datavorous/entrance-exam-dataset huggingface dataset:
Let fx=∫0xgtloge1-t1+tdt, where g is a continuous odd function. If ∫-π2π2fx+x2cosx1+exdx=πα2-α, then α is equal to _____.

Options:

The correct answer is:
2

Solution:
Given: fx=∫0xgtlog1-t1+tdt

⇒f-x=∫0-xgtlog1-t1+tdt

⇒f-x=-∫0xg-ylog1+y1-ydy

=-∫0xg(y)ln1-y1+ydy (g is odd)

f(-x)=-f(x)⇒f is also odd.

Now,

I=∫-π2π2f(x)+x2cosx1+exdx ...i

⇒I=∫-π2π2f(-x)+x2excosx1+exdx ...ii

⇒2I=∫-π2π2x2cosxdx

⇒2I=2∫0π2x2cosxdx

⇒I=∫0π2x2cosxdx

⇒I=x2sinx0π2-∫0π22xsinxdx

⇒I=π24-2-xcosx+sinx0π2

⇒I=π24-2(0+1)

⇒I=π24-2

⇒I=π22-2

∴α=2

Topic: Mathematics, Definite Integration, JEE Main
Let fx=∫0xgtloge1-t1+tdt, where g is a continuous odd function. If ∫-π2π2fx+x2cosx1+exdx=πα2-α, then α is equal to _____.

Options:

The correct answer is:
2

Solution:
Given: fx=∫0xgtlog1-t1+tdt

⇒f-x=∫0-xgtlog1-t1+tdt

⇒f-x=-∫0xg-ylog1+y1-ydy

=-∫0xg(y)ln1-y1+ydy (g is odd)

f(-x)=-f(x)⇒f is also odd.

Now,

I=∫-π2π2f(x)+x2cosx1+exdx ...i

⇒I=∫-π2π2f(-x)+x2excosx1+exdx ...ii

⇒2I=∫-π2π2x2cosxdx

⇒2I=2∫0π2x2cosxdx

⇒I=∫0π2x2cosxdx

⇒I=x2sinx0π2-∫0π22xsinxdx

⇒I=π24-2-xcosx+sinx0π2

⇒I=π24-2(0+1)

⇒I=π24-2

⇒I=π22-2

∴α=2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let fx =â«0x(t-1)(t-2)2dt. If fxâ¥k for all x and for some k, then the set of exhaustive values of k is

Options:
A 0,∞
B 0,2
C 1,∞
D –∞,–1712

The correct answer is:
–∞,–1712

Solution:
x=1 is a point of global minima of fx
Minimum value of
fx= ∫01t-1t-22dt
=∫01t3-5t2+8t-4dt
=-1712

Topic: Mathematics, Definite Integration, JEE Main
Let $f(x)=\int_0^x\left(t+\sin \left(1-e^{\prime}\right)\right) d t, x \in \mathbb{R}$. Then, $\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

Options:
A $-\frac{1}{6}$
B $\frac{2}{3}$
C $-\frac{2}{3}$
D $\frac{1}{6}$

The correct answer is:
$-\frac{1}{6}$

Solution:
$\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$
Using L Hopital Rule.
$\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}$ (Again L Hopital)
Using L.H. Rule
$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \\
& =-\frac{1}{6}
\end{aligned}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let fx=12-tanπx2, -1<x<1 and gx=3+4x-4x2, then the domain of f+g is

Options:
A 12, 1
B -12, 12
C -12, 1
D -12, -1

The correct answer is:
-12, 1

Solution:
Given, fx=12-tanπx2, -1<x<1 and gx=3+4x-4x2

Since, we are given with the domain of fx, then the domain f+g can be obtained by finding the domain of gx and taking the intersection with the domain of fx.

Now, for gx to be defined 3+4x-4x2≥0

⇒4x2-4x-3≤0

⇒2x+12x-3≤0

Thus, the domain of gx is -12, 32.

