datavorous/entrance-exam-dataset huggingface dataset:
If fx=11-x, then the points of discontinuity of the function f30x where fnx=fof......of (n times) are

Options:
A x=2,1
B x=0,1
C x=1,2
D no points of discontinuity

The correct answer is:
x=0,1

Solution:
Clearly, x=1 is a point of discontinuity of the function fx=11-x.
if xâ 1, then
fofâx=ffx=f11-x=x-1x, which is discontinuous at x=0.
If xâ 0 and xâ 1, then
foâfofx=ffofx=fx-1x=x
Which is continuous everywhere.
Hence, f30x=x, which is continuous everywhere.
So, the only points of discontinuity are x=0 and x=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $\mathrm{f}(x)=\left(\frac{2^{x}-1}{1-3^{x}}\right)$, for $\mathrm{x} \neq 0$ is continuous at $x=0$, then $\mathrm{f}(0)=$

Options:
A $\cdot \log 3$
B $\frac{-(\log 2)}{(\log 3)}$
C $\frac{(\log 2)}{(\log 3)}$
D $-\log 2$

The correct answer is:
$\frac{-(\log 2)}{(\log 3)}$

Solution:
$f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}$
$$
f(0)=\lim _{x \rightarrow 0} \frac{2^{x}-1}{-\left(3^{x}-1\right)}=-\frac{\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}}{\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}}=\frac{-\log 2}{\log 3}
$$

Topic: Mathematics, Limits, JEE Main
If fx=1+5+x-ax-40≤x<4bx≥4 is continuous at x=4,
then value of 1ab is equal to

Options:

The correct answer is:
12

Solution:
∵fx is continuous at x=4
∴limx→4-fx=limx→4+fx=f4
limx→4-1+5+x-ax-4=b
for numerator to be zero ⇒a=2
now after rationalization,
limx→4-1+5+x-21+5+x+2x-41+5+x+2=b
limx→4-1+5+x-4x-44=b
again on rationalization,

limx→4-5+x-9x-4×45+x+3=b
14×6=b⇒b=124
1ab=242=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\frac{e^x}{1+e^x}, l_1=\int_{f(-a)}^{f(a)} x g\{x(1-x)\} d x$ and $I_2=\int_{f(-a)}^{f(a)} g\{x(1-x)\} d x$ then the value of $\frac{l_2}{l_1}$ is

Options:
A
2

B
$-3$

C
$-1$

D
1

The correct answer is:
2

Solution:
$f(-a)+f(a)=1$
$I_1=\int_{f(-a)}^{f(a)} x g\{x(1-x)\} d x=\int_{f(-a)}^{f(a)}(1-x) g\{x(1-x)\} d x \quad\left(\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right)$
$2 I_1=\int_{f(-a)}^{f(a)} g\{x(1-x)\} d x=I_2 \Rightarrow I_2 / I_1=2$.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2004
If $f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$, for $x \neq \pi$ is continuous at $x=\pi$, then the value of $f(\pi)$ is

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

The correct answer is:
$-1$

Solution:
$f(x)=\frac{(1+\cos x)-(\sin x)}{(1+\cos x)+(\sin x)}$, is continuous at $x=\pi$
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=-1
\end{aligned}
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=1+sinxpsinâ¡x:-Ï6<x<0q:x=0etanâ¡3x.cotâ¡5x:0<x<Ï6 is continuous at x=0, then the value of 2p+10lnâ¡q is equal to

Options:

The correct answer is:
7.20

Solution:
limx→0+fx=q⇒limx→0+etan⁡3xtan⁡5x=q
=limx→0+etan⁡3x3x.5xtan⁡5x.3x5x⇒e35=q
Now, limx→0-fx=q⇒elimx→0-1-sin⁡x-psinx=q
=elimx→0--psin⁡x1-sin⁡x-1=ep
∴ep=e35⇒p=35, q=e35
2p+10.ln⁡q=65+1035=365

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{l}1+x^2, \text { when } 0 \leq x \leq 1 \\ 1-x, \text { when } x\gt1\end{array}\right.$, then

Options:
A $\lim _{x \rightarrow 1^{+}} f(x) \neq 0$
B $\lim _{x \rightarrow 1^{-}} f(x) \neq 2$
C $f(x)$ is discontinuous at $x=1$
D None of these

