datavorous/entrance-exam-dataset huggingface dataset:
If a→, b→ & c→ are non-coplanar vectors, then the point of intersection of the line passing through the points 2a→+3b→-c→, 3a→+4b→-2c→ with the line joining the points a→-2b→+3c→, a→-6b→+6c→ is

Options:
A a→+b→+c→
B a→+2b→
C a→+c→
D a→+2b→+c→2

The correct answer is:
a→+2b→

Solution:
Let, OA→=2a→+3b→-c→

OB→=3a→+4b→-2c→

OC→=a→-2b→+3c→

OD→=a→-6b→+6c→

Now, vector equation of line joining OA & OB

r→=OA→+tOB→-OA→, t∈R

=2a→+3b→-c→+ta→+b→-c→

=2+ta→+3+tb→+-1-tc→ ...i

Similarly, vector equation of line joining OC & OD

r→=OC→+sOD→-OC→, s∈R

=a→-2b→+3c→+s-4b→+3c→

=a→+-2-4sb→+3+3sc→ ...ii

Comparing Equations i & ii, we get

2+t=1=t=-1

And, -2-4s=3+t

⇒-2-4s=3-1

⇒s=-1

Substituting these values in i, we get

r→=a→+2b→

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If a→,b→,c→ are non-coplaner vectors such that b→×c→=a→; c→×a→=b→; a→×b→=c→, then which of the following is not TRUE?

Options:
A |a→|-|b→|=0
B |a→|=|b→|=|c→|=2
C a→b→c→=1
D |a→||b→||c→|=1

The correct answer is:
|a→|=|b→|=|c→|=2

Solution:
Given,

a→×b→=c→

b→×c→=a→

c→×a→=b→

(b→×c→)·a→=a→·a→

(c→×a→)·b→=b→·b→

a→2-b→2=(b→×c→)·a→-(c→×a→)·b→

=[c→a→b→]-[b→c→a→]=0

∴ |a→|2-|b→|2=0

|a→|-|b→|=0

⇒|a→|=|b→|

We know that,

a→×b→ b→×c→ c→×a→=[a→b→c→]2

a→b→c→=a→b→c→2

⇒a→b→c→=1

|a→||b→||c→|=1

∴ |a→|=|b→|=|c→|≠2

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-zero real numbers and if the equations $(a-1) x=y+z,(b-1) y=z+x,(c-1) z=x+y$ have a non-trivial solution, then $a b+b c+c a=$

Options:
A $a^2 b^2 c^2$
B 0
C $a b c$
D $a+b+c$

The correct answer is:
$a b c$

Solution:
We have system of equation,
$$
(a-1) x-y-z=0 ; x-(b-1) y+z=0
$$
and $x+y-(c-1) z=0$
It is a homogeneous system of equations.
Now, for non-trivial solution.
$$
\begin{aligned}
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & -(b-1) & 1 \\
1 & 1 & -(c-1)
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & 1-b & 1 \\
1 & 1 & 1-c
\end{array}\right|=0 \\
& (a-1)[(1-b)(1-c)-1]+1[1-c-1] \\
& \quad-[1-(1-b)]=0 \\
& (a-1)(1-b)(1-c)-a+1-c-b=0 \\
& (a-1)(1-c-b+b c)-a-b-c+1=0 \\
& a-a c-a b+a b c-1+c+b-b c-a-b-c+1=0 \\
& \therefore \quad a b+b c+c a=a b c
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
If a,b,c are non-zero real numbers, the system of equations y+z=a+2x, x+z=b+2y, x+y=c+2z is consistent and b=4a+c4, then the sum of the roots of the equation at2+bt+c=0 is

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

The correct answer is:
-3

Solution:
Given,

2x-y-z=-ax-2y+z=bx+y-2z=c

For consistent system of equations ââ 0 or â=â1=â2=â3=0
â=2-1-11-2111-2=0
â1=-a-1-1b-21c1-2=0
âa+b+c=0
â2=2-a-11b11c-2=0
âa+b+c=0
â3=2-1-a1-2b11c=0
âa+b+c=0
a+b+c=016a-4b+c=0âa5=b15=c-20
âa=Î»
b=3Î»
c=-4Î»
â Sum of roots =-ba=-3