Now, the domain of f+g is -1, 1∩-12, 32≡-12, 1

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
Let fx=1-cosx2Ï-x2sin2xlog1+4Ï2-4Ïx+x2:xâ 2Ï Î»:x=2Ï is continuous at x=2Ï, then the value of Î» is equal to

Options:

The correct answer is:
0.5

Solution:
For the function fx to be continuous at x=2Ï
limxâ2Ïfx=f2Ï
Now, limxâ2Ï1-cosâ¡x2Ï-x2sin2xlog1+2Ï-x2=Î»
Putting x=2Ï+t, we get,
limtâ01-costt2âsin2tlog1+t2=Î»
limtâ012âsin2tt2ât2log1+t2=Î»
12Ã1Ã1=Î»âÎ»=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=1−cosx2π−x2⋅tan2xln1+4π2−4πx+x2:x≠2π λ:x=2π is continuous at x=2π, then the value of λ is equal to

Options:

The correct answer is:
0.5

Solution:
For fx to be continuous at x=2π,
limx→2π⁡fx=f2π
Now, limx→2π⁡1-cos⁡xtan2⁡x2π-x2ln⁡1+2π-x2=λ
Putting x=2π+t, we get,
limt→01−costt2⋅tan2tln1+t2=λ
limt→012⋅tan2tt2⋅t2ln1+t2=λ
So, 12×1×1=λ⇒λ=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=1+cosxπ−x2⋅sin2xln1+π2−2πx+x2 x≠πλ x=π is continuous at x=π, then λ is equal to

Options:
A 1
B -1
C 12
D 14

The correct answer is:
12

Solution:
For fx to be continuous at x=π,
limx→π⁡fx=fπ
Now, limx→π⁡1+cos⁡xπ-x2·sin2⁡xln⁡1+π-x2=λ
Putting x=π+t, we get,
∴ lim t→0 1−cost t 2 ⋅ sin 2 t ln 1+ t 2 =λ
lim t→0 1 2 ⋅ sin 2 t t 2 ⋅ t 2 ln 1+ t 2 =λ
12×1×1=λ⇒λ=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=1+sin2xcos2xsin2xsin2x1+cos2xsin2xsin2xcos2x1+sin2x, x∈π6,π3 . If α and β respectively are the maximum and the minimum values of f, then

Options:
A β2-2α=194
B β2+2α=194
C α2-β2=43
D α2+β2=92

The correct answer is:
β2-2α=194

Solution:
Given:

fx=1+sin2xcos2xsin2xsin2x1+cos2xsin2xsin2xcos2x1+sin2x

Applying C1→C1+C2+C3

fx=2+sin2xcos2xsin2x2+sin2x1+cos2xsin2x2+sin2xcos2x1+sin2x

⇒fx=2+sin2x1cos2xsin2x11+cos2xsin2x1cos2x1+sin2x

Applying R2→R2-R1 and R3→R3-R1

fx=2+sin2x1cos2xsin2x010001

⇒fx=2+sin2x1=2+sin2x

Now, for

x∈π6,π3

⇒2x∈π3,2π3

⇒sin2x∈32,1

⇒2+sin2x∈2+32,3

Hence,

β=2+32

α=3

So,

β2-2α=4+34+23-23=194

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let fx=1+sin⁡xlsin⁡x,-π6<x<0etan⁡2xtan⁡3x,0<x<π6m,x=0 is continuous at x=0. Then, the values of l and m are

Options:
A l=-23,m=e23
B l=23,m=e-23
C l=23,m=e23
D None of these

The correct answer is:
l=23,m=e23

Solution:
We have,
limh→0+⁡f0-h=limh→0+⁡1+sin⁡-hlsin⁡-h
=limh→0+⁡1+sin⁡hlsin⁡h
=limh→0+⁡1+sin⁡h1sin⁡hl=el
limh→0+⁡f0+h=limh→0+⁡etan⁡2htan⁡3h
=elimh→0+⁡tan⁡2h2h×23×3htan⁡3h=e1×23×1=e23

Also, f0=m
For fx to be continuous at x=0, we must have
limh→0+⁡f0-h=limh→0+⁡f0+h=f0⇒el=e23=m
⇒l=23 and m=e23

