datavorous/entrance-exam-dataset huggingface dataset:
f(x)=72x−9x−8x+12−1+cos⁡x,x≠0klog⁡2log⁡3,x=0

Find the value of 'k' for which the function f is continuous.

Options:
A 2
B 24
C 183
D 242

The correct answer is:
242

Solution:
Given that f(x)=72x−9x−8x+12−1+cos⁡x,x≠0klog⁡2log⁡3,x=0

We know that fx is continuous ⇒limx→afx = fa

limx→072x-9x-8x+12-1+cosx =k log2log3

limx→09x-18x-12-1+cosx×2+1+cosx2+1+cosx =k log2log3

limx→09x-18x-12-1-cosx×2+1+cosx1 =k log2log3

⇒limx→09x-18x-12sin2x2×2+1+cosx1 =k log2log3

We know that limx→0sinaxx = a , limx→0ax-1x= loga

⇒limx→0 x22sin2x2×limx→0 9x-1x×limx→0 8x-1x× lim x→0 2+1+cosx=k log2log3

⇒12×14×log32×log232×22 = k log3×log2

⇒2×2log3×32log2×22 = k log3×log2

242log3×log2= k log3×log2

So required value k =242.

Topic: Mathematics, Continuity and Differentiability, JEE Main
$f(x)=\int_a^x t^3 e^t d t$, then $\frac{d}{d x} f(x)=$

Options:
A $e^x\left(x^3+3 x^2\right)$
B $x^3 e^x$
C $a^3 e^a$
D None of these

The correct answer is:
$x^3 e^x$

Solution:
$\begin{aligned} f(x) & =\int_a^x t^3 e^t d t=\int_a^0 t^3 \cdot e^t d t+\int_0^x t^3 e^t d t \\ & \Rightarrow \frac{d f(x)}{d x}=\frac{d}{d x}\left(\int_a^0 t^3 \cdot e^t d t\right)+\frac{d}{d x}\left(\int_0^x t^3 \cdot e^t d t\right)=x^3 e^x\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$f(x)$ and $g(x)$ are two differentiable functions on $[0,2]$ such that $f^{\prime \prime}(x)-g^{\prime \prime}(x)=0$
$f^{\prime}(1)=2 g^{\prime}(1)=4 f(2)=3 g(2)=9$ then $f(x)-g(x)$ at $x=3 / 2$ is

Options:
A
0

B
2

C
10

D
5

The correct answer is:
5

Solution:
$\because f^{\prime \prime}(x)-g^{\prime \prime}(x)=0$
Integrating, $f^{\prime}(x)-g^{\prime}(x)=c \Rightarrow f^{\prime}(1)-g^{\prime}(1)=c \Rightarrow 4-2=c \Rightarrow c=2$
$\therefore f^{\prime}(x)-g^{\prime}(x)=2 ;$ Integrating, $f(x)-g(x)=2 x+c_1$
$\Rightarrow f(2)-g(2)=4+c_1 \Rightarrow 9-3=4+c_1 \Rightarrow c_1=2 \quad \therefore f(x)-g(x)=2 x+2$
At $x=3 / 2, f(x)-g(x)=3+2=5$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2002
$f(x)=\cos x$ is monotonic decreasing under which one of the following conditions? $\quad$

Options:
A $0 < x < \frac{\pi}{2}$ only
B $\frac{\pi}{2} < x < \pi$ only
C $\quad 0 < x < \pi$
D $\quad 0^{2} < x < 2 \pi$

The correct answer is:
$\quad 0 < x < \pi$

Solution:
Given, $f(x)=\cos x$,
$\Rightarrow f^{\prime}(x)=-\sin x$
Now $f^{\prime}(x)=0$
$\Rightarrow-\sin x=0$
$\Rightarrow \sin x=\sin (0)$
$\Rightarrow x=n \pi$
Clearly, $f^{\prime}(x) < 0$, when $0 < x < _{x}$
Hence $f(x)$ is decreasing when $0 < x < \pi$

Topic: Mathematics, Application of Derivatives, JEE Main
fx=limn→∞⁡cos2n⁡πx2+x (where, ⋅ denotes the greatest integer function and n∈N) is

