datavorous/entrance-exam-dataset huggingface dataset:
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-\left(1-2 a^{2}\right) x+\left(1-2 a^{2}\right)=0$
Under what condition is $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} < 1 ?$

Options:
A $\mathrm{a}^{2} < \frac{1}{2}$
B $\mathrm{a}^{2}>\frac{1}{2}$
C $\mathrm{a}^{2}>1$
D $\mathrm{a}^{2} \in\left(\frac{1}{3}, \frac{1}{2}\right)$ only

The correct answer is:
$\mathrm{a}^{2} < \frac{1}{2}$

Solution:
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} < 1 \Rightarrow \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} < 1$
$\Rightarrow \frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}} < 1$
$\alpha+\beta=\frac{-b}{a}=\left(1-2 a^{2}\right) \& \alpha \beta=\frac{c}{a}=\left(1-2 a^{2}\right)$
On solving: $\frac{4 a^{4}-1}{4 a^{4}-4 a^{2}+1} < 1$
$\Rightarrow 4 a^{4}-1 < 4 a^{4}-4 a^{2}+1$
$4 a^{2} < 2 \Rightarrow a^{2} < \frac{1}{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)
Let $\alpha$ and $\beta(\alpha < \beta)$ be the roots of the equation $x^{2}+b x+c=0$, where $\mathrm{b}>0$ and $\mathrm{c} < 0 .$
Consider the following:
1.$\beta < -\alpha$
2.$\beta < \alpha \mid$
Which of the above is/are correct?

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:
Given quadratic equation, Hence roots of given quadratic equation are
$\beta=\frac{-b+\sqrt{b^{2}-4 c}}{2}$
$\alpha=\frac{-b-\sqrt{b^{2}-4 c}}{2} \quad(\because \alpha < \beta)$
$\Rightarrow-\alpha=\frac{b+\sqrt{b^{2}-4 c}}{2}$ and $\mid \alpha \models \frac{b+\sqrt{b^{2}-4 c}}{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0 .$ The equation whose roots are $\alpha^{19}$ and $\beta^{7}$ is

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:
Given Equation is $x^{2}+x+1=0$
$\Rightarrow x=\frac{-1 \pm \sqrt{3} i}{2}$
Thus, roots are $\omega$ and $\omega^{2}$ $\alpha=\omega$ and $\beta=\omega^{2}$
$\mathrm{S} o, \alpha^{19}=(\omega)^{19}=\left(\omega^{3}\right)^{6}, \omega=\omega \quad\left(\because \omega^{3}=1\right)$
and $\beta^{7}=\left(\omega^{2}\right)^{7}=\omega^{14}=\left(\omega^{3}\right)^{4} \cdot \omega^{2}=\omega^{2}$
Now, $\alpha^{19}+\beta^{7}=\omega+\omega^{2}=-1\left(\because 1+\omega+\omega^{2}=0\right)$
and $\left(\alpha^{19}\right)\left(\beta^{7}\right)=(\omega)\left(\omega^{2}\right)=1$
$\therefore$ Required Quadratic equation whose roots are $\alpha^{19}$
and $\beta^{7}$ is
$x^{2}+\left(\alpha^{19}+\beta^{7}\right) x+\left(\alpha^{19}\right)\left(\beta^{7}\right)=0$
$\Rightarrow x^{2}+(-1) x+1=0$
$\Rightarrow x^{2}-x+1=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
Let $\theta$ be a positive angle, If the number of degrees in $\theta$ is divided by the number of radians in $\theta$, then an irrational
number $\frac{180}{\pi}$ results. If the number of degrees in $\theta$ is
multiplied by the number of radians in $\theta$, then an irrational
number $\frac{125 \pi}{9}$ results. The angle $\theta$ must be equal to

