datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=x^2+1$, then $f^{-1}(17)$ and $f^{-1}(-3)$ will be

Options:
A 4,1
B 4,0
C 3,2
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned}
& \text { Let } y=x^2+1 \Rightarrow x= \pm \sqrt{y-1} \\
& \Rightarrow f^{-1}(y)= \pm \sqrt{y-1} \Rightarrow f^{-1}(x)= \pm \sqrt{x-1} \\
& \Rightarrow f^{-1}(17)= \pm \sqrt{17-1}= \pm 4
\end{aligned}$
and $f^{-1}(-3)= \pm \sqrt{-3-1}= \pm \sqrt{-4}$, which is not possible.

Topic: Mathematics, Functions, JEE Main
If $f(x)=\left\{\begin{array}{cl}\frac{x^{2}+3 x-10}{x^{2}+2 x-15} & \text { , when } x \neq-5 \\ \text { a }, & \text { when } x=-5\end{array}\right.$

is continuous at $x=-5$, then the value of 'a' will be

Options:
A $3 / 2$
B $7 / 8$
C $8 / 7$
D $2 / 3$

The correct answer is:
$7 / 8$

Solution:
$\lim _{x \rightarrow-5} f(x)=\frac{(x-2)(x+5)}{(x+5)(x-3)}=\frac{-7}{-8}=\frac{7}{8}$

Topic: Mathematics, Limits, JEE Main
If $f(x)=x^{2}-3 x+4$ and $f(x)=f(2 x+1)$, then $x=$

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

The correct answer is:
$-1, \frac{2}{3}$

Solution:
$\begin{array}{ll}\text { Here } \mathrm{f}(2 \mathrm{x}+1) & =(2 \mathrm{x}+1)^{2}-3(2 \mathrm{x}+1)+4 \\ & =4 \mathrm{x}^{2}+4 \mathrm{x}+1-6 \mathrm{x}-3+4 \\ \therefore \quad \mathrm{f}(2 \mathrm{x}+1) & =4 \mathrm{x}^{2}-2 \mathrm{x}+2 \\ \text { Given } \mathrm{f}(\mathrm{x})=\mathrm{f}(2 \mathrm{x}+1) \\ \therefore \mathrm{x}^{2}-3 \mathrm{x}+4=4 \mathrm{x}^{2}-2 \mathrm{x}+2 \\ \therefore \quad 3 \mathrm{x}^{2}+\mathrm{x}-2 \quad=0 \Rightarrow 3 \mathrm{x}^{2}+3 \mathrm{x}-2 \mathrm{x}-2=0 \\ \quad 3 \mathrm{x}(\mathrm{x}+1)-2(\mathrm{x}+1)=0 \Rightarrow(\mathrm{x}+1)(3 \mathrm{x}-2)=0 \\ \therefore \mathrm{x}=-1, \frac{2}{3}\end{array}$

Topic: Mathematics, Functions, JEE Main
If $f(x)=[x]^{2}-5[x]+6=0$, where $[x]$ denotes greatest integer function
then $x \in$

Options:
A $(2,4]$
B $[2,4]$
C $[2,4)$
D $(2,4)$

The correct answer is:
$[2,4)$

Solution:
We have, $[x]^{2}-5[x]+6=0$
$\therefore([x]-3)([x]-2)=0 \Rightarrow[x]=2,3$
For $[x]=2, x \in[2,3)$ and for $[x]=3, x \in[3,4)$
$\therefore \quad x \in[2,4)$

Topic: Mathematics, Functions, JEE Main
If $f(x)=\left\{\begin{array}{rr}\left(x^2 / a\right)-a, & \text { when } x \lt a \\ 0, & \text { when } x=a, \\ a-\left(x^2 / a\right), & \text {when } x \gt a\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow a} f(x)=a$
B $f(x)$ is continuous at $x=a$
C $f(x)$ is discontinuous at $x=a$
D None of these

