datavorous/entrance-exam-dataset huggingface dataset:
If $[x]$ denotes the greatest integer not exceeding the number $x$, then $f(x)$ defined by
$f(x)=\left\{\begin{array}{ll}{[x],} & \text { if } x < 2 \\ {[x]-1,} & \text { if } x \geq 2\end{array}\right.$ is continuous in the interval.

Options:
A $[1,2) \cup(2,3)$
B $[1,3)$
C $(1,3)$
D $R$

The correct answer is:
$[1,3)$

Solution:
Given function,
$$
\begin{aligned}
& \qquad f(x)=\left\{\begin{aligned}
{[x], } & \text { if } x < 2 \\
{[x]-1, } & \text { if } x \geq 2
\end{aligned}\right. \\
& \text { At } \quad x=2, f(2)=1 \\
& \text { LHL } \\
& \qquad(\text { at } x=2)=\lim _{h \rightarrow 0}[2-h]=1 \\
& \text { and RHL (at } x=2)=\lim _{h \rightarrow 0}([2+h]-1)=2-1=1
\end{aligned}
$$
So, $f(x)$ is continuous at $x=2$
And $[x]$ is continuous in $[n, n+1), \forall n \in$ integer.
So, given function continuous in the interval $[1,3)$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by
$f(x)= \begin{cases}\frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0\end{cases}$
is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

Options:
A (-2, 1)
B (-2,-1)
C $(-1, \sqrt{3})$
D $(-2,-\sqrt{3})$

The correct answer is:
(-2,-1)

Solution:
Given, $f(x)= \begin{cases}\frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0\end{cases}$
At $\mathrm{x}=0$
$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{a+2 \cos x}{x^2}$
$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{a+2\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\ldots\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{(a+2)+2\left(-\frac{x^2}{x !}+\frac{x^4}{4 !}-\ldots\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{0+2\left(\frac{-x^2}{2 !}+\frac{x^4}{4 !}-\ldots\right)}{x^2}=-1 \\ & {[\because f(x) \text { is continuous so we take } a+2=0 \text { ] }} \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} b \tan \frac{\pi}{[x+4]} \\ & =b \tan \frac{\pi}{4}=b \\ & \text { But } \quad \mathrm{LHL}=\text { RHL } \\ & \Rightarrow \quad-1=b \text { and } a=-2 \\ & \end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If [x] denotes the greatest integer $\leq x$, then the domain of the function fx=4-x2[x]+2 is

Options:
A (-∞,-2]∪[-1,2)
B (-∞,-2)∪[-1,2]
C (-∞,-2)∪(-1,2)
D (-∞,-1]∪[1,2]

The correct answer is:
(-∞,-2)∪[-1,2]

Solution:
fx=4-x2[x]+2

it will be defined if

4-x2x+2≥0⇒x2-4x+2≤0

x-2x+2x+2≤0

x∈-∞,-2∪-1,2

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2019 (03 May Shift 2)
If [x] denotes the greatest integer ≤x, then the system of linear equations [sinθ]x+[-cosθ]y=0, [cotθ]x+y=0

Options:
A has a unique solution if Î¸âÏ2,2Ï3âªÏ,7Ï6
B have infinitely many solution if θ∈π2,2π3∪π,7π6
C has a unique if Î¸âÏ2,2Ï3 and have infinitely many solutions if Î¸âÏ,7Ï6
D have infinitely many solutions if Î¸âÏ2,2Ï3 and has a unique solution if Î¸âÏ,7Ï6

The correct answer is:
have infinitely many solutions if θ∈π2,2π3 and has a unique solution if θ∈π,7π6

Solution:
sin⁡θx+-cos⁡θy=0 and cos⁡θx+y=0 for infinite many solution
sin⁡θ-cos⁡θcot⁡θ 1=0
i.e. sinθ= -cosθ cotθ ...i
when θ∈π2, 2π3 then sinθ∈0, 12 ⇒ sinθ=0 and -cosθ∈0, 12 ⇒ -cosθ=0 and

cotθ∈-13, 0 ⇒ cotθ=-1
When θ∈π, 7π6 then sinθ∈-12, 0 ⇒ sinθ=-1 and -cosθ∈32, 1 ⇒-cosθ=0 and
cotθ∈3, ∞ ⇒ cotθ∈2, 3, 4,...∞
When θ∈π2, 2π3 then the equation i satisfied therefore infinitely many solution.
When θ∈π, 7π6 then the equation i not satisfied therefore unique solution.