Options:
A continuous at x=1 but discontinuous at x=0
B continuous at x=1 and x=0
C discontinuous at x=1 and x=0
D discontinuous at x=1 but continuous at x=0

The correct answer is:
discontinuous at x=1 and x=0

Solution:
fx=limn→∞⁡cos2⁡πx2n+x=xcos2⁡πx2∈0,11+xcos2⁡πx2=1
=xx2∉I1+x x2∈I
For x=1
f1-=1-=0, f1+=1+=1, f1=1+1=2
∴f1+≠f1-

∴fx is discontinuous at x=1
For x=0
f0-=0-=-1, f0+=0+=0, f0=1+0=1
∴f0+≠f0-
∴fx is discontinuous at x=0

Topic: Mathematics, Continuity and Differentiability, JEE Main
$f(x)=\log \left[\frac{1+x}{1-x}\right]$, then $f\left[\frac{2 x}{1+x^2}\right]$ is equal to

Options:
A $[f(x)]^2$
B $[f(x)]^3$
C $2 f(x)$
D $3 f(x)$

The correct answer is:
$2 f(x)$

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

Topic: Mathematics, Functions, JEE Main
$f(x)=\log \frac{1+x}{1-x}$, then $f(x)$ is

Options:
A Even function
B $f\left(x_1\right) f\left(x_2\right)=f\left(x_1+x_2\right)$
C $\frac{f\left(x_1\right)}{f\left(x_2\right)}=f\left(x_1-x_2\right)$
D Odd function

The correct answer is:
Odd function

Solution:
Here,
$f(x)=\log \left(\frac{1+x}{1-x}\right)$
and
$\begin{aligned} f(-x)=\log \left(\frac{1-x}{1+x}\right)= & \log \left(\frac{1+x}{1-x}\right)^{-1} \\ =-\log \left(\frac{1+x}{1-x}\right)=-f(x) \end{aligned}$
$\Rightarrow f(x)$ is an odd function

Topic: Mathematics, Functions, JEE Main
$f(x)=\int \frac{d x}{\sin ^6 x}$ is a polynomial of degree

Options:
A
5 in $\cot x$

B
5 in $\tan x$

C
3 in $\tan x$

D
3 in $\cot x$

The correct answer is:
5 in $\cot x$

Solution:
Let $f(x)=\int \frac{d x}{\sin ^6 x}$
$$
f(x)=\int \operatorname{cosec}^6 x d x
$$
From reduction formula, we have
$$
\begin{aligned}
& \mathrm{I}_{\mathrm{n}}=\int \operatorname{cosec}^n x d x \\
& =-\frac{\operatorname{cosec}^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \mathrm{I}_{n-2}
\end{aligned}
$$
$$
\begin{gathered}
\therefore f(x)=-\frac{\operatorname{cosec}^4 x \cot x}{5}+\frac{4}{5}\left[\frac{-\operatorname{cosec}^2 x \cot x}{3}+\frac{2}{3} \mathrm{I}_2\right] \\
=-\frac{\operatorname{cosec}^4 x \cot x}{5}-\frac{4}{15} \operatorname{cosec}^2 x \cdot \cot x+\frac{8}{15}[-\cot x] \\
=\frac{-\left(1+\cot ^2 x\right)^2 \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^2 x\right) \cot x \\
=\frac{-1}{5}\left[1+\cot ^4 x+2 \cot ^2 x\right] \cot x-\frac{4}{15}[-\cot x)\left(\because \operatorname{cosec}^2 x=1+\cot ^2 x\right) \\
=\frac{-1}{5}\left[\cot x+\cot ^3 x\right] \\
\frac{-4}{15} \cot ^3 x-\frac{4}{15} \cot ^3 x-\frac{8}{15} \cot x \\
\cot x]
\end{gathered}
$$
$$
\begin{aligned}
& =\frac{-15}{15} \cot x-\frac{\cot ^5 x}{5}-\frac{10}{15} \cot ^3 x \\
& =\frac{-\cot ^5 x}{5}-\frac{2}{3} \cot ^3 x-\cot x
\end{aligned}
$$
It is a polynomial of degree 5 in $\cot x$.

