datavorous/entrance-exam-dataset huggingface dataset:
Let fx=ex-x and gx=x2-x, ∀ x ϵ R . Then the set of all x ϵ R , where the function hx=fogx is increasing, is:

Options:
A -1,-12⋃12,∞
B [0,∞)
C 0,12∪[1,∞)
D -12, 0∪[1,∞)

The correct answer is:
0,12∪[1,∞)

Solution:
hx=fogx=egx-gx
=ex2-x-x2-x
h'x=ex2-x×2x-1-2x+1
=2x-1ex2-x-1
∵hx is increasing ∴h'x≥0 ⇒2x-1ex2-x-1≥0
Case 1: 2x-1≥0 & x2-x≥0
⇒x≥12 & x∈-∞, 0∪1,∞⇒x∈1, ∞
Case 2: 2x-1≤0 and x2-x≤0
x<12 and x∈0, 1 ⇒x∈0,12
∴ x∈0,12∪[1,∞)

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Let $F(x)=f(x)+f\left(\frac{1}{x}\right)$, where $f(x)=\int_1^x \frac{\log t}{1+t} d t$. Then $F(e)$ equals

Options:
A
$\frac{1}{2}$

B
$0$

C
$1$

D
$2$

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

Solution:
$f(x)=\int_1^x \frac{\log t}{1+t} d t$
$F(e)=f(e)+f\left(\frac{1}{e}\right)$
$F(e)=\int_1^e \frac{\log t}{1+t} d t+\int_1^{1 / e} \frac{\log t}{1+t} d t$
$=\int_1^e \frac{\log t}{1+t}+\int_1^e \frac{\log t}{t(1+t)} d t$
$=\int_1^e \frac{\log t}{t} d t=\frac{1}{2}$.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2007
Let fx is a differentiable function on x∈R, such that fx+y=fxfy for all x,y∈R where f0≠0. If f5=10, f'0=6, then the value of f ′ 5 is equal to

Options:

The correct answer is:
60

Solution:
f ′ 5 = lim h→0 f 5+h −f 5 h
=limh→0f5+h-f5+0h
=limh→0f5.fh-f5.f0h
[∵ fx+y=fx.fy for all x,y]
=limh→0fh-f0h.f5
= f ′ 0 .f 5 =6×10
=60

Topic: Mathematics, Differentiation, JEE Main
Let fx=∑k=110k·xk, x∈ℝ, if 2f2+f'2=1192n+1 then n is equal to ______.

Options:

The correct answer is:
10

Solution:
Given,

fx=∑k=110k·xk, x∈ℝ

⇒fx=x+2x2+3x3+………+10x10

Now let S=x+2x2+3x3+………+10x10

So, S·x = x2+2x3+…………+9x10+10x11

Now subtracting above equations we get,

S1-x=x+x2+x3+……+x10-10x11

⇒S1-x=x1-x101-x-10x11

⇒S=x1-x101-x2-10x111-x=fx

Now finding, f2=21-210+10·211

⇒f2=2+18·210

Now finding f'x we get,

f'x=-10x111-x2-110x101-x-10x101-x2+2x1-x101-x3+1-x101-x2

⇒f'2=-10·2111-22-110·2101-2-10·2101-22+2·2·1-2101-23+1-2101-22

⇒f'2=-10·211+110·210-10·210-41-210+1-210

⇒f'2=83·210-3

2f(2)+f'(2)=22+18·210+83·210-3=119(2)10+1

∴ n=10

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let fx=limn→∞x2+2x+4+sin⁡πxn-1x2+2x+4+sin⁡πxn+1, then

Options:
A fx is continuous and differentiable for all x∈R.
B fx is continuous but not differentiable for all x∈R.
C fx is discontinuous at infinite number of points.
D fx is discontinuous at two points.

The correct answer is:
fx is continuous and differentiable for all x∈R.

