datavorous/entrance-exam-dataset huggingface dataset:
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, JEE Main 2020 (04 Sep Shift 1)
Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof $)(x)=(\operatorname{gogof})(x)$, where $(f \circ g)(x)=f(g(x))$ is

Options:
A
$\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$

B
$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$

C
$\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

D
$2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

The correct answer is:
$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$

Solution:
$f(x)=x^2, g(x)=\sin x$
$(g \circ f)(x)=\sin x^2$
go(gof) $(x)=\sin \left(\sin x^2\right)$
$(f \circ g \circ g \circ f)(x)=\left(\sin \left(\sin x^2\right)\right)^2$
Again, $(g \circ f)(x)=\sin x^2$
(gogof) $(x)=\sin \left(\sin x^2\right)$
Given, $($ fogogof $)(x)=($ gogof $)(x)$
$\Rightarrow \quad\left(\sin \left(\sin x^2\right)\right)^2=\sin \left(\sin x^2\right)$
$\Rightarrow \sin \left(\sin x^2\right)\left\{\sin \left(\sin x^2\right)-1\right\}=0$
$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$
$\Rightarrow \sin x^2=0$ or $\sin x^2=\frac{\pi}{2}$
$\therefore \quad x^2=n \pi$
(i.e. not possible as $-1 \leq \sin \theta \leq 1$ )

Topic: Mathematics, Trigonometric Equations, JEE Main
Let $f(x)= \begin{cases}x^2+k, & \text { when } x \geq 0 \\ -x^2-k, & \text { when } x \lt 0\end{cases}$. If the function $f(x)$ be continuous at $x=0$, then $k=$

Options:
A $0$
B $1$
C $2$
D $-2$

The correct answer is:
$0$

Solution:
We have,
$\begin{array}{ll}x^2+k & \text { when, } x \geq 0 \\
-x^2-k & \text { when, } x \leq 0\end{array}$
Given that,
L. H. L = R.H.L = f (x)
At point $x=0$
$\begin{aligned}& x^2+k=-x^2-k=0 \\
& \Rightarrow x^2+k=0 \\
& \Rightarrow k=-x^2 \\
& \Rightarrow k=0\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=x2sin1x; x≠00; x=0, then at x=0

Options:
A f is continuous but not differentiable
B f is continuous but f' is not continuous
C both f and f' are continuous
D f' is continuous but not differentiable

The correct answer is:
f is continuous but f' is not continuous

Solution:
Given:

fx=x2sin1x; x≠00; x=0

limx→0fx=limx→0x2sin1x

⇒limx→0fx=limx→0x2×number between -1 to 1

⇒limx→0fx=0=f0

Hence, fx is continuous at x=0.

Now,

RHD=f'0+=limh→0f0+h-f0h

⇒RHD=f'0+=limh→0h2sin1h-0h

⇒RHD=f'0+=limh→0hsin1h=0

And,

LHD=f'0-=limh→0f0-h-f0-h

⇒LHD=f'0-=limh→0-h2sin1h-0-h=0

Since, LHD=RHD=finite

So, fx is differentiable at x=0.

Now,

f'x=2xsin1x-cos1x; x≠00; x=0

Now,

limx→0f'x=limx→02xsin1x-cos1x

⇒limx→0f'x=limx→02x1x-13!1x3+15!1x5-....-1-12!1x2+14!1x4-....

⇒limx→0f'x=limx→01-23!1x2+25!1x4-....--12!1x2+14!1x4-....

