datavorous/entrance-exam-dataset huggingface dataset:
For any two vectors $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ consider the following
statement:
$1.$ $\quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}| \Leftrightarrow \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ are orthogonal.
$2.$ $\quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}|+|\overrightarrow{\mathbf{b}}| \Leftrightarrow \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ are orthogonal.
$3.$ $\quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2} \Leftrightarrow \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ are orthogonal.
Which of the above statements is/are correct?

Options:
A 1 and 2 only
B 1 and 3 only
C 2 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
(i) $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$
Squaring both the sides
$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}$
or, $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \vec{b}$
or, $4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \vec{a}$ and $\vec{b}$ are orthogonal.
(ii) $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}|$
Squaring both the sides
$|\vec{a}+\vec{b}|=(|\vec{a}|+|\vec{b}|)^{2}$
or, $|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|$
or, $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}|$
or, $|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cdot \cos \theta=|\overrightarrow{\mathrm{a}} \| \overrightarrow{\mathrm{b}}|$
$\cos \theta=1 \Rightarrow \theta=0$
$\vec{a}$ and $\vec{b}$ are parallel, and not orthogonal.
(iii) $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}$
$|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}$
$2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \vec{a}$ and $\overrightarrow{\mathrm{b}}$ are orthogonal. Statements 1 and 3 are correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
For any vector $\overrightarrow{\mathrm{a}}|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}|^{2}+|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}}|^{2}+|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^{2}$ is equal to

Options:
A $|\vec{a}|^{2}$
B $2|\overrightarrow{\mathrm{a}}|^{2}$
C $3|\overrightarrow{\mathrm{a}}|^{2}$
D $4|\overrightarrow{\mathrm{a}}|^{2}$

The correct answer is:
$2|\overrightarrow{\mathrm{a}}|^{2}$

Solution:
Let $\overrightarrow{\mathrm{a}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
$|\vec{a} \times \hat{i}|^{2}=y^{2}+z^{2}$
$|\vec{a} \times \hat{j}|^{2}=x^{2}+z^{2}$
$|\vec{a} \times \hat{k}|^{2}=x^{2}+y^{2}$
$\therefore|\vec{a} \times \hat{i}|^{2}+|\vec{a} \times j|^{2}+|\vec{a} \times \hat{k}|^{2}$
$=y^{2}+z^{2}+x^{2}+z^{2}+x^{2}+y^{2}$
$=2 x^{2}+2 y^{2}+2 z^{2}$
$=2\left(x^{2}+y^{2}+z^{2}\right)$
$=2|\vec{a}|^{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
For any vector $\vec{\alpha}$, what is $(\vec{\alpha} \cdot \hat{i}) \hat{i}+(\vec{\alpha} \cdot \hat{j}) \hat{j}+(\vec{\alpha} \cdot \hat{k}) \hat{k}$ equal
to ?

Options:
A $\vec{\alpha}$
B $3 \vec{\alpha}$
C $-\vec{\alpha}$
D $\overrightarrow{0}$

The correct answer is:
$\vec{\alpha}$

Solution:
Let $\vec{\alpha}=\vec{a} \hat{i}+\vec{b} \hat{j}+\vec{c} \hat{k}$
Now, $\vec{\alpha} \cdot \hat{\mathrm{i}}=(\overrightarrow{\mathrm{a}} \hat{\mathrm{i}}+\overrightarrow{\mathrm{b}} \hat{\mathrm{j}}+\overrightarrow{\mathrm{c}} \hat{\mathrm{k}}) \cdot \hat{\mathrm{i}}=\overrightarrow{\mathrm{a}}$
$\vec{\alpha} \cdot \hat{j}=(\vec{a} \hat{i}+\vec{b} \hat{j}+\vec{c} \hat{k}) \cdot \hat{j}=\vec{b}$
$\vec{\alpha} \cdot \hat{\mathrm{k}}=(\overrightarrow{\mathrm{a}} \hat{\mathrm{i}}+\overrightarrow{\mathrm{b}} \hat{\mathrm{j}}+\overrightarrow{\mathrm{c}} \hat{\mathrm{k}}) \cdot \hat{\mathrm{k}}=\overrightarrow{\mathrm{c}}$
Now, $\vec{a} \hat{i}+b \hat{j}+c \hat{k}=\vec{\alpha}$
Thus, Required expression $=\vec{\alpha}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 1)
For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$, suppose the points $\left(x_1, \frac{1}{x_1}\right),\left(x_2, \frac{1}{x_2}\right),\left(x_3, \frac{1}{x_3}\right)$ and $\left(x_4, \frac{1}{x_4}\right)$ lie on the boundary of a circle of radius 4 . Then, the value of $x_1 x_2 x_3 x_4$ is