Topic: Mathematics, Determinants, JEE Main
If $a, b, c$ are positive integers, then the determinant $\Delta=\left|\begin{array}{ccc}a^2+x & a b & a c \\ a b & b^2+x & b c \\ a c & b c & c^2+x\end{array}\right|$ is divisible by

Options:
A $x^3$
B $x^2$
C $\left(a^2+b^2+c^2\right)$
D None of these

The correct answer is:
$x^2$

Solution:
$\begin{aligned} & \Delta=\frac{1}{a b c}\left|\begin{array}{ccc}a^3+a x & a^2 b & a^2 c \\ a b^2 & b^3+b x & b^2 c \\ c^2 a & c^2 b & c^3+c x\end{array}\right| \\ & =\left|\begin{array}{ccc}a^2+x & a^2 & a^2 \\ b^2 & b^2+x & b^2 \\ c^2 & c^2 & c^2+x\end{array}\right| \\ & =\left(a^2+b^2+c^2+x\right) \times\left|\begin{array}{ccc}1 & 1 & 1 \\ b^2 & b^2+x & b^2 \\ c^2 & c^2 & c^2+x\end{array}\right|\end{aligned}$
\{Applying $R_1 \rightarrow R_1+R_2+R_3$ \}
$\begin{gathered}=\left(a^2+b^2+c^2+x\right)\left|\begin{array}{ccc}1 & 0 & 0 \\ b^2 & x & 0 \\ c^2 & 0 & x\end{array}\right|\left\{\begin{array}{ll}\text { Applying } & C_2 \rightarrow C_2-C_1 \\ C_3 \rightarrow C_3-C_1\end{array}\right\} \\ =x^2\left(a^2+b^2+c^2+x\right) .\end{gathered}$
Hence $\Delta$ is divisible by $x^2$ as well as by $x$.

Topic: Mathematics, Determinants, JEE Main
If $a, b, c$ are positive numbers, then least value of $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is

Options:
A 1
B 6
C 9
D None

The correct answer is:
9

Solution:
$\mathrm{AM} \geq \mathrm{HM}$

$\Rightarrow \frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$\Rightarrow(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9$

Topic: Mathematics, Sequences and Series, JEE Main
If a,b,c are real numbers satisfying the condition a+b+c=0, then the roots of the quadratic equation 3ax2+5bx+7c=0 are

Options:
A Positive
B negative
C real and equal
D distinct but not imaginary

The correct answer is:
distinct but not imaginary

Solution:
D=25b2-4×3a×7c
=25-a-c2-84ac
=25a2+c2+2ac-84ac
=25a2+c2-34ac
=17a2+c2-2ac+8a2+c2
=17a-c2+8a2+c2
D>0⇒ Roots are real and distinct

Topic: Mathematics, Quadratic Equation, JEE Main
If $a, b, c$ are real numbers such that $a+b+c=0$, then the quadratic equation $3 a x^2+2 b x+c=0$ has

Options:
A At least one root in $[0,1]$
B At least one root in $[1,2]$
C At least one root in $[-1,0]$
D None of these

The correct answer is:
At least one root in $[0,1]$

Solution:
Let $f(x)$ denotes the quadratic expression $\quad f(x) \equiv 3 a x^2+2 b x+c, \quad$ whose antiderivative be denoted by $f(x)=a x^3+b x^2+c x$
Now $f(x)$ being a polynomial in $R, f(x)$ is continuous and differentiable on $R$. To apply Rolle's theorem.
We observe that $f(0)=0$ and $f(1)=a+b+c=0$, by hypothesis. So there must exist at least one value of $x$, say $x=\alpha \in(0,1)$ such that
$f^{\prime}(\alpha)=0 \Leftrightarrow 3 a \alpha^2+2 b \alpha+c=0$
That is,
$f^{\prime}(x)=3 a x^2+2 b x+c=0$ has at least one root in [o, 1].

