datavorous/entrance-exam-dataset huggingface dataset:
Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, where $[.]$ is the greatest integer function and $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}$ be two real valued functions over R.
Which of the following statements is correct?

Options:
A $\operatorname{Both} \mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are continuous at $\mathrm{x}=0$.
B $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$, but $\mathrm{g}(\mathrm{x})$ is not continuous at $\mathrm{x}=0 .$
C $\mathrm{g}(\mathrm{x})$ is continuous at $\mathrm{x}=0$, but $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=0 .$
D Both $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are discontinuous at $\mathrm{x}=0$.

The correct answer is:
$\mathrm{g}(\mathrm{x})$ is continuous at $\mathrm{x}=0$, but $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=0 .$

Solution:
$f(x)=[x]$ and $g(x)=\sin x$ $\lim _{x \rightarrow 0^{+}} f(x)=[0+h]=0$
$\lim _{x \rightarrow 0^{-}} f(x)=[0-h]=-1$
$f(0)=0$
$\Rightarrow f(x)$ is not continuous at $x=0$ and also $g(x)$ is
continuous at $x=0$. (every trignometric function is continuous).

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, where $[.]$ is the greatest integer function and $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}$ be two real valued functions over R.
Which one of the following statements is correct?

Options:
A $\lim _{x \rightarrow 0}($ fog $)(\mathrm{x})$ exists
B $\lim _{x \rightarrow 0}$ (gof)(x)exists
C $\lim _{x \rightarrow 0+}($ fog $)(x)=\lim _{x \rightarrow 0-}(g o f)(x)$
D $\lim _{x \rightarrow 0+}($ fog $)(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}}(\mathrm{gof})(\mathrm{x})$

The correct answer is:
$\lim _{x \rightarrow 0+}($ fog $)(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}}(\mathrm{gof})(\mathrm{x})$

Solution:
At $x=0$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=|x|-2+|x-2|$
$\begin{aligned} g(x) &=x-2-(x-2)=0 \\ \mathrm{~g}(\mathrm{x}) &=0 \\ \text { For }(x=0): g(x) &=-2+|-2|=0 \end{aligned}$
LHL $=\lim _{x \rightarrow 0^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} g(x)=0$
$g(0)=0$
$\Rightarrow g(x)$ is continuous at $x=0$ At $x=2$
For LHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2)$
$g(x)=0$
For RHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2+x-2$
$2(x)=2 x-4$
For $(x=2): g(x)=|x|-2+|x-2|$
$\mathrm{LHL}=\lim _{x \rightarrow 2^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 2^{+}} g(x)=\lim _{x \rightarrow 2} 2 x-4=2(2)-4=0$
$g(2)=|2|-2+|2-2|$
$g(2)=2-2+2-2=0$
$\Rightarrow g(x)$ is continuous at $x=2$. At $x=-1$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=-2+|-2|=0$
For $(x=-1): g(x)=-2+|-2|=0$
$\therefore \mathrm{LHL}=\lim _{x \rightarrow-1^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow-1^{+}} g(x)=0$
$g(-1)=0$
$\Rightarrow g(\mathrm{x})$ is differentiable at $x=-1$.

Topic: Mathematics, Limits, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, where $[.]$ is the greatest integer function and $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}$ be two real valued functions over R.
Which one of the following statements is correct?

Options:
A $\lim _{x \rightarrow 0}($ fog $)(\mathrm{x})$ exists
B $\lim _{x \rightarrow 0}$ (gof)(x)exists
C $\lim _{x \rightarrow 0+}($ fog $)(x)=\lim _{x \rightarrow 0-}(g o f)(x)$
D $\lim _{x \rightarrow 0+}($ fog $)(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}}(\mathrm{gof})(\mathrm{x})$

The correct answer is:
$\lim _{x \rightarrow 0+}($ fog $)(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}}(\mathrm{gof})(\mathrm{x})$