Topic: Mathematics, Indefinite Integration, JEE Main
$f(x)=\int \frac{d x}{\sin ^6 x}$ is a polynomial of degree

Options:
A
5 in $\cot x$

B
5 in $\tan x$

C
3 in $\tan x$

D
3 in $\cot x$

The correct answer is:
5 in $\cot x$

Solution:
Let $f(x)=\int \frac{d x}{\sin ^6 x}$
$$
f(x)=\int \operatorname{cosec}^6 x d x
$$
From reduction formula, we have
$$
\begin{aligned}
& \mathrm{I}_{\mathrm{n}}=\int \operatorname{cosec}^n x d x \\
& =-\frac{\operatorname{cosec}^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \mathrm{I}_{n-2}
\end{aligned}
$$
$$
\begin{gathered}
\therefore f(x)=-\frac{\operatorname{cosec}^4 x \cot x}{5}+\frac{4}{5}\left[\frac{-\operatorname{cosec}^2 x \cot x}{3}+\frac{2}{3} \mathrm{I}_2\right] \\
=-\frac{\operatorname{cosec}^4 x \cot x}{5}-\frac{4}{15} \operatorname{cosec}^2 x \cdot \cot x+\frac{8}{15}[-\cot x] \\
=\frac{-\left(1+\cot ^2 x\right)^2 \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^2 x\right) \cot x \\
=\frac{-1}{5}\left[1+\cot ^4 x+2 \cot ^2 x\right] \cot x-\frac{4}{15}[-\cot x)\left(\because \operatorname{cosec}^2 x=1+\cot ^2 x\right) \\
=\frac{-1}{5}\left[\cot x+\cot ^3 x\right] \\
\frac{-4}{15} \cot ^3 x-\frac{4}{15} \cot ^3 x-\frac{8}{15} \cot x \\
\cot x]
\end{gathered}
$$
$$
\begin{aligned}
& =\frac{-15}{15} \cot x-\frac{\cot ^5 x}{5}-\frac{10}{15} \cot ^3 x \\
& =\frac{-\cot ^5 x}{5}-\frac{2}{3} \cot ^3 x-\cot x
\end{aligned}
$$
It is a polynomial of degree 5 in $\cot x$.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2012 (26 May Online)
$f(x)=|| x|-1|$ is not differentiable at

Options:
A $0$
B $\pm 1,0$
C $1$
D $\pm 1$

The correct answer is:
$\pm 1,0$

Solution:
$\begin{aligned} & =\left\{\begin{array}{l}|x|-1, \quad|x|-1 \geq 0 \\ -|x|+1, \quad|x|-1 \lt 0\end{array}\right. \\ & \quad= \begin{cases}|x|-1, \quad x \leq-1 \text { or } x \geq 1 \\ -|x|+1, & -1 \lt x \lt 1\end{cases} \end{aligned}$
$=\left\{\begin{array}{cc}-x-1, & x \leq-1 \\ x+1, & -1 \lt x \lt 0 \\ -x+1, & 0 \leq x \lt 1 \\ x-1, & x \geq 1\end{array}\right.$
From the graph. It is clear that $f(x)$ is not differentiable at $x=-1,0$ and 1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
$f(x)=\left\{\begin{array}{cc}\frac{\log x}{x-1}, & \text { if } x \neq 1 \\ k, & \text { if } x=1\end{array}\right.$
is continuous at
$x=1$, then the value of $k$ is

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

The correct answer is:
1

Solution:
Given, $f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, & \text { if } x \neq 1 \\ k & , \text { if } x=1\end{array}\right.$ at $x=1$ Since, the function is continuous at $x=1$. Then, $f(1)=\lim _{x \rightarrow 1} f(x)$
$$
k=\lim _{x \rightarrow 1} \frac{\log x}{x-1} \quad\left(\frac{0}{0} \text { from }\right)
$$
Apply L'hopital rule,
$$
\begin{aligned}
&k=\lim _{x \rightarrow 1} \frac{1 / x}{1}=\frac{1 / 1}{1} \\
&k=1
\end{aligned}
$$
$$
\Rightarrow \quad k=1
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
$f(x)=\frac{x-3}{(x-1) \sqrt{x^2-4}}$ Domain of the function

