datavorous/entrance-exam-dataset huggingface dataset:
If $a, b$ and $c$ are real numbers then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are always

Options:
A real
B imaginary
C positive
D negative

The correct answer is:
real

Solution:
Given equation is $(x-a)(x-b)+(x-b)(x-c)+(x-c$ $(x-a)=0$
$\Rightarrow 3 x^{2}-2(b+a+c) x+a b+b c+c a=0$
Now, here $A=3, B=-2(a+b+c)$
$C=a b+b c+c a$
$\therefore D=\sqrt{B^{2}-4 A C}$
$=\sqrt{(-2(a+b+c))^{2}-4(3)(a b+b c+c a)}$
$=\sqrt{4(a+b+c)^{2}-12(a b+b c+c a)}$
$=2 \sqrt{a^{2}+b^{2}+c^{2}-a b-b c-c a}$
$=2 \sqrt{\frac{1}{2}\left\{(a-b)^{2}-(b-c)^{2}+(c-a)^{2}\right\}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)
If A, B and C are subsets of a Universal set, then which one of the following is not correct?

Options:
A $\mathrm{A} \cup(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \cup \mathrm{B}) \cap(\mathrm{A} \cup \mathrm{C})$
B $\mathrm{A}^{\prime} \cup(\mathrm{A} \cup \mathrm{B})=\left(\mathrm{B}^{\prime} \cap \mathrm{A}\right) \cup \mathrm{A}$
C $\mathrm{A} \cup(\mathrm{B} \cup \mathrm{C})=\left(\mathrm{C}^{\prime} \cap \mathrm{B}\right)^{\prime} \cap \mathrm{A}$
D $(A \cap B) \cup C=(A \cup C) \cap(B \cup C)$
Where $\mathrm{A}^{\prime}$ is the complement of $\mathrm{A}$.

The correct answer is:
$\mathrm{A} \cup(\mathrm{B} \cup \mathrm{C})=\left(\mathrm{C}^{\prime} \cap \mathrm{B}\right)^{\prime} \cap \mathrm{A}$

Solution:
Checkingthrough option 'c' is incorrect

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are the angles of a triangle and
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A+\sin ^{2} A & \sin B+\sin ^{2} B & \sin C+\sin ^{2} C\end{array}\right|=0$
then which one of the following is correct?

Options:
A The triangle $\mathrm{ABC}$ is isosceles
B The triangle $\mathrm{ABC}$ is equilateral
C The triangle $\mathrm{ABC}$ is scalene
D No conclusion can be drawn with regard to the nature of the triangle

The correct answer is:
The triangle $\mathrm{ABC}$ is isosceles

Solution:
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A+\sin ^{2} A & \sin B+\sin ^{2} B & \sin C+\sin ^{2} C\end{array}\right|$
$\begin{array}{l}
\left|\begin{array}{ccc}
1 & 0 & 0 \\
1+\sin A & \sin A-\sin B & \sin A-\sin C \\
\sin A+\sin ^{2} A & \sin ^{2} A-\sin ^{2} B & \sin ^{2} A-\sin ^{2} C
\end{array}\right|+0=0 \\
\left\{\text { Applying } C_{2} \rightarrow C_{1}-C_{2} \text { and } C_{3} \rightarrow C_{1}-C_{3}\right\} \\
=(\sin A-\sin B) \times(\sin A-\sin C)
\end{array}$
$\left|\begin{array}{ccc}1 & 0 & 0 \\ 1+\sin \mathrm{A} & 1 & 1 \\ \sin \mathrm{A}+\sin ^{2} \mathrm{~A} & (\sin \mathrm{A}+\sin \mathrm{B}) & (\sin \mathrm{A}+\sin \mathrm{C})\end{array}\right|=0$
$(\sin A-\sin B)(\sin A-\sin C)(\sin B-\sin C)=0$
$\therefore \sin \mathrm{A}=\sin \mathrm{B}$ or $\sin \mathrm{A}=\sin \mathrm{C}$ or $\sin \mathrm{B}=\sin \mathrm{C}$ either $\mathrm{A}=\mathrm{B}$
or $\mathrm{B}=\mathrm{C}$ or $\mathrm{A}=\mathrm{C}$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 2)
If $\vec{a}, \vec{b}$ and $\vec{c}$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then which one of the following is correct?