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=\left\{\begin{array}{cl}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$ and

$g(x)=|\eta(x)|+f(x \mid) .$ Then, in the interval $(-2,2), g$ is:

Options:
A differentiable at all points
B not continuous
C not differentiable at two points
D not differentiable at one point

The correct answer is:
not differentiable at one point

Solution:
$f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$

$\text { Then, } f(x \mid)=\left\{\begin{array}{cc}

-1, & -2 \leq|x| < 0 \\

|x|^{2}-1, & 0 \leq|x| \leq 2

\end{array}\right.$

$\Rightarrow f(|x|)=x^{2}-1,-2 \leq x \leq 2$

$\Rightarrow g(x)=\left\{\begin{aligned} 1+x^{2}-1, &-2 \leq x < 0 \\\left(x^{2}-1\right)+\left|x^{2}-1\right|, & 0 \leq x \leq 2 \end{aligned}\right.$

$=\left\{\begin{array}{cc}x^{2}, & -2 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ 2\left(x^{2}-1\right), & 1 \leq x \leq 2\end{array}\right.$

$g^{\prime}\left(0^{-}\right)=0, g^{\prime}\left(0^{*}\right)=0, g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=4$

$\Rightarrow g(x)$ is non-differentiable at $x=1$ $\Rightarrow g(x)$ is not differentiable at one point.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Let $f(x)=-1+|x-2|$, and $g(x)=1-|x|$; then the set of all points where $f_{o g}$ is discontinuous is :

Options:
A
$\{0,2\}$

B
$\{0,1,2\}$

C
$\{0\}$

D
an empty set

The correct answer is:
an empty set

Solution:
$$
\text { } \begin{aligned}
& f \circ g=f(g(x))=f(1-|x|) \\
& =-1+|1-| x|-2| \\
& =-1+|-| x|-1|=-1+|| x|+1|
\end{aligned}
$$
Let $f \circ g=y$
$$
\begin{aligned}
& \therefore y=-1+|| x|+1| \\
& \Rightarrow y= \begin{cases}-1+x+1, & x \geq 0 \\
-1-x+1, & x < 0\end{cases} \\
& \Rightarrow y=\left\{\begin{array}{cc}
x, & x \geq 0 \\
-x, & x < 0
\end{array}\right.
\end{aligned}
$$
LHL at $(x=0)=\lim _{x \rightarrow 0}(-x)=0$
RHL at $(x=0)=\lim _{x \rightarrow 0}(x)=0$
When $x=0$, then $y=0$
Hence, LHL at $(x=0)=$ RHL at $(x=0)=$ value of $y$ at $(x=0)$
Hence $y$ is continuous at $x=0$.
Clearly at all other point $y$ continuous. Therefore, the set of all points where fog is discontinuous is an empty set.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2013 (22 Apr Online)
Let fx=1x2:x≥1αx2+β:x<1. If fx is continuous and differentiable at any point, then

Options:
A Î±=2, Î²=-1
B Î±=-1, Î²=2
C Î±=1, Î²=0
D Î±=-2, Î²=3

The correct answer is:
Î±=-1, Î²=2

Solution:
The given function is clearly continuous at all points except possibly at x=±1.
For fx to be continuous at x=1, we must have
limx→1-fx=limx→1+fx=f1
⇒limx→1-αx2+β=limx→1+1x2
⇒α+β=1 … 1
Now, for fx to be differentiable at x=1, we must have
limx→1-fx-f1x-1=limx→1+fx-f1x-1
⇒limx→1-αx2+β-1x-1=limx→1+1x2-1x-1 ∵α+β=1∴β-1=-α
⇒limx→1-αx2-αx-1=limx→1+1x2-1x-1
⇒limx→1-αx+1=limx→1+-2x3⇒2α=-2
⇒α=-1
Putting α=-1 in 1, we get β=2

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}}$ for $n \geq 2$ and $g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text { occurs } n \text { times }}(x)$. Then, $\int x^{n-2} g(x) d x$ equals

Options:
A
$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

B
$\frac{1}{n-1}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

C
$\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$

D
$\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$

The correct answer is:
$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$

Solution:
Here, $\begin{array}{rlrl} & \text { and } & f f(x) & =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ & \therefore \quad f f f(x) & =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \text { Let } & g(x) & =\underbrace{(f \circ f o \ldots o f)}_{n \text { times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I & =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ & & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array}$
$\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$.