Options:
A $30^{\circ}$
B $45^{\circ}$
C $50^{\circ}$
D $60^{\circ}$

The correct answer is:
$50^{\circ}$

Solution:
From going by the options, option (a), $\theta=30^{\circ}$, as we know that $180^{\circ}=\pi \mathrm{radian}$
$\therefore \quad 30^{\circ}=\frac{30 \pi}{180}$ radian
Now according to question,
$\frac{30^{\circ} \times 180^{\circ}}{30^{\circ} \pi}=\frac{180}{\pi}$
Now number of degree in $\theta$ is multiplied by number of radians in $\theta$.
$\therefore \quad 30^{\circ} \times \frac{30 \pi}{180}=\frac{900 \pi}{180}=\frac{10 \pi}{2}=5 \pi \neq \frac{125 \pi}{9}$
From option (b), $\theta=45^{\circ}$
$\therefore \quad 45^{\circ}=\frac{45^{\circ} \pi}{180}$ radian
Now according to question,
$\frac{45^{\circ} \times 180}{45^{\circ} \pi}=\frac{180}{\pi}$
Now number of degree in $\theta$ is multiplied by number of $\operatorname{radian}$ in $\theta$.
$\therefore \quad 45^{\circ} \times \frac{45^{\circ} \pi}{180^{\circ}}=\frac{45 \pi}{4} \neq \frac{125 \pi}{9}$
From option (c), $\theta=50^{\circ}$
As we know that $180^{\circ}=\pi$ radian
$\therefore \quad 50^{\circ}=\frac{50 \pi}{180}$ radian
Now according to question
$\frac{50^{\circ} \times 180^{\circ}}{50^{\circ} \pi}=\frac{180}{\pi}$
Now number of degree in ' $\theta$ ' is multiplied bynumber of $\operatorname{radian} \operatorname{in} \theta .$
$\therefore \quad 50^{\circ} \times \frac{50 \pi}{180}=\frac{2500 \pi}{180}=\frac{125 \pi}{9}$
$\therefore \quad$ Option (c) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 1)
Let θ be an angle in the standard position such that the point (-5,12) lies on its terminal side, then

Options:
A |sinθ|=-sinθ
B |cosθ|=cosθ
C |tanθ|=-tanθ
D |cosecθ|=-cosecθ

The correct answer is:
|tanθ|=-tanθ

Solution:
Clearly, the point -5,12 lies in second quadrant. Since, it's x-coordinate is negative and y-coordinate is positive.

We know that, in second quadrant sinθ & cosecθ are positive and cosθ, secθ, tanθ & cotθ are negative.

So, by using definition of modulus function, we get |sinθ|=sinθ, |cosθ|=-cosθ, |tanθ|=-tanθ, |cosecθ|=cosecθ.

Topic: Mathematics, Trigonometric Ratios & Identities, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 2)
Let $\alpha, \beta, \gamma$ be distinct real numbers. The points with position
vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \hat{\beta} \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$

Options:
A are collinear
B form an equilateral triangle
C form a scalene triangle
D form a right-angled triangle

The correct answer is:
form an equilateral triangle

Solution:
$\alpha, \beta$ and $\gamma$ be distinct real numbers $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=\beta \hat{\mathrm{i}}+\gamma \hat{\mathrm{j}}+\alpha \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=\gamma \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$
Let $\alpha=1, \beta=2$ and $\gamma=3$
then, $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
Now, $|\overrightarrow{a b}|=\sqrt{(2-1)^{2}}+(3-2)^{2}+(1-3)^{2}=\sqrt{6}$
Similarly $|\overrightarrow{\mathrm{bc}}|=|\overrightarrow{\mathrm{ac}}|=\sqrt{6}$
Hence, point for equilateral triangle.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
Let $\alpha$ be the angle which the vector $\overrightarrow{\mathrm{V}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ makes
with the z-axis, Then, what is the value of $\sin \alpha$ ?

Options:
A $\frac{2}{3}$
B $\frac{1}{3}$
C $\frac{\sqrt{5}}{3} \quad$
D $\frac{\sqrt{5}}{9}$

The correct answer is:
$\frac{\sqrt{5}}{3} \quad$

Solution:
The given vector is $\overrightarrow{\mathrm{V}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
and for z-axis $\mathrm{x}=0$ and $\mathrm{y}=0$, so the vector equation is $\overrightarrow{\mathrm{A}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\cos \alpha=\frac{\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{A}}}{|\overrightarrow{\mathrm{V}}| \cdot|\overrightarrow{\mathrm{A}}|}$
$\cos \alpha=\frac{2.0+(-1) \cdot(0)+2 \cdot(1)}{\sqrt{4+1+4} \sqrt{0+0+1}}=\frac{2}{3}$
Hence, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}$
$=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
Let $\alpha$ be the root of the equation $25 \cos ^{2} \theta+5 \cos \theta-12=0$,
where $\frac{\pi}{2} < \alpha < \pi$
What is $\sin 2 \alpha$ equal to?

