datavorous/entrance-exam-dataset huggingface dataset:
The function $f(x)=e^{x}, x \in R$ is

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

The correct answer is:
one-one but not onto

Solution:
Let $f(x)=f(y)$ To show that $f(x)$ is one-one We have to show that $x=y$ Now, $f(x)=f(y)$
$\Rightarrow e^{x}=e^{y} \Rightarrow \frac{e^{x}}{e^{y}}=1 \Rightarrow e^{x-y}=1$
Take log on both side $\log e^{x-y}=\log 1$
$\Rightarrow \quad x-y=0 \Rightarrow x=y$
Hence $f(x)$ one-one $\forall x \in R$
$-2$ is an element of the co-domain $\mathrm{R}$. There doesnot exist any element $\mathrm{X}$ in the domain $\mathrm{R}$ such that $-2=e^{x}=f(x)$
Hence, by definition, $\mathrm{f}$ is not a onto function.

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 1)
The function fx=limn→∞cos2nπx+x is (where, . denotes the greatest integer function and n∈N)

Options:
A continuous at x=1 but discontinuous at x=32
B continuous at x=1 and x=32
C discontinuous at x=1 and x=32
D discontinuous at x=1 but continuous at x=32

The correct answer is:
discontinuous at x=1 but continuous at x=32

Solution:
fx=limn→∞cos2πxn+x

=x:cos2πx∈0,11+x:cos2πx=1
=x:x∉Ι1+x:x∈Ι
For x=1,
f1-=1-=0, f1+=1+=1, f1=1+1=2
∵f1+≠f1-⇒fx is discontinuous at x=1
For x=32,

f32+=32+=1, f32-=32-=1, f32=32=1
∵f32-=f32+=f32
⇒fx is continuous at x=32

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=limn→∞x-22n-1x-22n+1 ∀n∈N is discontinuous at

Options:
A x=1 only
B x=3 only
C x=1 and 3
D x=0,1 and 2

The correct answer is:
x=1 and 3

Solution:
Given, fx=limn→∞x-22n-1x-22n+1
⇒fx=limn→∞1-1x-22n1+1x-22n
⇒fx=-1,0,1,0≤x-22<1x-22=1x-22>1
⇒fx=1,x<10,x=1-1,1<x<30,x=31,x>3
Thus, fx is discontinuous at x=1,3

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

Options:
A $(0, \infty)$
B $(-\infty, 0)$
C $(-\infty, \infty)$
D None of these

The correct answer is:
$(0, \infty)$

Solution:
Given, $f(x)=\log (1+x)-\frac{2 x}{2+x}$
$$
\begin{aligned}
\therefore \quad f^{\prime}(x) &=\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \\
&=\frac{x^{2}}{(1+x)(x+2)^{2}}
\end{aligned}
$$
Clearly, $f^{\prime}(x)>0$ for all $x>0$. Hence, $f(x)$ is increasing on $(0, \infty)$.

Topic: Mathematics, Application of Derivatives, JEE Main
The function $f(x)=\log \left(x+\sqrt{x^2+1}\right)$, is

Options:
A An even function
B An odd function
C A Periodic function
D Neither an even nor odd function

The correct answer is:
An odd function

Solution:
$f(x)=\log \left(x+\sqrt{x^2+1}\right)$
and $f(-x)=-\log \left(x+\sqrt{x^2+1}\right)=-f(x)$
$f(x)$ is odd function.

Topic: Mathematics, Functions, JEE Main
The function $f(x)=\log \left(x+\sqrt{x^2+1}\right)$, is

Options:
A
neither an even nor an odd function

B
an even function

C
an odd function

D
a periodic function

The correct answer is:
an odd function

Solution:
$f(x)=\log \left(x+\sqrt{x^2+1}\right)$
$f(-x)=-\log \left(x+\sqrt{x^2+1}\right)$
$f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.