The correct answer is:
$f(x)$ is discontinuous at $x=1$

Solution:
$\lim _{x \rightarrow 1^+} f(x)=0$ and $\lim _{x \rightarrow 1^-} f(x)=1+1=2$.
Hence $f(x)$ is discontinuous at $x=1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x$ and $f(0)=0$, then $f(1)$ equals

Options:
A
$\tan 1-\frac{\pi}{4}$

B
$\tan 1+1$

C
$\frac{\pi}{4}$

D
$1-\frac{\pi}{4}$

The correct answer is:
$\tan 1-\frac{\pi}{4}$

Solution:
$$
\begin{aligned}
& \text { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\
& =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2} d x
\end{aligned}
$$
$$
\begin{aligned}
& =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2} d x \\
& =\int \frac{x^2}{1+x^2} d x+\int \tan ^2 x d x \\
& =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right. \\
& =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \\
& =-\tan -1 x+\tan ^{-1} x+\mathrm{c} \\
& \text { Given: } f(0)=0 \\
& \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\
& \Rightarrow c=0 \\
& \therefore f(x)=-\tan ^{-1} x+\tan ^2 x
\end{aligned}
$$
Now,
$$
f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-\frac{\pi}{4}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If $f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x$ and $f(0)=0$, then $f(1)$ equals

Options:
A
$\tan 1-\frac{\pi}{4}$

B
$\tan 1+1$

C
$\frac{\pi}{4}$

D
$1-\frac{\pi}{4}$

The correct answer is:
$\tan 1-\frac{\pi}{4}$

Solution:
$$
\begin{aligned}
& \text { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\
& =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2} d x
\end{aligned}
$$
$$
\begin{aligned}
& =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2} d x \\
& =\int \frac{x^2}{1+x^2} d x+\int \tan ^2 x d x \\
& =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right. \\
& =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \\
& =-\tan -1 x+\tan ^{-1} x+\mathrm{c} \\
& \text { Given: } f(0)=0 \\
& \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\
& \Rightarrow c=0 \\
& \therefore f(x)=-\tan ^{-1} x+\tan ^2 x
\end{aligned}
$$
Now,
$$
f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-\frac{\pi}{4}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2012 (19 May Online)
If $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ is continuous at $\mathrm{x}=0$, then what is the value of $\mathrm{f}(0)$ ?

Options:
A 0
B 1
C $\infty$
D $\mathrm{e}^{5}$

The correct answer is:
$\mathrm{e}^{5}$

Solution:
Given that $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ and $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$. Value of function at $\mathrm{x}=0$ is same as limit of the function at $\mathrm{x}=0$.
$\mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{5 / \mathrm{x}}=\left\{\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{1 / \mathrm{x}}\right\}^{5}=\mathrm{e}^{5}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=1|x|;|x|≥1ax2+b;|x|<1 is differentiable at every point of the domain, then the values of a and b are respectively:

Options:
A 12,12
B 12,-32
C 52,-32
D -12,32

The correct answer is:
-12,32

Solution:
fx=1|x|,|x|≥1ax2+b,|x|<1

at x=1 function must be continuous

So, 1=a+b …1

differentiability at x=1

-1x2x=1=(2·ax)x=1

⇒-1=2a⇒a=-12

Put in (1)⇒b=1+12=32

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (18 Mar Shift 1)
If $f(x)=\frac{x}{1+x}$ and $g(x)=f(f(x))$, then $g^{\prime}(x)$ is equal to

Options:
A $\frac{1}{(2 x+3)^2}$
B $\frac{1}{(x+1)^2}$
C $\frac{1}{x^2}$
D $\frac{1}{(2 x+1)^2}$

The correct answer is:
$\frac{1}{(2 x+1)^2}$

Solution:
Given, $f(x)=\frac{x}{1+x}$
and
$$
\begin{array}{rlrl}
\text { and } & g(x) & =f(f(x)) \\
\therefore & g(x) & =f\left(\frac{x}{x+1}\right) \\
& =\frac{\frac{x}{1+x}}{1+\frac{x}{x+1}} \\
\Rightarrow \quad & g(x) & =\frac{x}{2 x+1}
\end{array}
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
g^{\prime}(x) & =\frac{(2 x+1) 1-x(2)}{(2 x+1)^2} \\
& =\frac{1}{(2 x+1)^2}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\left\{\begin{array}{r}2^{1 / x}, \text { for } x \neq 0 \\ 3, \text { for } x=0\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 0^+} f(x)=0$
B $\lim _{x \rightarrow 0^-} f(x)=\infty$
C $f(x)$ is continuous at $x=0$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0^+} f(x)=\lim _{h \rightarrow 0} 2^{1 / h}=\infty \\ & \lim _{x \rightarrow 0^-} f(x)=\lim _{h \rightarrow 0} 2^{-1 / h}=\lim _{h \rightarrow 0} \frac{1}{2^{1 / h}}=0\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=2+2x, -1≤x<01-x3, 0≤x≤3; gx=-x, -3≤x≤0x, 0<x≤1, then range of (f∘g(x)) is