Topic: Mathematics, Indefinite Integration, JEE Main
Let fx=21x-121x+1:x≠00:x=0, then fx is

Options:
A continuous and differentiable at x=0
B continuous but not differentiable at x=0
C differentiable but not continuous at x=0
D None of these

The correct answer is:
None of these

Solution:
For function fx at x=0
L.H.L. =limh→0+f0-h=limh→0+2-1h-12-1h+1=0-10+1=-1
R.H.L. =limh→0+f0+h=limh→0+21h-121h+1=limh→0+1-2-1h1+2-1h=1-01+0=1
Since, limh→0+f0-h≠limh→0+f0-h,

therefore, fx is not continuous and not differentiable at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=210x+1 and gx=310x-1. If fogx=x, then x is equal to:

Options:
A 210-1210-3-10
B 1-2-10310-2-10
C 310-1310-2-10
D 1-3-10210-3-10

The correct answer is:
1-2-10310-2-10

Solution:
Given fgx=x

Where,fx=210x+1 ,gx=310x-1.
⇒f310x-1=210310x-1+1
⇒210310x-1+1=x
⇒x610-1=210-1
⇒x=210-1610-1=210-1210(2×3)10-1210
=1-2-10310-2-10

Topic: Mathematics, Functions, JEE Main, JEE Main 2017 (08 Apr Online)
Let fx=2tan3⁡x-6tan2⁡x+1+sgnex, ∀x∈-π4,π4. Then the positive difference between the least value and the local maximum value of the function is (where sgnf(x) represents the signum function)

Options:
A 7
B 8
C 9
D 10

The correct answer is:
8

Solution:
Here, sgnex=1

Let tan⁡x=t⇒t∈-1,1 ∵x∈-π4,π4
∴ft=2t3-6t2+2
⇒f't=23t2-6t
=6tt-2
∴f-1=2-1-3+1=-6
f1=21-3+1=-2
f0=2
∴ Least value =-6
Local maximum value =2
∴ the required positive difference =8

Topic: Mathematics, Application of Derivatives, JEE Main
Let fx=2x+1 and gx=∫fxx2x+12dx. If 6g2+1=0, then g-12 is equal to

Options:
A 4
B -4
C 3
D 2

The correct answer is:
4

Solution:
gx=∫x+12-x2x2x+12dx=∫1x2-1x+12dx=-1x+1x+1+c
=-1xx+1+c
g2=-16+c=-16⇒c=0
Hence, g-12=4

Topic: Mathematics, Indefinite Integration, JEE Main
Let $f(x)=2 x^{2}+5 x+1$. If we write $f(x)$ as $f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)$
for real numbers $a, b, c,$ then

Options:
A there are infinite number of choices for $a, b, c$
B only one choice for a but infinite number of choices for $b$ and $c$
C exactly one choice for each of $a, b, c$
D more than one but finite number of choices for $a, b, c$

The correct answer is:
exactly one choice for each of $a, b, c$

Solution:
$$
\begin{array}{l}
\text { Given, } f(x)=2 x^{2}+5 x+1 \\
\text { Also, } f(x)=a(x+1)(x-2)+b(x-2)(x-1) \\
+c(x-1)(x+1) \\
\qquad \begin{array}{r}
=a\left(x^{2}-x-2\right)+b\left(x^{2}-3 x+2\right) \\
+c\left(x^{2}-1\right) \\
f(x)=(a+b+c) x^{2}+(-a-3 b) x \\
+(-2 a+2 b-c)
\end{array}
\end{array}
$$
On equating the coefficients of $x^{2}$, $x$ and constant term we get
$$
\begin{array}{l}
a+b+c=2,-a-3 b=5 \\
\text { and }-2 a+2 b-c=1
\end{array}
$$
On solving above equations, we get $a=-\frac{35}{4}, b=\frac{5}{4}$ and $c=\frac{38}{4}$
Hence, exactly one choice for each of b and c