Options:
A $1$
B $2$
C $4$
D $\frac{1}{4}$

The correct answer is:
$1$

Solution:
Let the equation of the circle be $x^2+y^2+2 g x+2 f y+c=0$
$\left(x_i, \frac{1}{x_i}\right)$ lies on the above circle.
$\therefore \quad x_i^2+\frac{1}{x_i^2}+2 g x_i+\frac{2 f}{x_i}+C=0$
$\begin{aligned} & \Rightarrow \quad x_i^4+2 g x_i^3+c x_i^2+2 f x_i+1=0 \\ & \therefore \quad x_1 x_2 x_3 x_4=1\end{aligned}$

Topic: Mathematics, Circle, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 1)
If $(1+\tan \theta)(1+\operatorname{tan\phi})=2$, then what is $(\theta+\phi)$ equal to?

Options:
A 30^{\circ}
B 45^{\circ}
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
45^{\circ}

Solution:
Given, $(1+\tan \theta)(1+\tan \phi)=2$
$\Rightarrow 1+\tan \theta+\tan \phi+\tan \theta \tan \phi=2$
$\Rightarrow \tan \theta+\tan \phi=1-\tan \theta \tan \phi$
$\Rightarrow \frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=1$
$\Rightarrow \tan (\theta+\phi)=\tan 45^{\circ}$
$\Rightarrow \quad \theta+\phi=45^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 1)
If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a$, then what is $\frac{d y}{d x}$ equal to?

Options:
A $\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}$
B $\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
C $\sqrt{\frac{1-x^{2}}{1-y^{2}}}$
D None of these

The correct answer is:
None of these

Solution:
Let $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a$
On differentiating w.r.t. $x$, we get
$\frac{1}{2 \sqrt{1-x^{2}}}(-2 x)+\frac{1(-2 y)}{2 \sqrt{1-y^{2}}} \frac{d y}{d x}=0$
$\Rightarrow \frac{-x}{\sqrt{1-x^{2}}}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0$
$\Rightarrow \frac{-x}{\sqrt{1-x^{2}}}=\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=-\frac{x}{y} \sqrt{\frac{1-y^{2}}{1-x^{2}}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
If $1, x, y, z, 16$ are in geometric progression, then what is the value of $x+y+z$ ?

Options:
A 8
B 12
C 14
D 16

The correct answer is:
14

Solution:
As given $1, \mathrm{x}, \mathrm{y}, \mathrm{z} 16$ are in geometric progression. Let common ratio be $\mathrm{r}$, $\mathrm{x}=1 . \mathrm{r}=\mathrm{r}$
$\mathrm{y}=1 \cdot \mathrm{r}^{2}=\mathrm{r}^{2}$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=\mathrm{r}^{3}$
and $16=1 \cdot r^{4} \Rightarrow 16=r^{4}$
$\Rightarrow \mathrm{r}=2$
$\therefore \quad \mathrm{x}=1 . \mathrm{r}=2, \mathrm{y}=1 \cdot \mathrm{r}^{2}=4$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=8$
$\therefore \quad x+y+z=2+4+8=14$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
If $\left|\begin{array}{ccc}2 & 4 & 0 \\ 0 & 5 & 16 \\ 0 & 0 & 1+p\end{array}\right|=20$, then what is the value of $p$?

