datavorous/entrance-exam-dataset huggingface dataset:
If $\alpha$ and $\beta$ are positive angles such that $\alpha+\beta=\frac{\pi}{4}$, then what is $(1+\tan \alpha)(1+\tan \beta)$ equal to?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
Let $\alpha+\beta=\frac{\pi}{4}$
$\Rightarrow \quad \tan (\alpha+\beta)=\tan \frac{\pi}{4}$
$\Rightarrow \quad \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=1$
$\Rightarrow \quad \tan \alpha+\tan \beta=1-\tan \alpha \tan \beta$
By adding 1 on both sides, we get $1+\tan \alpha+\tan \beta+\tan \alpha \tan \beta=2$
$\Rightarrow \quad(1+\tan \alpha)(1+\tan \beta)=2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
If $\alpha$ and $\beta$ are such that $\tan \alpha=2 \tan \beta$, then what is $\sin (\alpha+\beta)$ equal to ?

Options:
A 1
B $2 \sin (\alpha-\beta)$
C $\sin (\alpha-\beta)$
D $3 \sin (\alpha-\beta)

The correct answer is:
$3 \sin (\alpha-\beta)

Solution:
As given :
$\tan \alpha=2 \tan \beta$
$\Rightarrow \frac{\tan \alpha}{\tan \beta}=2 \Rightarrow \frac{\sin \alpha / \cos \alpha}{\sin \beta / \cos \beta}=2$
$\Rightarrow \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=2$
Using componendo and dividendo we get
$\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{2+1}{2-1}=3$
$\Rightarrow \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$
$\sin (\alpha+\beta)=3 \sin (\alpha-\beta)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
If $\alpha, \beta$ and $\gamma$ are the angles which the vector $\overrightarrow{\mathrm{OP}}$ (O being the origin) makes with positive direction of the coordinate axes, then which of the following are correct?
$1.$ $\quad \cos ^{2} \alpha+\cos ^{2} \beta=\sin ^{2} \gamma$
$2.$ $\quad \sin ^{2} \alpha+\sin ^{2} \beta=\cos ^{2} \gamma$
$3.$ $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2$
Select the correct answer using the code given below.

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

The correct answer is:
1 and 3 only

Solution:
We know, $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta=1-\cos ^{2} \gamma=\sin ^{2} \gamma$
$\therefore$ Statement 1 is correct.
Now, $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow 1-\sin ^{2} \alpha+1-\sin ^{2} \beta+1-\sin ^{2} \gamma=1$
$\Rightarrow 3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1 \Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$
$=2 .$
$\therefore$ Statement 3 is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
If $\alpha$ and $\beta$ are the complex cube roots of unity, then what is the value of $(1+\alpha)(1+\beta)\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right) ?$

Options:
A -1
B 0
C 1
D 4

The correct answer is:
1

Solution:
Since, $\alpha$ and $\beta$ are the complex cube roots of unity therefore $1+\alpha+\alpha^{2}=0=1+\beta+\beta^{2}$
and $\alpha^{3}=1=\beta^{3}$.
Consider $(1+\alpha)(1+\beta)\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right)$
$=(1+\alpha)\left(1+\alpha^{2}\right)(1+\beta)\left(1+\beta^{2}\right)$
$=\left(1+\alpha^{2}+\alpha+\alpha^{3}\right)\left(1+\beta^{2}+\beta+\beta^{3}\right)$
$=\left(0+\alpha^{3}\right)\left(0+\beta^{3}\right)$
$=\left(\alpha^{3}\right)\left(\beta^{3}\right)=(1)(1)=1$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
If $\alpha$ and $\beta$ are the roots of the equation $1+\mathrm{x}+\mathrm{x}^{2}=0$, then
the matrix product $\left[\begin{array}{cc}1 & \beta \\ \alpha & \alpha\end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ 1 & \beta\end{array}\right]$ is equal to

Options:
A $\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]$
B $\left[\begin{array}{cc}-1 & -1 \\ -1 & 2\end{array}\right]$
C $\left[\begin{array}{lr}1 & -1 \\ -1 & 2\end{array}\right]$
D $\left[\begin{array}{ll}-1 & -1 \\ -1 & -2\end{array}\right]$