Topic: Mathematics, Functions, JEE Main, JEE Main 2003
The function fx=max1-x, 1+x, 2 âxâR is

Options:
A discontinuous at exactly two points
B differentiable âxâR
C differentiable âxâR--1,1
D continuous ∀x∈R-0,1,-1

The correct answer is:
differentiable ∀x∈R--1,1

Solution:
fx=1-x:x≤-12:-1<x≤11+x:x>1
Continuity at x=-1
f-1=1--1=2
f-1-=1--1=2
f-1+=2
∵f-1=f-1-=f-1+
∴ continuous at x=-1
f1=2, f1-=2
f1+=1+1=2
∵f1-=f1=f1+

∴ continuous at x=1
For differentiability,
f ′ x = −1 x<−1 0 −1<x<1 1 x>1
at x=-1,

f'-1-=-1, f-1+=0
∵f'−1−≠f'−1+⇒non-differentiable
at x=1,

f'1-=0, f'1+=1
∵f'1−≠f'1+⇒non-differentiable

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=sin-12x-x2+2-1x+1x2is defined in the interval (where · is the greatest integer function)

Options:
A x∈1-2,1
B x∈1,1+2
C x∈1-2,1+2
D x∈1-2,2

The correct answer is:
x∈1,1+2

Solution:
(i) -1≤2x-x2≤1 (for sin-1 to be defined)
⇒-1≤x2-2x≤1
i.e. x2-2x+1≥0 and x2-2x-1≤0
x-12≥0 and x-12-22≤0
x∈R and x-1-2x-1+2≤0
⇒x∈1-2,1+2 … 1
(ii) 2-1x≥0⇒1x≤2⇒x≥12⇒x∈-∞,-12∪12,∞ … 2
(iii) x2≠0⇒x2∉0,1
⇒x∉-1,1⇒x∈-∞,-1∪1,∞ … 3
Hence, 1∩2∩3
⇒x∈1,1+2

Topic: Mathematics, Functions, JEE Main
The function f(x)=seclogx+1+x2 is

Options:
A an odd function
B an even function
C neither an odd nor an even function
D a constant function

The correct answer is:
an even function

Solution:
f âx =sec log âx+ 1+ x 2
=secâ¡logâ¡x+ 1+x2-1
âµ1+x2-x= 11+x2+x
=sec âlog x+ 1+ x 2
=seclog x+1+x2 âµsecâÎ¸=secÎ¸
=f(x)
Hence f(x) is an even function.

Topic: Mathematics, Functions, JEE Main
The function fx=sin4x+cos2x, is a periodic function with a fundamental period

Options:
A π
B 2 π
C π 4
D π 2

The correct answer is:
π 2

Solution:
Given fx=sin4x+cos2x

We know that the period of sinx is π and also if fx is a periodic function with fundamental period T, then the fundamental period of fkx is Tk.

⇒T1=period of sin4x=π4

Also, we know that the period of cosx is π

⇒T2=Period of cos2x=π2

Again, we know that the period of the sum or difference of two periodic functions is the L.C.M. of their periods.

So, the period of f(x) is L.C.M. of the periods of sin4x and cos2x i.e. the L.C.M. T1, T2

Now, L.C.M. π4, π2=L.C.M. π, πH.C.F. 4, 2=π2.

Topic: Mathematics, Functions, JEE Main, JEE Main 2014 (19 Apr Online)
The function $f(x)=\sin \left(\log \left(x+\sqrt{x^2+1}\right)\right)$ is

Options:
A Even function
B Odd function
C Neither even nor odd
D Periodic function

The correct answer is:
Odd function

Solution:
$\begin{aligned}
f(x) & =\sin \left(\log \left(x+\sqrt{1+x^2}\right)\right) \\
& \Rightarrow f(-x)=\sin \left[\log \left(-x+\sqrt{1+x^2}\right)\right] \\
& \Rightarrow f(-x)=\sin \log \left(\left(\sqrt{1+x^2}-x\right) \frac{\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}\right) \\
& \Rightarrow f(-x)=\sin \log \left[\frac{1}{\left(x+\sqrt{1+x^2}\right)}\right] \\
& \Rightarrow f(-x)=\sin \left[\log \left(x+\sqrt{1+x^2}\right)^{-1}\right] \\
& \Rightarrow f(-x)=\sin \left[-\log \left(x+\sqrt{1+x^2}\right)\right] \\
& \Rightarrow f(-x)=-\sin \left[\log \left(x+\sqrt{1+x^2}\right)\right] \Rightarrow f(-x)=-f(x)
\end{aligned}$
$\therefore f(x)$ is odd function.