Options:
A 0
B 1
C 2
D 5

The correct answer is:
1

Solution:
$\left|\begin{array}{ccc}2 & 4 & 0 \\ 0 & 5 & 16 \\ 0 & 0 & 1+p\end{array}\right|=20$
On expanding along $\mathrm{C}_{1}$, $2\{5(1+p)-0\}=20$
$\Rightarrow 1+p=2$
$\Rightarrow p=1$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 2)
If $(2,0)$ is the vertex and the $y$ -axis is the directrix of a parabola, then where is its focus?

Options:
A $(0,0)$
B $(-2,0)$
C $(4,0)$
D $(-4,0)$

The correct answer is:
$(4,0)$

Solution:
Vertec is $(2,0)$. Since, $y$ -axis is the directrix of a parabola. Equation directrix is $x=0 .$ So, axis of parabolais $x$ -axis. Let the focus be $(\mathrm{a}, 0)$
Distance of the vertex of a parabola from directrix = its distance from focus
So, $\mathrm{OV}=\mathrm{VF} \Rightarrow(2-0)^{2}=(\mathrm{a}-2)^{2}$
$a^{2}=4 a b a=4$
$\Rightarrow$ Focus is $(4,0)$

Topic: Mathematics, Parabola, NDA, NDA 2006 (Phase 1)
If $\alpha=\frac{1+\mathrm{i} \sqrt{3}}{2}$, then what is the value of $1+\alpha^{8}+\alpha^{16}+$ $\alpha^{24}+\alpha^{32} ?$

Options:
A 0
B 1
C $\omega$
D $-\omega^{2}$

The correct answer is:
$-\omega^{2}$

Solution:
$\frac{-1+\mathrm{i} \sqrt{3}}{2}$ and $\frac{-1-\mathrm{i} \sqrt{3}}{2}$
are complex cube roots of unity $\omega$ and $\omega^{2}$
$\begin{aligned} & \alpha=\frac{1+\mathrm{i} \sqrt{3}}{2}=-\omega^{2} \\ \therefore \quad & 1+\alpha^{8}+\alpha^{16}+\alpha^{24}+\alpha^{32} \\=& 1+\omega^{16}+\omega^{32}+\omega^{48}+\omega^{64} \\=& 1+\omega+\omega^{2}+1+\omega=0+1+\omega=-\omega^{2} \end{aligned}$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 2)
If $\left(\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right) \times\left(\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ 17 & \lambda\end{array}\right)$, then what is $\lambda$ equal to?

Options:
A 7
B $-7$
C 9
D $-9$

The correct answer is:
$-7$

Solution:
$\left(\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right) \times\left(\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ 17 & \lambda\end{array}\right)$
$\left(\begin{array}{cc}1 & -1 \\ 17 & -7\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ 17 & \lambda\end{array}\right)$
$\lambda=-7$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 2)
If $\int_{-2}^{5} \mathrm{f}(\mathrm{x}) \mathrm{dx}=4$ and $\int_{0}^{5}\{1+\mathrm{f}(\mathrm{x})\} \mathrm{dx}=7$, then what is
$\int_{-2}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to?

Options:
A $-3$
B 2
C 3
D 5

The correct answer is:
2

Solution:
$\int_{-2}^{5} f(x) d x=4$ and $\int_{0}^{5}\{1+f(x)\} d x=7$
$\int_{0}^{5} f(x) d x=7-\int_{0}^{5} 1 d x=7-5=2$
$\int_{-2}^{0} f(x) d x=?$
$\int_{-2}^{0} f(x) d x+\int_{0}^{5} f(x) d x=4$
$\int_{-2}^{0} f(x) d x=4-\int_{0}^{5} f(x) d x$
$=4-2=2 . \quad[$ Using eqn. $(1)]$
$\int_{-2}^{0} f(x) d x=2$

Topic: Mathematics, Definite Integration, NDA, NDA 2016 (Phase 1)
If $2 A=\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right)$, then what is $A^{-1}$ equal to?

Options:
A $\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
B $\frac{1}{2}\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
C $\frac{1}{4}\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
D None of these

The correct answer is:
None of these

Solution:
Given, $2 \mathrm{~A}=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]$
$\mathrm{A}=\left(\begin{array}{cc}1 & \frac{1}{2} \\ \frac{3}{2} & 1\end{array}\right)$
$|\mathrm{A}|=\frac{1}{4}$
$\operatorname{adj} \mathrm{A}=\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)$
$\mathrm{A}^{-1}=4\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)=\left(\begin{array}{cc}4 & -2 \\ -6 & 4\end{array}\right)$

Topic: Mathematics, Matrices, NDA, NDA 2013 (Phase 2)
If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$; then which one of the following is correct?