This limit does not exists finitely, hence f'x is discontinuous at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let fx=x2-x+1, âxâ¥12, then the solution of the equation fx=f-1x is

Options:
A x=1
B x=2
C x=12
D x=3

The correct answer is:
x=1

Solution:
f(x) and f−1(x) intersect only on the line y=x
∴ x 2 −x+1=x
⇒ x 2 −2x+1=0
⇒ ( x−1 ) 2 = 0
⇒x=1

Topic: Mathematics, Functions, JEE Main
Let fx=x2-x+1, ∀x≥12, then the solution of the equation fx=f-1x is

Options:
A x=1
B x=2
C x=12
D x=3

The correct answer is:
x=1

Solution:
f(x) and f−1(x) intersect only on the line y=x
∴ x 2 −x+1=x
⇒ x 2 −2x+1=0
⇒ ( x−1 ) 2 = 0
⇒x=1

Topic: Mathematics, Functions, JEE Main
Let fx=-x2+x+p, where p is a real number. If gx=fx and gx is discontinuous at x=12, then p cannot be (where . represents the greatest integer function)

Options:
A 12
B 34
C 74
D -14

The correct answer is:
12

Solution:
fx=-x-122+p+14
If gx is discontinuous at x=12, then f12 should be an integer.

Hence, p cannot be 12

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=x2, x∈R . For any A⊆R, define gA={x∈R :f(x)∈A} . If S=[0, 4] , then which one of the following statements is not true?

Options:
A g(f(S))≠S
B f(g(S))≠f(S)
C fgS=S
D gfS=gS

The correct answer is:
gfS=gS

Solution:
From the given information we can conclude
gS=-2,2 , So, f(g(S))=0, 4=S
And f(S)=0, 16⇒fgS≠fS
Also, gfS=-4, 4≠gS
So, gfS≠S
Hence, gfS=gS is not true.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Let fx=x2-x+-x+x, where x∈ℝ and t denotes the greatest integer less than or equal to t. Then, f is

Options:
A continuous at x=0, but not continuous at x=1
B continuous at x=1, but not continuous at x=0
C continuous at x=0 and x=1
D not continuous at x=0 and x=1

The correct answer is:
continuous at x=1, but not continuous at x=0

Solution:
Given,

fx=x2-x+-x+x

⇒fx=x2-x+x-x, as -A=A

⇒fx=x2-x+x, as x+x=x

⇒fx=x2-x+x (∵x≥0)

Now at x=0, f0=0

And f0+=-1, as x2-x<0 for x→0+

So, function is discontinuous at x=0

Now at x=1, f1=0,

f1+=0+0=0 and f1-=-1+1=0

Hence, the function is continuous at x=1.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let fx=x+32x-23,x∈[-4,4]. If M and m are the maximum and minimum values of f, respectively in [-4,4], then the value of M-m is :

Options:
A 600
B 392
C 608
D 108

The correct answer is:
608

Solution:
Given: fx=x+32x-23

⇒f'x=x+32×3x-22+x-232x+3

⇒f'x=x+3x-22x+3×3+x-22

⇒f'x=x+3x-223x+9+2x-4

⇒f'x=x+3x-225x+5

⇒f'x=5x+3x-22x+1

For critical points, we put f'x=0.

⇒5x+3x-22x+1=0

⇒x=-3, 2, -1

⇒f-3=f2=0

⇒f-1=22-33=-108

Also, x∈-4,4

⇒f(-4)=-216

⇒f(4)=49×8=392

⇒M=392, m=-216

⇒M-m=392+216

⇒M-m=608

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let $f(x)=\left|\begin{array}{ccc}x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|$, where $p$ is a constant. Then $\frac{d^3}{d x^3}\{f(x)\}$ at $x=0$ is

Options:
A $p$
B $p+p^2$
C $p+p^3$
D Independent of $p$

The correct answer is:
Independent of $p$

Solution:
$f^{\prime \prime \prime}(x)=\left|\begin{array}{ccc}\frac{d^3}{d x^3} x^3 & \frac{d^3}{d x^3} \sin x & \frac{d^3}{d x^3} \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=\left|\begin{array}{ccc}6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|$
$\therefore f^{\prime \prime \prime}(0)=\left|\begin{array}{ccc}6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=0$, which is independent of p.