Topic: Mathematics, Quadratic Equation, JEE Main
Let fx=2x2+5x-3, x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to:

Options:
A 5
B 2
C 0
D 3

The correct answer is:
3

Solution:
Given: fx=2x2+5x-3

Now, let y=2x2+5x−3

⇒y=2x2+6x-x−3

⇒y=2xx+3-x+3

⇒y=2x-1x+3

Graph of ⇒y=2x-1x+3 is given below:

Graph of fx=2x2+5x-3 is given below:

So, fx is continuous for x∈R and non-differentiable at x=±12,0

So, number of points of discontinuity=0=m

Also, number of points of non-discontinuity=3=n

⇒m+n=3

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Let fx=2-x+4sin2x, x≠0. In order that fx is continuous at x=0, f0 is to be defined as

Options:
A -18
B 12
C 1
D 18

The correct answer is:
-18

Solution:
Given f(x) is continuous at x=0.

Therefore, f(0) =limx→0 2-x+4sin 2x ...i

f(x) forms a 00 form at x=0.

Hence, using L'Hospital's Rule,

limx→0 -12x+42cos 2x

-12(4)2(1) =-18

Using i, f0=-18

Topic: Mathematics, Limits, JEE Main
Let fx=2x+tan-1x and gx=loge1+x2+x, x∈0,3. Then

Options:
A There exists x∈0,3 such that f'x<g'x
B max fx>max gx
C There exist 0<x1<x2<3 such that fx<gx, ∀x∈x1,x2
D min f'x=1+max g'x

The correct answer is:
max fx>max gx

Solution:
Given:

fx=2x+tan-1x

⇒f'x=2+11+x2

And,

gx=ln1+x2+x

⇒g'x=11+x2

Now,

0≤x≤3

⇒0≤x2≤9

⇒1≤1+x2≤10

So,

110≤11+x2≤1

⇒2+110≤2+11+x2≤3

⇒2+110≤f'x≤3

⇒2110≤f'x≤3

And,

110≤g'x≤1

So, min f'x=2110≠1+max g'x

Option 4 is incorrect

From above,

g'x<f'x∀x∈0,3

Option 1 is incorrect.

Since, f'x and g'x are both positive so fx and gx both are increasing

So,

maxfx at x=3 is 6+tan-13

maxgx at x=3 is ln3+10

And, 6+tan-13>ln3+10

Option 2 is correct

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Let fx=2+x-x-1+x+1, x∈R.

Consider

S1:f'-32+f'-12+f'12+f'32=2

S2: ∫-22 fxdx=12

Then,

Options:
A both S1 and S2 are correct
B both S1 and S2 are wrong
C only S1 is correct
D only S2 is correct

The correct answer is:
only S2 is correct

Solution:
Given fx=2+x-x-1+x+1, x∈R ⇒fx=-x,x<-1x+2,-1≤x<03x+2,0≤x<1x+4,x≥1

⇒f'x=-1,x<-11,-1<x<03,0<x<11,x>1

∴ f'-32+f'-12+f'12+f'32

=-1+1+3+1=4

i.e. S1 is incorrect.

Now ∫-22fxdx=∫-2-1fxdx+∫-10fxdx+∫01fxdx+∫12fxdx

=-x22-2-1+x+222-10+3x+22601+x+42212

-12+2+2-12+256-46+18-252

=32+32+72+112=12

i.e. S2 is correct.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Let fx=∫2xx2+1x2+3dx. If f3=12loge5-loge6, then f4 is equal to