Options:
A $2 \tan \beta+\tan \gamma=\tan \alpha$
B $\tan \beta+2 \tan \gamma=\tan \alpha$
C $\tan \beta+2 \tan \gamma=\tan \alpha$
D $2(\tan \beta+\tan \gamma)=\tan \alpha$

The correct answer is:
$\tan \beta+2 \tan \gamma=\tan \alpha$

Solution:
As given, $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$
$\Rightarrow \tan (\beta+\gamma)=\tan \alpha$
$\Rightarrow \frac{\tan \beta+\tan \gamma}{1-\tan \beta \tan \gamma}=\tan \alpha$
$\Rightarrow \tan \beta+\tan \gamma=\tan \alpha-\tan \alpha \tan \beta \tan \gamma$
$\Rightarrow \tan \beta+\tan \gamma=\tan \alpha-\tan \alpha \cot \alpha \tan \gamma$
$\left(\because \beta+\alpha=\frac{\pi}{2} \Rightarrow \beta=\pi / 2-\alpha \Rightarrow \tan (\pi / 2-\alpha)=\cot \alpha\right)$
$\Rightarrow \tan \beta+\tan \gamma=\tan \alpha-\tan \gamma$
$\Rightarrow \tan \beta+2 \tan \gamma=\tan \alpha$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
If $\overline{ }_{-2}$ is one of the roots of $a x^{2}+b x+c=0$, where $a$,
$b, c$ are real, then what are the values of $a, b, c$ respectively?

Options:
A 6,-4,1
B 4,6,-1
C $3,-2,1$
D $6,4,1$

The correct answer is:
6,-4,1

Solution:
Given quadratic equation is $a x^{2}+b x+c=0$ whose
one root is $\frac{1}{2-\sqrt{-2}}$
Consider $\frac{1}{2-\sqrt{-2}}=\frac{1}{2-\sqrt{2} i} \times \frac{2+\sqrt{2} i}{2+\sqrt{2} i}$
$=\frac{2+\sqrt{2} i}{4+2}=\frac{2+\sqrt{2} i}{6}$
$\therefore$ Another root will be $\frac{2-\sqrt{2} i}{6}$
$(\because$ complex roots always occurs in pairs)
Thus, sum of roots $=\frac{2+\sqrt{2} i}{6}+\frac{2-2 \sqrt{2} i}{6}=\frac{4}{6}$
and product of roots $=\left(\frac{2+\sqrt{2} i}{6}\right)\left(\frac{2-\sqrt{2} i}{6}\right)$
$=\frac{4+2}{36}=\frac{1}{6}$
$\therefore$ Requiredequation is $x^{2}-($ sum of roots $) x+($ product of roots $)=0$
$x^{2}-\frac{4}{6} x+\frac{1}{6}=0$
$\Rightarrow 6 x^{2}-4 x+1=0$
Thus, the values of $a, b, c$ are $6,-4,1$ respectively

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If $2 p+3 q=18$ and $4 p^{2}+4 p q-3 q^{2}-36=0$, then what $(2 p+q)$ equal to?

Options:
A 6
B 7
C 10
D 20

The correct answer is:
10

Solution:
$4 p^{2}+4 p q-3 q^{2}-36=0$
$\Rightarrow\left(4 p^{2}+4 p q+q^{2}\right)-4 q^{2}-36=0$
$(2 p+q)^{2}=4 q^{2}+36$
$2 p+3 q=18$
$p=\frac{18-3 q}{2}$
$\Rightarrow\left[\frac{18-3 \mathrm{q}}{2} \times 2+\mathrm{q}\right]^{2}=4 \mathrm{q}^{2}+36$
$\Rightarrow(18-3 \mathrm{q}+\mathrm{q})^{2}$
$\begin{aligned} &(18-2 q)^{2}=4 q^{2}+36 \\ & 324+4 q^{2}-72 q=4 q^{2}+36 \\ 72 q=& 324-36=288 \end{aligned}$
$\mathrm{q}=\frac{288}{72}=4$
Putting the value of $\mathrm{q}$ in $2 \mathrm{p}+3 \mathrm{q}=18$ $2 p+3 \times 4=18$
$2 p=18-12$
$\mathrm{p}=\frac{6}{2}=3$
$(2 p+q)=2 \times 3+4=10$
Option (c) is correct