Options:
A $\quad \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
B $\vec{a}+\vec{b}+\vec{c}=$ unit vector
C $\quad \vec{a}+\vec{b}=\vec{c}$
D $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$

The correct answer is:
$\quad \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

Solution:
Position vectors of vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $\vec{a}, \vec{b}$ and $\vec{c}$
$\because$ triangle is equilateral.
$\therefore$ Centroid and orthocenter will coincide. Centroid $\equiv$ orthocenter position vector
$=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$
$\because$ given in question orthocenter is at origin.
Hence $\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})=0$
$\vec{a}+\vec{b}+\vec{c}=0$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 1)
If $A, B$ and $C$ are three finite sets, then what is $[(A \cup B) \cap C]$ equal to?

Options:
A $A^{\prime} \cup B^{\prime} \cap C^{\prime}$
B $A^{\prime} \cap B^{\prime} \cap C^{\prime}$
C $A^{\prime} \cap B^{\prime} \cup C^{\prime}$
D $A \cap B \cap C$

The correct answer is:
$A^{\prime} \cap B^{\prime} \cup C^{\prime}$

Solution:
We know that
$[(A \cup B) \cap C]^{\prime}=A^{\prime} \cap B^{\prime} \cup C$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
If $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are three positive numbers in an arithmetic progression, then:

Options:
A $\mathrm{ac}>\mathrm{b}^{2}$
B $b^{2}>a+c$
C $a b+b c \leq 2 a c$
D $\mathrm{ab}+\mathrm{bc} \geq 2 \mathrm{ac}$

The correct answer is:
$\mathrm{ab}+\mathrm{bc} \geq 2 \mathrm{ac}$

Solution:
Since, $\mathrm{a}, \mathrm{b}, \mathrm{c}$, are in $\mathrm{AP}$
$\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}$ are in $\mathrm{HP}$.
Since, $\mathrm{AM} \geq \mathrm{HM}$
$\Rightarrow \quad \mathrm{b} \geq \frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}$
[since $\mathrm{AM}=\mathrm{b}$ and $\left.\mathrm{HM}=\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}\right]$
$\Rightarrow \quad \mathrm{ab}+\mathrm{bc} \geq 2 \mathrm{ac}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three sets and $\mathrm{U}$ is the universal set such thatn $(\mathrm{U})=700, \mathrm{n}(\mathrm{A})=200, \mathrm{n}(\mathrm{B})=300$ and $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=100$,
then what is the value of $\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right) ?$

Options:
A 100
B 200
C 300
D 400

The correct answer is:
300

Solution:
From the given data $\mathrm{n}(\mathrm{U})=700, \mathrm{n}(\mathrm{A})=200, \mathrm{n}(\mathrm{B})=300$ and
$n(A \cap B)=100$
We know that, $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=200+300-100=400$
Now, $\mathrm{n}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\sim(\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{n}(\mathrm{U})-\mathrm{n}(\mathrm{A} \cup \mathrm{B})$
$=700-400=300$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
If $(\mathrm{A}-\mathrm{B}) \cup(\mathrm{B}-\mathrm{A})=\mathrm{A}$ for subsets $\mathrm{A}$ and $\mathrm{B}$ of the universal set $\mathrm{U}$, then which one of the following is correct?