Topic: Mathematics, Determinants, JEE Main
If [x] denotes the greatest integer ≤x, then the system of linear equations [sinθ]x+[-cosθ]y=0, [cotθ]x+y=0

Options:
A has a unique solution if Î¸âÏ2,2Ï3âªÏ,7Ï6
B have infinitely many solution if θ∈π2,2π3∪π,7π6
C has a unique if Î¸âÏ2,2Ï3 and have infinitely many solutions if Î¸âÏ,7Ï6
D have infinitely many solutions if Î¸âÏ2,2Ï3 and has a unique solution if Î¸âÏ,7Ï6

The correct answer is:
have infinitely many solutions if θ∈π2,2π3 and has a unique solution if θ∈π,7π6

Solution:
sin⁡θx+-cos⁡θy=0 and cos⁡θx+y=0 for infinite many solution
sin⁡θ-cos⁡θcot⁡θ 1=0
i.e. sinθ= -cosθ cotθ ...i
when θ∈π2, 2π3 then sinθ∈0, 12 ⇒ sinθ=0 and -cosθ∈0, 12 ⇒ -cosθ=0 and

cotθ∈-13, 0 ⇒ cotθ=-1
When θ∈π, 7π6 then sinθ∈-12, 0 ⇒ sinθ=-1 and -cosθ∈32, 1 ⇒-cosθ=0 and
cotθ∈3, ∞ ⇒ cotθ∈2, 3, 4,...∞
When θ∈π2, 2π3 then the equation i satisfied therefore infinitely many solution.
When θ∈π, 7π6 then the equation i not satisfied therefore unique solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (12 Apr Shift 2)
If $x=e^{y+e^{y+. .10 \infty}}, x>0$, then $\frac{d y}{d x}$ is

Options:
A
$\frac{\mathrm{x}}{1+\mathrm{x}}$

B
$\frac{1}{x}$

C
$\frac{1-x}{x}$

D
$\frac{1+x}{x}$

The correct answer is:
$\frac{1-x}{x}$

Solution:
$x=e^{y+e^{y+e^{y+\ldots \ldots \infty}}} \Rightarrow x=e^{y+x}$
$\Rightarrow \ln x-x=y \Rightarrow \frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2004
If $X$ follows a binomial distribution with parameters $n=6$ and $\quad p$ and $4(P(X=4))=P(X=2)$, then $p=$

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{6}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:
$4 P(X=4)=P(X=2) \Rightarrow 4 .{ }^6 C_4 p^4 q^2={ }^6 C_2 p^2 q^4$
$\Rightarrow 4 p^2=q^2 \Rightarrow 4 p^2=(1-p)^2$
$\Rightarrow 3 p^2+2 p-1=0 \Rightarrow p=\frac{1}{3}$

Topic: Mathematics, Probability, JEE Main
If $x$ gram of gas is adsorbed by $m$ gram of adsorbent at pressure $P$ the plot of $\log \frac{x}{\mathrm{~m}}$ versus $\log \mathrm{P}$ is linear. The slope of the plot is: ( $\mathrm{m}$ and $\mathrm{k}$ are constants and $n>1$ )