Options:
A $(1,2)$
B $(-\infty,-2) \cup(2, \infty)$
C $(-\infty,-2) \cup(1, \infty)$
D $(-\infty, \infty)-\{1, \pm 2\}$

The correct answer is:
$(-\infty,-2) \cup(2, \infty)$

Solution:
Obviously, here $|x|\gt2$ and $x \neq 1$ i.e., $\quad x \in(-\infty,-2) \cup(2, \infty)$.

Topic: Mathematics, Functions, JEE Main
f $\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c \quad$, then value of k is

Options:
A $\frac{2}{3}$
B $\frac{3}{2}$
C $\frac{5}{2}$
D $\frac{2}{5}$

The correct answer is:
$\frac{3}{2}$

Solution:
(C)
$\begin{array}{l}
\text { Let } \mathrm{I}=\int \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}=\frac{2}{3}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{\mathrm{k}}+\mathrm{c} \\
\text { Put } \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}}\left(1+\frac{1}{\mathrm{x}^{2}}\right) \mathrm{dx}=\mathrm{dt} \\
\therefore\left(\frac{\mathrm{x}^{2}+1}{\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}=2 \mathrm{t} \mathrm{dt} \\
\therefore \mathrm{I}=\int \mathrm{t}(2 \mathrm{t}) \mathrm{dt}=2 \int \mathrm{t}^{2} \mathrm{dt}=\frac{2 \mathrm{t}^{3}}{3}=\frac{2}{3}\left[\sqrt{\mathrm{x}-\frac{1}{\mathrm{x}}}\right]^{3} \\
=\frac{2}{3}\left[\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{\frac{1}{2}}\right]^{3}=\frac{2}{3}\left[\mathrm{x}-\frac{1}{\mathrm{x}}\right]^{\frac{3}{2}} \Rightarrow \mathrm{k}=\frac{3}{2}
\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
$f(x)=\frac{x-3}{x+1}$, then $f[f\{f(x)\}]$ equals

Options:
A $X$
B $-X$
C $\frac{x}{2}$
D $-\frac{1}{x}$

The correct answer is:
$X$

Solution:
$f[f(x)]=\frac{f(x)-3}{f(x)+1}$
$= \frac{\left(\frac{x-3}{x+1}\right)-3}{\left(\frac{x-3}{x+1}\right)+1}=\frac{x-3-3 x-3}{x-3+x+1}=\frac{3+x}{1-x}$
$\text {Now } f[f(f(x))] =f\left(\frac{3+x}{1-x}\right)$
$=\frac{\left(\frac{3+x}{1-x}\right)-3}{\left(\frac{3+x}{1-x}\right)+1}=\frac{3+x-3+3 x}{3+x+1-x}=x$

Topic: Mathematics, Functions, JEE Main
$f(x)=\frac{x}{x-1}$, then $\frac{f(a)}{f(a+1)}=$

Options:
A $f(-a)$
B $f\left(\frac{1}{a}\right)$
C $f\left(a^2\right)$
D $f\left(\frac{-a}{a-1}\right)$

The correct answer is:
$f\left(a^2\right)$

Solution:
Step 1. Find the value of $f(a)$ and $f(a+1)$. In the function $f\left({ }^x\right)=\frac{x}{x-1}$, substitute $a$ for $x$.
$f(a)=\frac{a}{a-1}$
In the function $f\left({ }^x\right)=\frac{x}{x-1}$, substitute $a+1$ for $x$.
$\begin{aligned}
f(a+1) & =\frac{a+1}{a+1-1} \\
& =\frac{a+1}{a}
\end{aligned}$
Step 2. Find the value of $\frac{f\left({ }^{(a)}\right.}{f(a+1)}$.
The value of $\frac{f\left({ }^{(a)}\right.}{f(a+1)}$ is given as:
$\frac{f\left({ }^{(a)}\right.}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a}}$
$\begin{aligned}
& =\frac{a \times a}{(a+1)(a-1)} \\
& =\frac{a^2}{a^2-1}
\end{aligned}$
The value $\frac{a^2}{a^2-1}$ is the function value of $f\left({ }^x\right)=\frac{x}{x-1}$ when $x=a^2$.
So, $\frac{f\left({ }^a\right)}{f(a+1)}=f\left(a^2\right)$.