Topic: Mathematics, Differentiation, JEE Main
Let f(x)=x3+ax2+bx+c be polynomial with integer coefficients. If the roots of f(x) are integer and are in Arithmetic Progression, then 'a' cannot take the value

Options:
A −642
B 1214
C 1323
D 1626

The correct answer is:
1214

Solution:
Given that, f(x)=x3+ax2+bx+c

Now, since the roots of the polynomial are in Arithmetic Progression,

Let the roots are s-t, s, s+t

⇒s-t+s+s+t=-a1

⇒-3s=a

Hence, a must be multiple of 3.

From given options we see that only 1214 is not multiple of 3.

Topic: Mathematics, Sequences and Series, JEE Main
Let fx=x3x-11-cos⁡x for x≠0. Then value of f0, which make fx continuous at x=0, is

Options:
A log⁡3
B 12log⁡3
C 12log⁡3
D 2log⁡3

The correct answer is:
2log⁡3

Solution:
limx→0⁡fx=limx→0⁡x3x-11-cos⁡x
=limx→0⁡3x-1x⋅x21-cos⁡x
=log⁡3⋅112=2log⁡3

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $\mathrm{f}(\mathrm{x})=\sqrt{\frac{x+1}{x+3}}$ and $\mathrm{g}(\mathrm{x})=\sqrt{\frac{2-x}{x+3}}$ be two real valued functions. Then the domain of $f / g$ is

Options:
A $(-\infty,-3) \cup[-1, \infty)$
B $[-1,2)$
C $(-3,2)$
D $(-\infty,-3) \cup[2, \infty)$

The correct answer is:
$[-1,2)$

Solution:
$f(x)=\sqrt{\frac{x+1}{x-3}}$ and $g(x)=\sqrt{\frac{2-x}{x+3}}$
Domain for $\mathrm{f} / \mathrm{g}$
$\frac{x+1}{x+3} \geq 0 \frac{+-+}{-\infty-3-1}$
$x \in\left(-{ }^{\circ},-3\right) \cup\left[-1,{ }^{\circ}\right)$ ........(i)
$\frac{2-x}{x+3}>\Rightarrow \frac{x-2}{x+3} < 0 \frac{+-+}{-\infty-3-2 \infty}$
$x \in(-3,2)$ ........(ii)
from (i) and (ii)
$x \in[-1,2)$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
Let fx=x3-x2+10x-7,x≤1-2x+log2b2-4,x>1 Then the set of all values of b, for which fx has maximum value at x=1, is:

Options:
A -6,-2
B 2,6
C -6,-2)∪(2,6
D -6,-2)∪(2,6

The correct answer is:
-6,-2)∪(2,6

Solution:
Given, fx=x3-x2+10x-7,x≤1-2x+log2b2-4,x>1

Now f1=3

For x<1,f'x=3x2-2x+10>0

⇒ fx is increasing

For x>1,f'x<0

⇒ function is decreasing.

So, at x=1 there is a point of maxima

So, limx→1+fx=-2+log2b2-4

For maximum value at x=1

⇒3≥-2+log2 b2-4

⇒32≥b2-4>0

⇒b∈[-6,-2)∪(2,6]

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let f(x)=x3+x2f'(1)+xf"(2)+f'''(3), x∈R. Then f'(10) is equal to

Options:

The correct answer is:
202

Solution:
Given: f(x)=x3+x2f'(1)+xf"(2)+f'''(3)

⇒f'(x)=3x2+2xf'(1)+f"(2) ...i

⇒f"(x)=6x+2f'(1) ...ii

⇒f'''(x)=6

⇒f'''(3)=6 ...iii

Using ii,

⇒f"(2)=12+2f'(1)

Putting this value in equation i,

⇒f'(1)=3+2f'(1)+12+2f'(1)

⇒0=15+3f'(1)

⇒f'(1)=-5 ...iv

⇒f"(2)=12+2-5

⇒f"(2)=2 ...v

⇒f(x)=x3+x2·(-5)+x·(2)+6

⇒f'(x)=3x2-10x+2

⇒f'(10)=300-100+2

⇒f'(10)=202

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Let fx=x3+x2+x+1, then the area (in sq. units) bounded by y=fx, x=0, y=0 and x=1 is equal to