Options:
A $\mathrm{B}$ is proper non-empty subset of $\mathrm{A}$
B $\mathrm{A}$ and $\mathrm{B}$ are non-empty disjoint sets
C $\quad B=\phi$
D None of the above

The correct answer is:
$\quad B=\phi$

Solution:
For subsets $\mathrm{A}$ and $\mathrm{B}$ of $\mathrm{U}$. If $(A-B) \cup(B-A)=A$
$\Rightarrow \mathrm{B}=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
If $a+b+c=0$, then one of the solution of
$\left|\begin{array}{ccc}\mathrm{a}-\mathrm{x} & \mathrm{c} & \mathrm{b} \\ \mathrm{c} & \mathrm{b}-\mathrm{x} & \mathrm{a} \\ \mathrm{b} & \mathrm{a} & \mathrm{c}-\mathrm{x}\end{array}\right|=0$ is

Options:
A $x=a$
B $\mathrm{x}=\sqrt{\frac{3\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)}{2}}$
C $x=\sqrt{\frac{2\left(a^{2}+b^{2}+c^{2}\right)}{3}}$
D $\mathrm{x}=0$

The correct answer is:
$\mathrm{x}=0$

Solution:
$\left|\begin{array}{ccc}a-x & c & b \\ c & b-x & a \\ b & a & c-x\end{array}\right|=0$
$-\left|\begin{array}{ccc}x-(a+b+c) & c & b \\ x-(a+b+c) & b-x & a \\ x-(a+b+c) & a & c-x\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}x & c & b \\ x & b-x & a \\ x & a & c-x\end{array}\right|=0$
$\left\{\right.$ Applying $\left.\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right\}$
$\left|\begin{array}{ccc}x & c & b \\ 0 & c+x-b & b-a \\ 0 & c-a & b+x-c\end{array}\right|=0$
$\left\{\right.$ Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3}\right\}$
$x\{(c+x-b)(b+x-c)-(b-a)(c-a)\}=0$
$x\left\{\left(x^{2}-(b-c)^{2}-b c+a c+a b-a^{2}\right\}=0\right.$
$x\left(x^{2}-a^{2}-b^{2}-c^{2}+a b+b c+c a\right)=0$
$x\left\{x^{2}-(a-b)^{2}-(b-c)^{2}-(c-a)^{2}\right\}=0$
$\therefore x=0$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 2)
If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, then which of the following is/are correct?
$1.$ $\vec{a}, \vec{b}, \vec{c}$ are coplanar.
$2.$ $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$
Select the correct answer using the code given below.

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
According to statements (1) and (2), when $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$, then both statements are correct. If $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=0$, then $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are
collinear. Therefore, option (c) is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
If $A+B+C=\pi / 2$, then what is the value of $\tan A \tan B+\tan B \tan C+\tan C \tan A ?

Options:
A 0
B 1
C $-1$
D $\tan A \tan B \tan C$

The correct answer is:
1

Solution:
Given $A+B+C=\frac{\pi}{2}$ Take tan on both sides,
$\Rightarrow \tan (A+B+C)=\tan \left(\frac{\pi}{2}\right)$
$\Rightarrow \frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}=\frac{1}{0}$
$\left(\because \frac{1}{0}=\infty\right)$
$\Rightarrow \tan A \tan B+\tan B \tan C+\tan C \tan A=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$, then what is $\sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}-\sin 2 \mathrm{C}$ equal
to?

Options:
A $-4 \sin A \sin B \sin C$
B $-4 \cos A \sin B \cos C$
C $-4 \cos A \cos B \sin C$
D $-4 \sin A \cos B \cos C$

The correct answer is:
$-4 \sin A \cos B \cos C$

Solution:
$\begin{aligned} & \sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}-\sin 2 \mathrm{C} \\ &=2 \cos (\mathrm{A}+\mathrm{B}) \cdot \sin (\mathrm{A}-\mathrm{B})+\sin (2 \mathrm{~A}+2 \mathrm{~B}) \\ &=2 \cos (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})+2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}+\mathrm{B}) \\ &=2 \cos (\mathrm{A}+\mathrm{B})[\sin (\mathrm{A}-\mathrm{B})+\sin (\mathrm{A}+\mathrm{B})] \\ &=-2 \cos \mathrm{C}[2 \sin \mathrm{A} \cdot \cos \mathrm{B}] \\ &=-4 \sin \mathrm{A} \cos \mathrm{B} \cos \mathrm{C} \end{aligned}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2018 (Phase 2)
If $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}$, then which one of the following
is correct?