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

Solution:
$\alpha, \beta$ are roots of the equation $1+x+x^{2}=0$. $1+x+x^{2}=0 \Rightarrow x^{2}+x+1=0$
Solving for $x, x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2(1)}$
$=\frac{-1 \pm \sqrt{-3}}{2}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2}$
$\therefore$ roots are $\frac{-1 \pm \sqrt{3} \mathrm{i}}{2}$ and $\frac{-1-\sqrt{3} \mathrm{i}}{2}$
i.e., $\alpha=\mathrm{a}, \beta=\omega^{2}$
$\left[\begin{array}{ll}1 & \beta \\ \alpha & \alpha\end{array}\right]\left[\begin{array}{ll}\alpha & \beta \\ 1 & \beta\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & \beta+\beta^{2} \\ \alpha^{2}+\alpha & \alpha \beta+\alpha \beta\end{array}\right]$
$=\left[\begin{array}{ll}\omega+\omega^{2} & \omega^{2}+\omega^{4} \\ \omega^{2}+\omega & \omega^{3}+\omega^{3}\end{array}\right]$
$=\left[\begin{array}{cc}-1 & \omega^{2}+\omega \\ -1 & 2 \omega^{3}\end{array}\right]$
$=\left[\begin{array}{cc}-1 & -1 \\ -1 & 2\end{array}\right]$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 2)
If $\alpha$ and $\beta$ are the roots of the equation $4 x^{2}+3 x+7=0$, then what is the value of $\left(\alpha^{-2}+\beta^{-2}\right)$ ?

Options:
A 47 / 49
B 49 / 47
C $-47 / 49$
D $-49 / 47$

The correct answer is:
$-47 / 49$

Solution:
Let $\alpha$ and $\beta$ be the roots of the equation. $4 x^{2}+3 x+7=0$
$\therefore \quad$ Sum $=\alpha+\beta=-\frac{3}{4}$ and Product $=\alpha \beta=\frac{7}{4}$
Consider, $\alpha^{2}+\beta^{-2}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}}$
$=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{\frac{9}{16}-\frac{7}{2}}{\frac{49}{16}}$
$=\frac{\frac{9-56}{16}}{\frac{49}{16}}=\frac{-47}{16} \times \frac{16}{49}=-\frac{47}{49}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2 x+4=0$, then what is the value of $\alpha^{3}+\beta^{3} ?$

Options:
A 16
B $-16$
C 8
D -8

The correct answer is:
$-16$

Solution:
Let $\alpha$ and $\beta$ are the roots of $x^{2}-2 x+4=0$ sum of roots $=\alpha+\beta=2$, product $=\alpha \beta=4$
Now, $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)=2^{3}-3 \times 4 \times 2$
$=8-24=-16$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 2)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+6 x+1=0$, then what is $|\alpha-\beta|$ equal to ?

Options:
A 6
B $3 \sqrt{2}$
C $4 \sqrt{2}$
D 12

The correct answer is:
$4 \sqrt{2}$

Solution:
$\alpha$ and $\beta$ are the roots ofthe equation $x^{2}+6 x+1=0$
Now, $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$
$=(-6)^{2}-4=36-4=32$
$\Rightarrow|\alpha-\beta|=\sqrt{32}=4 \sqrt{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+b x+c=0$, then what is the value of $\alpha^{-1}+\beta^{-1} ?$

Options:
A $-\frac{b}{c}$
B $\frac{\mathrm{b}}{\mathrm{c}}$
C $\frac{\mathrm{c}}{\mathrm{b}}$
D $-\frac{\mathrm{c}}{\mathrm{b}}$

The correct answer is:
$-\frac{b}{c}$

Solution:
Consider $\alpha^{-1}+\beta^{-1}=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}$
Equation is $x^{2}+b x+c=0$
Now, sum of roots $=\alpha+\beta=-b$
and product of roots $=\alpha \cdot \beta=\mathrm{c}$
$\frac{\alpha+\beta}{\alpha \beta}=-\frac{\mathrm{b}}{\mathrm{c}}$
Hence, $\alpha^{-1}+\beta^{-1}=-\frac{\mathrm{b}}{\mathrm{c}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 1)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-q(1+x)-r=0$, then what is $(1+\alpha)(1+\beta)$ equal to?