Options:
A $\frac{24}{25}$
B $\frac{-24}{25}$
C $\frac{-5}{12}$
D $\frac{-21}{25}$

The correct answer is:
$\frac{-24}{25}$

Solution:
$\sin 2 \alpha=2 \sin \alpha \cdot \cos \alpha$
$=2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)$
$=\frac{6}{5} \times \frac{-4}{5}=\frac{-24}{25}$
$\therefore \quad$ Option (b) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 1)
Let $\alpha$ be the root of the equation $25 \cos ^{2} \theta+5 \cos \theta-12=0$,
where $\frac{\pi}{2} < \alpha < \pi$
What is tan $\alpha$ equal to?

Options:
A $\frac{-3}{4}$
B $\frac{3}{4}$
C $\frac{-4}{3}$
D $\frac{-4}{5}$

The correct answer is:
$\frac{-3}{4}$

Solution:
Here $\alpha$ is the root of equation $25 \cos ^{2} \theta+5 \cos \theta-12=0$
$\Rightarrow \quad 25 \cos ^{2} \alpha+5 \cos \alpha-12=0$
$\Rightarrow \quad 25 \cos ^{2} \alpha+20 \cos \alpha-15 \cos \alpha-12=0$
$\Rightarrow \quad 5 \cos \alpha(5 \cos \alpha+4)-3(5 \cos \alpha+4)=0$
$(5 \cos \alpha-3)(5 \cos \alpha+4)=0$
$\cos \alpha=\frac{3}{5}$ or $\cos \alpha=\frac{-4}{5}$
Here, $\frac{\pi}{2} < \alpha < \pi$
$\therefore \quad \cos \alpha=\frac{-4}{5}$
$\left(\because\right.$ In $2^{\text {nd }}$ quadrant, $\cos \alpha$ value is negative)
Now, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\frac{16}{25}}$
$\therefore \quad \sin \alpha=\frac{3}{5}$
$\therefore \quad \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{3}{5} \times \frac{-5}{4}=\frac{-3}{4}$
$\therefore \quad$ Option (a) is correct.

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2015 (Phase 1)
Let $\alpha, \gamma$ be the roots of $A x^{2}-4 x+1=0$ and $\beta, \delta$ be the
rootsof $B x^{2}-6 x+1=0 .$ If $\alpha, \gamma, \beta, \delta$ are in $\mathrm{HP}$, then what are the values of $A$ and $B$ respectively?

Options:
A 3,8
B $-3,-8$
C $3 .$
D $-3,8$

The correct answer is:
$-3,8$

Solution:
Sum and product of roots $=0$ will be $\alpha+\gamma=\frac{4}{A}$ and $\alpha \gamma=\frac{1}{A}$ respectively
Sum and product of roots of $\mathrm{Bx}^{2}-6 \mathrm{x}+1=0$ will be
$\beta+\delta=\frac{6}{B}$ and $\beta \delta=\frac{1}{B}$ respectively
Since, $\alpha, \beta, \gamma$ and $\delta$ are in HP.
Then, $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ and $\frac{1}{\delta}$ will be in AP.
$\Rightarrow \frac{1}{\beta}-\frac{1}{\alpha}=\frac{1}{\delta}-\frac{1}{\gamma} \Rightarrow \frac{1}{\beta}-\frac{1}{\delta}=\frac{1}{\alpha}-\frac{1}{\delta}$
$\Rightarrow \frac{\delta-\beta}{\beta \delta}=\frac{\gamma-\alpha}{\alpha \gamma}$
$\Rightarrow \frac{\sqrt{(\delta+\beta)^{2}-4 \beta \delta}}{\beta \delta}=\frac{\sqrt{(\gamma+\alpha)^{2}-4 \alpha \gamma}}{\alpha \gamma}$
$\Rightarrow 36-4 B=16-4 A$
$\Rightarrow 4 A+4 B=20$
$\Rightarrow A+B=5$
It is possible only, if\
$A=-3$ and $B=8$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
Let $\alpha, \beta$ be the roots of the equation $(x-a)(x-b)=c, c \neq$ $0 .$ Then the roots of the equation $(x-\alpha)(x-\beta)+\mathrm{c}=0$ are