Solution:
$(f \circ g)(x)=[\sin x]$
$\lim _{x \rightarrow 0^{+}}(f o g)(x)=\lim _{x \rightarrow 0^{+}}[\sin x]=[h]$ where $h>0$
$\Rightarrow \lim _{x \rightarrow 0^{+}}(f \circ g)(x)=0$
$\lim _{x \rightarrow 0^{-}}(f o g)(x)=\lim _{x \rightarrow 0^{-}}[\sin x]=[h]$ where $h < 0$
$\Rightarrow \lim _{x \rightarrow 0^{-}}(f \circ g)(x)=-1$
$\Rightarrow \lim _{x \rightarrow 0}(f \circ g)(x)$ does not exist.
Now, $(g o f)(x)=\sin [x]$
$\lim _{x \rightarrow 0^{+}}(g o f)(x)=\lim _{x \rightarrow 0^{+}} \sin [x]=\sin 0=0$
$\lim _{x \rightarrow 0^{-}}($ gof $)(x)=\lim _{x \rightarrow 0^{-}} \sin [x]=\sin (-1)=-0.01745$
$\Rightarrow \lim _{x \rightarrow 0}($ gof $)(x)$ does not exist.
$\Rightarrow \lim _{x \rightarrow 0^{+}}(f o g)(x)=\lim _{x \rightarrow 0^{+}}(g o f)(x)$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[|\mathrm{x}|-\mid \mathrm{x}-1]]^{2}$
What is $\mathrm{f}^{\prime}(\mathrm{x})$ equal to when $0 < \mathrm{x} < 1 ?$

Options:
A 0
B $2 x-1$
C $4 x-2$
D $8 x-4$

The correct answer is:
$8 x-4$

Solution:
Given $f(x)=(|x|-|x+1|)^{2}$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
When $0 < x < 1$
$f(x)=(2 x-1)^{2}$
$f^{\prime}(x)=2(2 x-1) \cdot 2=4(2 x-1)$
$f^{\prime}(x)=8 x-4$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[|\mathrm{x}|-\mid \mathrm{x}-1]]^{2}$
What is $\mathrm{f}^{\prime}(\mathrm{x})$ equal to when $\mathrm{x}>1$?

Options:
A 0
B $2 x-1$
C $4 x-2$
D $8 x-4$

The correct answer is:
0

Solution:
Given $f(x)=(|x|-|x+1|)^{2}$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
When $x>1$ $f(x)=1$
$f^{\prime}(x)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 2)
Let $\mathrm{f}(\mathrm{x})=[|\mathrm{x}|-\mid \mathrm{x}-1]]^{2}$
Which of the following equations is/are correct?
$1 \quad \mathrm{f}(-2)=\mathrm{f}(5)$
$2 \quad f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=4$
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:
1 only

Solution:
Given $f(x)=(|x|-|x+1|)^{2}$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
For $x=-2$
$f(x)=1$ so $f(-2)=1$
For $x=5$ $f(x)=1 \Rightarrow f(5)=1$
Hence $f(-2)=f(5)$
Now, for $x=-1$
$f^{\prime \prime}(x)=0$
$f^{\prime \prime}(-2)=0$
For $x=0.5$
$f^{\prime \prime}(x)=8 \Rightarrow f^{\prime \prime}(0.5)=8$
For $x=3$
$f^{\prime \prime}(x)=0 \Rightarrow f^{\prime \prime}(3)=0$
$\Rightarrow f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=8 \neq 4$
Only statement 1 is correct.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 2)
Let $f(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}$, when $\mathrm{x} \in(0,1)$. Then which one of the following is correct?

Options:
A $f(\mathrm{x})$ fluctuates in the interval
B $f(\mathrm{x})$ increases in the interval
C $f(\mathrm{x})$ decreases in the interval
D None of the above

The correct answer is:
$f(\mathrm{x})$ decreases in the interval

Solution:
$f(x)=x+\frac{1}{x}$
$f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}=\frac{(x-1)(x+1)}{x^{2}}$
For $x \in(0,1), f^{\prime}(x) < 0$
$\Rightarrow \mathrm{f}(\mathrm{x})$ decreases

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
$\quad$ Let $f(x+y)=f(x) \cdot f(y)$ and $f(1)=2$ for all $x, y \in R$, where $f(x)$ is continuous function. What is $\mathrm{f}^{\prime}(1)$ equal to ?