Options:
A
$\log \mathrm{k}$

B
$\frac{1}{n}$

C
$2 \mathrm{k}$

D
$\mathrm{n}$

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

Solution:
According to Freundlich adsorption isotherm.
$$
\begin{aligned}
&\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{\frac{1}{\mathrm{n}}} \\
&\log _{10} \frac{\mathrm{x}}{\mathrm{m}}=\frac{1}{\mathrm{n}} \log _{10} \mathrm{P}+\log _{10} \mathrm{k}
\end{aligned}
$$
This is the equation of straight line of type $y=m x+c$
Hence slope is $1 / \mathrm{n}(\mathrm{m})$ and intercept is $\log _{\mathrm{e}} \mathrm{k}$.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If $X$ has a binomial distribution, $B(n, p)$ with parameters $n$ and $p$ such that $P(X=2)=P(X=3)$, then $\mathrm{E}(\mathrm{X})$, the mean of variable $\mathrm{X}$, is

Options:
A
$2-\mathrm{p}$

B
$3-p$

C
$\frac{\mathrm{p}}{2}$

D
$\frac{p}{3}$

The correct answer is:
$3-p$

Solution:
Since $\mathrm{X}$ has a binomial distribution, $\mathrm{B}(n$,
p)
$$
\begin{aligned}
&\therefore \mathrm{P}(X=2)={ }^n \mathrm{C}_2(p)^2(1-p)^{n-2} \\
&\text { and } \mathrm{P}(X=3)={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\text { Given } \mathrm{P}(X=2)=\mathrm{P}(X=3) \\
&\Rightarrow{ }^n \mathrm{C}_2 p^2(1-p)^{n-2}={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\
&\Rightarrow \frac{n !}{2 !(n-2) !} \cdot \frac{p^2(1-p)^n}{(1-p)^2} \\
&=\frac{n !}{3 !(n-3) !} \cdot \frac{p^3(1-p)^n}{(1-p)^3}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{1}{n-2}=\frac{1}{3} \cdot \frac{p}{1-p} \\
&\Rightarrow 3(1-p)=p(n-2) \\
&\Rightarrow 3-3 p=n p-2 p \\
&\Rightarrow n p=3-p \\
&\Rightarrow \mathrm{E}(X)=\text { mean }=3-p \\
&(\because \text { mean of } \mathrm{B}(n, p)=n p)
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (11 Apr Online)
If $X$ is a random variable such that $\sigma(X)=2.6$ then $\sigma(1-4 x)$ is equal to

Options:
A 7.8
B $-10.4$
C 13
D 10.4

The correct answer is:
10.4

Solution:
Given that,
$$
\begin{aligned}
\sigma(X) &=2.6=\sqrt{\operatorname{var}(x)} \\
\sigma(1-4 x) &=\sqrt{\operatorname{var}(1-4 x)} \\
&=\sqrt{16 \operatorname{var}(x)} \\
&=4 \times \sqrt{\operatorname{var}(x)}=4 \times 2.6=10.4
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
If x is a solution of the equation 2x+1- 2x-1=1, x≥12 , then 4x2-1 is equal to :

Options:
A 34
B 12
C 22
D 2

The correct answer is:
34

Solution:
2x+1- 2x-1=1 ...1

⇒2x+1+2x-1-24x2-1=1

⇒4x-1=24x2-1

⇒16x2-8x+1=16x2-4

⇒8x=5

⇒x=58 which satisfies equation 1.

So, 4x2-1=34

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2016 (10 Apr Online)
If $x$ is amount of adsorbate and $m$ is amount of adsorbent, which of the following relations is not related to adsorption process?

Options:
A $\frac{x}{m}=f(T)$ at constant $p$
B $p=f(T)$ at constant $\left(\frac{x}{m}\right)$
C $\frac{x}{m}=p \times T$
D $\frac{x}{m}=f(p)$ at constant $T$

The correct answer is:
$\frac{x}{m}=p \times T$

Solution:
$\frac{x}{m}=p \times T$ is the incorrect relation. The correct relation is $\frac{x}{m} \propto \frac{p}{T}$.

Topic: Chemistry, Solutions, JEE Main
If $\mathrm{x}$ is positive, the first negative term in the expansion of $(1+x)^{27 / 5}$ is

Options:
A
6th term

B
7th term

C
5th term

D
8th term

The correct answer is:
8th term

Solution:
$T_{r+1}=\frac{n(n-1)(n-2) \ldots \ldots \ldots(n-r+1)}{r !}(x)^r$
For first negative term, $n-r+1 < 0$ or $r>\frac{32}{5}$
$\therefore \mathrm{r}=7$. Therefore, first negative term is $\mathrm{T}_8$.

