datavorous/entrance-exam-dataset huggingface dataset:
If the function fx=sin⁡3x+asin⁡2x+bx3, x≠0 is continuous at x=0 and f0=K, ∀K∈R, then b-a is equal to

Options:
A 4
B 52
C 5
D 32

The correct answer is:
32

Solution:
∵fx is continuous at x=0
∴limx→0⁡fx=f0=finite
limx→0⁡sin⁡3x+asin⁡2x+bx3=finite
For limx→0⁡3x-27x33!+…+a2x-8x33!+…+bx3 to be finite
b=0,3+2a=0
So, a=-32

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx=tan⁡π4+x1xforx≠0Kforx=0 is continuous at x=0, then k=?

Options:
A e
B e-1
C e2
D e-2

The correct answer is:
e2

Solution:
Since f( x ) is continuous at x=0
âf0=limxâ0â¡fx
=limxâ0â¡tanâ¡Ï4+Ï1x
=limxâ0â¡1+tanâ¡x1-tanâ¡x1x
limxâ0â¡1+tanâ¡x1tanâ¡xtanâ¡xx1-tanâ¡x-1tanâ¡x-tanâ¡xx
Taking limits
= e+1e-1=e1âe1=e2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function $f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$ attains the maximum value at $x=\frac{1}{\mathrm{e}}$ then :

Options:
A $\mathrm{e}^\pi < \pi^{\mathrm{e}}$
B $\mathrm{e}^\pi>\pi^{\mathrm{e}}$
C $(2 e)^\pi>\pi^{(2 e)}$
D $\mathrm{e}^{2 \pi} < (2 \pi)^{\mathrm{e}}$

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

Solution:
Let $y=\left(\frac{1}{x}\right)^{2 x}$
$\begin{aligned}
& \ell \text { ny }=2 x \ell n\left(\frac{1}{x}\right) \\
& \ell \text { ny }=-2 x \ell \ln x \\
& \frac{1}{y} \frac{d y}{d x}=-2(1+\ell n x)
\end{aligned}$
for $\mathrm{x}>\frac{1}{\mathrm{e}} \mathrm{f}^{\mathrm{n}}$ is decreasing
$\text {so, } \mathrm{e} < \pi$
$\begin{aligned}
& \left(\frac{1}{\mathrm{e}}\right)^{2 \mathrm{e}}>\left(\frac{1}{\pi}\right)^{2 \pi} \\
& \mathrm{e}^\pi>\pi^{\mathrm{e}}
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the function $f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is

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

The correct answer is:
$1 / 2$

Solution:
We have,
$$
f(x)=\left\{\begin{array}{cl}
\frac{1-\cos x}{x^{2}}, & x \neq 0 \\
k, & x=0
\end{array}\right.
$$
Since, $f(x)$ is continuous at $x=0$, then
$f(0)=\lim _{x \rightarrow 0} f(x) \Rightarrow k=\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$
$\Rightarrow \quad k=\lim _{x \rightarrow 0} \frac{\sin x}{2 x} \quad$ [using L' Hospital Rule]
$\Rightarrow \quad k=\frac{1}{2}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx=x3-3ax has a local minimum at x=Î» Î»â¥4 and aâ10,18, then the sum of all the possible integral values of a is

Options:
A 50
B 112
C 51
D 16

The correct answer is:
51

Solution:
f â² x =3 x 2 â3a
=3x-ax+a
â´fx has a local minimum at x=a
âaâ¥4
âaâ¥16
â´ âaâ can be 16,17,18

Topic: Mathematics, Application of Derivatives, JEE Main
If the function fx=x3-6x2+ax+b satisfies all conditions of Rolle's theorem in xâ1,3, then the value of a is equal to

Options:
A 1
B 11
C 22
D 2

The correct answer is:
11

Solution:
For Rolle’s theorem,
f1=f3
⇒1-6+a+b=27-54+3a+b
⇒2a=22
⇒a=11

Topic: Mathematics, Application of Derivatives, JEE Main
If the function $f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}$, is continuous at $x=0$, then $f(0)$ is equal to :