Topic: Mathematics, Functions, JEE Main
The function $f(x)=\sin |x|$ is

Options:
A Continuous for all $X$
B Continuous only at certain points
C Differentiable at all points
D None of these

The correct answer is:
Continuous for all $X$

Solution:
To analyze the function $f(x)=\sin (|x|)$ with respect to its continuity and differentiability, let's go through the details step-by-step:
1. Continuity:
- The function $\sin (x)$ is known to be continuous for all $x \in \mathbb{R}$.
- The absolute value function $|x|$ is also continuous for all $x \in \mathbb{R}$.
- Since the composition of continuous functions is continuous, the function $\sin (|x|)$ is continuous for all $x$.
Therefore, $f(x)=\sin (|x|)$ is continuous for all $x$. This makes option (a) "Continuous for all $x^{\prime \prime}$ correct.
2. Differentiability;
- For $x\gt0, f(x)=\sin (x)$. The derivative in this region is $f^{\prime}(x)=\cos (x)$.
- For $x \lt 0, f(x)=\sin (-x)=-\sin (x)$. The derivative in this region is $f^{\prime}(x)=-\cos (x)$.
At $x=0$ :
- The left-hand derivative (LHD) is calculated as:
$\lim _{h \rightarrow 0^{-}} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0^{-}} \frac{-\sin (h)-0}{h}=\lim _{h \rightarrow 0^{-}} \frac{-\sin (h)}{h}=-\cos (0)=-1$
- The right-hand derivative (RHD) is calculated as:
$\lim _{h \rightarrow 0^{+}} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{\sin (h)-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{\sin (h)}{h}=\cos (0)=1$
Since the left-hand derivative -1 and the right-hand derivative 1 at $x=0$ are not equal, $f(x)$ is not differentiable at $x=0$.
Therefore, option (c) "Differentiable at all points" is incorrect.
Based on this analysis, the correct answer is:
(a) Continuous for all $x$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=sin⁡xtan2x is not defined at x=π2. The value of fπ2 such that f is continuous at x=π2 is

Options:
A e
B 1e
C 2
D None of these

The correct answer is:
1e

Solution:
f is continuous at x=π2, if
limx→π2fx=fπ2
Now, L=limx→π2sin⁡xtan2x
⇒log⁡L=limx→π2tan2xlog⁡sin⁡x
=limx→π2log⁡sin⁡xcot2x00
=limx→π2cotx2cot⁡xcosec2x=-12
or L=e-12

∴fπ2=e-12=1e

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in

Options:
A
$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

B
$\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$

C
$\left(0, \frac{\pi}{2}\right)$

D
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

The correct answer is:
$\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$

Solution:
$f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^2}(\cos x-\sin x)$
$=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
$f(x)$ is increasing if $-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2}$
$-\frac{3 \pi}{4} < x < \frac{\pi}{4}$
hence $f(x)$ is increasing when $x \in\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2007
The function fx=tan⁡x+1x, ∀x∈0,π2 has

Options:
A one local maximum
B one local minimum
C one local maximum and one minimum
D no local maximum or minimum

The correct answer is:
one local minimum

Solution:
dydx=sec2⁡x–1x2
=x-cos⁡xx+cos⁡xcos2⁡x.x2
Hence, at α sign of dydx changes from negative to positive
i.e., y=fx has a local minimum

Topic: Mathematics, Application of Derivatives, JEE Main
The function $f(x)=\tan x-x$

Options:
A Always increases
B Always decreases
C Never decreases
D Sometimes increases and sometimes decreases

The correct answer is:
Always increases

Solution:
$y=\tan x-x \Rightarrow \frac{d y}{d x}=\sec ^2 x-1=\tan ^2 x \geq 0$

Topic: Mathematics, Application of Derivatives, JEE Main
The function f(x), that satisfies the condition f(x)=x+∫0π/2sinxcosyf(y)dy, is :

Options:
A x+π2sinx
B x+(π+2)sinx
C x+23(π-2)sinx
D x+(π-2)sinx

The correct answer is:
x+(π-2)sinx

Solution:
Given the function fx=x+∫0π2sinx·cosyfydy

fx=x+ksinx ...i

fy=y+ksiny ...ii

k=∫0π2cosyfydy

From equations i & ii

fx=x+∫0π2sinxcosyy+ksinydy

fx=x+sinx∫0π2ycosydy+k2sinx∫0π2sin2ydy

fx=x+sinxy∫0π2cosydy-∫0π2ddyy∫0π2cosydydy+k2sinx∫0π2sin2ydy

fx=x+sinxπ2-1+k2sinx×1

fx=x+(π-2)sinx2+ksinx2 ...iii

from (i) and (iii)

k=π2-1+k2

k=π-2

Hence, f(x)=x+(π-2)sinx

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The function fx=x1+|x| is

Options:
A strictly increasing
B strictly decreasing
C neither increasing nor decreasing
D not differentiable at x=0