Solution:
∵ x2+2x+4+sinπx=x+12+3+sin⁡πx≥2,∀x∈R.
∴ fx=limn→∞1-x+12+3+sin⁡πx-n1+x+12+3+sin⁡πx-n
=1-01+0=1
Clearly, fx=1, ∀x∈R
Hence, f(x) is continuous and differentiable ∀x∈R

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=logesinx, 0<x<π and gx=sin-1(e-x), (x≥0). If α is a positive real number such that a=fog'(α) and b=fog(α), then

Options:
A aα2+bα+a=0
B aα2+bα-a=-2α
C aα2-bα-a=0
D aα2-bα-a=1

The correct answer is:
aα2-bα-a=1

Solution:
Given fx=logesinx, 0<x<π and gx=sin-1(e-x), (x≥0), then we have

fogx=fgx

=fsin-1e-x

=logesinsin-1e-x

=logee-x=-x

⇒fogx=-x

⇒b=fgα=-α

Now (fg(x))'=-1

⇒a=fgα'=-1

∴ aα2-bα-a=-1α2--αα--1

=-α2+α2+1=1.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Let fx=max tan⁡x,cot⁡x. Then the number of roots of the equation fx=12 in 0,2π is

Options:
A 0
B 1
C 2
D 4

The correct answer is:
0

Solution:
From the graph of tanx and cotx, we see that the range of fx is -1,0∪1,∞.
Hence, fx=max tan⁡x, cot⁡x and gx=12 do not cut each other
⇒ The number of solutions is zero.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let fx=maxx, x2,x≤28-2x,2<x≤4. Let S be the set of points in the interval -4, 4 at which f is not differentiable. Then S

Options:
A equals -2, -1, 0, 1, 2
B equals -2, 2
C is an empty set
D equal -2, -1, 1, 2

The correct answer is:
equals -2, -1, 0, 1, 2

Solution:
Given function is fx=maxx, x2,x≤28-2x,2<x≤4

The graph of the function, is

We know that, a function is not differentiable at a point where its graph has a sharp corner.

And, from the graph we can easily conclude that fx is non-derivable at x=-2, -1, 0,1, 2

Hence, the set S=-2, -1, 0,1, 2.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let fx=maxâ¡x,x2xâ¤312-x 3<xâ¤12. If S is the set of points in the interval -12,12 at which f is not differentiable, then S is

Options:
A equal to -3,3
B equal to -3,-1,1,3
C an empty set
D equal to -3,-1,0,1,3

The correct answer is:
equal to -3,-1,0,1,3

Solution:
Here, fx=12+x-12≤x<-3x2-3≤x≤-1x-1<x<1x21≤x≤312-x3<x≤12
Hence, fx is not differentiable at x=-3,-1,0,1,3
⇒S=-3,-1,0,1,3

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=\max \{x+|x|, x-[x]\}, \quad$ where $\quad \mid x]$ denotes the greateat integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) d x$ is

Options:
A 0
B $51 / 2$
C $21 / 2$
D 1

The correct answer is:
$21 / 2$

Solution:
Given, $f(x)=\max \{x+|x|, x-[x]\}$
$$
\begin{aligned}
=\left\{\begin{array}{ll}
2 x, & x \geq 0 \\
x-[x], & x \leq 0
\end{array}\right.\\
\begin{aligned}
\int_{-3}^{3} f(x) d x &=\int_{-3}^{0} x-[x] d x+\int_{0}^{3} 2 x d x \\
&=3 \int_{-1}^{0}(1+x) d x+2 \int_{0}^{3} x d x
\end{aligned}
\end{aligned}
$$
[$\because$ $x-[x]$ is a periodic function at $x=1]$
$$
\begin{array}{l}
=3\left[x+\frac{x^{2}}{2}\right]_{-1}^{0}+2\left[\frac{x^{2}}{2}\right]_{0}^{3} \\
=3\left[0-0-\left(-1+\frac{1}{2}\right)\right]+3^{2}-0 \\
=3\left[\frac{1}{2}\right]+9=\frac{3}{2}+9 \\
=\frac{21}{2}
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main
Let $f(x)=$ Maximum $\left\{x^{2},(1-x)^{2}, 2 x(1-x)\right\}$, where $0 \leq x \leq 1$. Determine the area of the region bounded by the curves $y=f(x), x$-axis, $x=0 \& x=1$