Options:
A 12loge17-logc19
B loge17-loge18
C 12logc19-logc17
D logc19-logc20

The correct answer is:
12loge17-logc19

Solution:
Let

I=∫2xx2+1x2+3dx

Put x2=t⇒2xdx=dt

I=∫1t+1t+3dt

⇒I=12∫2t+1t+3dt

⇒I=12∫1t+1-1t+3dt

⇒I=12lnt+1-lnt+3+C

⇒fx=12lnx2+1-lnx2+3+C

Put x=3, then

12ln5-ln6=12ln10-ln12+C

12ln5-ln6=12ln2+ln5-ln2-ln6+C

⇒C=0

So,

fx=12lnx2+1-lnx2+3

⇒f4=12ln17-ln19 or f4=12loge17-loge19

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let $f(x)=\sqrt{2-x-x^{2}}$ and $g(x)=\cos x$. Which of the following statements are true?
(I) Domain of $f\left((g(x))^{2}\right)=$ Domain of $f(g(x))$
(II) Domain of $f(g(x))+g(f(x))=$ Domain of $g(f(x))$
(III) Domain of $f(g(x))=$ Domain of $f(g(x))$ (V) Domain of $g\left((f(x))^{3}\right)=$ Domain of $f(g(x))$
(IV) Domain of $g\left((f(\mathrm{x}))^{3}\right)=$ Domain of $f(\mathrm{~g}(\mathrm{x}))$

Options:
A Only (I)
B Only (I) and (II)
C Only (III) and (IV)
D Only (I) and (IV)

The correct answer is:
Only (I) and (II)

Solution:
Domain of $f(g(x))$ is $R$
$$
\begin{array}{l}
\because 2-\cos x-\cos ^{2} x \geq 0 \\
(\cos x+2)(\cos x-1) \leq 0 \Rightarrow-2 \leq \cos x \leq 1 \\
x \in R
\end{array}
$$
Domain of $g(f(x))$ is $[-2,1]$
$$
\begin{array}{c}
\because \cos \left(\sqrt{2-x-x^{2}}\right) \\
2-x-x^{2} \geq 0
\end{array}
$$
Domain of $f\left(g(x)^{2}\right)$ is $R$
$$
\begin{array}{l}
\because 2-\cos ^{2} x-\cos ^{4} x \geq 0 \\
\left(\cos ^{2} x+2\right)\left(\cos ^{2} x-1\right) \leq 0 \\
-1 \leq \cos x \leq 1 \\
\quad x \in R
\end{array}
$$
Domain of $\mathrm{g}\left(\mathrm{f}^{\beta}(\mathrm{x})\right.$ is Domain of $\mathrm{g}(\mathrm{f}(\mathrm{x})$
i.e., $[-2,1]$

Topic: Mathematics, Functions, JEE Main
Let $f(x)=\sqrt{2-x-x^{2}}$ and $g(x)=\cos x$. Which of the following statements are true?
(I) Domain of $f\left((g(x))^{2}\right)=$ Domain of $f(g(x))$
(II) Domain of $f(g(x))+g(f(x))=$ Domain of $g(f(x))$
(III) Domain of $f(g(x))=$ Domain of $f(g(x))$ (V) Domain of $g\left((f(x))^{3}\right)=$ Domain of $f(g(x))$
(IV) Domain of $g\left((f(\mathrm{x}))^{3}\right)=$ Domain of $f(\mathrm{~g}(\mathrm{x}))$

Options:
A Only (I)
B Only (I) and (II)
C Only (III) and (IV)
D Only (I) and (IV)

The correct answer is:
Only (I) and (II)

Solution:
Domain of $f(g(x))$ is $R$
$$
\begin{array}{l}
\because 2-\cos x-\cos ^{2} x \geq 0 \\
(\cos x+2)(\cos x-1) \leq 0 \Rightarrow-2 \leq \cos x \leq 1 \\
x \in R
\end{array}
$$
Domain of $g(f(x))$ is $[-2,1]$
$$
\begin{array}{c}
\because \cos \left(\sqrt{2-x-x^{2}}\right) \\
2-x-x^{2} \geq 0
\end{array}
$$
Domain of $f\left(g(x)^{2}\right)$ is $R$
$$
\begin{array}{l}
\because 2-\cos ^{2} x-\cos ^{4} x \geq 0 \\
\left(\cos ^{2} x+2\right)\left(\cos ^{2} x-1\right) \leq 0 \\
-1 \leq \cos x \leq 1 \\
\quad x \in R
\end{array}
$$
Domain of $\mathrm{g}\left(\mathrm{f}^{\beta}(\mathrm{x})\right.$ is Domain of $\mathrm{g}(\mathrm{f}(\mathrm{x})$
i.e., $[-2,1]$