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

The correct answer is:
$a, b$

Solution:
Given equation is $(x-a)(x-b)=c, c \neq 0$
$\Rightarrow x^{2}-(a+b) x+a b-c=0$
Let $\alpha, \beta$ be the roots of this equation. $\alpha+\beta=a+b, \alpha \beta=a b-c$
Consider $(x-\alpha)(x-\beta)+c=0$
$\Rightarrow x^{2}-(\alpha+\beta) x+\alpha \beta+c=0$
Roots of this equation is
$x=\frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^{2}-4(\alpha \beta+c)}}{2}$
$=\frac{(a+b) \pm \sqrt{(a+b)^{2}-4(a b-c+c)}}{2}$
$=\frac{(a+b) \pm \sqrt{a^{2}+b^{2}-2 a b}}{2}$
$=\frac{(a+b) \pm \sqrt{(a-b)^{2}}}{2}=\frac{(a+b) \pm(a-b)}{2}$
$=\frac{a+b+a-b}{2}, \frac{a+b-a+b}{2}=a, b$.
Hence, roots of the equation $(x-\alpha)(x-\beta)+\mathrm{c}=0$ are $a$ and $b$.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
Let be the set of all numbers, which are perfect squares, $\mathrm{B}$ be the set of all numbers which are multiples of 5 and $C$ be the set of all numbers, which are divisible by 2 and 3 . Consider
the following statements :
I. $A, B, C$ are mutually exdusive
II. $A, B, C$ are mutually exhaustive. III. The number of elements in the complement set of $\mathrm{A} \cup \mathrm{B}$ is 12
Which of the statements given above are the correct?

Options:
A I and II only
B I and III only
C II and III only
D I, II and III

The correct answer is:
I and III only

Solution:
Let, $\mathrm{U}=\{1,2,3, \ldots, 20\}$
$\mathrm{A}=$ Set of all numbers which are perfect square $=\{1,4,9,16\}$
$\mathrm{B}=$ Set of all numbers which are multiples of $5=\{5,10,15,20\}$
$\mathrm{C}=$ Set of all numbers which are divisible by 2 and 3 $=\{6,12,18\}$
$\mathrm{A} \cup \mathrm{B}=\{1,4,9,16,5,10,15,20\}$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})=8$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})^{\prime}=20-8=12$
Also, $\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})$
$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\{1,4,9,16,5,10,15,20,6,12,18\}$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=11$
and $\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})=4+4+3=11$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
Let $C$ be the set of complex number and $z_{1}, z_{2}$ are in $C$
1- agr $\left(z_{1}\right)=\operatorname{agr}\left(z_{2}\right) \Rightarrow z_{1}=z_{2}$
2- $\left|z_{1}\right|=\left|z_{2}\right| \Rightarrow z_{1}=z_{2}$
Which of the statements given above is/are correct?

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

The correct answer is:
Neither 1 nor 2

Solution:
Both statements are not correct.

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 1)
Let \( \Delta=\left|\begin{array}{ccc}A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1\end{array}\right| \) and
\( \Delta_{1}=\left|\begin{array}{cccc}A & B & C & \\ x & y & z & \\ z y & z x & x y & \end{array}\right| \) then

Options:
A \( \Delta_{1}=-\Delta \)
B \( \Delta_{1}=\Delta \)
C \( \Delta_{1} \neq \Delta \)
D \( \Delta_{1}=2 \Delta \)

The correct answer is:
\( \Delta_{1}=\Delta \)