Options:
A $1-\mathrm{r}$
B $\mathrm{q}-\mathrm{r}$
C $1+r$
D $\mathrm{q}+r$

The correct answer is:
$1-\mathrm{r}$

Solution:
Givenequation is $x^{2}-q(1+x)-r=0$
$\Rightarrow x^{2}-q x+(-q-r)=0$
Now, $\alpha+\beta=q$
$\alpha \beta=-q-r$
Consider $(1+\alpha)(1+\beta)$ $+\alpha+\beta+\alpha \beta=1+q-q-$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+x+1=0$, then what is the equation whose roots are $\alpha^{19}$ and $\beta^{7}$ ?

Options:
A $x^{2}-x-1=0$
B $x^{2}-x+1=0$
C $x^{2}+x-1=0$
D $x^{2}+x+1=0$

The correct answer is:
$x^{2}+x+1=0$

Solution:
If $\alpha$ and $\beta$ are the roots of the equations $x^{2}+x+1=0$
$\Rightarrow \alpha=\omega$ and $\beta=\omega^{2}$
or, $\alpha=\omega^{2}$ and $\beta=\omega$
$\alpha^{19}+\beta^{7}=\omega^{19}+\omega^{14}=\omega+\omega^{2}=-1$
$\alpha^{19}+\beta^{7}=\omega^{38}+\omega^{7}=\omega^{2}+\omega=-1$
$+\beta^{7}=-1$
and $-\beta^{7}=\omega^{19} \cdot \omega^{14}=\omega^{33}=1$
or $\omega^{38} \cdot \omega^{7}=\omega^{45}=$
and $\beta$ The is
$x^{2}-\left(\alpha^{19}+\beta^{7}\right) x+\alpha^{19} \beta^{7}=0 \Rightarrow \quad x^{2}+x+1$
$=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+x+1=0$. then which of the following are the roots of the equation $x^{2}-x+1=0 ?$

Options:
A $\alpha^{7}$ and $\beta^{13}$
B $a^{13}$ and $\beta^{7}$
C $\alpha^{20}$ and $\beta^{20}$
D None of these

The correct answer is:
None of these

Solution:
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0$ $\begin{array}{ll}\therefore & \alpha+\beta=-1 \text { and } \alpha \beta=1 \\ \Rightarrow & \alpha=\omega \text { and } \beta=\omega^{2} \\ (\because & \left.\omega^{3}=1,1+\omega+\omega^{2}=0\right)\end{array}$
option a, $b, c$ does not sat isfies the eqn $x^{2}-x+1=0$ Hence, option (d) is correct.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 2)
If $\alpha$ and $\beta$ are the roots of $x^{2}+4 x+6=0$, then what is the
value of $\alpha^{3}+\beta^{3} ?$

Options:
A $-2 / 3$
B $2 / 3$
C 4
D 8

The correct answer is:
8

Solution:
$\alpha$ and $\beta$ are theroots of $x^{2}+4 x+6=0$ $\therefore \alpha+\beta=-4$ and $\alpha \beta=6$
Now, $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$=(-4)^{3}-3 \times 6(-4)=-64+72=8$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 2)
If $\alpha$ and $\beta$ are the roots ofthe equation $a x^{2}+b x+c=0$, then what are the roots of the equation $c x^{2}+b x+a=0 ?$

Options:
A $\beta, \frac{1}{\alpha}$
B $\alpha, \frac{1}{\beta}$
C $-\alpha,-\beta$
D $\frac{1}{\alpha}, \frac{1}{\beta}$

The correct answer is:
$\frac{1}{\alpha}, \frac{1}{\beta}$

Solution:
As given, $\alpha$, and $\beta$ are the roots of $a x^{2}+b x+c=0$, then the roots of $\mathrm{cx}^{2}+\mathrm{bx}+\mathrm{a}=0$ will be reciprocal of $\alpha$ and $\beta$, i.e., $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $\alpha$ and $\beta$ are the roots ofthe equation $a x^{2}+b x+c=0$, where $a \neq 0$, then $(a \alpha+b)(a \beta+b)$ is equal to:

Options:
A $a b$
B $b c$
C $c$
D $a b c$

The correct answer is:
$c$

Solution:
Given equation $a x^{2}+b x+c=0 \quad$ (where $a \neq 0$ ) $\alpha$ and $\beta$ are roots of given equation. $(a \alpha+b)(a \beta+b)=a^{2} \alpha \beta+a b \alpha+a b \beta+b^{2}$
$=a^{2} \alpha \beta+a b(\alpha+\beta)+b^{2}$
From the given quadratic equation
$\alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$a^{2} \times \frac{c}{a}+a b \times-\frac{b}{a}+b^{2}=a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2014 (Phase 1)
If $\alpha$ and $\beta$ are theroots of the equation $3 x^{2}+2 x+1=0$, then the equation whose roots are $\alpha+\beta^{-1}$ and $\beta+\alpha^{-1}$ is

Options:
A 3 x^{2}+8 x+16=0
B 3 x^{2}-8 x-16=0
C $3 x^{2}+8 x-16=0$
D $x^{2}+8 x+16=0$

The correct answer is:
3 x^{2}+8 x+16=0

Solution:
$3 \mathrm{x}^{2}+2 \mathrm{x}+1=0$.
Sum of the roots $=\alpha+\beta=\frac{-2}{3}$...(1)
Product of the roots $=\alpha \cdot \beta=\frac{1}{3}$...(2)
We have to find the equation with the roots $\alpha+\beta^{-1}$ and $\beta+\alpha^{-1} .$
Sum of the roots $(\mathrm{S})=\alpha+\beta^{-1}+\beta+\alpha^{-1}$
$=\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}$
$=(\alpha+\beta)+\left(\frac{\alpha+\beta}{\alpha \beta}\right)$
$=\frac{-2}{3}+\left(\frac{-2}{3}\right) \quad($ from $(1),(2))$
$=\frac{-2}{3}-2=\frac{-2-6}{3}=\frac{-8}{3}$
Product of the roots $(\mathrm{P})=\left(\alpha+\beta^{-1}\right)\left(\beta+\alpha^{-1}\right)$
$=\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$
$=\alpha \beta+1+1+\frac{1}{\alpha \beta}=\alpha \beta+2+\frac{1}{\alpha \beta}$
$=\frac{1}{3}+2+\frac{1}{\frac{1}{3}}$
$=\frac{1}{3}+2+3=\frac{1}{3}+5=\frac{16}{3}$.
so the required equation is $\mathrm{x}^{2}-\mathrm{s} . \mathrm{x},+\mathrm{P}=0$
$=x^{2}+\frac{8}{3} x+\frac{16}{3}=0 \Rightarrow 3 x^{2}+8 x+16=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 2)
If angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in $\mathrm{AP}$, then what is $\sin A+2 \sin B+\sin C$ equal to?

Options:
A $4 \sin B \cos ^{2}\left(\frac{A-C}{2}\right)$
B $4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$
C $4 \sin (2 B) \cos ^{2}\left(\frac{A-C}{2}\right)$
D $4 \sin (2 B) \cos ^{2}\left(\frac{A-C}{4}\right)$

The correct answer is:
$4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$

Solution:
Since, $A, B, C$ are in $A P$. $\therefore B-A=C-B$
$\Rightarrow 2 B=A+C$
But we know $A+B+C=180^{\circ}$
$\Rightarrow \quad 3 B=180^{\circ} \Rightarrow B=60^{\circ}$
Consider $\sin A+2 \sin B+\sin C$
$=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}+2 \sin B$
$=2 \sin B\left[\cos \frac{A-C}{2}+1\right] \quad(\because A+C=2 B)$
$=2 \sin B\left[2 \cos ^{2}\left(\frac{A-C}{4}\right)\right]$
$=4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2010 (Phase 1)
If $\alpha, \beta$ are real, what is $\left|\frac{\alpha+i \beta}{\beta+i \alpha}\right|$ equal to ?