Topic: Mathematics, Binomial Theorem, JEE Main
If $\mathrm{x}$ is positive, the first negative term in the expansion of $(1+x)^{27 / 5}$ is

Options:
A
6th term

B
7th term

C
5th term

D
8th term

The correct answer is:
8th term

Solution:
$T_{r+1}=\frac{n(n-1)(n-2) \ldots \ldots \ldots(n-r+1)}{r !}(x)^r$
For first negative term, $n-r+1 < 0$ or $r>\frac{32}{5}$
$\therefore \mathrm{r}=7$. Therefore, first negative term is $\mathrm{T}_8$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2003
If x is rational and 4x2+1x2+16x+1x-57=0,â¡ then the product of all possible values of x is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
Given equation,
4 x 2 + 1 x 2 + 1 6 x + 1 x - 5 7 = 0
Let, x+1x=y; x 2 + 1 x 2 = y 2 - 2
â 4 y 2 + 1 6 y - 6 5 = 0
â y=-132 orâ¡ 52
When, y = 5 2
x+1x=52âx=2 or 12
When, y=-132
â x + 1 x = - 1 3 / 2
â 2 x 2 + 1 3 x + 2 = 0
â x = - 1 3 Â± 1 5 3 4
Since x is rational, x=2 or 1 2

Topic: Mathematics, Quadratic Equation, JEE Main
If $x$ is real, the maximum value of $\frac{3 x^2+9 x+17}{3 x^2+9 x+7}$ is

Options:
A
1/4

B
41

C
1

D
17/7

The correct answer is:
41

Solution:
$y=\frac{3 x^2+9 x+17}{3 x^2+9 x+7}$
$3 x^2(y-1)+9 x(y-1)+7 y-17=0$
$D \geq 0 \quad \because x$ is real
$81(y-1)^2-4 x 3(y-1)(7 y-17) \geq 0$
$\Rightarrow(y-1)(y-41) \leq 0 \Rightarrow 1 \leq y \leq 41$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2006
If $x$ is real, then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$, is

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

The correct answer is:
2

Solution:
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{\left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)}{\left(x^2+x+1\right)^2}$
for maximum or minimum, put $f^{\prime}(x)=0$
$\begin{array}{lc}
\Rightarrow & \left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)=0 \\
\Rightarrow & x^2+x-1-\left(-x^2+x+1\right)=0 \\
\Rightarrow & 2 x^2-2=0 \Rightarrow x= \pm 1
\end{array}$
Now, $f^{\prime}(c)=\frac{2 x^2-2}{\left(x^2+x+1\right)^2}$
Again differentiating, we get
$\begin{aligned}
& \left(x^2+x+1\right)^2(4 x)-\left(2 x^2-2\right) \\
& f^{\prime \prime}(x)=\frac{2\left(x^2+x+1\right)(2 x+1)}{\left(x^2+x+1\right)^4}
\end{aligned}$
at $x=1, f^{\prime \prime}(x)>0$
Therefore it is minimum at $x=1$
Put $x=1$ in equation (i), we get
$f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
$\therefore$ The minimum value is $\frac{1}{3}$.

Topic: Mathematics, Application of Derivatives, JEE Main
If $x$ is small, so that $x^2$ and higher powers can be neglected, then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is

Options:
A $1-2 x$
B $1-3 x$
C $1-4 x$
D $1-5 x$

The correct answer is:
$1-2 x$

Solution:
Given expression is
$$
\begin{aligned}
& E=\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}} \\
&=\frac{\left(1+2 x+2 x^2+\ldots\right)(1+6 x+\ldots)}{(1+12 x+\ldots)} \\
&=\frac{(1+2 x+6 x+\ldots)}{(1+12 x)} \\
&=(1+8 x)(1+12 x)^{-1} \\
&=(1+8 x)(1-12 x \ldots) \\
&=(1+8 x-12 x \ldots) \\
&=(1-4 x) \\
& \quad(\because \text { neglecting the higher term })
\end{aligned}
$$
( $\because$ neglecting the higher term)