Solution:
Given that, \( \Delta=\left|\begin{array}{ccc}A x & x^{2} & 1 \\ B y & y^{2} & 1 \\ C z & z^{2} & 1\end{array}\right| \rightarrow(1) \)
\[
\text { and } \Delta_{1}=\left|\begin{array}{ccc}
A & B & C \\
x & y & z \\
z y & z x & x y
\end{array}\right| \rightarrow(2)
\]
Take out common \( x, y \) and \( z \) in Eq. (1) from \( R_{1}, R_{2} \) and \( R_{3} \) respectively we get
\( \Delta=x y z\left|\begin{array}{ccc}A & x & \frac{1}{x} \\ B & y & \frac{1}{y} \\ C & z & \frac{1}{z}\end{array}\right| \)
\( C_{3} \rightarrow x y z C_{3} \), we get
\( \Delta=\left|\begin{array}{ccc}A & x & y z \\ B & y & z x \\ C & z & x y\end{array}\right| \)
\( =\left|\begin{array}{ccc}A & B & C \\ x & y & z \\ y z & z x & x y\end{array}\right|=\Delta_{1} \)
Therefore, \( \Delta=\Delta_{1} \)

Topic: Mathematics, Indefinite Integration, KCET, KCET 2017
Let E be the ellipe $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and $C$ be the circle $x^{2}+y^{2}=9$. Let $P=(1,2)$ and $Q=(2,1) .$ Which one of the following

Options:
A $Q$ lies inside $C$ but outside $E$
B $Q$ lies outside both $C$ and $E$
C $P$ lies inside both $C$ and $E$
D $P$ lies inside $C$ but outside $E$.

The correct answer is:
$P$ lies inside $C$ but outside $E$.

Solution:
Given equation of ellipse $\mathrm{E}$ is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\Rightarrow \frac{4 x^{2}+9 y^{2}}{36}=1 \Rightarrow 4 x^{2}+9 y^{2}=36$
$\Rightarrow 4 x^{2}+9 y^{2}-36=0$ $\ldots(1)$
and $C:$ Eqn of circle is $x^{2}+y^{2}=9$ Which can be rewritten as
$x^{2}+y^{2}-9=0$ $\ldots(2)$
For a point $P(1,2)$ we have
$4(1)^{2}+9(2)^{2}-36=40-36>0 \quad[$ from $(1)]$
and $1^{2}+2^{2}-9=5-9 < 0 \quad[$ from $(2)]$
Point $P$ lies outside of $E$ and inside of $C$.

Topic: Mathematics, Ellipse, NDA, NDA 2010 (Phase 1)
Let $f:[-6,6] \rightarrow \mathrm{R}$ be defined by $f(\mathrm{x})=\mathrm{x}^{2}-3$. Consider the $\begin{array}{ll}\text { following : } & {[2017-I]}\end{array}$
1- $\left(f^{\circ} f^{\circ} f\right)(-1)=\left(f^{\circ} f^{\circ} f\right)(1)$
2- $\left(f^{\circ} f^{\circ} f\right)(-1)-4\left(f^{\circ} f^{\circ} f\right)(1)=\left(f^{\circ} f\right)(0)$
Which of the above is/are correct?

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:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-3$
$\operatorname{fof}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))=\mathrm{f}\left(\mathrm{x}^{2}-3\right)=\left(2^{2}-3\right)^{2}-3$
$=2^{4}-6 \mathrm{x}^{2}+9-3=\mathrm{x}^{4}-6 \mathrm{x}^{2}+6$
$\mathrm{fof}(0)=0-0+6=6$
$\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))=\mathrm{f}\left(\mathrm{x}^{4}-6 \mathrm{x}^{2}+6\right)=\left(\mathrm{x}^{4}-6 \mathrm{x}^{2}+6\right)^{2}-3$
$\operatorname{fofof}(1)=\left((-1)^{4}-6(-1)^{2}+6\right)^{2}-3$
$=(1-6+6)-3=-2$
$\quad \mathrm{fofof}(1)=\left(1^{4}-6(1)^{2}+6\right)^{2}-3$
$=(1-6+6)-3=-2$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
Let f $[-100 \pi, 100 \pi] \rightarrow[-1,1]$ be defined by $f(\theta)=\sin \theta$.
Then what is the number of values of $\theta \in[-100 \pi, 1000 \pi]$ for which $\mathrm{f}(\theta)=0$ ?