Options:
A $2 \ln 2$
B $\ln 2$
C 1
D 0

The correct answer is:
$2 \ln 2$

Solution:
Given that $\mathrm{f}(1)=2$ and $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})$
These are value for all values of $x$ and $y$ Puting $\mathrm{x}=1$ and $\mathrm{y}=1$, we get $\mathrm{f}(2)=\mathrm{f}(1) \cdot \mathrm{f}(1)=2.2=2^{2}$
Similarly, $\mathrm{f}(3)=\mathrm{f}(1) \cdot \mathrm{f}(2)=2 \cdot 2^{2}=2^{3}$
$\Rightarrow \quad \mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$
$\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=2^{\mathrm{x}} \log _{\mathrm{e}} 2$
Hence, $\mathrm{f}^{\prime}(1)=2 \log _{\mathrm{e}} 2$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2006 (Phase 1)
Let $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})$ and $\mathrm{f}(\mathrm{x})=1+\mathrm{xg}(\mathrm{x}) \phi(\mathrm{x})$,
where $\lim _{x \rightarrow 0} g(x)=a$ and $\lim _{x \rightarrow 0} \phi(x)=b$. Then what is $f^{\prime}(x)$
equal to?

Options:
A $1+\mathrm{ab} \mathrm{f}(\mathrm{x})$
B $1+\mathrm{ab}$
C ab
D $\operatorname{abf}(\mathrm{x})$

The correct answer is:
$\operatorname{abf}(\mathrm{x})$

Solution:
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=\lim _{h \rightarrow 0} \frac{f(x)[h g(h) \phi(h)]}{h}$
$=a b f(x)$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 2)
Let $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})$ for all $\mathrm{x}$ and $\mathrm{y}$. Then what is $f^{\prime}(5)$
equal to [where $f^{\prime}(x)$ is the derivative of $\left.f(x)\right] ?$

Options:
A $f(5) f^{\prime} 0$
B $f(5)-f^{\prime}(0)$
C $f(5) f(0)$
D $f(5)+f^{\prime}(0)$

The correct answer is:
$f(5) f^{\prime} 0$

Solution:
$\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})$
$\operatorname{Let} \mathrm{f}(\mathrm{x})=\mathrm{a}^{\mathrm{x}}$
$\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{a}^{\mathrm{x}+\mathrm{y}}=\mathrm{a}^{\mathrm{x}} \cdot \mathrm{a}^{\mathrm{y}}=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) .$
$\mathrm{f}(5)=\mathrm{a}^{5}$
$\mathrm{f}^{\prime}(5)=\mathrm{a}^{5} \cdot \log \mathrm{a}$
$=\mathrm{f}(5) \cdot \mathrm{f}(0)$
$\mathrm{f}^{\mathrm{l}}(0)=\mathrm{a}^{0} \cdot \log \mathrm{a}$
$=\log \mathrm{a}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 1)
Let \(f(x)=\left\{\begin{array}{cc}x+1, & -1 \leq x \leq 0 \\ -x, & 0 \leq x \leq 1\end{array}\right.\)

Options:
A \(f(x)\) is discontinuous in \([-1,1]\) and so has no maximum value or minimum value in \([-1,1]\)
B \(f(x)\) is continuous in \([-1,1]\) and so has maximum value and minimum value
C \(f(x)\) is discontinuous in \([-1,1]\) but still has the maximum and minimum value
D \(f(x)\) is bounded in \([-1,1]\) and does not attain maximum or minimum value

The correct answer is:
\(f(x)\) is discontinuous in \([-1,1]\) but still has the maximum and minimum value

Solution:
Hint : \(f(x)= \begin{cases}x+1 & -1 \leq x \leq 0 \\ -x & 0 < x \leq 1\end{cases}\)
\(\left.\begin{array}{l}f\left(-1^{-}\right)=0 \\ f(-1)=0 \\ f\left(-1^{+}\right)=0\end{array}\right]\) continuous at \(x=-1\)
\(\left.\begin{array}{l}\mathrm{f}(0)=1 \\ \mathrm{f}\left(0^{-}\right)=1 \\ \mathrm{f}\left(0^{+}\right)=0\end{array}\right]\) discontinuous at \(\mathrm{x}=0\)
obviously \(f(x)\) has local maximum at \(x=0\)