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 1)
If $(0.2)^{x}=2$ and $\log _{10} 2=0.3010$, then what is the value of x to the nearest tenth?

Options:
A $-10.0$
B $-0.5$
C $-0.4$
D $-0.2$

The correct answer is:
$-0.4$

Solution:
$(0.2)^{x}=2$
Taking log on both sides,
$x \log _{10} \frac{2}{10}=\log _{10} 2$
$x\left[\log _{10} 2-\log _{10} 10\right]=\log _{10} 2$
$x[0.3010-1]=0.3010$
$x=-\frac{0.3010}{0.6990}=-0.43$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
If $2 \mathrm{x}^{3}-3 \mathrm{y}^{2}=7$, what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to $(\mathrm{y} \neq 0)?$

Options:
A $\frac{\mathrm{x}^{2}}{2 \mathrm{y}}$
B $\frac{\mathrm{x}}{2 \mathrm{y}}$
C $\frac{\mathrm{x}^{2}}{\mathrm{y}}$
D None of the above

The correct answer is:
$\frac{\mathrm{x}^{2}}{\mathrm{y}}$

Solution:
Let $2 x^{3}-3 y^{2}=7$
Differentiate both side, w.r.t. 'x'
$6 x^{2}-6 y \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{x^{2}}{y}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
If $2 x=3+5 i$, then what is the value of $2 x^{3}+2 x^{2}-7 x+72$ ?

Options:
A 4
B -4
C 8
D -8

The correct answer is:
4

Solution:
Given 2 x=3+5i
$\Rightarrow \quad x=\frac{3+5 i}{2}$
Consider $x^{3}=\frac{27+125 i^{3}+225 i^{2}+135 i}{8}$
$=\frac{27-125 i-225+135 i}{8} \quad\left(\begin{array}{r}\because i^{2}=-1 \\ i^{3}=-i\end{array}\right)$
$=\frac{-198+10 i}{8}=\frac{-99+5 i}{4}$
and $x^{2}=\frac{9+25 i^{2}+30 i}{4}$
$=\frac{9-25+30 i}{4}=\frac{-8+15 i}{2}$
Now, Consider $2 x^{3}+2 x^{2}-7 x+72$
$=\left(\frac{-99+5 i}{2}\right)+(-8+15 i)-\frac{7(3+5 i)}{2}+72$
$=-\frac{99}{2}+\frac{5 i}{2}-8+15 i-\frac{21}{2}-\frac{35}{2} i+72$
$=\left(-\frac{99}{2}-8-\frac{21}{2}+72\right)+\left(\frac{5}{2}+15-\frac{35}{2}\right) i$
$=\frac{-99-16-21+144}{2}=\frac{8}{2}=4$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 1)
If $2^{x}=3^{y}=12^{2}$, then what is $(x+2 y) /(x y)$ equal to ?

Options:
A $\mathrm{z}$
B $\frac{1}{z}$
C $2 \underline{2}$
D $\frac{\mathrm{z}}{2}$

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

Solution:
Given that $2^{\mathrm{x}}=3^{\mathrm{y}}=12^{\mathrm{z}}=\mathrm{k}$
Taking $\log _{2}$ on both the sides $x=\log _{2} k, y=\log _{3} k$ and $z=\log _{12} k$
$\frac{x+2 y}{x y}=\frac{\log _{2} k+2 \log _{3} k}{\log _{2} k \log _{3} k}$
$=\frac{1}{\log _{3} \mathrm{k}}+\frac{2}{\log _{2} \mathrm{k}}$
$=\log _{\mathrm{k}} 3+2 \log _{\mathrm{k}} 2=\log _{\mathrm{k}} 3+\log _{\mathrm{k}} 4$
$=\log _{\mathrm{k}} 12=\frac{1}{\log _{12} \mathrm{k}}=\frac{1}{\mathrm{z}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
If $2^{x}+3^{y}=17$ and $2^{x+2}-3^{y+1}=5$, then what is the value of $x ?$