Topic: Mathematics, Functions, KVPY, KVPY 2013 (SB/SX)
Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{1-\sin ^{3} \mathrm{x}}{3 \cos ^{2} \mathrm{x}}, \mathrm{x}<\frac{\pi}{2} \\ \mathrm{p}, \quad \mathrm{x}=\frac{\pi}{2} \\ \frac{\mathrm{q}(1-\sin \mathrm{x})}{(\pi-2 \mathrm{x})^{2}}, \mathrm{x}>\frac{\pi}{2}\end{array}\right.$

If $f(x)$ is continuous at $x=\frac{\pi}{2},(p, q)=$

Options:
A (1,4)
B $\left(\frac{1}{2}, 2\right)$
C $\left(\frac{1}{2}, 4\right)$
D None of these

The correct answer is:
$\left(\frac{1}{2}, 4\right)$

Solution:
$\mathrm{f}\left[(\pi / 2)^{-}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\sin ^{3}[(\pi / 2)-\mathrm{h}]}{3 \cos ^{2}[(\pi / 2)-\mathrm{h}]}$

$\quad=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos ^{3} \mathrm{~h}}{3 \sin ^{2} \mathrm{~h}}=\frac{1}{2}$

$\mathrm{f}\left[(\pi / 2)^{+}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}[1-\sin \{(\pi / 2)+\mathrm{h}\}]}{[\pi-2\{(\pi / 2)+\mathrm{h}\}]^{2}}$

$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}(1-\cosh )}{4 \mathrm{~h}^{2}}=\frac{\mathrm{q}}{8}$

$\therefore \mathrm{p}=\frac{1}{2}=\frac{\mathrm{q}}{8} \Rightarrow \mathrm{p}=\frac{1}{2}, \mathrm{q}=4$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=3x2-23+4,x∈R. Then which of the following statements are true?

P: x=0 is a point of local minima of f

Q: x=2 is a point of inflection of f

R:f' is increasing for x>2

Options:
A Only P and Q
B Only P and R
C Only Q and R
D All P,Q and R

The correct answer is:
All P,Q and R

Solution:
Given,

fx=3x2-23+4

Now on rearranging we get,

fx=81.3x2-23

Now differentiating both side we get,

f'x=81.3x2-23·ln3·3x2-22·2x

=81×63x2-23×x2-22ln3

So by using first derivate test we get x=0 is point of local minima

Now differentiating the function f'x=486·ln3⏟k3x2-23xx2-22⏟gx

We get,g'x=3x2-23x2-22+x·3x2-23·4x·x2-2

+x·x2-22·3x2-23ln3·3x2-22·2x

=3x2-23x2-2x2-2+4x2+6x2ln3x2-23

g'x=3x2-23x2-25x2-2+6x2ln3x2-23

f''x=k·g'x

f''2=0,f''2+>0,f''2-<0

Since double derivate is changing the sign at given point so we can say that x=2 is point of inflection

Also, f''x>0 for x>2 so f'x is increasing

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha^2+2 \beta^2$ is equal to

Options:
A 42
B 38
C 24
D 44

The correct answer is:
42

Solution:
$\begin{aligned}
& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\
& x-2 \geq 0 \& 4-x \geq 0 \\
& \therefore x \in[2,4]
\end{aligned}$
Let $x=2 \sin ^2 \theta+4 \cos ^2 \theta$
$\begin{aligned}
& \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \\
& \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} \\
& \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} \\
& \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} \\
& \alpha^2+2 \beta^2=2+40=42
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let $f(x)=4$ and $f^{\prime}(x)=4$. Then $\operatorname{Lim}_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is given by