The correct answer is:
$f(x)$ is continuous at $x=a$

Solution:
$f(a)=0$
$\lim _{x \rightarrow a-} f(x)=\lim _{x \rightarrow a}^{-}\left(\frac{x^2}{a}-a\right)=\lim _{h \rightarrow 0}\left\{\frac{(a-h)^2}{a}-a\right\}=0$
and $\lim _{x \rightarrow a}^{+} f(x)=\lim _{h \rightarrow 0}\left\{a-\frac{(a+h)^2}{a}\right\}=0$
Hence it is continuous at $x=a$

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

Options:
A $\frac{90}{\pi}$
B $\frac{180}{\pi}$
C $\frac{\pi}{90}$
D $\frac{\pi}{180}$

The correct answer is:
$\frac{\pi}{90}$

Solution:
$\begin{aligned} \mathrm{f}(0) &=\lim _{\mathrm{x} \rightarrow 0} \frac{\left(\mathrm{e}^{2 \mathrm{x}}-1\right) \sin \frac{x \pi}{180}}{\mathrm{x}^{2}} \quad\quad\cdots\left[\because x^{\circ}=\left(\frac{x \pi}{180}\right)^{c}\right] \\ &=\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\mathrm{e}^{2 \mathrm{x}}-1}{\mathrm{x}}\right)\left(\frac{\sin \frac{\pi \mathrm{x}}{180}}{\mathrm{x}}\right) \\ &=\left(2 \lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{e}^{2 \mathrm{x}}-1}{2 \mathrm{x}}\right) \frac{\pi}{180}\left[\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\sin \mathrm{x} \pi}{180}}{\frac{\mathrm{x} \pi}{180}}\right]=2(\log \mathrm{e}) \times\left(\frac{\pi}{180}\right)=\frac{\pi}{90} \end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x2+g'1x+g"2 and gx=f1x2+xf'x+f"x, then the value of f4-g4 is equal to _____ .

Options:

The correct answer is:
14

Solution:
Given:

fx=x2+g'1x+g"2

⇒f'x=2x+g'1

⇒f"x=2

Now,

gx=f1x2+xf'x+f"x

⇒gx=f1x2+x2x+g'1+2

⇒gx=f1+2x2+g'1x+2

⇒g'x=2f1x+4x+g'1 ....1

⇒g"x=2f1+4 ...2

Put x=1 in 1, then we get

2f1+4=0

⇒f1=-2

∴g"2=2f1+4=0

Now,

fx=x2+g'1x+g"2

Put x=1, then we get

f1=1+g'1+g"2

⇒g'1=-3

So,

f'x=2x-3

fx=x2-3x+c

Since, f1=-2⇒c=0

So,

fx=x2-3x

And,

gx=f1x2+xf'x+f"x

⇒gx=-2x2+x2x-3+2

⇒gx=-3x+2

Hence, f4-g4=4+10=14

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (01 Feb Shift 1)
If fx=x2lncosxln1+x2, x≠00, x=0, then fx is

Options:
A discontinuous at zero
B continuous but not differentiable at zero
C differentiable at zero
D not continuous and not differentiable at zero

The correct answer is:
differentiable at zero

Solution:
f'0+=limh→0f0+h-f0h

=limh→0fhh

=limh→0lncoshln1+h2×h2h2 ....1 {Multiplying Numerator & denominator by h}

=limh→0h2ln1+h2×limh→0lncoshh2

=limh→0h2ln1+h2×limh→0-tanh2h {Using L'Hospital's Rule}
= 1 × - 1 2
= - 1 2
and f'0-=limh→0f0-h-f0-h

=limh→0f-h-h

=-limh→0-hlncos-hln1+-h2-h
=limh→0logcoshlog1+h2
= - 1 2 {from Eq. 1}
∴f'0+=f'0-
∴ fx is differentiable and hence continuous at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x2sin1x, when x≠0 and f0=0, then limx→0f'x

Options:
A Does not exist
B 0
C ∞
D 1

The correct answer is:
Does not exist

Solution:
limx→0f'x

=limx→02xsin1x+x2-1x2cos1x

=2limx→0xsin1x-limx→0cos1x

=0-limx→0cos1x

Limit does not exist as value of cos1x oscillatesy between -1 & 1 as x approaches 0.