Topic: Mathematics, Quadratic Equation, JEE Main
If $a, b, c$ are sides of a scalene triangle, then the value of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is :

Options:
A
non - negative

B
negative

C
positive

D
non-positive

The correct answer is:
negative

Solution:
$$
\begin{aligned}
&\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{array}\right| \\
&=(a+b+c)\left|\begin{array}{lll}
1 & 1 & 1 \\
b & c & a \\
c & a & b
\end{array}\right| \\
&=(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
b-c & c-a & a \\
c-a & a-b & b
\end{array}\right| \\
&=(a+b+c)\left[a b+b c+c a-a^2-b^2-c^2\right] \\
&=-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]
\end{aligned}
$$
Since $a, b, c$ are sides of $a$ scalene triangle, therefore at least two of the $a, b, c$ will be unequal.
$\begin{array}{ll}\therefore \quad & (a-b)^2+(b-c)^2+(c-a)^2>0 \\ & \text { Also } a+b+c>0 \\ \therefore \quad & -(a+b+c)\left[(a-b)^2+\right. \\ & \left.(b-c)^2+(c-a)^2\right] < 0\end{array}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2013 (09 Apr Online)
If a,b,c are sides of the triangle ABC and 1ab1ca1bc=0, then the value of cos⁡2A+cos⁡2B+cos⁡2C is equal to

Options:
A -32
B 32
C 332
D -1

The correct answer is:
-32

Solution:
By expansion of the determinant, we get,
1ab1ca1bc=1c2-ab-1ac-b2+1a2-bc=0
⇒a2+b2+c2-ab-bc-ca=0

which is true for a=b=c

Hence, ∆ABC must be an equilateral triangle
⇒A=B=C=60o
⇒cos⁡2A+cos⁡2B+cos⁡2C=-32

Topic: Mathematics, Determinants, JEE Main
If A, B, C are the angles of a triangle and 1111+sinâ¡A1+sinâ¡B1+sinâ¡Csinâ¡A+sin2â¡Asinâ¡B+sin2â¡Bsinâ¡C+sin2â¡C = 0, then triangle ABC is

Options:
A right angled isosceles
B isosceles
C equilateral
D scalene

The correct answer is:
isosceles

Solution:
R2â R2- R1
111sinâ¡Asinâ¡Bsinâ¡Csinâ¡A+sin2â¡Asinâ¡B+sin2â¡Bsinâ¡C+sin2â¡C
R3â R3- R2
= 111sinâ¡Asinâ¡Bsinâ¡Csin2â¡Asin2â¡Bsin2â¡C
= sinâ¡A-sinâ¡B Ã sinâ¡B-sinâ¡C Ã sinâ¡C-sinâ¡A
â Either A= B or B=C or C= A
So Î is isosceles

Topic: Mathematics, Determinants, JEE Main
If A, B, C are the angles of a triangle then the system of equations -x+ycosC+zcosB=0, xcosC-y+zcosA=0, xcosB+ycosA-z=0 have

Options:
A only zero solution
B a non-zero solution for all triangles ∆ABC
C only zero solution but for certain values of A, B, C
D a non-zero solution if ∆ABC is an equilateral triangle & not for all triangles

The correct answer is:
a non-zero solution if ∆ABC is an equilateral triangle & not for all triangles

Solution:
A,B,C are angles of triangle.

-x+ycosC+zcosB=0

xcosC-y+zcosA=0

xcosB+ycosA-z=0

A=-1cosCcosBcosC-1cosAcosBcosA-1

A=-11-cos2A-cosC-cosC-cosA.cosB+cosBcosA.cosC+cosB

A=-1+cos2A+cos2C+cosA.cosB.cosC+cosA.cosB.cosC+cos2B

∵A,B, C are angles of triangle.

So, A=0 for equilateral triangle and A≠0 for all other triangles.

Hence, a non-zero solution if △ABC is an equilateral triangle and not for all triangle.

Topic: Mathematics, Determinants, JEE Main
If $a, b, c$ are the positive integers, then $(a+b)(b+c)(c+a)$ is

Options:
A $ \lt 8 a b c$
B $\geq 8 a b c$
C $=8 a b c$
D None of these

The correct answer is:
$\geq 8 a b c$

Solution:
We know that,
$A M \geq G M$
$\Rightarrow \frac{a+b}{2} \geq \sqrt{a b}$\ldots(i)
Similarly,
$\frac{b+c}{2} \geq \sqrt{b c}$\ldots(ii)
$\frac{c+a}{2} \geq \sqrt{c a}$\ldots(iii)
Multiplying these inequalities, we will get
$\frac{(a+b)(b+c)(c+a)}{2 \times 2 \times 2} \geq \sqrt{a^2 b^2 c^2}$
$\Rightarrow(a+b)(b+c)(c+a) \geq 8 a b c$
Hence, $\geq$ 8abc is the correct answer.