Options:
A 3
B 2
C 1
D 0

The correct answer is:
3

Solution:
$A=-3$ and $B=8$
Given, $2^{x}+3 y=17$
and $2^{x+2}-3^{y+1}=5$
$\Rightarrow 4.2^{x}-3.3^{y}=5$
From equation (i) and (ii), we get $2^{x}=8$ and $3 y=9$
$\Rightarrow x=3$ and $y=2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
If \( 2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right] \) then the value of \( x \) and \( y \) are

Options:
A \( x=3, y=3 \)
B \( x=-3, y=3 \)
C \( x=3, y=-3 \)
D \( x=-3, y=-3 \)

The correct answer is:
\( x=3, y=3 \)

Solution:
Given that,
$2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
$\left[\begin{array}{cc}2+y & 6 \\ 1 & 2 x+2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
Equating first elements, we have
$2+y=5 \Rightarrow y=3$
Equating last elements, we have
$2 x+2=8 \Rightarrow x=3$
Therefore, $x=3, y=3$.

Topic: Mathematics, Determinants, KCET, KCET 2017
If $\int_{-3}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{7}{3}$ and $\int_{-3}^{9} \mathrm{f}(\mathrm{x}) \mathrm{dx}=-\frac{5}{6}$, then what is the
value of $\int_{2}^{9} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ ?

Options:
A $-\frac{19}{6}$
B $\frac{19}{6}$
C $\frac{3}{2}$
D $-\frac{3}{2}$

The correct answer is:
$-\frac{19}{6}$

Solution:
Value of the integral $\int_{2}^{9} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$
$=\int_{-3}^{9} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}-\int_{-3}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
Given, $\int_{-3}^{9} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{-5}{6}$ and $\int_{-3}^{2} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{7}{3}$
Putting these values in equation (i)
$\int_{2}^{9} f(x) d x=\frac{-5}{6}-\frac{7}{3}=-\frac{19}{6}$

Topic: Mathematics, Definite Integration, NDA, NDA 2007 (Phase 1)
If $(3 \vec{a}-\vec{b}) \times(\vec{a}+3 \vec{b})=k \vec{a} \times \vec{b}$ then what is the value of $k$ ?

Options:
A 10
B 5
C 8
D $-8$

The correct answer is:
10

Solution:
$\quad \begin{aligned} (3 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}+3 \overrightarrow{\mathbf{b}}) \\ &=(3 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{a}}+(3 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times 3 \overrightarrow{\mathbf{b}} \\ &=3 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+3 \overrightarrow{\mathbf{a}} \times 3 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{b}} \times 3 \overrightarrow{\mathbf{b}} \\ &=0-(-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+9 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-0 \\ &=10 \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \\ & \therefore k=10 \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
If \( 3 A+4 B^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right] \) and \( 2 B-3 A^{\prime}=\left[\begin{array}{cc}-1 & 18 \\ 4 & 0 \\ 5 & -7\end{array}\right] \) then \( B= \)

Options:
A \( \left[\begin{array}{cc}1 & -3 \\ -1 & 1 \\ 2 & 4\end{array}\right] \)
B \( \left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & -4\end{array}\right] \)
C \( \left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right] \)
D \( \left[\begin{array}{cc}-1 & -18 \\ 4 & -16 \\ -5 & -7\end{array}\right] \)

The correct answer is:
\( \left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right] \)