The correct answer is:
strictly increasing

Solution:
Given, fx=x1+|x|
∴ f′x=1+x.1-x.xx1+x2
=11+x2>0 ∀x∈R
⇒ f(x) is strictly increasing

Topic: Mathematics, Application of Derivatives, JEE Main
The function, $f(x)=x \sqrt{1-x}$, where $x \in(0,1)$, has local maximum at $x=$

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

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

Solution:
$\begin{aligned}
& f(x)=x \sqrt{1-x} \\
& \Rightarrow f^{\prime}(x)=1 \sqrt{1-x}+\frac{x}{2 \sqrt{1-x}} x(-1)=\frac{2-3 x}{2 \sqrt{1-x}}
\end{aligned}$
sign scheme of $f^{\prime}(x)$
Hence, local maxima at $x=\frac{2}{3}$

Topic: Mathematics, Application of Derivatives, JEE Main
The function fx=x2-2x-3·e9x2-12x+4 is not differentiable at exactly :

Options:
A Four points
B Two points
C three points
D one point

The correct answer is:
Two points

Solution:
Given thatfx=x2-2x-3e9x2-12x+4

Differentiating on both sides

f'x=x2-2x-3.e9x2-12x+4.18x-12 + x2-2x-3x2-2x-3.2x-2.e9x2-12x+4

⇒f'(x) = x2-2x-3.e9x2-12x+4 18x-12+2x-2x2-2x-3

⇒f'x = x2-2x-3.e9x2-12x+4.18x3-36x2-54x-12x2+24x+36+2x-2x2-2x-3

⇒f'x = x2-2x-3.e9x2-12x+4.18x3-48x2-28x+34x-3x+1

Here f'x is not defined at x = -1 & x=3.

fx is not differentiable at two points.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}$ increases for all $\quad$

Options:
A $\mathrm{x}>-1$ only
B $\mathrm{x} < -1$ only
C $\mathrm{x}>1$ only
D $x < 1$ only

The correct answer is:
$\mathrm{x}>1$ only

Solution:
Given $f(x)=x^{2}-2 x$ On differentiating w.r.t $^{\circ} x^{\prime}$, we get $f^{\prime}(x)=2 x-2$ $f(x)$ is increasing, if $f^{\prime}(x)>0$ $\Rightarrow 2 x-2>0$
$\Rightarrow x>1$

Topic: Mathematics, Application of Derivatives, JEE Main
The function $f(x)=|x-2|+x$ is

Options:
A differentiable at both $x=2$ and $x=0$
B differentiable at $x=2$ but not at $x=0$
C continuous at $\mathrm{x}=2$ but not at $\mathrm{x}=0$
D continuous at both $\mathrm{x}=2$ and $\mathrm{x}=0$

The correct answer is:
continuous at both $\mathrm{x}=2$ and $\mathrm{x}=0$

Solution:
$$
f(x)=|x-2|+x
$$
First we check the continuity
At $\mathrm{x}=0$,
$\begin{aligned} \text { RHL } f(0+h) &=\lim _{h \rightarrow 0}|0+h-2|+(0+h) \\ &=|-2|=2 \\ \text { LHL } f(0-h) &=\lim _{h \rightarrow 0}|0-h-2|+(0-h) \\ &=|-2|=2 \end{aligned}$
At $\mathrm{x}=2$,
$$
\begin{aligned}
\operatorname{RHL} \mathrm{f}(2+\mathrm{h}) &=\lim _{\mathrm{h} \rightarrow 0}|2+\mathrm{h}-2|+(2+\mathrm{h}) \\
&=0+2+0=2
\end{aligned}
$$
LHL $f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)$
$$
=0+2-0=2
$$
and $\quad f(0)=2, f(2)=2$
Hence, $f(x)$, is continuous at $x=0,2$
Now, we check differentiability
$$
f(x)=\left\{\begin{array}{c}
(-x+2-x), x < 0 \\
(-x+2+x), 0 \leq x \leq 2 \\
(x-2+2), 2 \leq x
\end{array}\right.
$$
$$
f(x)=\left\{\begin{array}{c}
2-2 x, x < 0 \\
2,0 \leq x \leq 2 \\
2 x-2,2 \leq x
\end{array}\right.
$$
Now,$f^{\prime}(x)=\left\{\begin{array}{c}-2, x < 0 \\ 0,0 \leq x \leq 2 \\ 2,2 \leq x\end{array}\right.$
Hence, $f(x)$ is not differentiable at $x=0,2$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=πx3-3π2a+bx2+3πabx has a local minimum at x=a, then the values a and b can take are