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2023
Let $\mathrm{g}$ be the greatest integer function. Then the function $f(\mathrm{x})=(\mathrm{g}(\mathrm{x}))^{2}-\mathrm{g}(\mathrm{x})$ is discontinuous at

Options:
A all integers
B all integers except 0 and 1
C all integers except 0
D all integers except 1

The correct answer is:
all integers except 1

Solution:
$\mathrm{f}(\mathrm{x})=(\mathrm{g}(\mathrm{x}))^{2}-\mathrm{g}(\mathrm{x})$
Given, $\mathrm{g}(\mathrm{x})$ is greatest integer function. So, $\mathrm{g}(\mathrm{x})=[\mathrm{x}]$. $\therefore \mathrm{f}(\mathrm{x})=[\mathrm{x}]^{2}-[\mathrm{x}]$
$\mathrm{f}(\mathrm{x})$ is discontinuous at every integer except $\mathrm{x}=1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
Let $g: R \rightarrow R$ be a function such that, $g(x)=2 x+5$. Then, what is $g^{-1}(x)$ equal to?

Options:
A $\frac{x-5}{2}$
B $2 x-5$
C $x-\frac{5}{2}$
D $\frac{x}{2}+\frac{5}{2}$

The correct answer is:
$\frac{x-5}{2}$

Solution:
Let $y=2 x+5$
$\Rightarrow y-5=2 x$

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 2)
Let $G, W, E$ and $S$ be relative strength of gravitational, weak-nuclear, electromagnetic and strong-nuclear forces, respectively. Which of the correct statement?

Options:
A $S>W>E>G$
B $E>W>S>G$
C $S>E>W>G$
D $S>E>G>W$

The correct answer is:
$S>E>W>G$

Solution:
We know that, among all inverse forces based on inverse square law's gravitational force is the weakest and nuclear force is the strongest.
Hence, $F_{\text {Strongnuclear }}>E_{\text {Electromagnetic }}>F_{\text {Gravitational }}$ $\ldots(\mathrm{i})$
Weak nuclear force is a fundamental force in nature that underlies some forms of radioactivity, governs the decay of unstable subatomic particles such as meson and initiates the nuclear fusion reaction that fuels the sun.
These forces are considered weaker than electromagnetic forces and stronger than gravitational forces, i.e.
$F_{\text {Electromagnetic }}>F_{\text {Weak nuclear }}>F_{\text {Gravitational }}$ $\ldots(\mathrm{ii})$
Given, $\quad F_{\text {Electromagnetic }}=E \Rightarrow F_{\text {Strongnuclear }}=S$
$F_{\text {Weak nuclear }}=W$ and $F_{\text {Gravitational }}=G$
Using Eqs. (i) and (ii), we get
$S>E>W>G$

Topic: Physics, Mathematics in Physics, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
Let $g(x)=x^{3}-4 x+6$. If $f^{\prime}(x)=g^{\prime}(x)$ and $f(1)=2$, then what is $f(x)$ equal to?

Options:
A $x^{3}-4 x+3$
B $x^{3}-4 x+6$
C $x^{3}-4 x+1$
D $x^{3}-4 x+5$

The correct answer is:
$x^{3}-4 x+5$

Solution:
Now, (gof)' $(x)=1$ Given, $g(x)=x^{3}-4 x+6$
$\operatorname{But} f^{\prime}(x)=g^{\prime}(x)$
$\Rightarrow \int f^{\prime}(x) d x=\int g^{\prime}(x) d x$
$\Rightarrow f(x)=g(x)+c$
$\therefore f(x)=x^{3}-4 x+6+c$ where ' $\mathrm{c}$ ' is a constant. Now, $f(1)=2$ $\Rightarrow 2=f(1)=(1)^{3}-4(1)+6+c$
$\Rightarrow 2=1-4+6+c$
$\Rightarrow c=-1$
$\therefore f(x)=x^{3}-4 x+6-1=x^{3}-4 x+5$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
Let $\vec{\alpha}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \vec{\beta}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\vec{\gamma}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ be
three vectors. If $\vec{\alpha}$ and $\vec{\beta}$ are both perpendicular to the
vector $\vec{\delta}$ and $\vec{\delta} \cdot \vec{\gamma}=10$, then what is the magnitude of $\vec{\delta}$ ?

