix A respectively)

Options:
A 7353⋅46
B 7353⋅43
C 7453⋅212
D 5673⋅210

The correct answer is:
7353⋅46

Solution:
AAT=5Ι⇒A2=53 ….(1)
3A-1=2AT-adj4B
Multiplying by A on both sides, we get
3Ι=2AAT-Aadj4B
⇒A2adj4B=7Ι
A2adj4B=73
5346B2=73 (using (1))
B2=7353⋅46

Topic: Mathematics, Determinants, JEE Main
If A & B are two sets such that nA×B=60 & nA=12 also nA∩B=K, then the sum of maximum & minimum possible value of K is

Options:
A 17
B 12
C 5
D 7

The correct answer is:
5

Solution:
nB=nA×BnA=5
∴ nA∩B has minimum value =0
and nA∩B has maximum value =nB=5

Topic: Mathematics, Sets and Relations, JEE Main
If $\vec{A} \times \vec{B}=\vec{B} \times \vec{A}$, then the angle between $A$ and $B$ is

Options:
A
$\pi$

B
$\pi / 3$

C
$\pi / 2$

D
$\pi / 4$

The correct answer is:
$\pi$

Solution:
No Solution Available

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2004
If a-+b-,b-+c- and c-+a- are coterminous edges of a parallelepiped, then its volume is ________

Options:
A 3a-c-b-
B 0
C 2a-b-c-
D 4b-a-c-

The correct answer is:
2a-b-c-

Solution:
∵ Volume of a parallelelopiped whose, Coterminous edges are a-,b-,c-  is
v=a-b-c- …  (1)
Then, required volume = a-+b-,b-+c-,c-+a-=2a-b-c-

Topic: Mathematics, Vector Algebra, JEE Main
If a→×b→ b →×c→ c →×a→=λ a→ b → c →2 then λ is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
a→×b→b→×c→c→×a→=λa→b→c→2
a→×b→b→×c→c→×a→ =a→×b→×b→×c→.c→×a→
=a→b→c→b→-a→b→b→c→·c→×a→

=a→b→c→b→. c→×a→ ∵a→b→b→=0
=a→b→c→b→c→a→
=a→b→c→[a→b→c→]
=a→b→c→2
∴ λ = 1.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2014 (06 Apr)
If $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$ then $\vec{a}+\vec{b}+\vec{c}=$

Options:
A
abc

B
-1

C
0

D
2

The correct answer is:
0

Solution:
No Solution Available

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
If $a+b+c=0,|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is

Options:
A
$\frac{\pi}{3}$

B
$\frac{\pi}{4}$

C
$\frac{\pi}{6}$

D
$\frac{\pi}{2}$

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

Solution:
Let $a+b+c=0 \Rightarrow(a+b)=-c$ $\Rightarrow(a+b)^2=c^2$
$$
\begin{aligned}
& \Rightarrow a^2+b^2+2 a \cdot b=c^2 \\
& \Rightarrow 9+25+2.3 .5 \cos \theta=49 \\
& (\because|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7) \\
& \therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2012 (19 May Online)
If a+b+c=0 and a2+b2+c2-ab-bc-caâ 0, âa,b,câR, then the system of equations ax+by+cz=0, bx+cy+az=0 and cx+ay+bz=0 has

Options:
A A unique solution
B Infinite solutions
C No solution
D Exactly two solutions

The correct answer is:
Infinite solutions

Solution:
Note that abcbcacab=-a+b+c2a-b2+b-c2+c-a2
Since, D=0 & 0,0,0 is a solution, hence the system has infinite solutions.

Topic: Mathematics, Determinants, JEE Main
If $a+b+c=0$, then the roots of the equation $4 a x^2+3 b x+2 c=0$ are

Options:
A Equal
B Imaginary
C Real
D None of these

The correct answer is:
Real

Solution:
We have $4 a x^2+3 b x+2 c=0$ Let roots are $\alpha$ and $\beta$
Let $D=B^2-4 A C=9 b^2-4(4 a)(2 c)=9 b^2-32 a c$
Given that, $\quad(a+b+c)=0 \Rightarrow b=-(a+c)$
Putting this value, we get
$=9(a+c)^2-32 a c=9(a-c)^2+4 a c$
Hence roots are real

Topic: Mathematics, Quadratic Equation, JEE Main
If $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\mathbf{0}$, then

Options:
A $|\mathbf{a}|=|\mathbf{b}| .|\mathbf{c}|=1$
B $\mathbf{b} \| \mathbf{c}$
C $\mathbf{a} \| \mathbf{b}$
D $\mathbf{b} \perp \mathbf{c}$

The correct answer is:
$\mathbf{b} \| \mathbf{c}$

Solution:
$\quad \mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\mathbf{0} \Rightarrow \mathbf{a} \|(\mathbf{b} \times \mathbf{c})$ or $\mathbf{b} \times \mathbf{c}=\mathbf{0}$ i.e., $\mathbf{b} \| \mathbf{c}$ or $\mathbf{a}=\mathbf{0}$.