Options:
A a=π,b=e
B a=e,b=π
C a=b=π
D a=b=e

The correct answer is:
a=e,b=π

Solution:
f ′ x =3π x 2 − a+b x+ab
=3πx-ax-b
i.e. fx has a local maximum at x=a, if a<b

Topic: Mathematics, Application of Derivatives, JEE Main
The function fx=x3-6x2+ax+b is such that f2=f4=0.

Consider two statements:

S1 there exists x1,x2∈2,4, x1<x2, such that f'x1=-1 and f'x2=0.

S2 there exists x3,x4∈2,4, x3<x4, such that f is decreasing in 2,x4, increasing in x4,4 and 2f'x3=3fx4 then

Options:
A S1 is true and S2 is false
B both S1 and S2 are false
C both S1 and S2 are true
D S1 is false and S2 is true

The correct answer is:
both S1 and S2 are true

Solution:
Given:

fx=x3-6x2+ax+b

Given that:

f2=0⇒2a+b=16 ....i

f4=0⇒4a+b=32 ....ii

Solving both equations, we get

a=8, b=0

∴ fx=x3-6x2+8x

⇒ fx=xx-2x-4

Now,

f'x=3x2-12x+8

Now,

If f'x=-1

⇒3x2-12x+8=-1

⇒3x2-12x+9=0

⇒x2-4x+3=0

⇒x2-3x-x+3=0

⇒x-1x-3=0

⇒x=1, 3

Since, x1∈2,4, therefore x1=3.

And,

f'x=0

⇒3x2-12x+8=0

⇒x=12±144-966

⇒x=12±486

⇒x=12±436=6±233

⇒x=2+233, 2-233

Then, 2+233∈2,4

Hence, x2=2+233

So, S1 is true.

Now,

f'x=x-2+233x-2-233

Sign scheme for f'x is as follows:

f'x>0 ∀x∈-∞,2-233∪2+233,∞, hence it is increasing.

f'x<0 ∀x∈2-233,2+233, hence it is decreasing.

So, x4=2+233.

Now,

2f'x3=3fx4

⇒23x32-12x3+8=32+2332+233-22+233-4

⇒23x32-12x3+8=3233+2233233-2

⇒23x32-12x3+8=2129-4

⇒3x32-12x3+8=-83

⇒9x32-36x3+32=0

⇒x3=36±1296-115218=36±1218

⇒x3=83, 43

So, S2 is true.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The function fx=x3-ax has a local minimum at x=k, where kâ¥2, then a possible value of a is

Options:
A 9
B 11
C 13
D 8

The correct answer is:
13

Solution:
f'x=3x2-a=0⇒x=±a3
∴ f(x) has local minimum at x=+a3 ⇒a3≥2
⇒a≥12

Topic: Mathematics, Application of Derivatives, JEE Main
The function $f(x)=|x|$ at $x=0$ is

Options:
A Continuous but non-differentiable
B Discontinuous and differentiable
C Discontinuous and non-differentiable
D Continuous and differentiable

The correct answer is:
Continuous but non-differentiable

Solution:
Since this function is continuous at $x=0$
Now for differentiability
$\begin{aligned} & f(x)=|x|=|0|=0 \text { and } f(0+h)=f(h)=|h| \\ & \therefore \lim _{h \rightarrow 0-} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0-} \frac{|h|}{h}=-1 \\ & \text { and } \lim _{h \rightarrow 0+} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0+} \frac{|h|}{h}=1 .\end{aligned}$
Therefore it is continuous and non-differentiable.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=xex1-x,x∈R, is

Options:
A increasing in -12,1
B decreasing in 12,2
C increasing in -1,-12
D decreasing in -12,12

The correct answer is:
increasing in -12,1

Solution:
Let fx=xex1-x then fx is:

Now fx=xex1-x

Differentiating both side w.r.t x we get,

f'x=xex1-x1-2x+ex1-x

=-ex1-x2x2-x-1

=-ex1-x2x+1x-1

Now finding critical point by equating f'x=0

We get, x=-12 & x=1

∴ fx is decreasing in -∞,-12∪1,∞

And increasing in -12,1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The function $f(x)=\left\{\begin{array}{l}x, \text { if } 0 \leq x \leq 1 \\ 1, \text { if } 1 \lt x \leq 2\end{array}\right.$ is

Options:
A Continuous at all $x, \quad 0 \leq x \leq 2$ and differentiable at all $x$, except $x=1$ in the interval $[0,2]$
B Continuous and differentiable at all $x$ in $[0,2]$
C Not continuous at any point in $[0,2]$
D Not differentiable at any point $[0,2]$

The correct answer is:
Continuous at all $x, \quad 0 \leq x \leq 2$ and differentiable at all $x$, except $x=1$ in the interval $[0,2]$

Solution:
$\begin{aligned}& f(x)= \begin{cases}x, & 0 \leq x \leq 1 \\
1, & 1 \lt x \leq 2\end{cases} \\
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)=1 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\end{aligned}$
Hence function is continuous in $(0,2)$.
$\begin{aligned}& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=0=f(0) \\
& \text { Now } \\
& \lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=1=f(2)\end{aligned}$
Hence function is continuous in $[0,2]$
Clearly, from graph it is not differentiable at $x=1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $f(x)=x+\sin x$ has

Options:
A A minimum but no maximum
B A maximum but no minimum
C Neither maximum nor minimum
D Both maximum and minimum

The correct answer is:
Neither maximum nor minimum

Solution:
$f(x)=x+\sin x \Rightarrow f^{\prime}(x)=1+\cos x$
Now $f^{\prime}(x)=0 \Rightarrow 1+\cos x=0 \Rightarrow \cos x=-1 \Rightarrow x=\pi$
Now $f^{\prime \prime}(x)=-\sin x, f^{\prime \prime}(\pi)=0, f^{\prime \prime}(x)=-\cos x$,
$f^{\prime \prime \prime}(\pi)=1 \neq 0$
$\therefore$ Neither maximum nor minimum.

Topic: Mathematics, Application of Derivatives, JEE Main
The function f x = x sin Ï x , where [.] denotes the greatest integer function and {.} is the fractional part function, is discontinuous at

Options:
A all x
B all integer points
C no x
D x which is not an integer

The correct answer is:
no x

Solution:
f x = x sin π x
=xsin (integral multiple of π )
= 0
Hence, f(x) is continuous for all x.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $f(x)=[x]$, where $[x]$ denotes the greatest integer not greater than $x$, is

Options:
A continuous for all non-integral values of $x$
B continuous only at positive integral values of $\mathrm{x}$
C continuous for all real values of $x$
D continuous only at rational values of $x$

The correct answer is:
continuous for all non-integral values of $x$

Solution:
Given, $f(x)=[x]$ from the graph we observe that $f(x)=[x]$ discontinuous at every integral value of $x$. That means, $f(x)=[x]$. Continuous only for all non-integral values of $\mathrm{x}$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=xx2-6x-16,x∈ℝ--2,8

Options:
A decreases in (-2,8) and increases in (-∞,-2)∪(8,∞)
B decreases in (-∞,-2)∪(-2,8)∪(8,∞)
C decreases in (-∞,-2) and increases in (8,∞)
D increases in (-∞,-2)∪(-2,8)∪(8,∞)

The correct answer is:
decreases in (-∞,-2)∪(-2,8)∪(8,∞)

Solution:
Given: fx=xx2-6x-16

Now differentiating the above function we get,

⇒f'x=x2-6x-16·1-x·2x-6x2-6x-162

⇒f'x=-x2-16x2-6x-162

⇒f'x<0 ...i

Now, x2-6x-16≠0

⇒x2-8x+2x-16≠0

⇒x+2x-8≠0

⇒x≠8,-2 ...ii

Using i and ii,

So, the function is always decreasing inx∈(-∞,-2)∪(-2,8)∪(8,∞).