Options:
A 253
B 2512
C 125
D 53

The correct answer is:
2512

Solution:
f'x=3x2+2x+1>0 ∀x∈R
⇒fx is increasing
⇒f0=1, so fx is positive in 0,1
Required area =∫01x3+x2+x+1dx
=14+13+12+1
=2512 sq.units

Topic: Mathematics, Area Under Curves, JEE Main
Let $f(x)=|x-3|+|x+5|$ and
$A=\left\{a \in \mathbb{R} / \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\right.$ exists $\}$
Then the number of real numbers which are in $(-\infty,-3)$ $\cup(5, \infty)$ but not in $A$ is

Options:
A 2
B 0
C 1
D 3

The correct answer is:
1

Solution:
$A=I R /\{3,-5\}$
Now, real numbers which are in $(-\infty,-3) \cup(5, \infty)$ but not in $A$ is only 3 .
$\therefore$ No. of required real no. is 1 .

Topic: Mathematics, Sets and Relations, JEE Main
Let $f(x)=x^5+2 \mathrm{e}^{x / 4}$ for all $x \in \mathbf{R}$. Consider a function $g(x)$ such that $(g \circ f)(x)=x$ for all $x \in \mathbf{R}$. Then the value of $8 g^{\prime}(2)$ is :

Options:
A 2
B 8
C 4
D 16

The correct answer is:
16

Solution:
$\begin{aligned} & f(x)=2 \\ & \text { when } x=0 \\ & \because g^{\prime}(f(x)) f^{\prime}(x)=1 \\ & g^{\prime}(2)=\frac{1}{f^{\prime}(0)} \\ & \because f^{\prime}(x)=5 x^4+\frac{2}{4} e^{x / 4} \\ & g^{\prime}(2)=2 \\ & \text { Ans }=16 \\ & \text { Option (1) }\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Let $f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R}$, and $g(x)$ be a function such that $g(f(x))=x$ for all $x \in \mathbf{R}$. Then $\frac{g(7)}{g^{\prime}(7)}$ is equal to :

Options:
A 14
B 42
C 7
D 1

The correct answer is:
14

Solution:
$\begin{aligned} & f(x)=x^5+2 x^3+3 x+1 \\ & f^{\prime}(x)=5 x^4+6 x^2+3 \\ & f^{\prime}(1)=5+6+3=14 \\ & g(f(x))=x \\ & g^{\prime}(f(x)) f^{\prime}(x)=1 \\ & \text { for } f(x)=7 \\ & \Rightarrow x^5+2 x^3+3 x+1=7 \\ & \Rightarrow x=1 \\ & g^{\prime}(7) f^{\prime}(1)=1 \Rightarrow g^{\prime}(7)=\frac{1}{f^{\prime}(1)}=\frac{1}{14}\end{aligned}$
$\begin{aligned} & x=1, f(x)=7 \Rightarrow g(7)=1 \\ & \frac{g(7)}{g^{\prime}(7)}=\frac{1}{1 / 14}=14\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Let f(x)=x6+2x4+x3+2x+3, x∈R. Then the natural number n for which limx→1xnf(1)-f(x)x-1=44 is _____ .

Options:

The correct answer is:
7

Solution:
Given that: f(x)=x6+2x4+x3+2x+3, x∈R

So, f(1)=9

limx→19xn-x6-2x4-x3-2x-3x-1=44

⇒limx→19xn-9-x6+1-2x4+2-x3+1-2x+2x-1=44

⇒limx→19xn-1x-1-x6-1x-1-2x4-1x-1-x3-1x-1-2x-1x-1=44

As we know that, limx→axn-anx-a=nan-1

⇒9n-6-8-3-2=44

⇒n=7

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let fx=x6-2x5+x3+x2-x-1 and gx=x4-x3-x2-1 be two polynomials. Let a,b,c and d be the roots of gx=0. Then, the value of