Options:
A $\vec{a}, \vec{b}, \vec{c}$ are orthogonal in pairs and $|\vec{a}|=|\vec{c}|$ and
$|\overrightarrow{\mathrm{b}}|=1$
B $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are non-orthogonal to each other
C $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are orthogonal in pairs but $|\overrightarrow{\mathrm{a}}| \neq|\overrightarrow{\mathrm{c}}|$
D $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are orthogonal is pairs but $|\overrightarrow{\mathrm{b}}| \neq 1$

The correct answer is:
$\vec{a}, \vec{b}, \vec{c}$ are orthogonal in pairs and $|\vec{a}|=|\vec{c}|$ and
$|\overrightarrow{\mathrm{b}}|=1$

Solution:
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}, \quad \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}$
$\Rightarrow \vec{c} \perp \vec{a} \& \vec{c} \perp \vec{b}$
$\Rightarrow \vec{a} \perp \vec{b} \perp \vec{c}$
$\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are orthogonal.
Now, $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|$
$\Rightarrow|\vec{a}||\vec{b}| \sin 90^{\circ}=|\vec{c}|$
$\Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|$
$\Rightarrow|\vec{b}||\vec{\phi}| \vec{b}|=| \vec{\not} \mid$
$\Rightarrow|\overrightarrow{\mathrm{b}}|^{2}=1$
$\Rightarrow|\vec{b}|=1$...(i)
$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}$
$\Rightarrow \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}} \perp \overrightarrow{\mathrm{c}}$
Also, $|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{a}}|$
$\Rightarrow|\vec{b}||\vec{c}| \sin 90^{\circ}=|\vec{a}|$
$\Rightarrow|\overrightarrow{\mathrm{b}}| \cdot|\overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{a}}|$
$\Rightarrow|\vec{c}|=|\vec{a}| \quad$ (from $\quad$ (i)

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 1)
If $a \neq b \neq c$ are all positive, then the value of the determinant
$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is

Options:
A non-negative
B non-positive
C negative
D positive

The correct answer is:
negative

Solution:
$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=\left|\begin{array}{lll}a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b\end{array}\right|$
$\left(\right.$ Applying $\left.\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right)$
$=(a+b+c)\left|\begin{array}{lll}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{array}\right|$
[on taking $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\left.\mathrm{C}_{1}\right]$
$=(a+b+c)\left[1\left(b c-a^{2}\right)-b(b-a)+c(a-c)\right]$
$=(a+b+c)\left[b c-a^{2}-b^{2}+a b+a c-c^{2}\right]$
$=(a+b+c)\left[-\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$
$=-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$
Hence, the determinant value is Negative

Topic: Mathematics, Determinants, NDA, NDA 2014 (Phase 2)
If $A, B, C$ are any three arbitrary events, then which one of the following expressions shows that both $A$ and $B$ occur but not $C$ ?

Options:
A $A \cap B \cap \bar{C}$
B $A \cap B \cap \bar{C}$
C $A \cap B \cap C$
D $A \cap B \cap C$

The correct answer is:
$A \cap B \cap \bar{C}$

Solution:
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are any three arbitary events then only expression $A \cap B \cap \bar{C}$ will show that both $A$ and $B$ occur but not $\mathrm{C}$.