Options:
A $\frac{17}{27}$
B $\frac{15}{25}$
C $\frac{13}{23}$
D $\frac{14}{29}$

The correct answer is:
$\frac{17}{27}$

Solution:
$$
f(x)=\operatorname{Max}\left\{x^{2},(1-x)^{2}, 2 x(1-x)\right\}
$$
coordinate of $A,(1-x)^{2}=2 x(1-x)$
$(1-x)(1-x-2 x)=0$
$x=\frac{1}{3}, A\left(\frac{1}{3}, 0\right)$ coordinate of $B$
$2 x(1-x)=x^{2}, 2(1-x)=x, 2-3 x=0, x=\frac{2}{3}$
$B\left(\frac{2}{3}, 0\right)$ Req. Area $=\int_{0}^{1 / 3}(1-x)^{2} d x+\int_{1 / 3}^{2 / 3} 2 x(1-x) d x+\int_{\frac{2}{3}}^{1} x^{2} d x$
$=\frac{17}{27}$

Topic: Mathematics, Area Under Curves, JEE Main
Let fx=minâ¡4-x2,1+x2 âxâ-2,2, then the number of points where fx is non-differentiable is

Options:
A 1
B 0
C 4
D 2

The correct answer is:
4

Solution:
Clearly, from the graph we can see that fx is non-differentiable at 4 points

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=min⁡x+1,1-x ∀x≤1. Then, the area (in sq. units) bounded by y=fx, y=0 and x=0 from x=0 to x=1 is equal to

Options:
A 13
B 23
C 49
D 1

The correct answer is:
23

Solution:
fx=x+1;x≤01-x;x≥0
∴ Required area is
A=∫011-xdx
=-231-x3201
=-230-1=23 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
Let fx=sin-1x1-x-x1-x2, ∀0≤x≤1, then fx is

Options:
A negative
B positive
C non-negative
D non-positive

The correct answer is:
non-positive

Solution:
fx=sin-1x1-x2-x1-x2
∵ sin-1a-sin-1b=sin-1a1-b2-b1-a2
⇒fx=sin-1x-sin-1x≤0 as x≤x

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Let $f(x)=\int_{\sin x}^{\cos x} e^{-t^2} d t$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

Options:
A $\sqrt{1 / e}$
B $-\sqrt{2 / e}$
C $\sqrt{2 / e}$
D $-\sqrt{1 / e}$

The correct answer is:
$-\sqrt{2 / e}$

Solution:
$f^{\prime}(x)=-e^{-\cos ^2 x} \sin x-e^{-\sin ^2 x} \cos x$
$\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}$

Topic: Mathematics, Definite Integration, JEE Main
Let $f(x)$ satisfy all the conditions of the mean value theorem in $[0,2]$. If $f(0)=0$ and $|f(x)| \leq \frac{1}{2}$ for all $x$, in $[0,2]$ then

Options:
A $f(x) \leq 2$
B $|f(x)| \leq 1$
C $f(x)=2 x$
D $f(x)=3$ for at least one $x$ in $[0,2]$

The correct answer is:
$|f(x)| \leq 1$

Solution:
$\begin{aligned} & \frac{f(2)-f(0)}{2-0}=f^{\prime}(x) \Rightarrow \frac{f(2)-0}{2}=f^{\prime}(x) \\ & \Rightarrow \frac{d f(x)}{d x}=\frac{f(2)}{2} \Rightarrow f(x)=\frac{f(2)}{2} x+c \\ & \therefore f(0)=0 \Rightarrow c=0 . \therefore f(x)=\frac{f(2)}{2} x...(i)\end{aligned}$
Given
$|f(x)| \leq \frac{1}{2} \Rightarrow\left|\frac{f(2)}{2}\right| \leq \frac{1}{2}$
(i) $\Rightarrow$
$|f(x)|=\left|\frac{f(2)}{2} x\right|=\left|\frac{f(2)}{2}\right||x| \leq \frac{1}{2}|x|$
[from (ii)]
In $[0,2]$, for maximum $x(x=2)$
$\left.|f(x)| \leq \frac{1}{2} \cdot 2 \Rightarrow f(x) \right\rvert\, \leq 1$

Topic: Mathematics, Application of Derivatives, JEE Main
Let fx=sin-1x and gx=x2-x-22x2-x-6. If g2=limx→2gx, then the domain of the function fog is

Options:
A (-∞,-1]∪[2,∞)
B (-∞,-2]∪-32,∞
C (-∞,-2]∪-43,∞
D (-∞,-2]∪[-1,∞)