Topic: Mathematics, Vector Algebra, JEE Main
If a+b+c=1,ab+bc+ca=2 and abc=3, then the value of a4+b4+c4 is equal to:

Options:

The correct answer is:
13

Solution:
We have, a+b+c=1,ab+bc+ca=2 and abc=3

Now, a2+b2+c2=a+b+c2-2Σab=-3

⇒ab+bc+ca2=Σab2+2abcΣa

⇒Σab2=-2

So, a4+b4+c4=a2+b2+c22-2Σab2

=9-2-2

=13.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

$=(a+b+c)(x+a+b+c)^{2}, x \neq 0$ and $a+b+c \neq 0,$ then $x$

is equal to

Options:
A abc
B $-(a+b+c)$
C $2(a+b+c)$
D $-2(a+b+c)$

The correct answer is:
$-2(a+b+c)$

Solution:
$\Delta=\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

$R_{1} \rightarrow R_{1}+R_{2}+R$

$\Delta=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

$\begin{array}{l}

C_{1} \rightarrow C_{1}-C_{y}, C_{2} \rightarrow C_{2}-C_{3} \\

\Delta=(a+b+c)\left|\begin{array}{ccc}

0 & 0 & 1 \\

0 & -b-c-a & 2 b \\

c+a+b & c+a+b & c-a-b

\end{array}\right|

\end{array}$

$=(a+b+c)(a+b+c)^{2}$

Hence, $x=-2(a+b+c)$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$ then

Options:
A $\tan A \tan B+\tan B \tan C+\tan C \tan A=1$
B $\cot A+\cot B+\cot C=\cot A \cot B \cot C$
C $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}=1+4 \sin \mathrm{A} \sin \mathrm{B} \sin C$
D All three are correct

The correct answer is:
All three are correct

Solution:
$\mathrm{B}+\mathrm{C}=\frac{\pi}{2}-\mathrm{A} \Rightarrow \tan (\mathrm{B}+\mathrm{C})=\cot \mathrm{A}$

$\Rightarrow \frac{\tan \mathrm{B}+\tan \mathrm{C}}{1-\tan \mathrm{B} \tan \mathrm{C}}=\cot \mathrm{A}$

$\Rightarrow \tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan \mathrm{A}=1$

$\Rightarrow \tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}(\cot \mathrm{C}+\cot \mathrm{A}+\cot \mathrm{B})=1$

$\Rightarrow \cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}=\cot \mathrm{A} \cot \mathrm{B} \cot \mathrm{C}$

Again $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}$

$=2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2 \mathrm{C}$

$=2 \cos \left(\frac{\pi}{2}-C\right) \cos (A-B)+1-2 \sin ^{2} C$

$=2 \sin C[\cos (A-B)-\sin C]+1$

$=2 \sin \mathrm{C}\left[\cos (\mathrm{A}-\mathrm{B})-\sin \frac{\pi}{2}-(\mathrm{A}+\mathrm{B})\right]+1$

$\begin{array}{l}

=2 \sin C[\cos (A-B)-\cos (A+B)]+1 \\

=4 \sin A \sin B \sin C+1

\end{array}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If a+b+c=3 where a,b,c>0, then the greatest value of a2b3c2 is

Options:
A 3102477
B 392477
C 382477
D 392376

The correct answer is:
3102477

Solution:
Given, a+b+c=3
⇒2⋅a2+3⋅b3+2⋅c2=3
a2+a2+b3+b3+b3+c2+c27≥a22b33c2217
a+b+c7≥a2b3c2243317
377≥a2b3c22433
⇒a2b3c2≤3777×24×33=24×31077
Hence, maximum value =24×31077

Topic: Mathematics, Sequences and Series, JEE Main
If $[\bar{a} \bar{b} \bar{c}]=4$, then the volume (in cubic units) of the parallelepiped with $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}, \overline{\mathrm{b}}+2 \overline{\mathrm{c}}$ and $\overline{\mathrm{c}}+2 \overline{\mathrm{a}}$ as coterminal edges, is

Options:
A 32
B 16
C 9
D 36

The correct answer is:
36

Solution:
We have $\bar{a} \cdot(\bar{b} \times \bar{c})=4$
Volume of required parallelepiped is
$$
\begin{aligned}
& (\overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}}+2 \overline{\mathrm{c}}) \times \overline{\mathrm{c}}+2 \overline{\mathrm{a}}] \\
& =(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+2(\overline{\mathrm{c}} \times \overline{\mathrm{c}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+4(\overline{\mathrm{c}} \times \overline{\mathrm{a}})] \\
& =\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+\overline{\mathrm{a}}(0)+2 \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+4 \overline{\mathrm{a}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+2 \overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\
& +4 \overline{\mathrm{b}}(0)+4 \overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+8 \overline{\mathrm{b}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}}) \\
& =\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+0+0+0+0+0+0+8 \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\
& =9[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]=9(4)=36
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If a+b+c > 9c4 and the equation ax2+2bx - 5c=0 has non-real complex roots, then

Options:
A aÂ >Â 0,Â cÂ >Â 0
B a > 0, c < 0
C aÂ <Â 0,Â cÂ <Â 0
D aÂ <Â 0,Â cÂ >Â 0

The correct answer is:
a > 0, c < 0

Solution:
4a+4b+4c - 9c > 0
4a+4b - 5c > 0
fx=ax2+2bx - 5c, D < 0
f2=4a+4b - 5c > 0
a>0 and
-5c>0âc<0âa>0c<0

Topic: Mathematics, Quadratic Equation, JEE Main