Options:
A 0
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
1

Solution:
Let $z_{1}=\alpha+i \beta$ and $z_{2}=\beta+i \alpha$ Since, $\left|\frac{z_{1}}{z_{2}}\right|=\frac{\left|z_{1}\right|}{\left|z_{2}\right|}$
$\therefore\left|\frac{\alpha+\mathrm{i} \beta}{\beta+\mathrm{i} \alpha}\right|=\frac{|\alpha+\mathrm{i} \beta|}{|\beta+\mathrm{i} \alpha|}=\frac{\sqrt{\alpha^{2}+\beta^{2}}}{\sqrt{\alpha^{2}+\beta^{2}}}=1$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 1)
If $\alpha, \beta$ are the roots of $a x^{2}+b x+b=0$, then what is
$\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{b}}{\sqrt{a}}$ equal to?

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Let $\alpha$ and $\beta$ are the roots of $a x^{2}+b x+b=0$
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}$ and $\alpha \beta=\frac{\mathrm{b}}{\mathrm{a}}$
Consider, $\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}$
$=\frac{-b / a}{\sqrt{b / a}}+\sqrt{\frac{b}{a}}=-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}$ (by rationalizing)

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)
If $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0$ and $a+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$, then what is $h$ equal to ?

Options:
A $\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)$
B $\frac{1}{2}\left(-\frac{b}{a}+\frac{q}{p}\right)$
C $\frac{1}{2}\left(\frac{b}{p}+\frac{q}{a}\right)$
D $\frac{1}{2}\left(-\frac{b}{p}+\frac{q}{a}\right)$

The correct answer is:
$\frac{1}{2}\left(\frac{b}{a}-\frac{q}{p}\right)$

Solution:
$\because \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}$ and $\alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
Also, $\alpha+h+\beta+h=-\frac{q}{p}$
$\Rightarrow \alpha+\beta+2 h=-\frac{q}{p}$
$\Rightarrow 2 \mathrm{~h}=-\frac{\mathrm{q}}{\mathrm{p}}+\frac{\mathrm{b}}{\mathrm{a}}$
$\left(\because \alpha+\beta=-\frac{b}{a}\right)$
$\Rightarrow \mathrm{h}=\frac{1}{2}\left[\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{q}}{\mathrm{p}}\right]$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2014 (Phase 2)
If $\alpha, \beta$ are the roots of the equation $2 x^{2}-2\left(1+n^{2}\right) x+(1+$
$\left.n^{2}+n^{4}\right)=0$, then what is the value of $\alpha^{2}+\beta^{2} ?$

Options:
A $2 n^{2}$
B $2 n^{4}$
C 2
D $n^{2}$

The correct answer is:
$n^{2}$

Solution:
Since, $\alpha$ and $\beta$ be the roots of
$2 \mathrm{x}^{2}-2\left(1+n^{2}\right) x+\left(1+n^{2}+n^{4}\right)=0$
$\therefore \quad \alpha+\beta=-\left[\frac{-2\left(1+n^{2}\right)}{2}\right]=\left(n^{2}+1\right)$
and $\alpha \beta=\frac{1+n^{2}+n^{4}}{2}$
Now, Consider $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$=\left(n^{2}+1\right)^{2}-\left(1+n^{2}+n^{4}\right)$
$n^{4}+1+2 n^{2}-1-n^{2}-n^{4}=n^{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+b=0$, then
what is the value of $\sqrt{\frac{\alpha}{o}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\underline{b}} ?$

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

The correct answer is:
0

Solution:
$\alpha$ and $\beta$ are the roots of the given equation, then
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}$ and $\alpha \beta=\frac{\mathrm{b}}{\mathrm{a}}$
$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}$
$=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\alpha \beta}$
$=\frac{\alpha+\beta+\alpha \beta}{\sqrt{\alpha \beta}}=\frac{-\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{a}}}{\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}}=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then what is the value of $(a \alpha+b)^{-1}+(a \beta+b)^{-1} ?$

Options:
A $\mathrm{a} / (\mathrm{bc}$)
B bac
C $-\mathrm{b} /(\mathrm{ac})$
D $-a /(b c)$