The correct answer is:
(-∞,-2]∪-43,∞

Solution:
Domain of fogx=sin-1gx

⇒g(x)≤1,g2=37

x2-x-22x2-x-6≤1

x+1x-22x+3x-2≤1

x+12x+3≤1 and x+12x+3≥-1

x+1-2x-32x+3≤0 and x+1+2x+32x+3≥0

x+22x+3≥0 and 3x+42x+3≥0

x∈(-∞,-2]∪-43,∞

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let $f(x)=\sin 2 x+\cos 2 x$ and $g(x)=x^2-1$ then $g(f(x))$ is invertible in the domain

Options:
A $x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$
B $x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
C $x \in\left[0, \frac{\pi}{4}\right]$
D $x \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$

The correct answer is:
$x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$

Solution:
$f(x)=\sin 2 x+\cos 2 x$ and $g(x)=x^2-1$
$\begin{aligned} & g[f(x)]=\left[(\sin 2 x+\cos 2 x)^2-1\right] \\ & g[f(x)]=\sin ^2 2 x+\cos ^2 2 x+\sin 4 x-1 \\ & g[f(x)]=\sin 4 x \\ & \text { Let } g f(x)=y\end{aligned}$
$y=\sin 4 x \quad\left[\because\right.$ Domain of $\sin x$ is $\left.-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Domain of $\sin 4 x=\left[-\frac{\pi}{8}, \frac{\pi}{8}\right]$

Topic: Mathematics, Functions, KCET, KCET 2023
Let fx=sin3⁡x-3sin⁡x+6, ∀x∈0,π. The number of local maximum/maxima of the function f(x) is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
As f ′ x =3 sin 2 x.cosx−3cosx
=3cosâ¡xsin2â¡x-1
Hence, fx shows no local maximum.

Topic: Mathematics, Application of Derivatives, JEE Main
Let fx=sin4⁡x+cos4⁡x. Then, f is an increasing function in the interval:

Options:
A ] 5 π 8 , 3 π 4 [
B ] π 2 , 5 π 8 [
C ] π 4 , π 2 [
D ] 0 , π 4 [

The correct answer is:
] π 4 , π 2 [

Solution:
fx=sin4⁡x+cos4⁡x
f′x=4sin3⁡xcos⁡x+4cos3⁡x-sin⁡x
=4sin⁡xcos⁡xsin2⁡x-cos2⁡x
= -2sin⁡2xcos⁡2x
= -sin⁡4x

fx, is increasing when f′x>0
⇒ -sin⁡4x>0
⇒ sin⁡4x<0

⇒π<4x<2π

⇒π4<x<π2
⇒ x ∈π4,π2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (10 Apr Online)
Let fx=sintan-1x+sincot-1x2-1, x>1. If dydx=12ddxsin-1fx and y3=π6, then y-3 is equal to:

Options:
A 2π3
B -π6
C 5π6
D π3

The correct answer is:
5π6

Solution:
Let tan-1x=θ

⇒x=tanθ ⇒sinθ=x1+x2

So, y=x1+x2+11+x22-1

⇒y=x+121+x2-1

⇒y=2x1+x2=fx

Now, dydx=1211-f2×f'x

=1211-4x21+x22fx

=1+x22x2-1f'x

=1+x22x2-1×21+x2-2x21+x22

⇒dydx=1-x2x2-11+x2

⇒dy=1-x2x2-11+x2dx

Integrating both sides with respect to x

∫dy=∫1-x2x2-11+x2dx

⇒y=-tan-1x+c

Given, y3=π6⇒π6=-π3+c⇒c=π2

y=-tan-1x+π2=cot-1x

Now, y-3=cot-1x-3=5π6

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let fx=sinâ¡x3+cosâ¡3x10 for all real x. The least natural number n such that f10nÏ+x=fx for all real values of x is

Options:

The correct answer is:
6

Solution:
fx=sinâ¡x3+cosâ¡3x10
Period of sinâ¡x3=6Ï
Period of cosâ¡3x10=20Ï3
âµ sinâ¡aÎ¸,cosâ¡aÎ¸ are periodic with period2Ïa
LCM of 6Ï,20Ï3=60Ï
Therefore, the period of fx=60Ï
Hence, n=6.