Options:
A 1000
B 1101
C 1100
D 1110

The correct answer is:
1101

Solution:
$f(\theta)=0$, if $\theta$ is an integral multiple of $\pi$. From $-100$ m $\operatorname{to} 0 \pi$ there are 101 values of $\theta$ for which $f(\theta)=0$. From $\pi$ to $1000 \pi$, there are 1000 values for which $\mathrm{f}(\theta)$ $=0$
so, total values number is $101+1000=1101$

Topic: Mathematics, Functions, NDA, NDA 2007 (Phase 1)
Let $f(\mathrm{a})=\frac{\mathrm{a}-1}{\mathrm{a}+1} . \quad[2017-\mathrm{I}]$
Consider the following :
1- $ f(2 \mathrm{a})=f(\mathrm{a})+1$
2- $f\left(\frac{1}{\mathrm{a}}\right)=-f(\mathrm{a})$
Which of the above is/are correct?

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

The correct answer is:
2 only

Solution:
$\mathrm{f}(\mathrm{a})=\frac{\mathrm{a}-1}{\mathrm{a}+1}$
$\mathrm{f}(2 \mathrm{a})=\frac{2 \mathrm{a}-1}{2 \mathrm{a}+1}$
$\mathrm{f}(\mathrm{a})+1=\frac{\mathrm{a}-1}{\mathrm{a}+1}+1=\frac{2 \mathrm{a}}{\mathrm{a}+1}$
So, $\mathrm{f}(2 \mathrm{a}) \neq \mathrm{f}(\mathrm{a})+1$
Now, $f\left(\frac{1}{a}\right)=\frac{\frac{1}{a}-1}{\frac{l}{a}+1}=\frac{1-a}{1+a}=-\left(\frac{a-1}{a+1}\right)=-f(a)$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
Let $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{R}$, where $\mathrm{A}=\mathrm{R} \backslash\{0\}$ is such that $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}+|\mathrm{x}|}{\mathrm{x}}$.
On which one of the following sets is $\mathrm{f}(\mathrm{x})$ continuous?

Options:
A A
B $\mathrm{B}=\{\mathrm{x} \in \mathrm{R}: \mathrm{x} \geq 0\}$
C $\quad C=\{x \in R: x \leq 0\}$
D $\mathrm{D}=\mathrm{R}$

The correct answer is:
A

Solution:
For $x \geq 0$
$f(x)=\frac{x+x}{x}=2$
For $x < 0$
$f(x)=\frac{x-x}{x}=0$
$\lim _{x \rightarrow 0^{+}} f(x)=2$
$\lim _{x \rightarrow 0^{-}} f(x)=0$
$f(0)=2$
$\Rightarrow$ It is discontinuous at $x=0$. Option (a) is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
Let $f(\mathrm{n})=\left[\frac{1}{4}+\frac{\mathrm{n}}{1000}\right]$, where $[\mathrm{x}]$ denote the integral part
of $\mathrm{x}$. Then the value of $\sum f(\mathrm{n})$ is

Options:
A 251
B 1
C 250
D 0

The correct answer is:
251

Solution:
$\mathrm{f}(\mathrm{n})=\left[\frac{1}{4}+\frac{\mathrm{n}}{1000}\right]$
$\sum_{\mathrm{n}=1}^{1000} \mathrm{f}(\mathrm{n})=\left[\frac{1}{4}+\frac{1}{1000}\right]+\left[\frac{1}{4}+\frac{2}{1000}\right]+\ldots .+\left[\frac{1}{4}+\frac{1000}{1000}\right]$
$=[0.25+0.001]+[0.25+0.002]+\ldots .+[0.25+1]$
We get ' 0 ' for all values of $\mathrm{n}$ from 1 to 750 . From $\mathrm{n}=750$, we get all the values as 1 . So,
$\begin{array}{rl}\sum_{n=1}^{1000} & f(n)=0+0+0+\ldots .+\left[\frac{1}{4}+\frac{750}{1000}\right]+\left[\frac{1}{4}+\frac{751}{1000}\right]+\ldots .[1.25] \\ & =1+1+1+\ldots .(251 \text { times }) \\ & =251 .\end{array}$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)
Let \( f: N \rightarrow N \) defined by \( f(n)=\left\{\begin{array}{cl}\frac{n+1}{2} ; & \text { if } n \text { is odd } \\ \frac{n}{2} ; & \text { if } n \text { is even }\end{array}\right. \)
then \( \mathrm{f} \) is