Options:
A 2
B -2
C 4
D -4

The correct answer is:
-4

Solution:
$f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}$
is continuous at $\mathrm{x}=0$
$\lim _{x \rightarrow 0}=\frac{3 x-\frac{(3 x)^3}{\underline{|3}}+\ldots+\alpha\left(x-\frac{x^3}{\underline{|3}} \ldots\right)-\beta\left(1-\frac{(3 x)^2}{\lfloor 2} \ldots\right)}{x^3}=f(0)$
$\begin{aligned}
& \lim _{x \rightarrow 0}= \frac{-\beta+x(3+\alpha)+\frac{9 \beta x^2}{\underline{|2}}+\left(\frac{-27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}\right) x^3 \ldots}{x^3}=f(0) \\
& \text { for exist } \\
& \beta=0,3+\alpha=0,-\frac{27}{\underline{|3}}-\frac{\alpha}{\underline{|3}}=f(0)
\end{aligned}$
$\begin{aligned} & \alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0) \\ & f(0)=\frac{-27+3}{6}=-4\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If the function fx=x+a2 sin x0≤x<π42x cot x+b π4≤x≤π2a cos 2x-b sin x π2<x≤π is continuous in 0 , π , then the values of a and b respectively are

Options:
A π6, -π12
B -π6, π4
C -π3, π12
D None of these

The correct answer is:
π6, -π12

Solution:
∵ f x is continuous in 0 , π so it is continuous at x=π4 and x=π2
∴ limx→π4-fx=limx→π4+fx
⇒ π4+a=π2+b .....(1)
and limx→π2-fx=limx→π2+fx
⇒ 0 + b = - a - b .....(2)
By solving (1) and (2), we get,
a=π6, b=-π12

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx= -x,x<1a+cos-1⁡x+b,1 ≤x ≤2 is differentiable at x=1, then ab is equal to

Options:
A π+22
B π-22
C -π-22
D -1-cos-1⁡(2)

The correct answer is:
π+22

Solution:
fx= -xx<1a+cos-1⁡x+b1 ≤x ≤2
fx is continuous at x=1 as fx is differentiable at x=1.
⇒ lim x→1 f( x )= lim x→ 1 + ( a+ cos −1 ( x+b ) )=f( 1 )
⇒ -1=a+cos-1⁡1+b
cos-1⁡1+b=-1-a ........i
fx is differentiable at x=1.
⇒ LHD=RHD
⇒ -1=-11-1+b2
⇒ 1-1+b2=1
⇒ b= -1 ......ii
From i ⇒ cos-1⁡(0)= -1-a
∴ -1-a=π2
a= -1-π2
a=-π-22 ......iii
∴ ab=π+22

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2016 (09 Apr Online)
If the function gx={kx+1 , 0≤x≤3mx+2 , 3<x≤5 is differentiable, then the value of k+m is

Options:
A 4
B 2
C 165
D 103

The correct answer is:
2

Solution:
LHD at x=3
limx⟶3-⁡kx+1-2kx-3
= limx→3-⁡kx-3x-3.1x+1+2=k4
RHD at x=3
limx→3+⁡mx+2-2kx-3
Since this limit exists if 3m+2-2k=0 ...i (00 form)
So,limx→3+⁡mx+2-3m+2x-3=m
Hence m=k4 …ii
From i & ii
m=25 & k=85 ⇒k+m=2

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2015 (04 Apr)
If the functions are defined as f(x)=x and g(x)=1-x, then what is the common domain of the following functions: f+g, f-g, f/g, g/f, g-f, where f±gx=fx±gx,f/gx=f(x)g(x)

Options:
A 0≤x≤1
B 0≤x<1
C 0<x<1
D 0<x≤1

The correct answer is:
0<x<1

Solution:
f(x)+g(x)=x+1-x, domain 0, 1

f(x)-g(x)=x-1-x, domain 0, 1

g(x)-f(x)=1-x-x, domain 0, 1

f(x)g(x)=x1-x, domain [0, 1)

g(x)f(x)=1-xx, domain (0, 1]

So, common domain is 0, 1

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (18 Mar Shift 1)
If the functions $f$ and $g$ are defined by $f(x)=3 x-4, \quad g(x)=2+3 x \quad$ for $\quad x \in R$ respective, then $g^{-1}\left(f^{-1}(5)\right)$ is equal to

Options:
A $1$
B $\frac{1}{2}$
C $\frac{1}{3}$
D $\frac{1}{4}$

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

Solution:
We have,
$$
\begin{aligned}
& f(x)=3 x-4=y \\
& \Rightarrow \quad 3 x=y+4 \Rightarrow x=\frac{y+4}{3} \\
& \Rightarrow \quad f^{-1}(y)=\frac{y+4}{3} \\
& \text { and } \quad g(x)=2+3 x=z \\
& \Rightarrow \quad 3 x=z-2 \\
& \Rightarrow \quad x=\frac{z-2}{3} \\
& \Rightarrow \quad g^{-1}(z)=\frac{z-2}{3} \\
& \therefore \quad g^{-1}\left(f^{-1}(5)\right)=g^{-1}\left[\frac{5+4}{3}\right]=g^{-1}[3] \\
& =\frac{3-2}{3}=\frac{1}{3} \\
&
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main
If the functions fx=x33+2bx+ax22 and gx=x33+ax+bx2,a≠2b have a common extreme point, then a+2b+7 is equal to