Topic: Mathematics, Functions, JEE Main
$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x^2}, & \text { if } x \leq 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text { if } x>0\end{array}\right.$ is continuous at $x=0$, then the value of $\mathrm{k}$ is

Options:
A 4
B 2
C -1
D -3

The correct answer is:
4

Solution:
$\mathrm{f}(x)$ is continuous at $x=0$
$\begin{aligned}
\therefore \quad \text { L.H.L. } & =\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{\mathrm{k} x}{2}}{x^2} \\
& =2 \lim _{x \rightarrow 0} \frac{\sin ^2 \frac{\mathrm{k} x}{2}}{\frac{\mathrm{k}^2 x^2}{4} \times 4} \cdot \mathrm{k}^2 \\
& =\frac{1}{2} \mathrm{k}^2 \lim _{x \rightarrow 0}\left(\frac{\sin \frac{\mathrm{k} x}{2}}{\frac{\mathrm{k} x}{2}}\right)^2 \\
\therefore \quad \text { L.H.L. } & =\frac{1}{2} \mathrm{k}^2
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { R.H.L. } & =\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{x})(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} \\
& =\sqrt{16+\sqrt{0}+4}
\end{aligned}$
$\therefore \quad$ R.H.L. $=8$
$\mathrm{f}(x)$ is continuous at $x=0$... [Given]
$\begin{array}{ll}
\therefore \quad & \text { L.H.L. }=\text { R.H.L } \\
\therefore \quad & \frac{1}{2} \mathrm{k}^2=8 \\
\therefore \quad & \mathrm{k}^2=16 \\
\therefore \quad \mathrm{k}= \pm 4
\end{array}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
f(x)=x2+1;g(x)=x+1x2+1;h(x)=2x-3. Then the value of f'h'g'(x)=

Options:
A 5
B 25
C 52
D 15

The correct answer is:
25

Solution:
Given that

f(x)=x2+1;g(x)=x+1x2+1;h(x)=2x-3

By using Chain rule

f'x = 2x2x2+1 = xx2+1

By using Quotient rule

g'x = x2+11-x2xx2+12 = x3-2x2+1x2+12

h'x = 2

f'h'g'x = f'h'x3-2x2+1x2+12

=f'2

=222+1 = 25

Topic: Mathematics, Differentiation, JEE Main
$f(x)=\frac{x^2-4}{x^2+4}$ for $|x|\gt2$, then the function $f:(-\infty,-2] \cup[2, \infty) \rightarrow(-1,1)$ is

Options:
A One-one into
B One-one onto
C Many one into
D Many one onto

The correct answer is:
Many one into

Solution:
Let $f(x)=f(y) \Rightarrow \frac{x^2-4}{x^2+4}=\frac{y^2-4}{y^2+4}$
$\Rightarrow \frac{x^2-4}{x^2+4}-1=\frac{y^2-4}{y^2+4}-1 \Rightarrow x^2+4=y^2+4$
$\Rightarrow x= \pm y, \therefore f(x)$ is many-one.
Now for each $y \in(-1,1)$, there does not exist $x \in X$ such that $f(x)=y$. Hence $f$ is into.