Topic: Mathematics, Limits, JEE Main
If $f(x)=\left\{\begin{array}{rr}x^2 \sin \frac{1}{x}, & \text { when } x \neq 0 \\ 0, & \text { when } x=0\end{array}\right.$ then

Options:
A $f(0+0)=1$
B $f(0-0)=1$
C $f$ is continuous at $x=0$
D None of these

The correct answer is:
$f$ is continuous at $x=0$

Solution:
$\lim _{x \rightarrow 0^{+}} f(x)=x^2 \sin \frac{1}{x}$, but $-1 \leq \sin \frac{1}{x} \leq 1$ and $x \rightarrow 0$
Therefore, $\lim _{x \rightarrow 0^{+}} f(x)=0=\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
Hence $f(x)$ is continuous at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=|x-2|$, then

Options:
A $\lim _{x \rightarrow 2}^{+} f(x) \neq 0$
B $\lim _{x \rightarrow 2}^{-} f(x) \neq 0$
C $\lim _{x \rightarrow 2}^{+} f(x) \neq \lim _{x \rightarrow 2-} f(x)$
D $f(x)$ is continuous at $x=2$

The correct answer is:
$f(x)$ is continuous at $x=2$

Solution:
$\begin{aligned} & \text {Here } f(2)=0 \\ & \lim _{x \rightarrow 2}^{-} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}|2-h-2|=0 \\ & \lim _{x \rightarrow 2}^{+} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}|2+h-2|=0\end{aligned}$
Hence it is continuous at $x=2$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x2 :when x is rational 2-x :when x is irrational, then

Options:
A fx is continuous for all real x
B fx is discontinuous for all real x
C fx is continuous only at x=1,-2
D fx is discontinuous only at x=1,-2

The correct answer is:
fx is continuous only at x=1,-2

Solution:
Let a be any real number.
limxâaâ¡fx=limxâaâ¡x=a2 (when xâa through rational values)
limxâaâ¡fx=limxâaâ¡2-x=2-a (when xâa through irrational values)
limxâaâ¡fx will exist only when a2=2-a
i.e., when a=1,-2
Thus, if aâ 1 or -2 then limxâaâ¡fx will not exist
Hence, fx will be discontinuous at x=a where aâ 1 or -2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the
function because

Options:
A The function is not differentiable at every point in the $(0,4)$.
B $f(4) \neq f(0)$
C Function is not well-defined in the domain.
D The function is not continuous at every point in the $[0,4]$.

The correct answer is:
The function is not differentiable at every point in the $(0,4)$.

Solution:
Here $f(0)=|0-2|=-2$ and $f(4)=|4-2|=2$
Thus $\mathrm{f}(4) \neq \mathrm{f}(0)$
Hence Rolle's Theorem cannot be applied.

Topic: Mathematics, Application of Derivatives, MHT CET, MHT CET 2020 (12 Oct Shift 1)
If $f(x)=\int_{x^2}^{x^2+1} e^{-t^2} d t$ then $f(x)$ increases in

Options:
A $(2,2)$
B No value of $x$
C $(0, \infty)$
D $(-\infty, 0)$

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

Solution:
$\begin{aligned} & f(x)=e^{-\left(x^2+1\right)^2} \cdot 2 x-e^{-\left(x^2\right)^2} \cdot 2 x=2 x e^{-\left(x^4+1+2 x^2\right)}\left(1-e^{2 x^2+1}\right) \\ & \Rightarrow f^{\prime}(x)\gt0, \forall x \in(-\infty, 0) .\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
If fx=x2+x21+x2+x21+x22+.â¦â¦.. upto â, then