fa+fb+fc+fd is

Options:
A -5
B 0
C 4
D 5

The correct answer is:
0

Solution:
Given,

fx=x6-2x5+x3+x2-x-1

=x2x4-x3-x2-1-xx4-x3-x2-1+2x2-2x-1

⇒fx=x2-xgx+2x2-2x-1

∴ fa=2a2-2a-1

∴ Σfa=fa+fb+c+fd

=2Σa2-2Σa-Σ1

=2Σa2-2Σab-2Σa-4

∵ a,b,c,d are roots of equation

gx=x4-x3-x2-1=0

then Σa=1, Σab=-1

∴ Σfa=212-2-1-21-4

=6-2-4=0

Topic: Mathematics, Quadratic Equation, JEE Main
Let fx=x and gx=x, (where, . denotes the greatest integer function). Then, fog â² â1 is

Options:
A 0
B does not exist
C -1
D 1

The correct answer is:
does not exist

Solution:
fogx=fgx=fx=x
Now, LHL of fog ′ −1 = lim h→ 0 + fog −1−h − fog −1 −h
=limhâ0+â¡-1-h--1-h
=limhâ0+â¡-2--1-h=limhâ0+â¡1-hâ-â
∴ fog ′ −1 does not exist.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=xpsin1x,x≠00, x=0 then fx is continuous but not differentiable at x=0, if

Options:
A p<0
B p=0
C 0<p≤1
D p≥1

The correct answer is:
0<p≤1

Solution:
We have,

fx=xpsin1x,x≠00, x=0

For fx to be continuous at x=0

limx→0fx=f0

⇒limx→0xpsin1x=0

⇒limx→0xp×limx→0sin1x=0

⇒limx→0xp×value between -1 to 1=0

∴p>0

Now,

RHD at RHD at x=0=limh→0hpsin1h-0h=limh→0hp-1sin1h

LHD at x=0=limh→0-hpsin-1h-0-h=limh→0-1p-hp-1sin1h

Since, fx is not differentiable at x=0, therefore

p≤1

Hence, 0<p≤1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=\left\{\begin{array}{l}x^p \sin \frac{1}{x}, x \neq 0 \\ 0 \quad, x=0\end{array}\right.$ then $f(x)$ is continuous but not differential at $x=0$ if

Options:
A $0 \lt p \leq 1$
B $1 \leq p \lt \infty$
C $-\infty \lt p \lt 0$
D $p=0$

The correct answer is:
$0 \lt p \leq 1$

Solution:
$f(x)=x^p \sin \frac{1}{x}, x \neq 0$ and $f(x)=0, x=0$
Since at $x=0, f(x)$ is a continuous function
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)=0 \Rightarrow \lim _{x \rightarrow 0} x^p \sin \frac{1}{x}=0 \Rightarrow p\gt0$.
$f(x)$ is differentiable at $x=0$, if $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}$ exists
$\Rightarrow \lim _{x \rightarrow 0} \frac{x^p \sin \frac{1}{x}-0}{x-0}$ exists
$\Rightarrow \lim _{x \rightarrow 0} x^{p-1} \sin \frac{1}{x}$ exists
$\Rightarrow p-1\gt0$ or $p\gt1$
If $p \leq 1$, then $\lim _{x \rightarrow 0} x^{p-1} \sin \left(\frac{1}{x}\right)$ does not exist and at $x=0 \quad f(x)$ is not differentiable.
$\therefore$ for $0 \lt p \leq 1 f(x)$ is a continuous function at $x=0$ but not differentiable.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=xx2+xx2+xx2+xx2, âxâ0,10 (where Â· and Â· are the greatest integer and fractional part functions respectively). The number of points of discontinuity of fx is

Options:

The correct answer is:
0

Solution:
fx=xx2+x2+xx2+x2
=x+xx2+x2
=x.x2=x3
Hence, fx is continuous âxâ0,10

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x)=x|x|$ and $g(x)=\sin x$.
Statement-1 : gof is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-2 : gof is twice differentiable at $x=0$.