Topic: Mathematics, Functions, JEE Main
$f(x)=\left|\begin{array}{ccc}x^3 & x^2 & 3 x^2 \\ 1 & -6 & 4 \\ p & p^2 & p^3\end{array}\right|$, here $p$ is a constant, then $\frac{d^3 f(x)}{d x^3}$ is

Options:
A Proportional to $x^2$
B Proportional to $x$
C Proportional to $x^3$
D A constant

The correct answer is:
A constant

Solution:
$f(x)=\left|\begin{array}{ccc}x^3 & x^2 & 3 x^2 \\ 1 & -6 & 4 \\ p & p^2 & p^3\end{array}\right|$
$\begin{aligned} & \Rightarrow f(x)=x^3\left(-6 p^3-4 p^2\right)-x^2\left(p^3-4 p\right)+3 x^2\left(p^2+6 p\right) \\ & \Rightarrow f(x)=-6 p^3 x^3-4 p^2 x^3-x^2 p^3+4 p x^2+3 p^2 x^2+18 p x \\ & \frac{d}{d x} f(x)=-18 p^3 x^2-12 p^2 x^2-2 x p^3+8 p x+6 p^2 x+36 p \\ & \text { and } \frac{d^2}{d x^2} f(x)=-36 p^3 x-24 p^2 x-2 p^3+8 p+6 p^2+36 p \\ & \text { and } \frac{d^3 f(x)}{d x^3}=-36 p^3-24 p^2=\text { a constant. }\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
$f(x)=\left|\begin{array}{ccc}x-3 & 2 x^2-18 & 3 x^3-81 \\ x-5 & 2 x^2-50 & 4 x^3-500 \\ 1 & 2 & 3\end{array}\right|$ then $f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)=$

Options:
A $f(1)$
B $f(3)$
C $f(1)+f(3)$
D $f(1)+f(5)$

The correct answer is:
$f(3)$

Solution:
$f(x)=2(x-3)(x-5) ;\left|\begin{array}{ccc}1 & x+3 & 3\left(x^2+3 x+9\right) \\ 1 & x+5 & 4\left(x^2+5 x+25\right) \\ 1 & 1 & 3\end{array}\right|$
(Taking out $(x-3),(x-5)$ and 2 from 1st row, 2nd row and 3rd column respectively)
$f(x)=2(x-3)(x-5)$
$\left|\begin{array}{ccc}0 & (x+2) & 3\left(x^2+3 x+8\right) \\ 0 & 2 & x^2+11 x+73 \\ 1 & 1 & 3\end{array}\right|$ $\left(R_1 \rightarrow R_1-R_3\right.$ and $\left.R_2 \rightarrow R_2-R_1\right)$
$\begin{aligned} & =2(x-3)(x-5)\left[1(x+2)\left(x^2+11 x+73\right)-6\left(x^2+3 x+8\right)\right] \\ & =2\left(x^2-8 x+15\right)\left(x^3+13 x^2+95 x+146-6 x^2-18 x-48\right) \\ & =2\left(x^2-8 x+15\right)\left(x^3+7 x^2+77 x+98\right) \\ & \quad=2\left(x^5-x^4+36 x^3-413 x^2+371 x+1470\right)\end{aligned}$
$\begin{aligned} & f(1)=2928, f(3)=0, f(5)=0 \\ & \therefore f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)=0+0+0=0=f(3)\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
$f(x)=\frac{2 x-1}{x+5}(x \neq-5)$, then $f^{-1}(x)$ is equal to

Options:
A $\frac{x+5}{2 x-1}, x \neq \frac{1}{2}$
B $\frac{5 x+1}{2-x}, x \neq 2$
C $\frac{5 x-1}{2-x}, x \neq 2$
D $\frac{x-5}{2 x+1}, x \neq \frac{1}{2}$

The correct answer is:
$\frac{5 x+1}{2-x}, x \neq 2$

Solution:
$\begin{aligned} & \text { Let } f(x)=y \Rightarrow x=f^{-1}(y) . \text { Now, } y=\frac{2 x-1}{x+5},(x \neq-5) \\ & x y+5 y=2 x-1 \Rightarrow 5 y+1=2 x-x y . \\ & \Rightarrow x(2-y)=5 y+1 \Rightarrow x=\frac{5 y+1}{2-y} \Rightarrow f^{-1}(y)=\frac{5 y+1}{2-y} \\ & \quad f^{-1}(x)=\frac{5 x+1}{2-x}, x \neq 2\end{aligned}$

Topic: Mathematics, Functions, JEE Main
$f(x)=\left\{\begin{array}{rl}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$, then $\lim _{x \rightarrow 0} f(x)=$

Options:
A $1$
B $0$
C $-1$
D None of these

The correct answer is:
$0$

Solution:
Here $f(0)=0$
Since $-1 \leq \sin \frac{1}{x} \leq 1 \Rightarrow-|x| \leq x \sin \frac{1}{x} \leq|x|$
We know that $\lim _{x \rightarrow 0}|x|=0$ and $\lim _{x \rightarrow 0}-|x|=0$
In this way $\lim _{x \rightarrow 0} f(x)=0$.