Options:
A (0,1]
B [0,3)
C [0,1]
D [0,1)

The correct answer is:
[0,1]

Solution:
Given:

fx=2+2x, x∈[-1,0)1-x3, x∈0,3 and gx=x, x∈(0,1]-x, x∈-3,0

⇒fgx=2+2gx, gx∈[-1,0)1-gx3, gx∈0,3

⇒fgx=2+2x, x∈[-1,0)⇒x∈ϕ1-x3, x∈0,3

as x∈-3,3 for x≤3 but x∈-3,1 in gx

⇒fgx=1-x3, x∈-3,1

Now, maximum value of fgx=1-x3=1-03=1

And minimum value is given by fgx=1-33=1-1=0

Hence, the range of the function will be 0,1

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If fx=22x22x+2, x∈R, then f12023+f22023+f32023.........f20222023 is equal to

Options:
A 2011
B 1010
C 2010
D 1011

The correct answer is:
1011

Solution:
Given,

fx=22x22x+2,

so f1-x=22-2x22-2x+2=2222+22x+1=22+22x

Now fx+f1-x=22x22x+2+22+22x=1

So by using the relation fx+f1-x=1 we will solve the given equation

f12023+f22023+f32023.........f20222023

=f12023+f20222023+f22023+f20212023+f32023+f20202023.........f10112023+f10122023

=f12023+f20222023⏟1+f22023+f20212023⏟1+f32023+f20202023⏟1.........+f10112023+f10122023⏟1

=1+1+1.........1011 times=1011

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If $f(x)=\frac{2^{x}+2^{-x}}{2}$, then $f(x+y) \cdot f(x-y)$ is

Options:
A $\frac{1}{4}[f(2 x)-f(2 y)]$
B $\frac{1}{2}[f(2 x)-f(2 y)]$
C $\frac{1}{4}[f(2 x)+f(2 y)]$
D $\frac{1}{2}[f(2 x)+f(2 y)]$

The correct answer is:
$\frac{1}{2}[f(2 x)+f(2 y)]$

Solution:
Given $f(x)=\frac{2^{x}+2^{-x}}{2}$
Now, $f(x+y)=\frac{2^{x+y}+2^{-x-y}}{2}$
and $f(x-y)=\frac{2^{x-y}+2^{-x+y}}{2}$
$\therefore f(x+y) \cdot f(x-y)$
$=\frac{\left(2^{x+y}+2^{-x-y}\right)}{2} \cdot \frac{\left(2^{x-y}+2^{-x+y}\right)}{2}$
$=\frac{2^{2 x}+2^{-2 y}+2^{2 y}+2^{-2 x}}{4}$
$=\frac{1}{2}\left\{\left(\frac{2^{2 x}+2^{-2 x}}{2}\right)+\left(\frac{2^{2 y}+2^{-2 y}}{2}\right)\right\}$
$=\frac{1}{2}\{f(2 x)+f(2 y)\}$

Topic: Mathematics, Functions, JEE Main
If fx2-4x+3>0, âxâ2,3, then gx=fsinâ¡x is increasing in

Options:
A x∈0,π
B x∈0,π2
C x∈π,5π4
D x∈3π2,2π

The correct answer is:
x∈3π2,2π

Solution:
x2-4x+3=x-22-1
If x∈2,3, then x2-4x+3∈-1,0
Now, g ′ x = f ′ sinx .cosx
For gx to be increasing
sin⁡x∈-1,0 and cos⁡x>0
⇒x∈3π2,2π

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)=2^{100} x+1, g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f[g(x)\}=x$ is

Options:
A emply
B a singleton
C a finite set with more than one element
D infinite

The correct answer is:
a singleton

Solution:
Given, $\quad f(x)=2^{100} \cdot x+1$
$g(x)=3^{300} \cdot x+1$
Now, $f\{g(x)\}=x$
$\Rightarrow \quad f\left(3^{100}. x+1\right)=x$
$\Rightarrow \quad 2^{100}\left\{3^{100} \cdot x+1\right\}+1=x$
$\Rightarrow \quad 6^{100} \cdot x+2^{100}+1=x$
$\Rightarrow \quad x\left(1-6^{100}\right)=\left(1+2^{100}\right)$
$\Rightarrow \quad x=\frac{1+2^{100}}{1-6^{100}}$
Hence, fog $(x)=x$ represent a singleton set.