Solution:
(C)
\( 3 A+4 B^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right] \ldots .(1) \) \( \left(2 B-3 A^{\prime}\right)^{\prime}=(2 B)^{\prime}-\left(3 A^{\prime}\right)^{\prime}=2 B^{\prime}-3 A \) \( \Rightarrow 2 B^{\prime}-3 A=\left[\begin{array}{ccc}-1 & 4 & -5 \\ 18 & 0 & -7\end{array}\right] \ldots .(2) \)
$(1)$
(2)
Adding (1) and (2) we get, \( 6 \mathrm{~B}^{\prime}=\left[\begin{array}{ccc}6 & -6 & 12 \\ 18 & 6 & 24\end{array}\right] \)
\[
\begin{array}{l}
\Rightarrow B^{\prime}=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 1 & 4
\end{array}\right] \\
\therefore B=\left[\begin{array}{ccc}
1 & 3 & \\
-1 & 1 & \\
2 & 4 &
\end{array}\right]
\end{array}
\]

Topic: Mathematics, Matrices, KCET, KCET 2019
If 3 is the root of the equation $x^{2}-8 x+k=0$, then what is the val ue of $k$ ?

Options:
A $-15$
B 9
C 15
D 24

The correct answer is:
15

Solution:
Since, 3 is the root of the equation $x^{2}-8 x+k=0$ 3 satisfies the equation $x^{2}-8 x+k=0$ $9-24+k=0 \Rightarrow k=1 \mathrm{k}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
If $3 \tan \theta+4=0$, where $(\pi / 2) < \theta < \pi$, then what is the value of $2 \cot \theta-5 \cos \theta+\sin \theta ?$

Options:
A $-\frac{53}{10}$
B $\frac{7}{10}$
C $\frac{23}{10}$
D $\frac{37}{10}$

The correct answer is:
$\frac{23}{10}$

Solution:
As given, $3 \tan \theta+4=0 \Rightarrow \tan \theta=-\frac{4}{3}$
$\left[\theta\right.$ lies in second quadrant i.e., $\left.\frac{\pi}{2} < \theta < \pi\right]$
$\therefore \quad \cot \theta=-\frac{3}{4} \Rightarrow \cos \theta=-\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$
Now, $2 \cot \theta-5 \cos \theta+\sin \theta$
$=-\frac{6}{4}+\frac{15}{5}+\frac{4}{5}=\frac{-30+60+16}{20}

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
If $3^{(x-1)}+3^{(x+1)}=30$, then what is the value of $3^{(x+2)}+3^{x}$ ?

Options:
A 30
B 60
C 81
D 90

The correct answer is:
90

Solution:
Given: $3^{(x-1)}+3^{(x+1)}=30$
$\Rightarrow \frac{3^{\mathrm{x}}}{3}+3.3^{\mathrm{x}}=30$
Multiplying both the sides by 3 in equation (i) $3^{x}+3^{2} .3^{x}=90$
$\Rightarrow 3^{\mathrm{x}}+3^{\mathrm{x}+2}=90$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 1)
If $3^{\mathrm{x}}+3^{\mathrm{y}}=3^{\mathrm{x}+\mathrm{y}}$ then what is $\frac{d y}{d x}$ equal to?

Options:
A $\frac{3^{x+y}-3^{x}}{3^{y}}$
B $\frac{3^{x-y}\left(3^{y}-1\right)}{1-3^{x}}$
C $\frac{3^{x}+3^{y}}{3^{x}-3^{y}} \quad$
D $\frac{3^{x}+3^{y}}{1+3^{x+y}}$

The correct answer is:
$\frac{3^{x-y}\left(3^{y}-1\right)}{1-3^{x}}$

Solution:
$3^{x}+3^{y}=3^{x+y}$
On differentiating w.r.t. $x$, we get
$3^{x} \log 3+3^{y} \log 3 \frac{d y}{d x}=3^{x+y} \log 3\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \log 3\left[3^{x}+3^{y} \frac{d y}{d x}\right]=\log 3 \cdot 3^{x+y}\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \frac{d y}{d x}\left(-3^{x+y}+3^{y}\right)=3^{x+y}-3^{x}$
$=3^{x} \cdot 3^{y}-3^{x}=3^{x}\left(3^{y}-1\right)$
$\Rightarrow \frac{d y}{d x}=\frac{3^{x}\left(3^{y}-1\right)}{3^{y}\left(1-3^{x}\right)}=\frac{3^{x-y}\left(3^{y}-1\right)}{\left(1-3^{x}\right)}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)