Topic: Mathematics, Binomial Theorem, JEE Main
If $\mathrm{x}$ is so small that $\mathrm{x}^3$ and higher powers of $\mathrm{x}$ may be neglected, then $\frac{(1+x)^{3 / 2}-\left(1+\frac{1}{2} x\right)^3}{(1-x)^{1 / 2}}$

Options:
A
$1-\frac{3}{8} x^2$

B
$3 x+\frac{3}{8} x^2$

C
$-\frac{3}{8} x^2$

D
$\frac{x}{2}-\frac{3}{8} x^2$

The correct answer is:
$-\frac{3}{8} x^2$

Solution:
$$
\begin{aligned}
& (1-x)^{1 / 2}\left[1+\frac{3}{2} x+\frac{3}{2}\left(\frac{3}{2}-1\right) x^2-1-3\left(\frac{1}{2} x\right)-3(2)\left(\frac{1}{2} x\right)^2\right] \\
& =(1-x)^{1 / 2}\left[-\frac{3}{8} x^2\right]=-\frac{3}{8} x^2
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
If \([x]\) is the greatest integer less than or equal to \(x\) and \(|x|\) is the modulus of \(x\), then the system of three equations
\(\begin{aligned} & 2 x+3|y|+5[z]=0, x+|y|-2[z]=4, \\ & x+|y|+[z]=1 \text { has } \end{aligned}\)

Options:
A a unique solution
B finitely many solutions
C infinitely many solutions
D no solution

The correct answer is:
infinitely many solutions

Solution:
Given system of three equations
\(\begin{aligned}
2 x+3|y|+5[z] & =0 \\
x+|y|-2[z] & =4 \\
x+|y|+[z] & =1
\end{aligned}\)
and
According to Cramer's rule,
\(x=\frac{\Delta_1}{\Delta},|y|=\frac{\Delta_2}{\Delta} \text { and }[z]=\frac{\Delta_3}{\Delta}\)
where,
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
2 & 3 & 5 \\
1 & 1 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =2(1+2)-3(1+2)+5(1-1)=-3 \\
\Delta_1 & =\left|\begin{array}{ccc}
0 & 3 & 5 \\
4 & 1 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =0(1+2)-3(4+2)+5(4-1) \\
& =-18+15=-3 \\
\Delta_2 & =\left|\begin{array}{ccc}
2 & 0 & 5 \\
1 & 4 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =2(4+2)-0(1+2)+5(1-4)=-3
\end{aligned}\)
and \(\Delta_3=\left|\begin{array}{lll}2 & 3 & 0 \\ 1 & 1 & 4 \\ 1 & 1 & 1\end{array}\right|=2(l-4)-3(l-4)+0(l-1)=3\)
Now, \(x=\frac{-3}{-3}=1,|y|=\frac{-3}{-3}=1\) and \([z]=\frac{-3}{3}=-1\)
\(\therefore x=1,|y|=1 \Rightarrow y= \pm 1\) and \([z]=-1\)
\(\Rightarrow z \in[-1,0)\)
So, the given system of three equations has infinitely many solution. Hence, option (3) is correct.

Topic: Mathematics, Determinants, JEE Main
If $[x]$ is the greatest integer $\leq x$, then the value of the integral $\int_{-0.9}^{0.9}\left(\left[x^2\right]+\log \left(\frac{2-x}{2+x}\right)\right) d x$ is