Topic: Mathematics, Functions, JEE Main
If $f(x)=\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$ then $f^{\prime}(\pi)$ is equal to

Options:
A 0
B 2
C $\frac{\pi}{2}$
D $\pi-6$

The correct answer is:
2

Solution:
$f(x)=\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$
$f^{\prime}(x)=\left|\begin{array}{ccc}-2 \sin x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos x & 0 & 0 \\ x-\frac{\pi}{2} & -2 \sin x & 1 \\ 0 & 0 & 2 \cos x\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 0 \\ 0 & 1 & -2 \sin x\end{array}\right|$
$f^{\prime}(\pi)=\left|\begin{array}{ccc}-2 \sin \pi & 1 & 0 \\ 1 & 2 \cos \pi & 1 \\ 0 & 1 & 2 \cos \pi\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos \pi & 0 & 0 \\ \pi-\frac{\pi}{2} & -2 \sin \pi & 1 \\ 0 & 0 & 2 \cos \pi\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos \pi & 1 & 0 \\ \pi-\frac{\pi}{2} & 2 \cos \pi & 0 \\ 0 & 1 & -2 \sin \pi\end{array}\right|$
$f^{\prime}(\pi)=\left|\begin{array}{rrr}0 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -2\end{array}\right|$ $+\left|\begin{array}{ccc}-2 & 0 & 0 \\ \pi / 2 & 0 & 1 \\ 0 & 0 & -2\end{array}\right|+\left|\begin{array}{ccc}-2 & 1 & 0 \\ \pi / 2 & -2 & 0 \\ 0 & 1 & 0\end{array}\right|$
$\left[\begin{array}{l}\because \sin \pi=0 \\ \cos \pi=-1\end{array}\right]$
$f^{\prime}(\pi)=-(-2+0)=2$

Topic: Mathematics, Determinants, JEE Main
If fx+2f1x=3x, x≠0, and S=x∈R:fx=f-x, then S

Options:
A Contains exactly two elements
B Contains more than two elements
C Is an empty set
D Contains exactly one element

The correct answer is:
Contains exactly two elements

Solution:
fx+2. f1x=3x …1
Replace x by 1x, we get
f1x+2fx=3x …2
From 1 and 2
3fx=6x-3x
fx=2x-x
∵ fx=f-x
Therefore 2x-x= -2x+x
⇒ 4x=2x
⇒ 2=x2
⇒ x= ± 2
Contains exactly two elements

Topic: Mathematics, Functions, JEE Main, JEE Main 2016 (03 Apr)
If fx+2f1-x=6x âxâR, then the value of 34f8f'1 is equal to

Options:

The correct answer is:
5.5

Solution:
In given equation substituting x by 1-x, we get,
f1-x+2f1-1-x=61-x
f1-x+2fx=6-6x
From these two equations, we get,
fx+26-6x-2fx=6x
-3fx+12-12x=6x
3fx=12-18xâfx=4-6x
f8=-44, f'1=-6
f8f'1=-44-6=223

Topic: Mathematics, Functions, JEE Main
If fx=2sin⁡x-1-2cot⁡x, then the value of f'π3 is equal to

Options:
A 0
B -53
C 53
D 83

The correct answer is:
53

Solution:
In the neighbourhood of x=π3, we know,

2sin⁡x-1>0
⇒fx=2sin⁡x-1-2cot⁡x
Now, for the neighbourhood of x=π3
2sin⁡x-1-2cot⁡x=232-1-23=3-23-1
=13-1<0
∴fx=2cot⁡x-2sin⁡x-1
Now, f'x=2-cosec2x-2cos⁡x
f'π3=2-232-212
=243-1=53

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=2 \sin x, g(x)=\cos ^2 x$, then $(f+g)\left(\frac{\pi}{3}\right)=$