Options:
A limx→0⁡fx does not exist
B fx is continuous but not differentiable at x=0
C fx is discontinuous at x=0
D fx is differentiable at x=0

The correct answer is:
fx is discontinuous at x=0

Solution:
f0=0+0+0+.……..=0
limx→0⁡fx=limx→0⁡x2+x21+x2+x21+x22+.……..
=limx→0⁡x21+11+x2+11+x22+.……..
=limx→0⁡x211-11+x2 (as 0<11+x2<1)
=limx→0⁡1+x2=1
So, f0≠limx→0⁡f0

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x2-x21+x2-x2 (where . represents the greatest integer part of x ), then the range of fx is

Options:
A 0,1
B -1,1
C 0,∞
D 0,12

The correct answer is:
0,12

Solution:
x2-x2=x2
∴fx=x21+x2=1+x2-11+x2=1-11+x2

∵0≤x2<1⇒1≤x2+1<2
12<1x2+1≤1⇒-12>-11+x2≥-1
12>1+-11+x2≥0⇒ Range of fx∈0,12

Topic: Mathematics, Functions, JEE Main
If fx=x2-x2x2-x2-2 (where, â represents the greatest integer part of x), then the range of fx is

Options:
A 0,1
B -1,1
C 0,∞
D 0,13

The correct answer is:
0,13

Solution:
We know that, x2-x2=x2
∴fx=x22+x2=2+x2-22+x2=1-22+x2
∴0≤x2<1⇒2≤x2+2<3
23<2x2+2≤1⇒-23>-22+x2≥-1
⇒13>1+-22+x2≥0

Hence, the range of fx is 0,13

Topic: Mathematics, Functions, JEE Main
If $f(x)=\int_{x^2}^{x^4} \sin \sqrt{t} d t$, then $f^{\prime}(x)$ equals

Options:
A $\sin x^2-\sin x$
B $4 x^3 \sin x^2-2 x \sin x$
C $x^4 \sin x^2-x \sin x$
D None of these

The correct answer is:
$4 x^3 \sin x^2-2 x \sin x$

Solution:
We have $f(x)=\int_{x^2}^{x^4} \sin \sqrt{t} d t$
$\begin{aligned} \therefore \quad f^{\prime}(x) & =\frac{d}{d x}\left(x^4\right)\left(\sin \sqrt{x^4}\right)-\frac{d}{d x}\left(x^2\right)\left(\sin \sqrt{x^2}\right) \\ & =4 x^3 \sin x^2-2 x \sin x\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
If $f(x)=x^2-x+5, x>\frac{1}{2}$, and $\mathrm{g}(\mathrm{x})$ is its inverse function, then $\mathrm{g}^{\prime}(7)$ equals:

Options:
A
$-\frac{1}{3}$

B
$\frac{1}{13}$

C
$\frac{1}{3}$

D
$-\frac{1}{13}$

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

Solution:
$f(x)=y=x^2-x+5$
$$
\begin{aligned}
&x^2-x+\frac{1}{4}-\frac{1}{4}+5=y \\
&\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=y \\
&\left(x-\frac{1}{2}\right)^2=y-\frac{19}{4} \\
&x-\frac{1}{2}=\pm \sqrt{y-\frac{19}{4}}
\end{aligned}
$$
$$
\begin{aligned}
&x=\frac{1}{2} \pm \sqrt{y-\frac{19}{4}} \\
&\text { As } x>\frac{1}{2} \\
&x=\frac{1}{2}+\sqrt{y-\frac{19}{4}} \\
&g(x)=\frac{1}{2}+\sqrt{x-\frac{19}{4}} \\
&g^{\prime}(x)=\frac{1}{2 \sqrt{x-\frac{19}{4}}}
\end{aligned}
$$
$$
g^{\prime}(7)=\frac{1}{2 \sqrt{7-\frac{19}{4}}}=\frac{1}{2 \frac{\sqrt{28-19}}{2}}=\frac{1}{3}
$$