Topic: Mathematics, Functions, JEE Main
Let fx=sin⁡x+3sin⁡3x+5sin⁡5x+3sin⁡7xsin⁡2x+2sin⁡4x+3sin⁡6x, wherever defined. If x1+x2=π2, where fx is defined at x1 and x2, then f2x1+f2x2 is

Options:
A cos2⁡x
B sin2⁡x
C 4
D 1

The correct answer is:
4

Solution:
Numerator of fx=sin⁡x+sin⁡3x+2sin⁡3x+sin⁡5x+3sin⁡5x+sin⁡7x
=2cos⁡xsin⁡2x+2sin⁡4x+3sin⁡6x
Hence, fx=2cos⁡x, if fx1=2cos⁡x1 & fx2=2sin⁡x1

i.e. f2x1+f2x2=4

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let fx=sinx+cos-2sin x-cos x,x∈0, π-π4, then f7π12f"7π12 is equal to

Options:
A 29
B -23
C -133
D 233

The correct answer is:
29

Solution:
Give that:

fx=sinx+cosx-2sinx-cosx

Convert the numerator and denominator in the form of sinA±B.

⇒fx=2sinx+π4-22sinx-π4

⇒fx=sinx+π4-1sinx-π4

⇒fx+π4=cos x-1sin x

⇒fx+π4=-tanx2

⇒fx=-tanx2-π8

⇒f'x=-1 2sec2x2-π8

⇒f"x=-12sec2x2-x8tanx2-π8

⇒f7π12=-tan7π24-π8=-tan4π24

⇒-tanπ6=-13

Also,

⇒f"7π12=-12×232×13=-233

⇒f7π12f"7π12=29

Hence this is the correct option.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let $f(x)=\sin x+\cos x, g(x)=x^2-1$. Thus $g(f(x))$ is invertible for $x \in$

Options:
A $\left[-\frac{\pi}{2}, 0\right]$
B $\left[-\frac{\pi}{2}, \pi\right]$
C $\left[-\frac{\pi}{2}, \frac{\pi}{4}\right]$
D $\left[0, \frac{\pi}{2}\right]$

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

Solution:
By definition of composition of function,
$g(f(x))=(\sin x+\cos x)^2-1 \Rightarrow g(f(x))=\sin 2 x$
We know $\sin x$ is bijective only, when
$x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Thus $g(x)$ is bijective if $-\frac{\pi}{2} \leq 2 x \leq \frac{\pi}{2} \Rightarrow \frac{-\pi}{4} \leq x \leq \frac{\pi}{4}$

Topic: Mathematics, Functions, JEE Main
Let $f(x)=\sin x, g(x)=x$.
Statement 1: $f(x) \leqslant g x($ for $) \mathrm{x}$ in $(0, \infty)$
Statement 2: $f(x) \leq 1$ for $x$ in $(0, \infty)$ but $g(x) \rightarrow \infty$ as $x \rightarrow \infty$.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
Let $f(x)=\sin x$ and $g(x)=x$
Statement-1: $f(x) \leq g \quad x(\forall x) \in 0, \infty(\quad)$
i.e., $\sin x \leq x \forall x \in(0, \infty)$
which is true
Statement-2: $f(x) \leq 1 \forall x \in 0, \infty(\quad)$
i.e., $\sin x \leq 1 \forall x \in(0, \infty)$
It is true and
$g(x)=x \rightarrow \infty$ as $x \rightarrow \infty$ also true.

Topic: Mathematics, Functions, JEE Main, JEE Main 2012 (07 May Online)
Let fx=tan-1x3-1x2+x, then the value of 17 f ′ 2 is equal to

Options:

The correct answer is:
7.4

Solution:
fx=tan-1x2-1x1+x21x=tan-1x2-tan-11x
f ′ x = 2x 1+ x 4 + 1 1+ 1 x 2 1 x 2 ⇒ 2x 1+ x 4 + 1 1+ x 2
f'2=417+15⇒17f'2=4+175

Topic: Mathematics, Differentiation, JEE Main
Let fx=tan3⁡x-3tan2⁡x+1, ∀x∈-π4,π4. Then, the positive difference between the least value and the local maximum of the function is