Options:
A 4
B 32
C 3
D 6

The correct answer is:
6

Solution:
Given,

fx=x33+2bx+ax22 and gx=x33+ax+bx2

Now differentiating the above function we get,

f'x=x2+2b+ax

g'x=x2+a+2bx

Now given, f'(x)=0 and g'(x)=0 have a common extreme point,

So, x2+2b+ax=x2+a+2bx

⇒2b-a-x2b-a=0

∴ x=1 is the common extreme point

Put x=1 in f'x=0 or g'x=0 we get,

1+2b+a=0

⇒a+2b+7=6

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If the general solution of the differential equation $(y-x+1) d y-(y+x+2) d x=0$ is $f(x, y, c)=0$, then the value of $c$ such that $f(1,1, c)=0$ is

Options:
A 4
B -4
C 2
D 1

The correct answer is:
2

Solution:
(c) We have,
$$
\begin{array}{r}
(y-x+1) d y-(y+x+2) d x=0 \\
y d y-x d y+d y-y d x-x d x+2 d x=0 \\
y d y+d y-(x d y+y d x)-x d x+2 d x=0
\end{array}
$$
On integrating, we get
$$
\begin{gathered}
\frac{y^2}{2}+y-x y-\frac{x^2}{2}+2 x-C=0 \\
f(x, y, c)=\frac{y^2}{2}-\frac{x^2}{2}-x y+2 x+y-C \\
f(1,1, c)=0 \\
\therefore \quad \frac{1}{2}-\frac{1}{2}-1+2+1-C=0 \Rightarrow C=2
\end{gathered}
$$

Topic: Mathematics, Differential Equations, JEE Main
If the general solution of the differential equation $y^{\prime}=\frac{y}{x}+\Phi\left(\frac{x}{y}\right)$, for some function $\Phi$, is given by $y \ln |c x|=x$, where $\mathrm{c}$ is an arbitrary constant, then $\Phi(2)$ is equal to:

Options:
A
4

B
$\frac{1}{4}$

C
$-4$

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

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

Solution:
Given $\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{y}{x}\right)$
Let $\left(\frac{y}{x}\right)=v$ so that $y=x v$ or $\frac{d y}{d x}=x \frac{d v}{d x}+v$
from (1) \& (2), $x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$ or, $\frac{d v}{\phi\left(\frac{1}{v}\right)}=\frac{d x}{x}$
Integrating both sides, we get $\int \frac{d x}{x}=\int \frac{d v}{\phi\left(\frac{1}{v}\right)} \Rightarrow \ln x+c=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
(where $\mathrm{c}$ being constant of integration)
But, given $\mathrm{y}=\frac{x}{\ln |c x|}$ is the general solution
so that $\frac{x}{y}=\frac{1}{v}=\ln |c x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
Differentiating w.r.t $v$ both sides, we get $\phi\left(\frac{1}{v}\right)=\frac{-1}{v^2} \Rightarrow \phi\left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$
when $\frac{x}{y}=2$ i.e. $\phi(2)$
$$
=-\left(\frac{y}{x}\right)^2=-\left(\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)
$$

Topic: Mathematics, Differential Equations, JEE Main
If the general solution of the differential equation $y^{\prime}=\frac{y}{x}+\Phi\left(\frac{x}{y}\right)$, for some function $\Phi$, is given by $y \ln |c x|=x$, where $\mathrm{c}$ is an arbitrary constant, then $\Phi(2)$ is equal to:

Options:
A
4

B
$\frac{1}{4}$

C
$-4$

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

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

Solution:
Given $\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{y}{x}\right)$
Let $\left(\frac{y}{x}\right)=v$ so that $y=x v$ or $\frac{d y}{d x}=x \frac{d v}{d x}+v$
from (1) \& (2), $x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$ or, $\frac{d v}{\phi\left(\frac{1}{v}\right)}=\frac{d x}{x}$
Integrating both sides, we get $\int \frac{d x}{x}=\int \frac{d v}{\phi\left(\frac{1}{v}\right)} \Rightarrow \ln x+c=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
(where $\mathrm{c}$ being constant of integration)
But, given $\mathrm{y}=\frac{x}{\ln |c x|}$ is the general solution
so that $\frac{x}{y}=\frac{1}{v}=\ln |c x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
Differentiating w.r.t $v$ both sides, we get $\phi\left(\frac{1}{v}\right)=\frac{-1}{v^2} \Rightarrow \phi\left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$
when $\frac{x}{y}=2$ i.e. $\phi(2)$
$$
=-\left(\frac{y}{x}\right)^2=-\left(\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (11 Apr Online)
If the general solution of the differential equation y'=yx+Ïxy, for some function Ï , is given by yln|cx|=x, where c is an arbitrary constant, then Ï2 is equal to here, y'=dydx