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The function $f: X \rightarrow Y$ defined by $f(x)=\sin x$ is one-one but not onto if $X$ and $Y$ are respectively, is equal to

Options:
A $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and $[-1,1]$
B $\left[0, \frac{\pi}{2}\right]$ and $[-, 1,1]$
C $[0, \pi]$ and $[0,1]$
D $R$ and $R$

The correct answer is:
$\left[0, \frac{\pi}{2}\right]$ and $[-, 1,1]$

Solution:
We have,
$f: X \rightarrow Y$
$X:$ domain and $Y$ : codomain
In one-one function,
$x_{1} \neq x_{2}$ $f\left(x_{1}\right) \neq f\left(x_{2}\right)$
In onto function,
Range $=$ domain .
$f(x)=\sin x$
So, if $X=\left[0, \frac{\pi}{2}\right]$ and $Y=[-1,1]$
For each value of $x \in X$ and range $=[0,1]$, there exists a unique value for $y \in Y$
Hence, function is one-one but not onto.

Topic: Mathematics, Functions, COMEDK, COMEDK 2016
The function \( f(x)=\frac{x}{e^{x}-1}+\frac{x}{2}+1 \) is

Options:
A an odd function
B an even function
C neither an odd nor an even function
D a periodic function

The correct answer is:
an even function

Solution:
Given, fx=xex-1+x2+1

=2x+xex-x2ex-1+1=x+xex2ex-1+1

f-x=-x-xe-x2e-x-1+1=x+xex2ex-1+1

∴f(-x)=f(x) for all x.

∴fx is an even function.

Topic: Mathematics, Functions, JEE Main
The function $L(x)=\int_1^x \frac{d t}{t}$ satisfies the equation

Options:
A $L(x+y)=L(x)+L(y)$
B $L\left(\frac{x}{y}\right)=L(x)+L(y)$
C $L(x y)=L(x)+L(y)$
D None of these

The correct answer is:
$L(x y)=L(x)+L(y)$

Solution:
Given function $L(x)-\int_1^x \frac{1}{t} d t=[\log t]_1^x=\log x-\log 1$
$\Rightarrow L(x)-\log x$, Hence $L(x y)=L(x)+L(y)$.

Topic: Mathematics, Definite Integration, JEE Main
The function of $\mathrm{AlCl}_{3}$ in Friedel-Craft's reaction is

Options:
A to absorb $\mathrm{HCl}$
B to absorb water
C to produce nucleophile
D to produce electrophile

The correct answer is:
to produce electrophile

Solution:
The function of $\mathrm{AlCl}_{3}$ in Friedel-Craft's reaction is to produce electrophile.

Topic: Chemistry, Hydrocarbons, JEE Main
The function of $\mathrm{Fe}(\mathrm{OH})_{3}$ in the contact process is

Options:
A to remove arsenic impurity
B to detect colloidal impurity
C to remove moisture
D to remove dust particles

The correct answer is:
to remove arsenic impurity

Solution:
The function of $\mathrm{Fe}(\mathrm{OH})_{3}$ in the contact process is to remove arsenic impurity. $\mathrm{Fe}(\mathrm{OH})_{3}$ is a positive sol, hence it removes arsenic impurity which is a negative sol.

Topic: Chemistry, d and f Block Elements, JEE Main
The function of Sertoli cells is controlled by:

Options:
A Estrogen
B FSH
C Testosterone
D LH

The correct answer is:
FSH

Solution:
The function of Sertoli cells, also known as nurse cells is controlled by the Follicle Stimulating Hormone (FSH). The function of these cells is to provide nutrition to the growing germ cells i.e. spermatogonia. FSH stimulates the Sertoli cells to perform this function.
Estrogen is the hormone responsible for development of female sexual characters and development of female gametes.
Testosterone is the hormone responsible for development of male sexual characters and development of male gametes.
Luteinizing Hormone (LH) is the hormone responsible for induction of ovulation and maintenance of corpus luteum.

Topic: Biology, Human Reproduction, NEET
The function of "Sodium pump" is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a constituent of this pump

Options:
A Mg2+
B K+
C Fe2+
D Ca2+

The correct answer is:
K+

Solution:
K+ ion is a constituent of sodium pump. Sodium-potassium pump, in cellular physiology, a protein that has been identified in many cells that maintains the internal concentration of potassium ions [K+] higher than that in the surrounding medium (blood, body fluid, water) and maintains the internal concentration of sodium ions [Na+] lower than that of the surrounding medium.

Topic: Chemistry, Biomolecules, NEET