Topic: Mathematics, Sequences and Series, JEE Main
If A,B,C are three events such that PB=45, PA∩B∩Cc=14 and PAc∩B∩Cc=13, then PB∩C is equal to

Options:
A 1160
B 15
C 1360
D 14

The correct answer is:
1360

Solution:
From set theory
PB=PA∩B∩Cc+Ac∩B∩Cc+PB∩C
45=14+13+PB∩C
⇒PB∩C=45-14-13=48-15-2060
=1360

Topic: Mathematics, Probability, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors and $\mathbf{d}$ is any unit vector, then $|(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})|=$

Options:
A $2 \mid[\mathrm{abc}]$
B $\frac{1}{2} \mid[a b c]$
C $[\mathrm{abc}] \mid$
D $\frac{1}{6}[\mathbf{a b c}]$

The correct answer is:
$[\mathrm{abc}] \mid$

Solution:
a, b, c are non-coplanar.
$\therefore \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ and $\mathbf{a} \times \mathbf{b}$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let, $\mathbf{d}=\lambda(\mathbf{b} \times \mathbf{c})+\mu(\mathbf{c} \times \mathbf{a})+m(\mathbf{a} \times \mathbf{b})$
$\mathbf{a} \cdot \mathbf{d}=\lambda[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{gathered}
\mathbf{b} \cdot \mathbf{d}=\mu(\mathbf{b} \mathbf{c} \mathbf{a}) \\
\mathbf{c} \cdot \mathbf{d}=m(\mathbf{c} \mathbf{a}) \\
\therefore \mathrm{d}=\frac{(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})}{[\mathbf{a b} \mathbf{c}]} \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c}-\mathbf{d})(\mathbf{a} \times \mathbf{b})= \\
\mathbf{d}[\mathbf{a} \mathbf{b} \mathbf{c}] \\
\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c}+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b}) \mid \\
=[\mathbf{a} \mathbf{b} \mathbf{c}]
\end{gathered}$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d}$ and $\mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \mathbf{a}$, then $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$ is equal to

Options:
A 0
B $\alpha \mathbf{a}$
C $\beta \mathbf{b}$
D $(\alpha+\beta) \mathbf{c}$

The correct answer is:
0

Solution:
We have $\mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d}$ and $\mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \mathbf{a}$
$\begin{aligned}
& \therefore \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\alpha+1) \mathbf{d} \text { and } \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\beta+1) \mathbf{a} . \\
& \Rightarrow(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \\
& \text { If } \alpha \neq-1 \text {, then }(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \Rightarrow \mathbf{d}=\frac{\beta+1}{\alpha+1} \mathbf{a} \\
& \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d} \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha\left(\frac{\beta+1}{\alpha+1}\right) \mathbf{a} \\
& \Rightarrow\left(1-\frac{\alpha(\beta+1)}{\alpha+1}\right) \mathbf{a}+\mathbf{b}+\mathbf{c}=0
\end{aligned}$
$\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}$
are coplanar which is contradiction to the given condition,
$\therefore \alpha=-1$ and so $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$.

Topic: Mathematics, Vector Algebra, JEE Main
If a→, b→, c→ are three non-zero vectors and n^ is a unit vector perpendicular to c→ such that a→=αb→-n^,α≠0 and b→·c→=12, then c→×a→×b→ is equal to:

Options:
A 15
B 9
C 12
D 6

The correct answer is:
12

Solution:
Given:

a→=αb→-n^ where, α≠0

⇒a→·c→=αb→·c→-n^·c→

⇒a→·c→=12α-0

Since, n^⊥c→⇒n^·c→=0.

Now,

c→×a→×b→=c→·b→a→-c→·a→b→

⇒c→×a→×b→=12a→-12αb→

⇒c→×a→×b→=12a→-αb→

⇒c→×a→×b→=12-n^

⇒c→×a→×b→=12-n^

⇒c→×a→×b→=12-n^

⇒c→×a→×b→=12

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $\mathbf{a}=\mathbf{b}+\mathbf{c}$ and the angle between $\mathbf{b}$ and $\mathbf{c}$ is $\pi / 2$, then