Options:
A
$0.486$

B
$0.243$

C
$1.8$

D
0

The correct answer is:
0

Solution:
$$
\text { } \begin{aligned}
& \int_{-0.9}^{0.9}\left\{\left[x^2\right]+\log \left(\frac{2-x}{2+x}\right)\right\} d x \\
= & \int_{-0.9}^{0.9}\left[x^2\right] d x+\int_{-0.9}^{0.9} \log \left(\frac{2-x}{2+x}\right) d x
\end{aligned}
$$
$$
\begin{aligned}
& =0+\int_{-0.9}^{0.9} \log \left(\frac{2-x}{2+x}\right) d x \\
& \text { Put } x=-x \Rightarrow f(x)=\log \frac{2-x}{2+x} \\
& \text { and } f(-x)=\log \frac{2+x}{2-x} \\
& =-\log \frac{(2-x)}{2+x}=-f(x)
\end{aligned}
$$
So, it is an odd function, hence Required integral $=0$.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2012 (26 May Online)
If $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p$, Freundlich adsorption isotherm gives a straight line on plotting

Options:
A
$x / m$ vs $p$

B
$x / m$ vs $1 / p$

C
$\log x / m$ vs $\log p$

D
$\log x / m$ vs $p$

The correct answer is:
$\log x / m$ vs $\log p$

Solution:
Freundlich adsorption isotherm gives straight line on plotting $\log x / m$ vs $\log p$ as show below.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (07 May Online)
If $X$ is the specific resistance of the solution and $M$ is the molarity of the solution, the molar conductivity of the solution is given by

Options:
A $\frac{1000 X}{M}$
B $\frac{1000}{M X}$
C $\frac{1000 M}{X}$
D $\frac{M X}{1000}$

The correct answer is:
$\frac{1000}{M X}$

Solution:
Molar conductivity$=\frac{1000}{M X}$

Topic: Chemistry, Electrochemistry, JEE Main
If $x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0),$ then $\frac{d y}{d x}$ at $x=e$ is

equal to :

Options:
A $\frac{(1+2 e)}{2 \sqrt{4+e^{2}}}$
B $\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}$
C $\frac{(1+2 e)}{\sqrt{4+e^{2}}}$
D $\frac{e}{\sqrt{4+e^{2}}}$

The correct answer is:
$\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}$

Solution:
Consider the equation,

$x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4$

Differentiate both sides w.r.t. $x$

$\log _{e}\left(\log _{e} x\right)+x \cdot \frac{1}{x \cdot \log _{e} x}-2 x+2 y \frac{d y}{d x}=0$

$\log _{e}\left(\log _{e} x\right)+\frac{1}{\log _{e} x}-2 x+2 y \frac{d y}{d x}=0$

When $x=e, y=\sqrt{4+e^{2}}$. Put these values in (1),

$0+1-2 e+2 \sqrt{4+e^{2}} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{2 e-1}{2 \sqrt{4+e^{2}}}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If ϕx=log8⁡log3⁡x, then ϕ ′ e is equal to

Options:
A elog⁡8
B –elog⁡8
C 1elog⁡8
D None of these

The correct answer is:
1elog⁡8

Solution:
We have,
ϕx=log8⁡log3⁡x=log8⁡log⁡xlog⁡3
=log8⁡log⁡x-log8⁡log⁡3
=log⁡log⁡xlog⁡8-log8⁡log⁡3
ϕ ′ x = 1 log8 ⋅ 1 logx ⋅ 1 x −0
∴ ϕ ′ e = 1 log8 ⋅ 1 loge ⋅ 1 e = 1 elog8

Topic: Mathematics, Differentiation, JEE Main
If $x=m y+c$ is a normal to the parabola $x^2=4 a y$, then the value of $c$ is

Options:
A $-2 a m-a m^3$
B $2 a m+a m^3$
C $-\frac{2 a}{m}-\frac{a}{m^3}$
D $\frac{2 a}{m}+\frac{a}{m^3}$

The correct answer is:
$-2 a m-a m^3$

Solution:
The equation of the normal to $x^2=4 a y$ is of the form $x=m y-2 a m-a m^3$. Therefore $c=-2 a m-a m^3$.