Topic: Mathematics, Limits, JEE Main
$f(x)=\left\{\begin{array}{r}x, \text { when } 0 \leq x \leq 1 \\ 2-x, \text { when } 1 \lt x \leq 2\end{array}\right.$, then $\lim _{x \rightarrow 1} f(x)=$

Options:
A $1$
B $2$
C $0$
D Does not exist

The correct answer is:
$1$

Solution:
Hence $\lim _{x \rightarrow 1} f(x)=1$
Aliter : $\lim _{x \rightarrow 1-} f(x)=\lim _{h \rightarrow 0}(1-h)=1$
and $\lim _{x \rightarrow 1+} f(x)=\lim _{h \rightarrow 0} 2-(1+h)=1$
Hence limit of function is $1$

Topic: Mathematics, Limits, JEE Main
$f(x)=\left\{\begin{array}{l}x: x \lt 0 \\ 1: x=0, \\ x^2: x\gt0\end{array}\right.$ then $\lim _{x \rightarrow 0} f(x)=$

Options:
A $0$
B $1$
C $2$
D Does not exist

The correct answer is:
Does not exist

Solution:
\(\begin{aligned}
& f(x)=\left\{\begin{array}{l}
x: x < 0 \\
1: x=0 \\
x^2: x > 0
\end{array}\right. \\
& \lim _{x \rightarrow 0} f(x)=?
\end{aligned}\)
so first find
L.H.L \(\lim _{x \rightarrow 0^{-}} x=0\)
R.H.L \(\lim _{x \rightarrow 0^{+}} x^2=0\)
But given
\(\lim _{x \rightarrow 0} f(x)=1\)
So, we can say that limit at \(\mathrm{x}=\mathrm{o}\) does not exist.

Topic: Mathematics, Limits, JEE Main
$f(x)=x+\sqrt{x^2}$ is a function from $R \rightarrow R$, then $f(x)$ is

Options:
A Injective
B Surjective
C Bijective
D None of these

The correct answer is:
None of these

Solution:
We have $f(x)=x+\sqrt{x^2}=x+|x|$
Clearly $f$ is not one-one as $f(-1)=f(-2)=0$ but $-1 \neq 2$.
Also $f$ is not onto as $f(x) \geq 0, \forall x \in R$,
Also, range of $f=(0, \infty) \subset R$.

Topic: Mathematics, Functions, JEE Main
$f(x)=\left\{\begin{array}{ll}x, & \text { when } x\gt1 \\ x^2, & \text { when } x \lt 1\end{array}\right.$, then $\lim _{x \rightarrow 1} f(x)=$

Options:
A $x^2$
B $x$
C $-1$
D $1$

The correct answer is:
$1$

Solution:
$\lim _{x \rightarrow 1^{-}} f(x)=1=\lim _{x \rightarrow 1+} f(x)$.

Topic: Mathematics, Limits, JEE Main
f $y=c e^{\sin ^{-1} x}$ , then corresponding to this the differential equation is

Options:
A $\frac{d y}{d x}=\frac{y}{\sqrt{1-x^2}}$
B $\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}$
C $\frac{d y}{d x}=\frac{x}{\sqrt{1-x^2}}$
D None of these

The correct answer is:
$\frac{d y}{d x}=\frac{y}{\sqrt{1-x^2}}$

Solution:
$y=c e^{\sin ^{-1} x}$. Differentiate it w.r.t. $x$, we get
$\frac{d y}{d x}=\operatorname{ce}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}=\frac{y}{\sqrt{1-x^2}}$ or $\quad \frac{d y}{d x}=\frac{y}{\sqrt{1-x^2}}$.