Options:
A $1$
B $\frac{2 \sqrt{3}+1}{4}$
C $\sqrt{3}+\frac{1}{4}$
D None of these

The correct answer is:
$\sqrt{3}+\frac{1}{4}$

Solution:
$(f+g) \frac{\pi}{3}=\frac{2 \sqrt{3}}{2}+\frac{1}{4}=\sqrt{3}+\frac{1}{4}$

Topic: Mathematics, Functions, JEE Main
If fx=2sin⁡x-x2, then in x∈0,π

Options:
A fx has no local maximum
B fx has one local minimum
C fx has 2 local maxima
D fx has one local maximum

The correct answer is:
fx has one local maximum

Solution:
f â² x =2 cosxâx
Thus, x=Î± is a local maxima for fx and no local minimum exists

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=-2sinxπ≤x≤3π2pcosx+q3π2<x<2πcosx-22π≤x≤5π2rsinx5π2<x≤3π is continuous in π,3π, then the number of points at which fx is not differentiable is/are

Options:

The correct answer is:
2

Solution:
At 3π2, -2-1=p0+q⇒q=2
At 2π, p+q=1-2⇒p=–3
At 5π2, 0-2=r1⇒r=-2
f'x=-2cosx; π≤x<3π2ND; x=3π23sinx; 3π2<x<2π0; x=2π-sinx; 2π<x<5π2ND; x=5π2<x<3π-2cosx; 5π2<x<3π
⇒fx is not differentiable at 3π2 & 5π2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=2tan-1x+sin-12x1+x2, x>1, then f5 is equal to

Options:
A π2
B tan-165156
C π
D 4tan-15

The correct answer is:
π

Solution:
Substitute tan-1x=θ, θ∈π4,π2

Hence fx=2θ+sin-1sin2θ

=2θ+π-2θ, as θ>π4

=π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2015 (10 Apr Online)
If fx=2x-12tan⁡3xxsin2xλ::0<x<π6x=0 is continuous at x=0, then the value of 103λln2 is equal to

Options:

The correct answer is:
30

Solution:
∵ fx is continuous at x=0
∴ limx→0fx=f0
∵limx→0+2x-12tan⁡3xx.sin2x=λ
∵limx→0+2x-1x2tan⁡3x3x.3xxx2sin2x=λ
ln⁡22.1.3.1=λ
103λln⁡2=30

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\sqrt{2 x-1}+5 \cos ^{-1}\left(\frac{2 x-1}{3}\right)$ then the domain of the function $f(x)$ is

Options:
A $\left[-1, \frac{1}{2}\right]$
B $\left[\frac{1}{2}, 2\right]$
C $[-1,2]$
D $\left[\frac{1}{2}, \infty\right)$

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

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 2)
If f x = 2+x 1+x 1+x , then f ′ 0 is equal to

Options:
A 2log2
B log2
C 3log2−1
D 2log2â1

The correct answer is:
2log2−1

Solution:
By taking logarithm on both sides,
logf(x) = (1+x) log(2+x)−log(1+x)
On differentiating, we get,
1fx·f'x=1+x2+x+log(2+x)−1+x1+x−log(1+x)
Now, put x=0
f'(0)=f01+log22−1=f0log2−12
=2log2−1 ∵f 0 =2

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\left\{\begin{array}{ll}2 x^{2}+1, & x \leq 1 \\ 4 x^{3}-1, & x>1\end{array},\right.$ then $\int_{0}^{2} f(x) d x$ is

Options:
A $47 / 3$
B $50 / 3$
C $1 / 3$
D $47 / 2$

The correct answer is:
$47 / 3$

Solution:
Given, $f(x)=\left\{\begin{array}{ll}2 x^{2}+1, & x \leq 1 \\ 4 x^{3}-1, & x>1\end{array}\right.$
$\begin{aligned} \therefore \int_{0}^{2} f(x) d x &=\int_{0}^{1} f(x) d x+\int_{1}^{2} f(x) d x \\ &=\int_{0}^{1}\left(2 x^{2}+1\right) d x+\int_{1}^{2}\left(4 x^{3}-1\right) d x \\ &=\left[\frac{2 x^{3}}{3}+x\right]_{0}^{1}+\left[\frac{4 x^{4}}{4}-x\right]_{1}^{2} \\ &=\frac{2}{3}(1)^{3}+1-(0+0) \\ &+\left[(2)^{4}-2-\left\{(1)^{4}-1\right)\right.\end{aligned}$
$$
\begin{array}{l}
=\frac{2}{3}+1+[16-2-0] \\
=\frac{2}{3}+15=\frac{2+45}{3} \\
=\frac{47}{3} \mathrm{sq} \text { units }
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main
If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f\left(x^3\right)$, then find the value of $\frac{d y}{d x}$ at $x=1$