Options:
A $a^2=b^2+c^2$
B $b^2=c^2+a^2$
C $c^2=a^2+b^2$
D $2 a^2-b^2=c^2$

The correct answer is:
$a^2=b^2+c^2$

Solution:
Given that $\mathbf{a}=\mathbf{b}+\mathbf{c}$ and angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{2}$.
So,
$\begin{aligned}
& \mathbf{a}^2=\mathbf{b}^2+\mathbf{c}^2+2 \mathbf{b} \cdot \mathbf{c} \\or \\
& \mathbf{a}^2=\mathbf{b}^2+\mathbf{c}^2+2|\mathbf{b} \| \mathbf{c}| \cos \frac{\pi}{2}
\end{aligned}$
or $\quad \mathbf{a}^2=\mathbf{b}^2+\mathbf{c}^2+0, \quad \therefore \mathbf{a}^2=\mathbf{b}^2+\mathbf{c}^2$
i.e., $a^2=b^2+c^2$.

Topic: Mathematics, Vector Algebra, JEE Main
If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are three vectors such that $\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}$ and the angle between $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is $\frac{\pi}{2}$, then:

Options:
A $a^2=b^2+c^2$
B $b^2=c^2+a^2$
C $c^2=a^2+b^2$
D $2 a^2-b^2=c^2$

The correct answer is:
$a^2=b^2+c^2$

Solution:
Given that $\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}$
and $\overrightarrow{\mathbf{b}} \perp \overrightarrow{\mathbf{c}}$
then $(\overrightarrow{\mathbf{a}})^2=(\overrightarrow{\mathbf{b}})^2+(\overrightarrow{\mathbf{c}})^2$
$a^2=b^2+c^2$

Topic: Mathematics, Vector Algebra, JEE Main
If a→, b→, c→ are unit vectors such that a→+2b→+2c→=0→, then a→×c→ is equal to :

Options:
A 14
B 1516
C 154
D 1516

The correct answer is:
154

Solution:
Given a→+2b→+2c→=0

⇒a→+2c→=-2b→

Squaring both sides,

⇒a→2+4c→2+4a→.c→=4b→2

⇒1+4+4cos⁡θ=4⇒cos⁡θ=-14

∴sin⁡θ=1-cos2⁡θ=154

Now, a→×c→=a→c→sin⁡θ=154

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (15 Apr)
If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\vec{a}|=7,|\vec{b}|=5,|\vec{c}|=3$ then angle between vector $\vec{b}$ and $\vec{c}$ is

Options:
A
$60^{\circ}$

B
$30^{\circ}$

C
$45^{\circ}$

D
$90^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
$\vec{a}+\vec{b}+\vec{c}=0 \Rightarrow \vec{b}+\vec{c}=-\vec{a} \Rightarrow(\vec{b}+\vec{c})^2=(\vec{a})^2=5^2+3^2+2 \vec{b} \vec{c}=7^2$
$\Rightarrow 2|\vec{b}||\vec{c}| \cos \theta=49-34=15 \Rightarrow 2 \times 5 \times 3 \cos \theta=15 \Rightarrow \cos \theta=1 / 2 \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $|\vec{a} \vec{b} \vec{c}|=4$ then

Options:
A
16

B
64

C
4

D
8

The correct answer is:
16

Solution:

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be any three non-coplanar vectors, then $[\mathbf{a}+\mathbf{b} \mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a}]=$

Options:
A $|\mathbf{a b} \mathbf{c}|$
B $2|\mathbf{a b} \mathbf{c}|$
C $[\mathbf{a} \mathbf{b} \mathbf{c}]^2$
D $2[\mathbf{a} \mathbf{b} \mathbf{c}]^2$

The correct answer is:
$2|\mathbf{a b} \mathbf{c}|$

Solution:
$\begin{aligned} & {[\mathbf{a}+\mathbf{b} \mathbf{b}+\mathbf{c} \mathbf{c}+\mathbf{a}]=(\mathbf{a}+\mathbf{b}) .\{(\mathbf{b}+\mathbf{c}) \times(\mathbf{c}+\mathbf{a})\}} \\ & =(\mathbf{a}+\mathbf{b}) .(\mathbf{b} \times \mathbf{c}+\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}) \\ & =(\mathbf{a}+\mathbf{b}) .(\mathbf{b} \times \mathbf{c}+\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}), \quad[\mathbf{c} \times \mathbf{c}=0\} \\ & =\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}+\mathbf{a} \cdot \mathbf{b} \times \mathbf{a}+\mathbf{a} \cdot \mathbf{c} \times \mathbf{a}+\mathbf{b} . \mathbf{b} \times \mathbf{c}+\mathbf{b} \cdot \mathbf{b} \times \mathbf{a}+\mathbf{b} . \mathbf{c} \times \mathbf{a} \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]+[\mathbf{b} \mathbf{c} \mathbf{a}]=2[\mathbf{a} \mathbf{b} \mathbf{c}]\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c$ be in H.P., then