Topic: Mathematics, Functions, JEE Main, JEE Main 2014 (12 Apr Online)
If fx=x32x2+11+3x3x2+22xx3+6x3−x4x2−2 for all x∈ℝ, then 2f0+f'0 is equal to

Options:
A 48
B 24
C 42
D 18

The correct answer is:
42

Solution:
Given: fx=x32x2+11+3x3x2+22xx3+6x3-x4x2-2

⇒f0=01120604-2

⇒f0=0+4+8

⇒f0=12

⇒f'x=x32x2+11+3x3x2+22xx3+63x2-102x+x32x2+11+3x6x23x2x3-x4x2-2+3x24x33x2+22xx3+6x3-x4x2-2

⇒f'0=011206-100+01102004-2+00320604-2

⇒f'0=0-0+6+0+0+0+0+38

⇒f'0=-6+24=18

⇒2f0+f'0=42

Topic: Mathematics, Determinants, JEE Main
If fx=x32x2+11+3x3x2+22xx3+6x3−x4x2−2 for all x∈ℝ, then 2f0+f'0 is equal to

Options:
A 48
B 24
C 42
D 18

The correct answer is:
42

Solution:
Given: fx=x32x2+11+3x3x2+22xx3+6x3-x4x2-2

⇒f0=01120604-2

⇒f0=0+4+8

⇒f0=12

⇒f'x=x32x2+11+3x3x2+22xx3+63x2-102x+x32x2+11+3x6x23x2x3-x4x2-2+3x24x33x2+22xx3+6x3-x4x2-2

⇒f'0=011206-100+01102004-2+00320604-2

⇒f'0=0-0+6+0+0+0+0+38

⇒f'0=-6+24=18

⇒2f0+f'0=42

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If fx=x3+3x+1 and gx is the inverse function of fx, then the value of g'5 is equal to

Options:
A 3
B 13
C 16
D 6

The correct answer is:
16

Solution:
∵ fx, gx are inverse function of each other
∴ g'(fx)=1f'x
Now put x=1; g'f1=1f'1
Hence, g'5=13x2+3x=1=16

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\left\{\begin{array}{ll}x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 2} f(x)$ does not exist
B $f$ is not continuous at $x=2$
C $f$ is continuous but not differentiable at $x=2$
D $t$ is continuous and differentiable at $x=2$

The correct answer is:
$f$ is continuous but not differentiable at $x=2$

Solution:
Given. $f(x)=\left\{\begin{aligned} x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2 \end{aligned}\right.$
LHL
$$
\begin{array}{l}
=f(2-0)=\lim _{h \rightarrow 0}(2-h)^{3}-3(2-h)+2 \\
=(2)^{3}-6+2=8-6+2=4 \\
\begin{aligned}
LH L=f(2+0) &=\lim _{h \rightarrow 0}(2+h)^{3}-6(2+h)^{2} \\
&=(2)^{3}-6(2)^{2}+9(2)+2 \\
&=8-24+18+2=4
\end{aligned}
\end{array}
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 2} f(x)$ exist
and $\begin{aligned} f(2)=(2)^{3} &-6(2)^{2}+9(2)+2 \\ &=8-24+18+2=4 \end{aligned}$
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(2)$
So, $f(x)$ is continuous at $x=2$
Now, $\quad f^{\prime}(x)=\left\{\begin{array}{cc}3 x^{2}-3, & x < 2 \\ 3 x^{2}-12 x+9, & x \geq 2\end{array}\right.$
$\begin{array}{ll}\therefore \quad & L f^{\prime}(2)=3(2)^{2}-3=12-3=9 \\ \text { and } \quad & R f^{\prime}(2)=3(2)^{2}-12(2)+9 \\ & =12-24+9=-3\end{array}$
$\because \quad L f^{\prime}(2) \neq R f^{\prime}(2)$
$\therefore f(x)$ is not differentiable at $x=2$
Hence. $f$ is continuous but not differentiable at $x=2$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f(x)=x3+4x2+λx+1 is a monotonically decreasing function of x in the largest possible interval (-2, -23), then