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

The correct answer is:
3

Solution:
$$
\text { } \begin{aligned}
f^{\prime}(x) & =\sqrt{2 x^2-1} \\
y & =f\left(x^3\right)
\end{aligned}
$$
Differentiation w.r.t. $x$
$$
\begin{array}{rlr}
\frac{d y}{d x} & =f^{\prime}\left(x^3\right) \cdot 3 x^2 \quad \text { [Using chain rule] } \\
\left.\therefore \quad \frac{d y}{d x}\right|_{x=1} & =f^{\prime}\left(1^3\right) 3(1)^2 & \\
& =f^{\prime}(1) \cdot 3 \\
& =\sqrt{2(1)^2-1} \cdot 3 \quad\left[\because f^{\prime}(x)=\sqrt{2 x^2-1}\right] \\
& =\sqrt{2-1} \cdot 3=\sqrt{1} \cdot 3 \\
& =3
\end{array}
$$

Topic: Mathematics, Differentiation, JEE Main
If fx=2x2+3;x>3ax2+bx+1;x≤3 is differentiable everywhere, then ab2 is equal to

Options:
A 5
B 73
C 1
D 169

The correct answer is:
1

Solution:
∵fx is continuous at x=3
∴limx→3fx=f3
232+3=a32+b3+1
9a+3b=20 … 1
Now, f'x=4x;x>32ax+b;x≤3
∵fx is also differentiable at x=3
∴43=2a3+b
6a+b=12 … 2
From 1 & 2, we get,

9a=16⇒a=169, b=43

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=2-x2 and gx=ln1-x are two real valued functions then the domain of the function f+gx is

Options:
A -2,2
B −2,1
C -∞,1
D 1,2

The correct answer is:
−2,1

Solution:
Given fx=2-x2 and gx=ln1-x

f+gx=fx+gx=2-x2+ln1-x

Let domain of fx=2-x2 be D1,

then 2-x2≥0⇒x2-2≤0

i.e. D1≡-2,2

Let domain of gx=ln1-x be D2

then 1-x>0⇒x-1<0

i.e. D2≡-∞,1

Hence, domain of fx+gx will be D1∩D2≡[-2,1)

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
If $f\left(\frac{x-4}{x+2}\right)=2 x+1,(x \in R=\{1,-2\})$, then $\int f$ ( $x$ ) $d x$ is equal to (where $C$ is a constant of integration)

Options:
A
$12 \log _e|1-x|-3 x+c$

B
$-12 \log _e|1-x|-3 x+c$

C
$-12 \log _e|1-x|+3 x+c$

D
$12 \log _e|1-x|+3 x+c$

The correct answer is:
$-12 \log _e|1-x|-3 x+c$

Solution:
Suppose, $\frac{x-4}{x+2}=y \Rightarrow x-4=y(x+2)$
$$
\Rightarrow x(1-y)=2 y+4 \Rightarrow x=\frac{2 y+4}{1-y}
$$
So, $f(y)=2\left(\frac{2 y+4}{1-y}\right)+1$
Now, $f(x)=2\left(\frac{2 x+4}{1-x}\right)+1=\frac{3 x+9}{1-x}$
$$
\begin{aligned}
&=\frac{3(x+3)}{1-x}=\frac{3(x-1+4)}{1-x}=-3+\frac{12}{1-x} \\
&\therefore \int f(x) d x=-12 \log _e|1-x|-3 x+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If $\mathrm{f}(\mathrm{x})=2\{\mathrm{x}\}+5 \mathrm{x}$, where $\{\mathrm{x}\}$ is fractional part function, then $f(-1.4)$ is

Options:
A 8.2
B -8.2
C -5.8
D -5

The correct answer is:
-5.8

Solution:
We know that $\mathrm{x}=[\mathrm{x}]+\{\mathrm{x}\}$
When $\mathrm{x}=-1.4$, we get $[\mathrm{x}]=-2$ and $\{\mathrm{x}\}=0.6$
We have $f(x)=2\{x\}+5 x$
$\therefore \mathrm{f}(-1.4)=2(0.6)+5(-1.4)=-7+1.2=-5.8$