The correct answer is:
bac

Solution:
Since, $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$, then
Sum of the roots, $\alpha+\beta=-\frac{b}{a}$ and Product of the roots,
$\alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
The expression, $(a \alpha+b)^{-1}+(a \beta+b)^{-1}$
$=\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{\alpha \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$
$=\frac{\mathrm{a}(\alpha+\beta)+2 \mathrm{~b}}{\mathrm{a}^{2} \alpha \beta+\mathrm{ab}(\alpha+\beta)+\mathrm{b}^{2}}$
$=\frac{a(-b / a)+2 b}{a^{2}(c / a)+a b(-b / a)+b^{2}}=\frac{-b+2 b}{a c-b^{2}+b^{2}}=\frac{b}{a c}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If $\alpha, \beta$ are the roots of the equation $\ell \mathrm{x}^{2}-\mathrm{mx}+\mathrm{m}=0$, $\ell \neq \mathrm{m}, \ell \neq 0$, then which one of the following statements is
correct?

Options:
A $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$
B $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{\mathrm{m}}{\ell}}=0$
C $\sqrt{\frac{\alpha+\beta}{\alpha \beta}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$
D The arithmetic mean of $\alpha$ and $\beta$ is the same as their geometric mean

The correct answer is:
$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$

Solution:
As given : $\alpha$ and $\beta$ are the roots of the quadratio equation $\ell \mathrm{x}^{2}-\mathrm{mx}+\mathrm{m}=0$
So, sum of roots,
$\alpha+\beta=\frac{\mathrm{m}}{\ell}$ and product of roots, $\alpha \beta=\frac{\mathrm{m}}{\ell}$
$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}=\frac{\mathrm{m} / \ell}{\sqrt{\mathrm{m} / \ell}}$
$\Rightarrow \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\sqrt{\frac{\mathrm{m}}{\ell}}=0 \Rightarrow \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}-\sqrt{\frac{\mathrm{m}}{\ell}}=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 2)
If $\alpha, \beta$ are the roots of the equation $x^{2}+x+2=0$, then
what is $\frac{\alpha^{10}+\beta^{10}}{\alpha^{-10}+\beta^{-10}}$ equal to?

Options:
A 4096
B 2048
C 1024
D 512

The correct answer is:
1024

Solution:
Here, $\alpha$ and $\beta$ are the roots of the equation $\alpha+\beta=-1$
$\alpha \beta=2$
$\frac{\alpha^{10}+\beta^{10}}{\alpha^{-10}+\beta^{-10}}=\frac{\alpha^{10}+\beta^{10}}{\frac{1}{\alpha^{10}}+\frac{1}{\beta^{10}}}$
$=\frac{\left(\alpha^{10}+\beta^{10}\right)(\alpha \beta)^{10}}{\left(\alpha^{10}+\beta^{10}\right)}=(\alpha \beta)^{10}$
$\therefore(\alpha \beta)^{10}=2^{10}=1024$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}-x+1=0$, then which one of the following is correct?

Options:
A $\left(\alpha^{4}-\beta^{4}\right)$ is real
B $2\left(\alpha^{6}+\beta^{5}\right)=(\alpha \beta)^{5}$
C $\left(\alpha^{6}-\beta^{6}\right)=0$
D $\left(\alpha^{8}+\beta^{8}\right)=(\alpha \beta)^{8}$