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:
$$
\begin{aligned}
& f(x)=x|x| \text { and } g(x)=\sin x \\
& g \circ f(x)=\sin (x|x|)=\left\{\begin{array}{cc}
-\sin x^2 & , x < 0 \\
\sin x^2 & , x \geq 0
\end{array}\right. \\
& \therefore \text { (gof })^{\prime}(x)=\left\{\begin{array}{cc}
-2 x \cos x^2 & , x < 0 \\
2 x \cos x^2 & , x \geq 0
\end{array}\right.
\end{aligned}
$$
Clearly, L(gof )' $(0)=0=\mathrm{R}(\text { gof })^{\prime}(0)$
$\therefore$ gof is differentiable at $x=0$ and also its derivative is continuous at $x=0$
$$
\begin{aligned}
& \text { Now, (gof)" }(\mathrm{x})=\left\{\begin{array}{cc}
-2 \cos \mathrm{x}^2+4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} < 0 \\
2 \cos \mathrm{x}^2-4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} \geq 0
\end{array}\right. \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0)=-2 \text { and } \mathrm{R}(\text { gof })^{\prime \prime}(0)=2 \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0) \neq \mathrm{R}(\text { gof })^{\prime \prime}(0)
\end{aligned}
$$
$\therefore g o f(x)$ is not twice differentiable at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2009
Let $f(x)=x|x|, g(x)=\sin x$ and $h(x)=(g \circ f)(x)$. Then

Options:
A
$\mathrm{h}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

B
$\mathrm{h}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$, but $\mathrm{h}^{\prime}(\mathrm{x})$ is not continuous at $x=0$

C
$\mathrm{h}^{\prime}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but it is not differentiable at $x=0$

D
$\mathrm{h}^{\prime}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$

The correct answer is:
$\mathrm{h}^{\prime}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but it is not differentiable at $x=0$

Solution:
Let $f(x)=x|x|=x|x|, g(x)=\sin x$ and $h(x)=g o f(x)=g[f(x)]$
$$
\begin{aligned}
&\therefore h(x)= \begin{cases}\sin x^2, & x \geq 0 \\
-\sin x^2, & x < 0\end{cases} \\
&\text { Now, } h^{\prime}(x)=\left\{\begin{array}{cc}
2 x \cos x^2, & x \geq 0 \\
-2 x \cos x^2, & x < 0
\end{array}\right.
\end{aligned}
$$
Since, L.H.L and R.H.L at $x=0$ of $h^{\prime}(x)$ is equal to 0 therefore $h^{\prime}(x)$ is continuous at $x=0$
Now, suppose $h^{\prime}(x)$ is differentiable
$$
\begin{aligned}
&\therefore h^{\prime \prime}(x) \\
&= \begin{cases}2\left(\cos x^2+2 x^2\left(-\sin x^2\right),\right. & x \geq 0 \\
2\left(-\cos x^2+2 x^2 \sin x^2\right), & x < 0\end{cases}
\end{aligned}
$$
Since, L.H.L and R.H.L at $x=0$ of $h^{\prime \prime}(x)$ are different therefore $h^{\prime \prime}(x)$ is not continuous.
$\Rightarrow h^{\prime \prime}(x)$ is not differentiable
$\Rightarrow$ our assumption is wrong
Hence $h^{\prime}(x)$ is not differentiable at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2014 (11 Apr Online)
Let fx+y=fx⋅fy for all x,y∈R and fx=1+xϕxlog⁡3. If limx→0⁡ϕx=1, then f ′ x is equal to

Options:
A log⁡3fx
B log⁡fx3
C log⁡3
D None of these

The correct answer is:
log⁡3fx

Solution:
f ′ x = lim h→0 f x+h −f x h
=limh→0⁡fx⋅fh-fxh ∵fx+y=fx⋅fy
=fxlimh→0⁡fh-1h
=fxlimh→0⁡1+hϕhlog⁡3-1h ∵fx=1+xϕxlog⁡3
=fxlog⁡3limh→0⁡ϕh
=fx×log⁡3×1 ∵limh→0⁡ϕh=1
=log⁡3fx