Options:
A Î» = 4
B λ = 2
C λ = -1
D λ has no real value

The correct answer is:
λ = 4

Solution:
Here, f'(x)≤0
⇒3x2+8x+λ≤0, ∀x∈-2,-23
Then, situations for f'(x) is as follow :

Given that f(x) decreases in the largest possible interval
- 2 , - 2 3 then f'(x) = 0 must have roots -2 and -23
⇒ Product of roots is (-2) - 2 3 = λ 3 ⇒ λ = 4

Topic: Mathematics, Application of Derivatives, JEE Main
If $\mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d}$ and $0 < \mathrm{b}^2 < \mathrm{c}$, then in $(-\infty, \infty)$

Options:
A $\mathrm{f}(x)$ has a local maxima.
B $\mathrm{f}(x)$ is strictly increasing function.
C $\mathrm{f}(x)$ is bounded.
D $\mathrm{f}(x)$ is strictly decreasing function.

The correct answer is:
$\mathrm{f}(x)$ is strictly increasing function.

Solution:
$$
\begin{aligned}
& \mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d} \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{c}
\end{aligned}
$$
Now its discriminant $=4\left(b^2-3 c\right)$ $\Rightarrow 4\left(\mathrm{~b}^2-\mathrm{c}\right)-8 \mathrm{c} < 0$, as $\mathrm{b}^2 < \mathrm{c}$ and $\mathrm{c}>0$
$\Rightarrow \mathrm{f}^{\prime}(x)>0$ for all $x \in \mathrm{R}$
$\Rightarrow f$ is strictly increasing on $R$.

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)=\left\{\begin{array}{l}x^3 \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0 \quad, x=0\end{array}\right.$ then

Options:
A $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$
B $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}$
C $f^{\prime \prime}(0)=1$
D $f^{\prime \prime}(0)=0$

The correct answer is:
$f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$

Solution:
$\begin{aligned} & f^{\prime}(x)=3 x^2 \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right) \\ & f^{\prime \prime}(x)=6 x \sin \left(\frac{1}{x}\right)-3 \cos \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)-\frac{\sin \left(\frac{1}{x}\right)}{x} \\ & f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2 \pi}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (06 Apr Shift 1)
If $f(x)=|x-3|$, then $f$ is

Options:
A Discontinuous at $x=2$
B Not differentiable $x=2$
C Differentiable at $x=3$
D Continuous but not differentiable at $x=3$

The correct answer is:
Continuous but not differentiable at $x=3$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}|3-h-3|=0 \\ & \lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0}|3+h-3|=0 \\ & \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)\end{aligned}$
Hence $f$ is continuous at $x=3$
Now
$\begin{aligned}L f^{\prime}(3)= & \lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h} \\
& =\lim _{h \rightarrow 0} \frac{|3-h-3|-0}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1\end{aligned}$
$R f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h \rightarrow 0} \frac{|3+h-3|-0}{h}=1$
$L f^{\prime}(3) \neq R f^{\prime}(3)$. Hence $f$ is not differentiable at $x=3$.
Trick : Can be seen by graph it is continuous but tangent is not defined at $x=3$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=x3+x+1tanÏx (where, x represents the greatest integer part of x), then

Options:
A domain of fx is R-2n+1π2,n∈Ι
B range of fx∈R
C fx is an even function
D fx is a non-periodic function

The correct answer is:
fx is an even function

Solution:
∀x∈R⇒x is an integer
∴ πx=nπ, n∈Ι
and tanπx=tannπ=0
∴fx=0 ∀x∈R
∴fx is an even function