Options:
A $\sqrt{3}$ units
B $2 \sqrt{3}$ units
C $\frac{\sqrt{3}}{2}$ unit
D $\frac{1}{\sqrt{3}}$ unit

The correct answer is:
$2 \sqrt{3}$ units

Solution:
$\begin{aligned} \vec{\alpha} &=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \vec{\beta} &=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \vec{\gamma} &=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}} \end{aligned}$
Let $\vec{\delta}=a \hat{i}+b \hat{j}+c \hat{k}$
Since $\vec{\alpha}$ and $\vec{\delta}$ are perpendicular, $\vec{\alpha} \cdot \vec{\delta}=0$
$\Rightarrow \mathrm{a}+2 \mathrm{~b}-\mathrm{c}=0$
$\vec{\beta}$ and $\vec{\delta}$ are perpendicular, $\vec{\beta} \cdot \vec{\delta}=0$
$\Rightarrow 2 \mathrm{a}-\mathrm{b}+3 \mathrm{c}=0$
from $(1),(2), \frac{a}{5}=\frac{b}{-5}=\frac{c}{-5} \Rightarrow \frac{a}{1}=\frac{b}{-1}=\frac{c}{-1}=x($ say $)$
So, $a=x, b=-x, c=-x$
Also, it is given $\vec{\gamma} \cdot \vec{\delta}=10$
$\Rightarrow 2 \mathrm{a}+\mathrm{b}+6 \mathrm{c}=10$
$\Rightarrow 2 x-x-6 x=10$
$\Rightarrow-5 x=10$
$\Rightarrow x=-2$
$\therefore \vec{\delta}=-2 \hat{i}+2 \hat{j}+2 \hat{k} .$
$\therefore|\vec{\delta}|=\sqrt{4+4+4}=\sqrt{12}=2 \sqrt{3}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
Let $M$ be the set of men and $R$ is a relation 'is son of' defined on $M$. Then, $R$ is

Options:
A an equivalence relation
B a symmetric relation only
C a transitive relation only
D None of the above

The correct answer is:
None of the above

Solution:
Let $M=$ Set of men and $R$ is arelation'is son of defined on $M$. Reflexive : aRa (\because" $a$ can not be a son of $a$ ) Symmetric : $a R b \Rightarrow b R a$
which is not also possible (\because: If $a$ is a son of $b$ then $b$ can not be a son of $a$ )
Transitive : $a R b, b R c \Rightarrow a R c$
which is not possible.

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
Let matrix $\mathrm{B}$ be the adjoint of a square matrix $\mathrm{A}, \ell$ be the identitymatrix of same order as A. If $k(\neq 0)$ is the determinant of the matrix $\mathrm{A}$, then what is $\mathrm{AB}$ equal to?

Options:
A $\ell$
B $\mathrm{k} \ell$
C $\mathrm{k}^{2} \ell$
D $(1 / k) \ell$

The correct answer is:
$\mathrm{k} \ell$

Solution:
$\mathrm{B}=\mathrm{adj} \mathrm{A}, \mathrm{I}=$ Identity matrix, $|\mathrm{A}|=k$
$\mathrm{AB}=\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}=k \mathrm{I}$

Topic: Mathematics, Determinants, NDA, NDA 2018 (Phase 2)
Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)

Options:
A \(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
B \(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
C \(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
D \(\left[\begin{array}{cc}1 & 10 \\ 0 & 1\end{array}\right]\)

The correct answer is:
\(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)