Options:
A
2

B
-2

C
-4

D
3

The correct answer is:
-4

Solution:
We have, $\operatorname{Lim}_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\left(\frac{0}{0}\right)=\operatorname{Lim}_{x \rightarrow 2} f(2)-2 f^{\prime}(x)=f(2)-2 f^{\prime}(2)=4-2 \times 4=-4$

Topic: Mathematics, Limits, JEE Main, JEE Main 2002
Let fx=4x+1-cosâ¡x-sinâ¡x68sinâ¡Î±012sinâ¡Î±16sin2Î±1+4sinâ¡Î± and f0=0. If the sum of all possible values of Î± is kÏ for Î±â0,2Ï, then the value of k is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
f0=1-1068sinâ¡Î±012sinâ¡Î±16sin2Î±1+4sinâ¡Î±=0
â1+4sinâ¡Î±6+8sinâ¡Î±=0
âsinâ¡Î±=-14,-34
Let, sinâ¡a=-14,sinâ¡b=-34
(âµ both values are negative so the solution lies in 3rd or 4th quadrant)
sinâ¡Î±=sinâ¡a,sinâ¡Î±=sinâ¡b
Î±=Ï+a,2Ï-a,Ï+b,2Ï-b
Hence, sum of all possible values=6Ï

Topic: Mathematics, Determinants, JEE Main
Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is

Options:
A
1

B
$\frac{1}{2}$

C
$-\frac{1}{2}$

D
$-1$

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

Solution:
$f(x)=\frac{1-\tan x}{4 x-\pi} \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=-\frac{1}{2}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2004
Let f(x)=5-|x-2| and gx=x+1, x ∈ R. If f(x) attains maximum value at α and g(x) attains minimum value at β, then limx→-αβx-1x2-5x+6x2-6x+8 is equal to

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

The correct answer is:
12

Solution:
Maxima of fx will occur at x=2 i.e., α=2

Minima of g(x) will occur at x=-1 i.e. β=-1

∴ limx→2(x-1)(x-2)(x-3)(x-2)(x-4)=12.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let fx=9x25+c,c>0. If the curve y=f-1x passes through 14,-54 and gx is the antiderivative of f-1x such that g0=52, then the value of g1 is, (where [.] represents the greatest integer function)

Options:

The correct answer is:
2

Solution:
y=fx=9x25+c1
⇒x=259y-c1=f-1y
⇒f-1x=259x-c1
The curve y=f-1x passes through 14,-59
⇒-59=25914-c1
⇒c1=14+15=920
⇒f-1x=259x-920 =25x9-54
gx=∫f-1xdx=‍∫25x9-54dx‍
=259x22-5x4+c2
g0=52⇒c2=52
⇒gx=259x22-5x4+52
⇒g1=2518-54+52=50-45+9036=9536
⇒g1=2

Topic: Mathematics, Indefinite Integration, JEE Main
Let fx=10-x-5,x∈R, then the set of all values of x at which ffx is not differentiable is

Options:
A 0,5,10
B 5,10
C 0,5,10,15
D 5,10,15

The correct answer is:
0,5,10

Solution:
ffx=10-fx-5=10-10-x-5-5
=10-5-x-5â10-x-5-5
Points where ffx is non-differentiable are
x-5=0 & x-5=5
x=5 & x=0,10

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=25x25x+5, then the number of solution(s) of the equation fsin2Î¸+fcos2Î¸=tan2Î¸, Î¸â0,10Ï is/are

Options:
A 10
B 2
C 40
D 20

The correct answer is:
20

Solution:
fsin2Î¸+f1-sin2Î¸
Let, sin2Î¸=t
â´ft+f1-t
=25t25t+5+251-t251-t+5=25t25t+5+2525+525t
=25t+525t+5=1
Therefore, the given equation will become tan2Î¸=1
â´ tanâ¡Î¸=Â±1âÎ¸=nÏÂ±Ï4, nâZ
Hence, the number of solutions are
4Ã5=20
{âµ4 solutions in 0,2Ï}

Topic: Mathematics, Trigonometric Equations, JEE Main