Options:

The correct answer is:
4

Solution:
Let, tan⁡x=t⇒t∈-1,1 ∵x∈-π4,π4
∴ft=t3-3t2+1
⇒ f ′ t =3 t 2 −6t=3t t−2
∴f-1=-1-3+1=-3
f1=1-3+1=-1
f0=1
∴ Least value =-3
Local maximum value =1
∴ The positive difference =4

Topic: Mathematics, Application of Derivatives, JEE Main
Let $f(x)=(x+1)^2-1, x \geq-1$
Statement-1: The set $\left\{x: f(x)=f^{-1}(x)\right\}=\{0,-1\}$
Statement-2 : $\mathrm{f}$ is a bijection.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:
There is no information about co-domain therefore $f(x)$ is not necessarily onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2009
Let $f(x)=\left\{\begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k, & x=2\end{array}\right.$
The value of $k$ for which $f$ is continuous at $x=2$ is

Options:
A
$e^{-2}$

B
$e$

C
$e^{-1}$

D
1

The correct answer is:
$e^{-1}$

Solution:
Since $f(x)$ is continuous at $x=2$.
$$
\begin{aligned}
&\therefore \lim _{x \rightarrow 2} f(x)=f(2) \\
&\Rightarrow \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=k \quad\left(1^{\infty} \text { form }\right) \\
&\therefore e^{l=k}
\end{aligned}
$$
where
$$
\begin{aligned}
&l=\lim _{x \rightarrow 2}(x-1-1) \times \frac{1}{2-x}=\lim _{x \rightarrow 2} \frac{x-2}{2-x} \\
&=\lim _{x \rightarrow 2}\left(\frac{x-2}{x-2}\right) \\
&\Rightarrow k=e^{-1}
\end{aligned}
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Let $f(x)=\left\{\begin{array}{ll}(x-1) \sin \left(\frac{1}{x-1}\right), & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.$. Then which one of the following is true?

Options:
A
$f$ is neither differentiable at $x=0$ nor at $x=1$

B
$f$ is differentiable at $x=0$ and at $x=1$

C
$\mathrm{f}$ is differentiable at $\mathrm{x}=0$ but not at $\mathrm{x}=1$

D
$\mathrm{f}$ is differentiable at $\mathrm{x}=1$ but not at $\mathrm{x}=0$

The correct answer is:
$f$ is neither differentiable at $x=0$ nor at $x=1$

Solution:
$$
f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}
$$
$$
\begin{aligned}
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \left(\frac{1}{1+h-1}\right)-0}{h}=\lim _{h \rightarrow 0} \frac{h}{h} \sin \left(\frac{1}{h}\right) \\
& \Rightarrow f^{\prime}(1)=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)
\end{aligned}
$$
$\therefore \mathrm{f}$ is not differentiable at $x=1$.
Similarly, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$ $\Rightarrow f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{(h-1) \sin \left(\frac{1}{h-1}\right)-\sin (1)}{h}$
$\Rightarrow \mathrm{f}$ is also not differentiable at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2008
Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}+1}{\mathrm{x}-1}$ for all $\mathrm{x} \neq 1$. Let
$\mathrm{f}^{1}(\mathrm{x})=\mathrm{f}(\mathrm{x}), \mathrm{f}^{2}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$ and generally
$f^{n}(x)=f\left(f^{n-1}(x)\right)$ for $n>1$
Let $\mathrm{P}=\mathrm{f}^{1}(2) \mathrm{f}^{2}(3) \mathrm{f}^{3}(4) \mathrm{f}^{4}(5)$
Which of the following is a multiple of P -

Options:
A 125
B 375
C 250
D 147

The correct answer is:
375

Solution:
$\begin{array}{l}
f(x)=\frac{x+1}{x-1} \\
f^{2}(x)=f(f(x))=f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=x \\
f^{3}(x)=f(x)=\frac{x+1}{x-1} \\
f^{4}(x)=x \\
P=f(2) \cdot f^{3}(3) f^{3}(4) f^{4}(5) \\
P=3 \times 3 \times \frac{5}{3} \times 5=75
\end{array}$
Multiple of $\mathrm{P}$ is 375

Topic: Mathematics, Functions, JEE Main
Let fx=∫x1+x2dx x≥0. Then f3-f1 is equal to :