Options:
A one-one and onto
B one-one but not onto
C onto but not one-one
D neither one-one nor onto

The correct answer is:
onto but not one-one

Solution:
Given that, \( f: N \rightarrow N \) defined by
\( f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & n \text { is odd } \\ \frac{n}{2} & n \text { is even }\end{array}\right. \)
For \( n=1 \), we have \( f(1)=\frac{1+1}{2}=1 \)
and, if \( n=2 \), we have \( f(2)=\frac{2}{2}=1 \)
So, \( f(1)=f(2) \) but \( 1 \neq 2 \). Therefore, \( f(x) \) is not one-one.
Now, \( f(x)=\frac{n+1}{2} \) if \( \mathrm{n} \) is odd
if \( y=\frac{n+1}{2} \), then \( n=2 y-1, \forall y \)
Also, \( f(x)=\frac{n}{2} \) if \( n \) is even. That is,
\( y=\frac{n}{2} \) or \( n=2 y, \forall y \)
Therefore, \( f(x) \) is onto.

Topic: Mathematics, Differential Equations, KCET, KCET 2014
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
Consider the following:
$1 \quad f(2)=f(1)-f(0)$
$2 \quad \mathrm{f}^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
Which of the above is/are correct?

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:
1 $f(1)-f(0)=4-6$
$=-2$
$f(2)=8-20+4+6=-2$
$\quad$ Hence $f(2)=f(1)-f(0)$
$\therefore$ Statement $(1)$ is correct.
2 $f^{\prime \prime}(2)-2 f^{\prime}(1)=2-2(-5)$
$f^{\prime \prime}(2)-2 \mathrm{f}^{\prime}(1)=12$
$\therefore$ Statement $(2)$ is correct.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
What is $\mathrm{f}(1)$ equal to?

Options:
A $-2$
B $-1$
C 0
D 4

The correct answer is:
4

Solution:
$f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$ ...(1)
$f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...(2)
$f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$ ...(3)
$f^{\prime \prime \prime}(x)=6$ ...(4)
$f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)$ ...(5)
$f^{\prime}(2)=12+2 f^{\prime}(1)$ ...(6)
Using $(6)$ in $(5)$, we get
$f^{\prime}(1)=3+2 f^{\prime}(1)+12+2 f^{\prime \prime}(1)$
$-3 f^{\prime}(1)=15$
$f^{\prime}(1)=-5$
Using this value in eqn (6) we get
$f^{\prime}(2)=12+2 \times(-5)$
$f^{\prime \prime}(2)=2$
Using $x=3$ in eqn (4),
$f^{\prime \prime \prime}(3)=6$
Putting value of $f^{\prime}(1)+f^{\prime \prime}(2)$ and $\mathrm{f}^{\prime \prime \prime}(3)$ in eqn (1)
We get
$\begin{aligned} f(x) &=x^{3}+x^{2}(-5)+x(2)+6 \\ &=x^{3}-5 x^{2}+2 x+6 \end{aligned}$
Putting $\mathrm{x}=1$ $f(1)=(1)^{3}-5(1)^{2}+2(1)+6$
$f(1)=4$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
What is $\mathrm{f}^{\prime}(1)$ equal to?

Options:
A $-6$
B $-5$
C 1
D 0

The correct answer is:
$-5$

Solution:
$\mathrm{f}^{\prime}(1)=-5$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
What is $\mathrm{f}^{\prime \prime}(10)$ equal to?

Options:
A 1
B 5
C 6
D 8

The correct answer is:
6

Solution:
$f^{\prime \prime \prime}(10)=6$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)