Options:
A $a^2+c^2 \gt b^2$
B $a^2+b^2 \gt 2 c^2$
C $a^2+c^2 \gt 2 b^2$
D $a^2+b^2 \gt c^2$

The correct answer is:
$a^2+c^2 \gt 2 b^2$

Solution:
From the section of inequality, we know that
A.M. of $n^{\text {th }}$ powers $\gt n^{\text {th }}$ power of A.M.
i.e.$\frac{1}{2}\left(a^n+c^n\right) \gt \left(\frac{1}{2}(a+c)\right)^n$
Considering two quantities $a$ and $c$
or $\frac{1}{2}\left(a^n+c^n\right) \gt (A)^n$ or $\frac{1}{2}\left(a^n+c^n\right) \gt (H)^n$
Since A.M. $\gt H \cdot M$.
$\frac{1}{2}\left(a^n+c^n\right) \gt (b)^n$ $\Rightarrow a^n+c^n \gt 2 b^n$
Putting $n=2$, we have $a^2+c^2 \gt 2 b^2$.

Topic: Mathematics, Sequences and Series, JEE Main
If a → , b → , c → be three unit vectors perpendicular to each other, then 2 a → +3 b → +4 c → ⋅ a → × b → +5 b → × c → +6 c → × a → is equal to

Options:
A 18
B 0
C 4
D 32

The correct answer is:
32

Solution:
2 a → +3 b → +4 c → ⋅ a → × b → +5 b → × c → +6 c → × a →
= 10 a → b → c → +18 a → b → c → +4 a → b → c →
=32a→ b→ c→=32 as a→ b→ c→=1

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c, d$ and $p$ are distinct real numbers such that $\left(a^2+b^2+c^2\right) p^2-2 p(a b+b c+c d)+\left(b^2+\right.$ $\left.c^2+d^2\right) \leq 0$, then

Options:
A
$a, b, c, d$ are in A.P.

B
$a b=c d$

C
$a c=b d$

D
$a, b, c, d$ are in G.P.

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

Solution:
The given relation can be written as $\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2+c^2-2 b p c\right)+$ $\left(c^2 p^2+d^2-2 p c d\right) \leq 0$ or $\quad(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0$
...(1)
Since $a, b, \mathrm{c}, d$ and $p$ are all real, the inequality (1) is possible only when each of factor is zero. i.e., $a p-b=0, b p-c=0$ and $c p-d=0$ or $\quad p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ or $a, b, c, d$ are in G.P.

Topic: Mathematics, Quadratic Equation, JEE Main
If $a, b, c, d$ and $p$ are distinct real numbers such that $\left(a^2+b^2+c^2\right) p^2-2 p(a b+b c+c d)+\left(b^2+\right.$ $\left.c^2+d^2\right) \leq 0$, then

Options:
A
$a, b, c, d$ are in A.P.

B
$a b=c d$

C
$a c=b d$

D
$a, b, c, d$ are in G.P.

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

Solution:
The given relation can be written as $\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2+c^2-2 b p c\right)+$ $\left(c^2 p^2+d^2-2 p c d\right) \leq 0$ or $\quad(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0$
...(1)
Since $a, b, \mathrm{c}, d$ and $p$ are all real, the inequality (1) is possible only when each of factor is zero. i.e., $a p-b=0, b p-c=0$ and $c p-d=0$ or $\quad p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ or $a, b, c, d$ are in G.P.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (12 May Online)
If $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar vectors, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=$

Options:
A $|\mathbf{a} \times \mathbf{c}|^2$
B $|\mathbf{a} \times \mathbf{d}|^2$
C $|\mathbf{b} \times \mathbf{c}|^2$
D 0

The correct answer is:
0

Solution:
$\begin{aligned}
& (\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d} \\
& \text { a, b, c, }, \mathbf{d} \text { are coplanar vectors } \\
& \therefore[\mathbf{a} \mathbf{b} \mathbf{d}]=[\mathbf{a} \mathbf{b} \mathbf{c}]=0 . \text { So, }(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=\mathbf{0}
\end{aligned}$
So, $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=\mathbf{0}$