Topic: Mathematics, Parabola, JEE Main
If $x^m \cdot y^n=(x+y)^{m+n}$, then $\frac{d y}{d x}$ is

Options:
A
$\frac{y}{x}$

B
$\frac{x+y}{x y}$

C
xy

D
$\frac{x}{y}$

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

Solution:
$x^m \cdot y^n=(x+y)^{m+n} \Rightarrow m \ln x+n \ln y=(m+n) \ln (x+y)$
$\therefore \frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}\right)=\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x}$
$\quad \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}\right) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}$

Topic: Mathematics, Differentiation, JEE Main
If $x^m \cdot y^n=(x+y)^{m+n}$, then $\frac{d y}{d x}$ is

Options:
A
$\frac{y}{x}$

B
$\frac{x+y}{x y}$

C
xy

D
$\frac{x}{y}$

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

Solution:
$x^m \cdot y^n=(x+y)^{m+n} \Rightarrow m \ln x+n \ln y=(m+n) \ln (x+y)$
$\therefore \frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \Rightarrow\left(\frac{m}{x}-\frac{m+n}{x+y}\right)=\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x}$
$\quad \frac{m y-n x}{x(x+y)}=\left(\frac{m y-n x}{y(x+y)}\right) \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2006
If x=∑n=0∞-1ntan2⁡θ and y=∑n=0∞cos2n⁡θ, for 0<θ<π4, then:

Options:
A x1+y=1
B y1-x=1
C y1+x=1
D x1-y=1

The correct answer is:
y1-x=1

Solution:
y=1+cos2⁡θ+cos4⁡θ+…

Above series is an infinite G.P, apply sum of infinite G.P.

⇒y=11-cos2⁡θ

⇒1y=sin2θ

x=11--tan2⁡θ=1sec2⁡θ

⇒cos2⁡θ=x

⇒1y+x=1

⇒y1-x=1

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If x=∑n=0∞an, y=∑n=0∞bn, z=∑n=0∞cn, where a,b,c are in A.P. and a<1,b<1,c<1, abc≠0, then

Options:
A x,y,z are in A.P.
B x,y,z are in G.P.
C 1x,1y,1z are in A.P.
D 1x+1y+1z=1-a+b+c

The correct answer is:
1x,1y,1z are in A.P.

Solution:
x=1+a+a2=......

x=11-a⇒a=1-1x

y=11-b⇒b=1-1y

z=11-c⇒c=1-1z

a,b,c are in A.P.

⇒1-1x,1-1y,1-1z are in A.P.

⇒-1x,-1y,-1z are in A.P.

⇒1x,1y,1z are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If $x=\sum_{n=0}^{\infty} a^n, y=\sum_{n=0}^{\infty} b^n, z=\sum_{n=0}^{\infty} c^n$ where $a, b, c$ are in A.P. and $|a| < 1,|b| < 1,|c| < 1$, then $x, y, z$ are in

Options:
A
G.P.

B
A.P.

C
Arithmetic - Geometric Progression

D
H.P.

The correct answer is:
H.P.

Solution:
$$
\begin{array}{ll}
x=\sum_{n=0}^{\infty} a^n=\frac{1}{1-a} & \Rightarrow a=1-\frac{1}{x} \\
y=\sum_{n=0}^{\infty} b^n=\frac{1}{1-b} & \Rightarrow b=1-\frac{1}{y}
\end{array}
$$
$$
\begin{aligned}
& z=\sum_{n=0}^{\infty} c^n=\frac{1}{1-c} \quad \Rightarrow c=1-\frac{1}{z} \\
& a, b, c \text { are in A.P. } \\
& 2 b=a+c \\
& 2\left(1-\frac{1}{y}\right)=1-\frac{1}{x}+1-\frac{1}{y} \\
& \frac{2}{y}=\frac{1}{x}+\frac{1}{z} \\
& \Rightarrow x, y, z \text { are in H.P. }
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2005
If $x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}$, then $\prod_{n=1}^{\infty} x_n$ is equal to