Options:
A -π12+12+34
B π6+12-34
C -π6+12+34
D π12+12-34

The correct answer is:
π12+12-34

Solution:
fx=∫x1+x2dx

Let x=tan2θ

dx=2tanθsec2θ dθ

fx=∫tanθ1+tan2θ2.2tanθsec2θ dθ

fx=∫tanθsec4θ.2tanθsec2θ dθ

fx=∫2tan2θ.cos2θ dθ

fx=∫2sin2θ dθ

fx=∫1−cos2θ dθ

fx=θ−sin2θ2+C=θ−tanθ1+tan2θ+C

fx=tan−1x−x1+x+C

Now, f3−f1=tan−3−31+3−tan−11+11+1

=π3−π4+12−34=π12+12-34

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let fx=x2+1x2 and gx=x-1x, x∈R--1, 0, 1. If hx=fxgx , then the local minimum value of hx is:

Options:
A 22
B 3
C -3
D -22

The correct answer is:
22

Solution:
fx=x2+1x2=x-1x2+2
gx=x-1x
x∈R--1,0,1
Let x-1x=t
hx=fxgx=t2+2t=t+2t
t∈R-0
hx=t+2t
AM≥GM
Minimum of hx⇒ t+2t2 ≥ t×2t
t+2t≥22
Local minimum =22

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (08 Apr)
Let fx=x2+2px+2q2 and gx=-x2-2qx+p2 (where q≠0). If x∈R and the the minimum value of fx is equal to the maximum value of gx, then the value of p2q2 is equal to

Options:

The correct answer is:
0.5

Solution:
Since the minimum/maximum value of a quadratic expression ax2+bx+c is given by 4bc-b24a

So, fmin=2q2-p2, gmin=p2+q2
Now, 2q2-p2=p2+q2⇒q2=2p2

Hence, p2q2=12=0.5

Topic: Mathematics, Quadratic Equation, JEE Main
Let $f(x)=\left\{\begin{array}{rr}\frac{x^3+x^2-16 x+20}{(x-2)^2}, & \text { if } x \neq 2 \\ k, & \text { if } x=2\end{array}\right.$. If $f(x)$ be continuous for all $x$, then $k=$

Options:
A $7$
B $-7$
C $\pm 7$
D None of these

The correct answer is:
$7$

Solution:
For continuous $\lim _{x \rightarrow 2} f(x)=f(2)=k$
$\begin{aligned} & \Rightarrow k=\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4 x+4\right)(x+5)}{(x-2)^2}=7\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=\frac{x^2-x}{x^2+2 x} x \neq 0,-2$. Then $\frac{d}{d x}\left[f^{-1}(x)\right]$ (wherever it is defined) is equal to:

Options:
A
$\frac{-1}{(1-x)^2}$

B
$\frac{3}{(1-x)^2}$

C
$\frac{1}{(1-x)^2}$

D
$\frac{-3}{(1-x)^2}$

The correct answer is:
$\frac{3}{(1-x)^2}$

Solution:
Let $y=\frac{x^2-x}{x^2+2 x}$
$$
\begin{aligned}
& \Rightarrow \quad\left(x^2+2 x\right) y=x^2-x \\
& \Rightarrow \quad x(x+2) y=x(x-1) \\
& \Rightarrow \quad x[(x+2) y-(x-1)]=0 \\
& \because \quad x \neq 0, \quad \therefore \quad(x+2) y-(x-1)=0 \\
& \Rightarrow \quad x y+2 y-x+1=0 \\
& \Rightarrow \quad x(y-1)=-(2 y+1) \\
& \therefore \quad x=\frac{2 y+1}{1-y} \Rightarrow f^{-1}(x)=\frac{2 x+1}{1-x} \\
& \frac{d}{d x}\left(f^{-1}(x)\right)=\frac{2(1-x)-(2 x+1)(-1)}{(1-x)^2} \\
& \quad=\frac{2-2 x+2 x+1}{(1-x)^2}=\frac{3}{(1-x)^2}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2013 (09 Apr Online)
Let $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}^{2}-3 \mathrm{x}+2}$ and $\mathrm{g}(\mathrm{x})=\sqrt{\mathrm{x}}$ be two given functions. If $\mathrm{S}$ be the domain of $\mathrm{f} \circ \mathrm{g}$ and $\mathrm{T}$ be the domain of $\mathrm{g}$ of then