Topic: Mathematics, Functions, JEE Main
If $f(x)=x^4-2 x^3+3 x^2-a x+b$ is divided by $x-1$ and $x+1$, the remainders are 5 and 19, respectively. If $f(x)$ is divided by $x-2$. The remainder is

Options:
A 8
B 5
C 10
D 12

The correct answer is:
10

Solution:
$f(x)=x^4-2 x^3+3 x^2-a x+b$
Given that, $f(1)=5$ and $f(-1)=19$, then
$$
\begin{aligned}
& f(1)=1-2+3-a+b=2-a+b=5 \\
& \therefore \quad b-a=3 \\
& \text { and } f(-1)=1+2+3+a+b \\
& =b+a+6=19 \\
& \therefore \quad b+a=13 \\
&
\end{aligned}
$$
adding Eqs. (i) and (ii), we obtain
$$
\begin{aligned}
& \qquad 2 b=16 \Rightarrow b=8 \\
& \text { and } a=13-b \Rightarrow a=5 \\
& \therefore f(x)=x^4-2 x^3+3 x^2-5 x+8
\end{aligned}
$$
Thus,
$$
\begin{aligned}
f(2) & =(2)^4-2(2)^3+3(2)^2-5(2)+8 \\
& =16-16+12-10+8 \\
& =10
\end{aligned}
$$
$\therefore$ If $f(x)$ is divided by $x-2$, then remainder is 10 .

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
If $\mathrm{f}(\mathrm{x})=\frac{\sqrt{\mathrm{x}-1}}{\mathrm{x}-4}$ defines a function of $\mathrm{R}$, then what is its domain?

Options:
A $(-\infty, 4) \cup(4, \infty)$
B $[4, \infty]$
C $(1,4) \cup(4, \infty)$
D $[1,4) \cup(4, \infty)$

The correct answer is:
$[1,4) \cup(4, \infty)$

Solution:
$f(x)=\frac{\sqrt{x-1}}{x-4}$
$f(x)$ is defined for $(x-1) \geq 0 \Rightarrow x \geq 1$ and $x-4 \neq 0$ $\Rightarrow x \neq 4$
$\therefore \quad$ Domain of $f(x)=1 \leq x < \infty-\{4\}$
or, $\quad x=[1,4) \cup(4, \infty)$.

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 2)
If fx=x-4x-4+tan-11-2x2+x, ∀4<x<8, then the value of f ′ 5 is equal to

Options:
A -713
B 0
C 513
D -813

The correct answer is:
-713

Solution:
fx=x-42+22-2.2x-4+tan-112-x1+12.x
fx=x-4-22+tan-112-tan-1x
fx=2-x-4+tan-112-tan-1xâµx-4<2
f ′ x = −1 2 x−4 − 1 1+ x 2
f ′ 5 = −1 2 − 1 1+25 = −14 26 =− 7 13

Topic: Mathematics, Differentiation, JEE Main
If fx=x4+x-ax>0cx=04e2x+3e1xe2x+be1xx<0 is continuous at x=0 for some constants a, b and c, then the value of 50ba is equal to

Options:

The correct answer is:
18.75

Solution:
In limx→0+x4+x-a for 00 form a=2
⇒limx→0+x4+x-2×4+x+24+x+2
⇒limx→0+x4+x+24+x-4=limx→0+4+x+2=4
Now, limx→0-fx=limx→0-4e2x+3e1xe2x+be1x
=limx→0+e1x4e1x+3e1xe1x+b=limx→0+4e1x+3e1x+b=0+30+b
For fx to be continuous at x=0
4=3b⇒b=34

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{lll}x-5, & \text { for } x \leq 1 \\ 4 x^2-9, & \text { for } 1 < x < 2 \\ 3 x+4, & \text { for } x \geq 2,\end{array}\right.$ then $f^{\prime}\left(2^{+}\right)$is equal to