Solution:
\(\begin{aligned}
M & =\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\
M^2 & =\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\
M^2 & =\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
M^3 & =\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \\
& \vdots \\
M^{10} & =\left[\begin{array}{ll}
1 & 10 \\
0 & 1
\end{array}\right] \\
N & =\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right] \\
|N| & =2-0=2 \\
\operatorname{Adj}^N & =\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \\
N^{-1} & =\frac{1}{|N|} \text { Adj A } \\
N^{-1} & =\frac{1}{2}\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \\
N^{-1} & =\left[\begin{array}{ll}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right]
\end{aligned}\)
Consider,
\(\begin{aligned}
\mathbf{N} \cdot \mathbf{M}^{10} \cdot \mathbf{N}^{-1} & =\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 10 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 10 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right]=\left[\begin{array}{ll}
1 & 5 \\
0 & 1
\end{array}\right]
\end{aligned}\)
Hence, option (a) is correct.

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
Let $n \geq 3$. A list of numbers $0 < x_{1} < x_{2} < \ldots < x_{n}$ has mean $\mu$ and standard deviatiion $\sigma$. A new list of numbers is made as follows : $y_{1}=0, y_{2}=x_{2}, \ldots, y_{n-1}=x_{n-1}, y_{n}=x_{1}+x_{n}$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

Options:
A $\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$
B $\mu=\hat{\mu}, \sigma \geq \hat{\sigma}$
C $\sigma=\hat{\sigma}$
D $\mu$ may or may not be equal to $\hat{\mu}$

The correct answer is:
$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$

Solution:
$\hat{\mu}=\frac{0+x_{2}+x_{3} \ldots \ldots x_{n-1}+\left(x_{1}+x_{n}\right)}{n}=\mu$
$\hat{\sigma}^{2}=\frac{1}{n} \Sigma y_{i}^{2}-\hat{\mu}^{2}$
$=\frac{1}{n}\left[0^{2}+x_{2}^{2}+x_{3}^{2} \ldots\left(x_{n-1}\right)^{2}+\left(x_{1}+x_{n}\right)^{2}\right]-\mu^{2}$
$\hat{\sigma}^{2}-\sigma^{2}=\frac{2 x_{1} x_{n}}{n}>0$
so $\hat{\sigma}>\sigma$

Topic: Mathematics, Statistics, KVPY, KVPY 2013 (SB/SX)
Let $n \geq 3$.A list of numbers $x_{1}, x_{2}, \ldots \ldots \ldots x_{n}$ has mean $\mu$ and standard deviation $\sigma .$ A new list of numbers $y_{1}, y_{2}, \ldots \ldots \ldots . y_{n}$ is made as follows: $y_{1}=\frac{x_{1}+x_{2}}{2}, y 2=\frac{x_{1}+x_{2}}{2}$ and $y_{j}=x$ for $j=3,4, \ldots \ldots . n$. The mean and the standard deviation of the list are $\hat{\mu}$ and $\hat{\sigma}$. Then which of the following is necessarily true?

Options:
A $\mu=\hat{\mu}$ and $\sigma \leq \hat{\sigma}$
B $\mu=\hat{\mu}$ and $\sigma \geq \hat{\sigma}$
C $\sigma=\hat{\sigma}$
D $\mu \neq \hat{\mu}$

The correct answer is:
$\mu=\hat{\mu}$ and $\sigma \geq \hat{\sigma}$

Solution:
$\mu=\frac{\sum x_{i}}{n} \quad n \geq 3$
$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$
$\hat{\mu}=\frac{y_{1}+y_{2}+y_{3}+\ldots \ldots+y_{n}}{n}$
$=\frac{\frac{x_{1}+x_{2}}{2}+\frac{x_{1}+x_{2}}{2}+x_{3}+x_{4}+\ldots \ldots \ldots+x_{n}}{n}$
$=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots \ldots . .+\mathrm{x}_{\mathrm{n}}}{\mathrm{n}}=\mu$
$\hat{\sigma}=\sqrt{\frac{\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+x_{3}^{2}+\ldots \ldots+x_{n}^{2}}{n}-\left(\frac{\sum y_{1}}{n}\right)^{2}}$
$\begin{array}{l}=\sqrt{\frac{\frac{1}{4}\left(2 x_{1}^{2}+2 x_{2}^{2}+4 x_{1} x_{2}\right)+x_{3}^{2}+\ldots \ldots . .+x_{n}^{2}}{n}-\left(\frac{\sum y_{1}}{n}\right)^{2}} \\ =\sqrt{\frac{\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right)+x_{1} x_{2}+x_{3}^{2}+\ldots \ldots+x_{n}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}} \\ \leq \sqrt{\frac{x_{1}^{2}+x_{2}^{2}+\ldots \ldots+x_{2}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}} \\ \hat{\sigma} \leq \sigma & \ldots \ldots \ldots(2) \\ \text { by }(1) \text { and }(2)\end{array}$