Options:
A $S=T$
B $S \cap T=\phi$
C $\mathrm{S} \cap \mathrm{T}$ is a singleton
D $\mathrm{S} \cap \mathrm{T}$ is an interval

The correct answer is:
$\mathrm{S} \cap \mathrm{T}$ is an interval

Solution:
Hint:
$f(x)=\sqrt{(x-1)(x-2)}, g(x)=\sqrt{x}$
$f(g(x))=\sqrt{(\sqrt{x}-1)(\sqrt{x}-2)}$
$S=\{x: x \in[0,1] \cup[4, \infty)\}$
$g(f(x))=\sqrt{\sqrt{(x-1)(x-2)}}$
$T=\{x: x \in(-\infty, 1] \cup[2, \infty)\}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2020
Let fx=x2+4 : x<04-2x : x≥0, then the area bounded by y=fx and the x-axis from x=-1 to x=3 is equal to

Options:
A 9 sq. units
B 283 sq. units
C 293 sq. units
D 27 Sq. units

The correct answer is:
283 sq. units

Solution:
Required area =∫-10x2+4dx+12×4×2+1
=x33+4x-10+4+1
=13+4+5=283 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
Let fx=x2-4x-3,x>2 and g(x) be the inverse of f(x). Then the value of 1g'(2), where f(x)=2 , is (here, g' represents the first derivative of g)

Options:

The correct answer is:
6

Solution:
As g(f(x))=x (∵  g(x)=f−1(x))
Differentiating, we get,

g'(f(x))f'(x)=1
⇒ g'(f(x))=1f'(x)
when f(x)=2
⇒ x 2 −4x−3=2
⇒ x=5,  x=−1
∵x>2
∴  x=5
g'(2)=1(2x−4)x=5
= 1 6

Topic: Mathematics, Differentiation, JEE Main
Let $f(x)=x^{2}+a x+b$, where $a, b \in \mathrm{R}$. If $f(x)=0$ has all its roots imaginary, then the roots of $f(x)+f^{\prime}(\mathrm{x})+f^{\prime \prime}(\mathrm{x})=0$ are

Options:
A real and distinct
B imaginary
C equal
D rational and equal

The correct answer is:
imaginary

Solution:
$f(x)=x^{2}+a x+b$ has imaginary roots.
$$
\begin{array}{l}
\therefore \text { Discriminant, } D < 0 \Rightarrow a^{2}-4 b < 0 \\
f^{\prime}(x)=2 x+a \\
\Rightarrow f^{\prime \prime}(x)=2 \\
\text { Also, } f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0 ...(i)\\
\Rightarrow x^{2}+a x+b+2 x+a+2=0 \\
\Rightarrow x^{2}+(a+2) x+b+a+2=0 \\
\therefore x=\frac{-(a+2) \pm \sqrt{(a+2)^{2}-4(a+b+2)}}{2} \\
\quad=\frac{-(a+2) \pm \sqrt{a^{2}-4 b-4}}{2}
\end{array}
$$
Since, $a^{2}-4 b < 0$ $\therefore a^{2}-4 b-4 < 0$ Hence, Eq. (i) has imaginary roots.

Topic: Mathematics, Quadratic Equation, JEE Main
Let fx=x-2 and gx=ffx, x∈0,4. Then ∫03gx-fx dx is equal to

Options:
A 1
B 0
C 12
D 32

The correct answer is:
1

Solution:
fx=x−2=2−x;    x<2x−2;     x≥2

gx=ffx=2−fx;    fx<2fx−2;     fx≥2

=2−2−x;     2−x<2,    x<22−x−2;     2−x≥2,    x<22−x−2;     x−2<2,    x≥2x−2−2;     x−2≥2,    x≥2

=2−2−x;    0<x<22−x−2;  x<02−x−2;    2≤x<4x−2−2; x≥4

∫03gx−fxdx=∫02xdx+∫234−xdx−∫03x−2dx

=x2202+4x-x2223-2x-x2202-x22-2x23

=1.

Topic: Mathematics, Definite Integration, JEE Main