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c, d$ are in H.P., then $a b+b c+c d$ is equal to

Options:
A $3 a d$
B $(a+b)(c+d)$
C $3 \mathrm{ac}$
D None of these

The correct answer is:
$3 a d$

Solution:
Since $a, b, c, d$ are in H.P., therefore $b$ is the H.M. of $a$ and $c$ i.e. $b=\frac{2 a c}{a+c}$ and $c$ is the H.M. of $b$ and
d i.e. $c=\frac{2 b d}{b+d}$ $\therefore \quad(a+c)(b+d)=\frac{2 a c}{b} \cdot \frac{2 b d}{c}$
$\Rightarrow a b+a d+b c+c d=4 a d \Rightarrow a b+b c+c d=3 a d$.
Trick: Check for $a=1, b=\frac{1}{2}$, $c=\frac{1}{3}, d=\frac{1}{4}$

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b, c, d$ are in H.P., then

Options:
A $a+d \gt b+c$
B $a d \gt b c$
C Both (1) and (2)
D None of these

The correct answer is:
Both (1) and (2)

Solution:
As $a, b, c, d$ are in H.P. So $b$ is H.M. between $a$ and $c$. Also G.M. between $a$ and $c=\sqrt{a c}$.
Now, G.M.>H.M. so that $\sqrt{a c} \gt b \Rightarrow a c \gt b^2 \ldots(i)$
Again $a, b, c, d$ are in H.P. So $c$ is H.M. between $b$ and $d$.
Therefore $b d \gt c^2 \ldots(ii)$
Multiplying (i) and (ii), we get
$a b c d \gt b^2 c^2$ or $a d \gt b c$ Hence answer (2) is true.
Now A.M. between $a$ and $c=\frac{1}{2}(a+c)$
Now as A.M. > H.M. so here $\Rightarrow$ $a+c \gt 2 b \ldots(iii)$
And $C$ is H.M. between $b$ and $d \Rightarrow b+d \gt 2 c\ldots(iv)$
Adding (iii) and (iv), we get
$(a+c)+(b+d) \gt 2(b+c) \Rightarrow a+d \gt b+c$
Hence answer (1) is true. So both (1) and (2) are correct.

Topic: Mathematics, Sequences and Series, JEE Main
If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is:

Options:
A OR
B NAND
C NOT
D AND

The correct answer is:
OR

Solution:
a	b	c	d	X

0	0	0	0	0

0	0	0	1	1

When any input is one, output is one, hence the gate is 'OR' gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (03 Apr)
If $a, b, c, d$ are inputs to a gate and $x$ is its output, then as per the following time graph, the gate is

Options:
A NAND
B NOT
C AND
D OR

The correct answer is:
OR

Solution:
Output of OR gate is 0 when all inputs are $0 \&$ output is 1 when atleast one of the input is 1. Observing output $x$ :- It is 0 when all inputs are $0 \&$ it is 1 when atleast one of the inputs is $1 . \therefore$ OR gate

Topic: Physics, Semiconductors, JEE Main
If $a, b, c, d$ are positive, then $\lim _{x \rightarrow \infty}\left(1+\frac{1}{a+b x}\right)^{c+d x}=$

Options:
A $e^{d / b}$
B $e^{c / a}$
C $e^{(c+d) /(a+b)}$
D $e$

The correct answer is:
$e^{d / b}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \infty}\left(1+\frac{1}{a+b x}\right)^{c+c a x}=\lim _{x \rightarrow \infty}\left\{\left(1+\frac{1}{a+b x}\right)^{a+b x}\right\}^{\frac{c+c x}{a+b x}}=e^{d / b} \\ & \left\{\because \lim _{x \rightarrow \infty}\left(1+\frac{1}{a+b x}\right)^{a+b x}=e \text { and }\right. \left.\lim _{x \rightarrow \infty} \frac{c+d x}{a+b x}=\frac{d}{b}\right\} .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
If a,b,c,d,e,f are in arithmetic progression. Then e−c is equal to

Options:
A 2 câa
B 2 d−c
C 2 f−d
D d−c

The correct answer is:
2 d−c

Solution:
a, b, c, d, e, f are in A.P.
let common difference = D
∴e-c=a+4D-a+2D=2D=2(d-c)

Topic: Mathematics, Sequences and Series, JEE Main