Topic: Mathematics, Statistics, KVPY, KVPY 2014 (SB/SX)
Let $n$ be a positive integer and $(1+x)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$
What is $a_{0}+a_{1}+a_{2}+\ldots+a_{\mathrm{n}}$ equal to ?

Options:
A 1
B $2^{n}$
C $2^{n-1}$
D $2^{n+1}$

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

Solution:
$(1+\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1} \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}$
Putting $\mathrm{x}=1$ $2^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{0}+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}$
$\begin{aligned} \Rightarrow \mathrm{a}_{0}+\mathrm{a}_{1}+\ldots \mathrm{a}_{\mathrm{n}} &=2^{\mathrm{n}} \\
&\left[\text { here }{ }^{\mathrm{n}} \mathrm{C}_{0}=\mathrm{a}_{0},{ }^{\mathrm{n}} \mathrm{C}_{1}=\mathrm{a}_{1}, \ldots .,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}}\right] \end{aligned}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2013 (Phase 2)
Let $\mathrm{N}$ be the set of integers. A relation $\mathrm{R}$ on $\mathrm{N}$ is defined as $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{xy}>0, \mathrm{x}, \mathrm{y}, \in \mathrm{N}\}$. Then, which one of the following is correct?

Options:
A $\mathrm{R}$ is symmetric but not reflexive
B $\mathrm{R}$ is reflexive but not symmetric
C $\mathrm{R}$ is symmetric and reflexive but not transitive
D $\mathrm{R}$ is an equivalence relation

The correct answer is:
$\mathrm{R}$ is an equivalence relation

Solution:
A relation is equivalent if it is
(i) Reflexive
(ii)Symmetric and
(iii) Transitive
We check for the same, one by -one $\mathrm{x}, \mathrm{y} \in \mathrm{N} \Rightarrow \mathrm{x}>0, \mathrm{y}>0$
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{xy}>0, \mathrm{x}, \mathrm{y}, \in \mathrm{N}\}$
(i) Reflexive $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
$\therefore \quad x, x, \in N \Rightarrow x^{2}>0$
$\therefore \quad \mathrm{R}$ is reflexive
(ii) Symmetric $\mathrm{x}, \mathrm{y} \in \mathrm{N}$
and $x y>0 \Rightarrow y x>0$
$\mathrm{R}$ is also symmetric
(iii) Transitive
$\therefore \quad x, y, z \in N$
$\Rightarrow x y>0, y z>0 \quad \Rightarrow x z>0$
$\therefore \quad \mathrm{R}$ is also transtive. Conclusion: $\mathrm{R}$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
Let $N$ denote the set of all non-negative integers and $Z$ denote the set of all integers. The function $f: Z \rightarrow N$ given by
$f(x)=|x|$ is:

Options:
A One-one but not onto
B Onto but not one-one
C Both one-one and onto
D Neither one-one nor onto

The correct answer is:
Onto but not one-one

Solution:
$f: \mathrm{Z} \rightarrow \mathrm{N}$ and $f(\mathrm{x})=|\mathrm{x}|$
When we draw a parallel line to $\mathrm{x}$ -axis. It cuts the curve into more than one point. Therefore, $f(\mathrm{x})=|\mathrm{x}|$ is not one-one. but onto

Topic: Mathematics, Functions, NDA, NDA 2014 (Phase 1)