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f(x+y)=f(x) \cdot f(y), \forall x, y \in R$, suppose that $f(3)=3$ and $f^{\prime}(0)=11$, then $f^{\prime}(3)$ is given by

Options:
A 22
B 44
C 28
D 33

The correct answer is:
33

Solution:
$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$
$$
\begin{aligned}
&=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} \\
\Rightarrow \quad f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f(h)-1}{h}
\end{aligned}
$$
$[\because f(3)=3]$
Use L'Hospital rule,
$$
\begin{aligned}
\Rightarrow \quad f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1} \\
&=3 f^{\prime}(0)
\end{aligned}
$$
$\left[\because f^{\prime}(0)=11\right]$
$$
=3 \times 11=33
$$

Topic: Mathematics, Functions, JEE Main
Let for A=123α31112,A=2. If |2adj(2adj(2A))|=32n, then 3n+α is equal to

Options:
A 9
B 11
C 12
D 10

The correct answer is:
11

Solution:
Given that A=123α31112

⇒|A|=2

⇒16-1-α-7=2

⇒-α-2=2

⇒α=-4

⇒|2adj(2adj(2A))|=32n

We know that kA=knA, adj(adj(A))=An-12 and adjkA=kn-1adjA.

=23|adj(2adj(2A))|

=23|23-1adj(adj(2A))|

=23·(4)3|adj(adj(2A))|

=29|2A|(2)2

=29·|2A|4

=29·212|A|4

=221·24=225=(32)n=(2)5n

∴n=5

⇒3n+α=15-4=11

Hence this is the correct option.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let for a triangle ABC

AB→=-2i^+j^+3k^

CB→=αi^+βj^+γk^

CA→=4i^+3j^+δk^

If δ>0 and the area of the triangle ABC is 56 then CB→·CA→ is equal to

Options:
A 60
B 54
C 108
D 120

The correct answer is:
60

Solution:
Given,

A triangle ABC

AB→=-2i^+j^+3k^

CB→=αi^+βj^+γk^

CA→=4i^+3j^+δk^

Now plotting the diagram we get,

Now from triangle law of addition we get,

CA→+AB→=CB→

⇒4i^+3j^+δk^+-2i^+j^+3k^=αi^+βj^+γk^

⇒2i^+4j^+δ+3k^=αi^+βj^+γk^

Now on comparing both side we get,

α=2, β=4 & γ=δ+3

Now area of triangle is given by,

A=12AB→×BC→

⇒56=i^j^k^-21324γ

⇒562=γ-122+6+2γ2+100

⇒5γ2=320

⇒γ2=64⇒γ=8

So, δ=8-3=5

CB→·CA→=αi^+βj^+γk^·4i^+3j^+δk^

⇒CB→·CA→=2i^+4j^+8k^·4i^+3j^+5k^=8+12+40=60

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let for $\mathrm{i}=1,2,3, \mathrm{p}_{\mathrm{i}}(\mathrm{x})$ be a polynomial of degree 2 in $x, p^{\prime}{ }_i(x)$ and $p^{\prime \prime}{ }_i(x)$ be the first and second order derivatives of $p_i(x)$ respectively. Let,
$$
\left.\mathrm{A}(\mathrm{x})=\left[\begin{array}{lll}
\mathrm{p}_1(\mathrm{x}) & \mathrm{p}_1^{\prime}(\mathrm{x}) & \mathrm{p}_1^{\prime \prime} \mathrm{x}( \\
\mathrm{p}_2(\mathrm{x}) & \mathrm{p}_2^{\prime}(\mathrm{x}) & \mathrm{p}_2^{\prime \prime}( \\
\mathrm{p}_3(\mathrm{x}) & \mathrm{p}_3^{\prime}(\mathrm{x}) & \mathrm{p}_3^{\prime \prime}(\mathrm{x}
\end{array}\right]\right)
$$
and $\mathrm{B}(\mathrm{x})=[\mathrm{A}(\mathrm{x})]^{\mathrm{T}} \mathrm{A}(\mathrm{x})$. Then determinant of $\mathrm{B}(\mathrm{x})$ :

Options:
A
is a polynomial of degree 6 in $\mathrm{x}$.