Options:
A -4
B - 1 4
C 1 4
D 4

The correct answer is:
- 1 4

Solution:
given : y â² = y x + Ï x y
As y ln cx = x â y â² ln cx + y 1 cx c = 1
â y â² x y + y x = 1
â 1 - y x x y = y x + Ï x y
x = 2 , y = 1 â 1 - 1 2 2 1 = 1 2 + Ï 2 1
â Ï 2 = - 1 4

Topic: Mathematics, Differential Equations, JEE Main
If the given lines $y=m_1 x+c_1, y=m_2 x+c_2$ and $y=m_3 x+c_3$ be concurrent, then

Options:
A $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$
B $m_1\left(c_2-c_1\right)+m_2\left(c_3-c_2\right)+m_3\left(c_1-c_3\right)=0$
C $c_1\left(m_2-m_3\right)+c_2\left(m_3-m_1\right)+c_3\left(m_1-m_2\right)=0$
D None of these

The correct answer is:
$m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$

Solution:
$\begin{aligned}
& \left|\begin{array}{lll}
m_1 & -1 & c_1 \\
m_2 & -1 & c_2 \\
m_3 & -1 & c_3
\end{array}\right|=0 \\
& \Rightarrow m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0
\end{aligned}$
Note : Students should remember this question as a formula.

Topic: Mathematics, Straight Lines, JEE Main
If the graph of the function fx=ax3+x2+bx+c is symmetric about the line x=2, then the value of a+b is equal to

Options:
A 10
B -4
C 16
D -10

The correct answer is:
-4

Solution:
fx is symmetric about the line x=2
∴f2+x=f2-x
a2+x3+2+x2+b2+x+c=a2-x3+2-x2+b2-x+c
a2+x3-2-x3+2+x2-2-x2+b2+x-2-x=0
a8+12x+6x2+x3-8-12x+6x2-x3+24x+b2x=0
a24x+2x3+8x+2bx=0
2ax3+24a+2b+8x=0
Which must be true ∀x∈R
∴ it is an identity
∴2a=0, 24a+2b+8=0
a=0, 2b+8=0
b=-4

Topic: Mathematics, Functions, JEE Main
If the graph of the function \( f(x)=\frac{a^{x}-1}{x^{n}\left(a^{x}+1\right)} \) is symmetrical about \( Y \)-axis, then \( n \) equals

Options:
A \( 2 \)
B \( \frac{2}{3} \)
C \( \frac{1}{4} \)
D \( -\frac{1}{3} \)

The correct answer is:
\( -\frac{1}{3} \)

Solution:
fx=ax-1xnax+1

since, fx is symmetrical about Y-axis.

( i.e., it is a even function.)

∴ fx=f-x

⇒ax-1xnax+1=a-x-1-xna-x+1

⇒ax-1xnax+1=1-ax-xn1+ax

⇒xn=--xn

Hence, the value of n which satisfy this relation, is -13

Topic: Mathematics, Functions, JEE Main
If the graph of the function $y=(a-b)^2 x^2+2(a+b-2 c) x+1(\forall a \neq b)$ is strictly above the $x$-axis, then

Options:
A $\mathrm{a} \lt \mathrm{b} \lt \mathrm{c}$
B $\mathrm{a} \lt \mathrm{c} \lt \mathrm{b}$
C $\mathrm{b} \lt \mathrm{a} \lt \mathrm{c}$
D $\mathrm{c} \lt \mathrm{b} \lt \mathrm{a}$

The correct answer is:
$\mathrm{a} \lt \mathrm{c} \lt \mathrm{b}$

Solution:
$\mathrm{D}=\mathrm{B}^2-4 \mathrm{AC}$
$=(2(a+b-2 c))^2-4(a-b)^2$
$=4\left\{(\mathrm{a}+\mathrm{b}-2 \mathrm{c})^2-(\mathrm{a}-\mathrm{b})^2\right\}$
$=4(a+b-2 c-a+b)(a+b-2 c+a-b)$
$=4(2 b-2 c)(2 a-2 c)$
$=16(b-c)(a-c)$
$=16(c-b)(c-a)$
If $\mathrm{c}$ lies between $\mathrm{a}$ and $\mathrm{b}$, then $\mathrm{D}$ is negative. Hence, the roots will be imaginary and the graph will be entirely above the $\mathrm{x}$-axis as the coefficient of $x^2$ is positive.