Topic: Mathematics, Functions, JEE Main
If $f(x)=2 x^6+3 x^4+4 x^2$ then $f(x)$ is

Options:
A Even function
B An odd function
C Neither even nor odd
D None of these

The correct answer is:
An odd function

Solution:
$\begin{aligned}
& f(x)=2 x^6+3 x^4+4 x^2 \\
& \quad f(-x)=2(-x)^6+3(-x)^4+4(-x)^2=f(x)
\end{aligned}$
$\Rightarrow f(x)$ is an even function and derivative of an even function is always odd.

Topic: Mathematics, Functions, JEE Main
If fx=2-xcosx2+xcosx and g(x)=loge⁡x, then the value of the integral ∫-π4π4gfxdx is

Options:
A loge⁡e
B loge⁡2
C loge⁡1
D loge⁡3

The correct answer is:
loge⁡1

Solution:
Given,

fx=2-xcosx2+xcosx, g(x)=loge⁡x

⇒gfx=loge2-xcosx2+xcosx

⇒gf-x=⁡loge2-(-x)cos⁡(-x)2+(-x)cos⁡(-x)

⇒gf-x=loge2+xcosx2-xcosx

⇒gf-x=-log2-xcosx2+xcosx

⇒gf-x=–g(fx)

Hence, g(fx) is an odd function.

By using the property of definite integration, ∫-aafxdx=2∫0afxdx, f-x=fx0,f-x=-fx, we can write
∫-π4π4g(f(x))dx=0=loge⁡1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If f x =2xâsinx, f : R â R and g x = x 1 3 , g : R â R , then

Options:
A both f and g are onto
B gof is one-one
C both f and g are one - one
D All are true

The correct answer is:
All are true

Solution:
f'x=2-cos x âxâRâ one - one,
and g'x=13x-2/3>0âone - one
Since, both g and f are one - one, hence, gof is also one - one.
Range of fx is -â, â as fx is continuous and fxâÂ±â as xâÂ±â.

Also, since domain of gx is R, y=x13
âx=y3
ây3â-â, â
âyâ-â, â
â Range of gx is also R.

Topic: Mathematics, Functions, JEE Main
If fx=35x+45x-1, x∈R, then the equation fx=0 has :

Options:
A No solution
B More than two solutions
C One solution
D Two solutions

The correct answer is:
One solution

Solution:
f'x=35xln35+45xln45<0 ∀ x∈R

∵ddxax=axlna
Hence, fx is monotonically decreasing.
Also, limx→+∞fx→-1 ∵limx→∞ax=0, 0<a<1
and limx→-∞fx→∞ ∵limx→-∞ax=∞, 0<a<1
and there is only 1 solution for fx=0.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (09 Apr Online)
If $f(x)=3 x^{2}+6 x-9$, then

Options:
A $f(x)$ is increasing in $(-1,3)$
B $f(x)$ is decreasing in $(3, \infty)$
C $f(x)$ is increasing in $(-\infty,-1)$
D $f(x)$ is decreasing in $(-\infty,-1)$

The correct answer is:
$f(x)$ is decreasing in $(-\infty,-1)$

Solution:
Given $f(x)=3 x^{2}+6 x-9$ On differentiating w.r.t. $x$, we get $f^{\prime}(x)=6 x+6$
$\begin{aligned} & f^{\prime}(x) < 0 \\ \Rightarrow & 6 x+6 < 0 \end{aligned}$
$\Rightarrow 6 \mathrm{x} < -6$
$\Rightarrow \mathrm{x} < -1$
Hence $f(x)$ is decreasing in $(-\infty,-1)$

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=3x4+4x3-12x2+12, then f(x) is

Options:
A increasing in ââ,â2âª0, 1
B increasing in â2,0 âª 1,â
C decreasing in â2,0 âª 0,1
D decreasing in -â, -2âª1, â

The correct answer is:
increasing in −2,0 ∪ 1,∞

Solution:
Since, f(x)=3x4+4x3-12x2+12
fâ² x=12x3+12x2-24x
=12xx-1(x+2)
The above expression it is clear that f x is increasing in â2,â0âª1, â and decreasing in -â, -2âª0, 1

Topic: Mathematics, Application of Derivatives, JEE Main