Topic: Mathematics, Differential Equations, JEE Main
Factors affecting KC is -

Options:
A Increasing concentration of the reactant
B Presence of catalyst
C Method of writing balanced equation (or stoichiometry of reaction)
D Time taken by the chemical reaction

The correct answer is:
Method of writing balanced equation (or stoichiometry of reaction)

Solution:
KC depends on the expression that the coefficient of reactants and products in a given balanced equation.

Topic: Chemistry, Chemical Equilibrium, JEE Main
False statement is

Options:
A Chloroform is heavier than water
B $\mathrm{CCl}_4$ is non-inflammable
C Vinyl chloride is more reactive than allyl chloride
D $\mathrm{Br}^{-}$is a good nucleophile as compared to ${I^{-}}^{-}$

The correct answers are:
Vinyl chloride is more reactive than allyl chloride, $\mathrm{Br}^{-}$is a good nucleophile as compared to ${I^{-}}^{-}$

Solution:
Vinyl chloride is less reactive than allyl chloride due to resonance effect.
Order of nucleophilicity amongst the halide ion are as $\mathrm{I}^{-}\gt\mathrm{Br}^{-}\gt\mathrm{Cl}^{-}$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Faraday's laws of electrolysis will fail when

Options:
A Temperature is increased
B Inert electrodes are used
C A mixture of electrolytes is used
D In none of these cases

The correct answer is:
In none of these cases

Solution:
To check whether the faraday's law of electrolysis fails under the given situation first we have to define the faraday's law of electrolysis. The faraday's first law of electrolysis states that the amount of any substances deposited or liberated at any of the electrodes is directly proportional to the quantity of electricity passed through the electrolyte.
$W \propto Q$
Where $W$ is the amount of substance deposited or liberated in grams
And $Q$ is the quantity of electricity in coulombs.
$W=Z Q$
$\mathrm{Z}$ is the constant of proportionality and is known as the electrochemical equivalent of the substance.
We know that,
$Q= It$
Where ${I}$ is the current in amperes
And ' $t$ ' is the time in seconds.
Thus $\mathrm{W}=\mathrm{Zlt}$
From this, we understand that the faraday's first law of electrolysis does not depend on temperature, inert electrodes or the mixture of electrolytes used.
Now we can check the Faraday's second law of electrolysis
It states that when the same quantity of electricity is passed through solutions of different electrolytes connected in series, the amount of different substances deposited at the electrodes is directly proportional to their equivalent weights.
i.e., $\frac{W_1}{W_2}=\frac{E_1}{E_2}$
Where $W_1$ and $W_2$ are the weights of electrolyte 1 and electrolyte 2 deposited respectively and $E_1$ and $E_2$ are their respective equivalent weights.
From this also we can see that it does not depend on any of the temperatures, inert electrodes are used or a mixture of electrolytes is used.
Hence option (4) is correct.
Note: The faraday's law of electrolysis mainly depends only on the amount of electricity passed and the time it is passed. Since the amount of any substance deposited or liberated depends of quantity of electricity Is passed

Topic: Chemistry, Electrochemistry, JEE Main
Farmers in a particular region were concerned that pre-mature yellowing of leaves ora pulse crop might cause decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield?

Options:
A application of iron and magnesium to promote synthesis of chlorophyll
B frequent irrigation of the crop
C treatment of the plants with cytokinins along with a small dose of nitrogenous fertilizer
D removal of all yellow leaves and spraying the remaining green leaves with 2, 4, 5-trichlorophenoxy acetic acid

The correct answer is:
application of iron and magnesium to promote synthesis of chlorophyll

Solution:
Premature yellowing of leaves in pulse crops occurs because of degradation of chlorophyll. $\mathrm{Mg}$ and $\mathrm{Fe}$ are required for the biosynthesis of chlorophyll. Hence, $\mathrm{Mg}$ and Fe supplements will increase the yield of pulse crops.

Topic: Biology, Mineral Nutrition, NEET, NEET 2006