Options:
A $0$
B $2$
C $3$
D $4$

The correct answer is:
$3$

Solution:
Given, $\begin{aligned} f(x) & = \begin{cases}x-5 & \text { for } x \leq 1 \\ 4 x^2-9 & \text { for } 1 < x < 2 \\ 3 x+4 & \text { for } x \geq 2\end{cases} \\ f^{\prime}\left(2^{+}\right) & =\lim _{x \rightarrow 2^{+}}\left(\frac{f(x)-f(2)}{x-2}\right) \\ & =\lim _{x \rightarrow 2^{+}} \frac{3 x+4-(6+4)}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{3 x-6}{x-2} \\ & =3\end{aligned}$

Topic: Mathematics, Limits, JEE Main
if $f(x)=\left\{\begin{array}{r}\frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1, \text { when } x=a\end{array}\right.$ then

Options:
A $f(x)$ is continuous at $x=a$
B $f(x)$ is discontinuous at $x=a$
C $\lim _{x \rightarrow a} f(x)=1$
D None of these

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

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

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\sqrt{x}$ and $g(x)=x$ be two functions defined in the domain $R^{+} \cup\{0\}$, then find the value of.
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) $(f g)(x)$
(iv) $\left(\frac{f}{g}\right)(x)$

Options:

None

Solution:
Given $f(x)=\sqrt{x}$ and $g(x)=x$
(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$
(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$
(iii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$
(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$

Topic: Mathematics, Relations and Functions (Class 11)
If $f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \text { then } f(x) \text { is } \\ 0 & , x=0\end{cases}$

Options:
A
discontinuous every where

B
continuous as well as differentiable for all $x$

C
continuous for all $x$ but not differentiable at $x=0$

D
neither differentiable nor continuous at $x=0$

The correct answer is:
continuous for all $x$ but not differentiable at $x=0$

Solution:
$f(0)=0 ; f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}$
R.H.L. $\lim _{h \rightarrow 0}(0+h) \mathrm{e}^{-2 / h}=\lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^{2 / h}}=0$
L.H.L $\operatorname{Lim}_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0$
Therefore, $f(x)$ is continuous
R.H.D. $\lim _{h \rightarrow 0} \frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{h}=0$
L.H.D. $\operatorname{Lim}_{h \rightarrow 0} \frac{(0-h) \mathrm{e}^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{-h}=1$
Therefore, L.H.D. $\neq$ R.H.D.
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2003
If $f(x)=x e^{x(1-x)}, x \in R$, then $f(x)$ is

Options:
A
decreasing on $[-1 / 2,1]$

B
decreasing on $R$

C
increasing on $[-1 / 2,1]$

D
increasing on $R$

The correct answer is:
increasing on $[-1 / 2,1]$

Solution:
$$
\begin{aligned}
& f(x)=x e^{x(1-x)}, x \in R \\
& f^{\prime}(x)=e^{x(1-x)} \cdot\left[1+x-2 x^2\right]
\end{aligned}
$$
$$
\begin{aligned}
& =-e^{x(1-x} \cdot\left[2 x^2-x-1\right] \\
& =-2 e^{x(1-x} \cdot\left[\left(x+\frac{1}{2}\right)(x-1]\right) \\
& f^{\prime}(x)=-2 e^{x(1-x)} A \\
& \text { where } A=\left(x+\frac{1}{2}\right)(x-1)
\end{aligned}
$$
Now, exponential function is always $+$ ve and $f^{\prime}(x)$ will be opposite to the sign of $A$
which is -ve in $\left[-\frac{1}{2}, 1\right]$
Hence, $f^{\prime}(x)$ is $+$ ve in $\left[-\frac{1}{2}, 1\right]$
$\therefore f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$

Topic: Mathematics, Application of Derivatives, JEE Main