The correct answer is:
$\left(\alpha^{6}-\beta^{6}\right)=0$

Solution:
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-x+1=0$.
$\alpha+\beta=-(-1)=1$...(i)
$\alpha \beta=1$
Now, $\alpha-\beta=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$=\sqrt{(1)^{2}-4(1)}=\sqrt{-3}=\sqrt{3} i$...(ii)
On solving(i), (ii) we get
$\alpha=\frac{1+i \sqrt{3}}{2}$ and $\beta=\frac{1-i \sqrt{3}}{2}$
$\Rightarrow \alpha=\frac{1}{2}+i \cdot \frac{\sqrt{3}}{2}$ and $\beta=\frac{1}{2}-i \cdot \frac{\sqrt{3}}{2}$
$\Rightarrow \quad \alpha=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}$ and $\beta=\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}$
(a) $\alpha^{4}-\beta^{4}=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}-\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}$
$=2 i \sin \frac{4 \pi}{3}$
(By Demoiver's thm)
$\Rightarrow \alpha^{4}-\beta^{4}$ is not real.
(b) $2\left(\alpha^{5}+\beta^{5}\right)$
$=2\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}+\cos \frac{5 \pi}{3}-i \sin \frac{5 \pi}{3}\right)$
$=2.2 \cos \frac{5 \pi}{3}=4 \cdot \frac{1}{2}=2$
Now, $(\alpha \beta)^{5}=1$
$\Rightarrow 2\left(\alpha^{5}+\beta^{5}\right) \neq(\alpha \beta)^{5}$
(c) $\alpha^{6}-\beta^{6}=\cos \frac{6 \pi}{3}+i \sin \frac{6 \pi}{3}-\cos \frac{6 \pi}{3}+i \sin \frac{6 \pi}{3}$
$=2 i \sin 2 \pi=0$
Hence, option (c) is correct.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If $\alpha, \beta$ are the roots of $x^{2}+p x-q=0$ and $\gamma, \delta$ are the roots of $\mathrm{x}^{2}-\mathrm{px}+\mathrm{r}=0$ then what is $(\beta+\gamma)(\beta+\delta)$ equal to $?$

Options:
A $\mathrm{p}+\mathrm{r}$
B $\mathrm{p}+\mathrm{q}$
C $\mathrm{q}+\mathrm{r}$
D $\mathrm{p}-\mathrm{q}$

The correct answer is:
$\mathrm{q}+\mathrm{r}$

Solution:
Since, $\alpha \& \beta$ are the roots of $x^{2}+p x-q=0$, then, $\alpha+\beta=-\mathrm{p}$...(1)
and $\alpha \beta=-\mathrm{q}$...(2)
(2) put the value of $\alpha$ from $(2)$ in $(1)$
$-\frac{q}{\beta}+\beta=-p \Rightarrow-q+\beta^{2}=-p \beta$
$\Rightarrow \beta^{2}=q-p \beta$...(3)
Since $\gamma \& \delta$ are the roots of $x^{2}-p x+r=0$, then, $\gamma+\delta=\mathrm{p}, \gamma \delta=\mathrm{r}$
Now, $(\beta+\gamma)(\beta+\delta)=\beta^{2}+\beta \delta+\beta \gamma+\gamma \delta$.
$=\beta^{2}+\beta[\gamma+\delta]+\gamma \delta .=\mathrm{q}-\mathrm{p} \beta+\mathrm{p} \beta+\mathrm{r}=\mathrm{q}-\mathrm{r}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 2)
If $\alpha, \beta$ are the rootsofthe equations $x^{2}-2 x-1=0$, then what is the value of $\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}$

Options:
A $-2$
B 0
C 30
D 3

The correct answer is:
3

Solution:
Since, $\alpha$ and $\beta$ are the roots of the equation $\mathrm{x}^{2}-2 \mathrm{x}-1=0$, then
Sum of roots, $\alpha+\beta=2$ and
product of the roots $\alpha \beta=-1$ Since, $(\alpha+\beta)=\alpha^{2}+\beta^{2}+2 \alpha \beta$
$\Rightarrow 4=\alpha^{2}+\beta^{2}-2$
$\Rightarrow \alpha^{2}+\beta^{2}=6$
Now, $\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}=\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}=\frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}=6^{2}$
$\Rightarrow \alpha^{4}+\beta^{4}+2 \alpha^{2} \beta^{2}=36$
$\begin{aligned} \Rightarrow \alpha^{4}+\beta^{4}+2 &=36 &(\because \alpha \beta=-1) \end{aligned}$
$\Rightarrow \alpha^{4}+\beta^{4}=34$...(i)
$\Rightarrow \frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}=\frac{34}{(-1)^{2}}=34$
[Putting value of $\alpha^{4}+\beta^{4}=34$ from Equation (i)]

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)