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
If a, b, c are in APor GPHP, then $\frac{\mathrm{a}-\mathrm{b}}{\mathrm{b}-\mathrm{c}}$ is equal to

Options:
A $\frac{\mathrm{b}}{\mathrm{a}}$ or 1 or $\frac{\mathrm{b}}{\mathrm{c}}$
B $\frac{\mathrm{c}}{\mathrm{a}}$ or $\frac{\mathrm{c}}{\mathrm{b}}$ or
C 1 or $\frac{\mathrm{a}}{\mathrm{b}}$ or $\frac{\mathrm{a}}{\mathrm{c}}$
D 1 or $\frac{\mathrm{a}}{\mathrm{b}}$ or $\frac{\mathrm{c}}{\mathrm{a}}$

The correct answer is:
1 or $\frac{\mathrm{a}}{\mathrm{b}}$ or $\frac{\mathrm{a}}{\mathrm{c}}$

Solution:
From question,
$\begin{aligned} & a, b, c \in \text { A.P. } \Rightarrow a-b=b-c \Rightarrow \frac{a-b}{b-c}=1 \\ & a, b, c \in \text { G.P. } \Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{a-b}{b-c} \\ & a, b, c \in \text { H.P. } \Rightarrow b=\frac{2 a c}{a+c} \Rightarrow a b+b c=2 a c \\ \Rightarrow & a b-a c=a c-b c \\ \Rightarrow & \frac{a-b}{b-c}=\frac{a}{c} \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$, then what is the value of
$\left|\begin{array}{lll}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$

Options:
A 0
B 1
C $-1$
D None of these

The correct answer is:
0

Solution:
Since, $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$.
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
Explanding the determinant we get,
$\left|\begin{array}{ccc}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
$=a\left|\begin{array}{cc}c & b+c \\ b+c & 0\end{array}\right|-b\left|\begin{array}{cc}b & b+c \\ a+b & 0\end{array}\right|+(a+b)\left|\begin{array}{cc}b & c \\ a+b & b+c\end{array}\right|$
$=-a(b+c)^{2}+b(a+b)(b+c)+(a+b)\left(b^{2}+b c-a c-b c\right)$
$=-a\left(b^{2}+c^{2}+2 b c\right)+b\left(a b+a c+b^{2}+b c\right)$
$=-a b^{2}-a c^{2}-2 a b c+a b^{2}+2 a b c+b^{2} c \quad\left(\because b^{2}=a c\right)$
$=-a c^{2}+b^{2} c=-a c^{2}+a c \cdot c .$
$=-a c^{2}+a c^{2}=0$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 1)
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$, are non-zero real numbers and
$\left[\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right]=0$
then what is the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ ?

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

The correct answer is:
$-1$

Solution:
Given $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=0$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$
$\Rightarrow\left|\begin{array}{ccc}1+a & -a & -a \\ 1 & b & 0 \\ 1 & 0 & c\end{array}\right|=0$
$\Rightarrow$ Expanding along $R_{3}, 1(a b)+c(b+a b+a)=0$
$\Rightarrow a b+b c+c a+a b c=0$
$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 2)
If $a, b, c$ are non-zero real numbers, then the inverse of the
matrix $\mathrm{A}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$ is equal to

Options:
A $\left[\begin{array}{lll}a^{-1} & 0 & 0 \\ 0 & b^{-1} & 0 \\ 0 & 0 & c^{-1}\end{array}\right]$
B $\frac{1}{a b c}\left[\begin{array}{ccc}a^{-1} & 0 & 0 \\ 0 & b^{-1} & 0 \\ 0 & 0 & c^{-1}\end{array}\right]$
C $\frac{1}{a b c}\left[\begin{array}{llll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
D $\frac{1}{a b c}\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$

The correct answer is:
$\left[\begin{array}{lll}a^{-1} & 0 & 0 \\ 0 & b^{-1} & 0 \\ 0 & 0 & c^{-1}\end{array}\right]$

Solution:
$\mathrm{A}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$
$\mathrm{A}^{-1}=\left[\begin{array}{ccc}\frac{1}{\mathrm{a}} & 0 & 0 \\ 0 & \frac{1}{\mathrm{~b}} & 0 \\ 0 & 0 & \frac{1}{\mathrm{c}}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{a}^{-1} & 0 & 0 \\ 0 & \mathrm{~b}^{-1} & 0 \\ 0 & 0 & \mathrm{c}^{-1}\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 2)
If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are non-zero vectors and $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|$
$|\overrightarrow{\mathrm{c}}|$, then which one of the following is correct?