B
is a polynomial of degree 3 in $\mathrm{x}$.

C
is a polynomial of degree 2 in $\mathrm{x}$.

D
does not depend on $\mathrm{x}$.

The correct answer is:
is a polynomial of degree 6 in $\mathrm{x}$.

Solution:
Let $p_1 x=a_1 x^2+b_1 x+c_1$
$$
p_2 x=a_2 x^2+b_2 x+c_2
$$
and $p_3 x=a_3 x^2+b_3 x+c_3$
where $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3$ are real numbers.
$$
\begin{gathered}
\therefore \mathrm{A}(x)=\left[\begin{array}{ccc}
a_1 x^2+b_1 x+c_1 & 2 a_1 x+b_1 & 2 a_1 \\
a_2 x^2+b_2 x+c_2 & 2 a_2 x+b_2 & 2 a_2 \\
a_3 x^2+b_3 x+c_3 & 2 a_3 x+b_3 & 2 a_3
\end{array}\right] \\
=\left[\begin{array}{llll}
a_1 x^2+b_1 x+c_1 & a_2 x^2+b_2 x+c_2 & a_3 x^2+b_3 x+c_3 \\
2 a_1 x+b_1 & 2 a_2 x+b_2 & 2 a_3 x+b_2 \\
2 a_1 & 2 a_2 & 2 a_3
\end{array}\right] \\
\times\left[\begin{array}{ccc}
a_1 x^2+b_1 x+c_1 & 2 a_1 x+b_1 & 2 a_1 \\
a_2 x^2+b_2 x+c_2 & 2 a_2 x+b_2 & 2 a_2 \\
a_3 x^2+b_3 x+c_3 & 2 a_3 x+b_3 & 2 a_3
\end{array}\right]
\end{gathered}
$$
It is clear from the above multiplication, the degree of determinant of $\mathrm{B}(x)$ can not be less than 4 .

Topic: Mathematics, Matrices, JEE Main, JEE Main 2014 (11 Apr Online)
Let for some real numbers α and β,a=α-iβ. If the system of equations 4ix+1+iy=0 and 8cos2π3+isin2π3x+a¯y=0 has more than one solution then αβ is equal to

Options:
A 2-3
B 2+3
C -2+3
D -2-3

The correct answer is:
2-3

Solution:
The given system of equations has more than one solution, then it must have infinitely many solutions.

So, 4i8cos2π3+isin2π3=1+iα+iβ ∵a=α-iβ⇒a=α+iβ

⇒4i8-12+i32=1+iα+iβ

⇒αi-β=-1-i+3i-3

⇒-β+αi=-1-3+-1+3i

⇒α=-1+3 & -β=-1-3

∴αβ=-1+31+3=-1+31+3×1-31-3

=-1-321-3=1+3-232=2-3

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let for the 9th term in the binomial expansion of 3+6xn, in the increasing powers of 6x, to be the greatest for x=32, the least value of n is n0. If k is the ratio of the coefficient of x6 to the coefficient of x3, then k+n0 is equal to

Options:

The correct answer is:
24

Solution:
Given,

3+6xn=3n1+2xn

Now if T9 is numerically greatest term, then T8≤T9≥T10

So, C7n3n-76x7≤C8n3n-86x8≥C9n3n-96x9

⇒ n!n-7!7!9≤n!n-8!8!3·6x≥n!n-9!9!6x2

⇒9n-7n-8⏟≤1832n-88≥369.894⏟

⇒72≤27n-7 and 27≥9n-8

⇒293≤n and n≤11

So, n0=10

For 3+6x10

Now Tr+1=Cr10310-r6xr

For coefficient of x6

r=6⇒C61034·66

For coefficient of x3

r=3⇒C31037·63

So, k=C610C310·34·6637·63=10!7!3!6!4!10!·8

⇒k=14

∴ k+n0=24

Topic: Mathematics, Binomial Theorem, JEE Main