Options:
A $-1$
B $1$
C $\frac{1}{\sqrt{2}}$
D $\frac{i}{\sqrt{2}}$

The correct answer is:
$-1$

Solution:
$\begin{aligned} & \text { Given that, }=\frac{9+6 i}{13} \\ & \qquad \begin{aligned} X_n & =\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}=\operatorname{cis} \frac{\pi}{2^n} \\ \therefore \quad \prod_{n=1}^{\infty} X_n & =\prod_{n=1}^{\infty} \operatorname{cis} \frac{\pi}{2^n} \\ & =\operatorname{cis} \frac{\pi}{2} \cdot \operatorname{cis} \frac{\pi}{2^2} \cdot \operatorname{cis} \frac{\pi}{2^3} \ldots . \\ & =\operatorname{cis}\left(\frac{\pi}{2}+\frac{\pi}{2^2}+\frac{\pi}{2^3}+\ldots . .\right) \\ & =\operatorname{cis}\left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)=\operatorname{cis} \pi \\ & =-1\end{aligned}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If ∫xpq-p-1xp+1qdx =2(1+x-p)1-qλp(q-1)+c p,q∈N-{1}, then the value of λ is (here, c is an arbitrary constant)

Options:

The correct answer is:
2

Solution:
I=∫x-p-1(1+x-p)qdx

Let, 1+x-p=t
=-1p∫dttq
=t-q+1p(1-q)+c
∴2λ=1

Topic: Mathematics, Indefinite Integration, JEE Main
If $x_r=\cos \left(\frac{\pi}{2^r}\right)+i \sin \left(\frac{\pi}{2^r}\right)$, then $x_1 \cdot x_2 \cdots \cdots \infty$ is

Options:
A $-3$
B $-2$
C $-1$
D $0$

The correct answer is:
$-1$

Solution:
$x_1, x_2, x_3 \cdots \cdots$ upto $\infty=\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\left(\cos \frac{\pi}{2^2}+i \sin \frac{\pi}{2^2}\right)$ upto $\qquad$ $\infty$
$\begin{aligned} & =\cos \left(\frac{\pi}{2}+\frac{\pi}{2^2}+\ldots . .\right)+i \sin \left(\frac{\pi}{2}+\frac{\pi}{2^2}+\ldots . .\right) \\ & =\cos \left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)+i \sin \left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)=\cos \pi+i \sin \pi=-1 .\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $[x]$ represent greatest integer $\leq x$ and $[\alpha, \beta]$ is the set of all real values of $x$ for which the real function $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$ is defined, then $f^2(\alpha+1)+5 f^2(\beta)=$

Options:
A $0$
B $\frac{36}{5}$
C $12$
D $1$

The correct answer is:
$12$

Solution:
The given function $f(x)=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{[x]+2}}$,
then function $f$ is defined if
$3+x \geq 0,3-x \geq 0$ and $[x]+2>0$
$\begin{aligned} & \Rightarrow \quad x \geq-3, x \leq 3 \text { and }[x]>-2 \\ & \Rightarrow \quad x \geq-3, x \leq 3 \text { and } x \in[-1, \infty) \\ & \Rightarrow \quad x \in[-1,3]=[\alpha, \beta] \text { (given) } \\ & \text { So, } \alpha=-1 \text { and } \beta=3\end{aligned}$
$\therefore f^2(\alpha+1)+5 f^2(\beta)=f^2(0)+5 f^2(3)$
$\begin{aligned} & =\left(\frac{\sqrt{3}+\sqrt{3}}{\sqrt{2}}\right)^2+5\left(\frac{\sqrt{6}+\sqrt{0}}{\sqrt{5}}\right)^2=\left(\frac{4 \times 3}{2}\right)+5\left(\frac{6}{5}\right) \\ & =6+6=12\end{aligned}$
Hence, option (c) is correct.

Topic: Mathematics, Functions, JEE Main
If x=sin-1sin⁡10 and  y=cos-1 cos⁡10, then y-x is equal to:

Options:
A 10
B π
C 0
D 7π

The correct answer is:
π

Solution:
From the graph of y=sin-1⁡sin⁡x, it is clear that sin-1⁡(sin⁡10)=3π-10
From the graph of y=cos-1⁡cos⁡x
It is clear that cos-1⁡cos⁡10=4π-10
Hence y-x=4π-10 -3π-10=π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2019 (09 Jan Shift 2)