Options:
A $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a} \neq 0$
B $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$ only
C $\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0$ only
D $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0$

The correct answer is:
$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0$

Solution:
Refer to the figure. $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}|$ is triple dot product and is volume $(\mathrm{V})$ of
the parallelepiped whose adjacent edges are a, b, and c.
i.e., $V=|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})| \cdot \overrightarrow{\mathrm{OC}}$
$\Rightarrow \quad \mathrm{V}=|\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}|(\sin \theta)(\cos \phi)|\overrightarrow{\mathrm{c}}|$
where $0 \leq \theta \leq \pi$ is the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$.
As given $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{a}}\|\overrightarrow{\mathrm{b}}\| \overrightarrow{\mathrm{c}}|$
$\Rightarrow \quad \sin \theta \cos \phi \mid .=1$
$\Rightarrow \quad \sin \theta=1, \cos \phi=1$
$\Rightarrow \quad \theta=\frac{\pi}{2}, \phi=0$
$\Rightarrow \quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 2)
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are real numbers, then the value of the determinant $\left|\begin{array}{ccc}1-a & a-b-c & b+c \\ 1-b & b-c-a & c+a \\ 1-c & c-a-b & a+b\end{array}\right|$ is

Options:
A 0
B $(a-b)(b-c)(c-a)$
C $(a+b+c)^{2}$
D $(a+b+c)^{3}$

The correct answer is:
0

Solution:
$\left|\begin{array}{lll}1-a & a-b-c & b+c \\ 1-b & b-c-a & c+a \\ 1-c & c-a-b & a+b\end{array}\right|$
apply $C_{2} \rightarrow C_{2}+C_{3}$
$\left|\begin{array}{lll}1-a & a & b+c \\ 1-b & b & c+a \\ 1-c & c & a+b\end{array}\right|$
apply $C_{1} \rightarrow C_{1}+C_{2}$
$\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
apply $C_{3} \rightarrow C_{2}+C_{3}$
$(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|=0$

Topic: Mathematics, Determinants, NDA, NDA 2015 (Phase 2)
If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of corners $A, B, C$ of a parallelogram $\mathrm{ABCD}$, then what is the position vector of the corner $\mathrm{D}$ ?

Options:
A $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$
B $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$
C $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$
D $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$

The correct answer is:
$\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$

Solution:
Let O be the origin and $\mathrm{ABCD}$ be the parallelogram. $\operatorname{In} \Delta \mathrm{ODC}$,
$\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CD}}$
$\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}$
and $\operatorname{In} \Delta \mathrm{AOB}, \quad \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
Thus, $\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, $\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}[$ since, $\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}]$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
If $a, b, c$ are the sides of a triangle $A B C$, then $a^{\frac{1}{p}}+b^{\frac{1}{p}}-c^{\frac{1}{p}}$ where $\mathrm{p}>1$, is

Options:
A always negative
B always positive
C always zero
D positive if $1 < p < 2$ and negative if $p>2$

The correct answer is:
always positive

Solution:
Consider any equilateral triangle:
$c=b=a=1$ unit
Take value of $p$ between $1 \& 2$ i.e., $\frac{3}{2}$
$=1+1-1=1>0$
Take value of $\mathrm{p}$ greater than 2 i.e; $3 .$
$\therefore$ By considering all the options carefully; we came to
a conclusion that opton (b) is correct.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2016 (Phase 1)