Topic: Mathematics, Quadratic Equation, JEE Main
If the graph of the function y=a-b2x2+2a+b-2cx+1 ∀a≠b is strictly above the x-axis, then

Options:
A a<b<c
B a<c<b
C b<a<c
D c<b<a

The correct answer is:
a<c<b

Solution:
D=B2-4AC
=2a+b-2c2-4a-b2
=4a+b-2c2-a-b2
=4a+b-2c-a+ba+b-2c+a-b
=42b-2c2a-2c
=16b-ca-c
=16c-bc-a
If c lies between a and b, then D is negative. Hence, the roots will be imaginary and the graph will be entirely above the x-axis as the coefficient of x2 is positive.

Topic: Mathematics, Quadratic Equation, JEE Main
If the greatest value of the term independent of x in the expansion of xsinα+acosαx10 is 10!5!2, then the value of a is equal to:

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

The correct answer is:
2

Solution:
We have, rth term in the expansion of xsinα+acosαx10 is

Tr+1=Cr10xsinα10-racosαxr

r=0,1,2,…,10

Tr+1 will be independent of x, when 10-2r=0

⇒r=5

Now, T6=C510xsinα5×acosαx5

=C510×a5×125sin2α5

T6 will be greatest, when sin2α=1

Therefore, C510a525=C510

⇒a=2

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If the half cell reaction $A+e^{-} \rightarrow A^{-}$has a large negative reduction potential, it follows that

Options:
A $A$ is readily reduced
B $A$ is readily oxidised
C $A^{-}$is readily reduced
D $A^{-}$is readily oxidised

The correct answer is:
$A^{-}$is readily oxidised

Solution:
Since $E_{A / A^{-}}^o$ has large negative value, the tendency of $A$ to be reduced to $A^{-}$is very small. In other words tendency of $A^{-}$to be oxidized to $A$ is very large.

Topic: Chemistry, Electrochemistry, JEE Main
If the half life period of a reaction is inversely proportional to the initial concentration, the order of the reaction is

Options:
A Zero
B One
C Two
D Three

The correct answer is:
Two

Solution:
$t_{1 / 2} \propto \frac{1}{(a)}$ for II order reaction

Topic: Chemistry, Chemical Kinetics, JEE Main
If the half-life of a radioactive element is \( 10 \mathrm{~h} \), its average life \( = \)
h.

Options:
A \( 14.4 \)
B \( 6.93 \)
C \( 1.44 \)
D \( 0.693 \)

The correct answer is:
\( 14.4 \)

Solution:
The relation between average life and half-life,

τ=t1/2ln2=100.693

≃14.4 h.

Topic: Physics, Nuclear Physics, JEE Main
If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the third day? (Given, $\sqrt[3]{0.25} \approx 0.63)$

Options:
A 0.63
B 0.5
C 0.37
D 0.13

The correct answer is:
0.13

Solution:
Given, $t_{1 / 2}=3$ days
Number of active nuclei remaining after time $t$,
$N=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{t_{1 / 2}}}$
After $t=2$ days,
$N_{1}=N_{0}\left(\frac{1}{2}\right)^{2 / 3}=\frac{N_{0}}{2^{2 / 3}}=\frac{N_{0}}{4^{1 / 3}}=0.63 \mathrm{N}_{0}$
$\therefore \quad N_{1}=0.63 \mathrm{N}_{0}$
After $t=3$ days,
$N_{2}=\frac{N_{0}}{2}=0.5 \mathrm{N}_{0}$
Number of nuclei that will decay on the 3rd day. $N_{3}=N_{2}-N_{1}=0.63 \mathrm{N}_{0}-0.5 \mathrm{N}_{0}=013 \mathrm{N}_{0}$
In the term, fraction is 0.13.

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2018
If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{H-a}+\frac{1}{H-b}$ is

Options:
A $a+b$
B $a b$
C $\frac{1}{a}+\frac{1}{b}$
D $\frac{1}{a}-\frac{1}{b}$

The correct answer is:
$\frac{1}{a}+\frac{1}{b}$

Solution:
Putting $H=\frac{2 a b}{a+b}$, we have $\frac{1}{H-a}+\frac{1}{H-b}$
$=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}$ $=\frac{a+b}{a b-a^2}+\frac{a+b}{a b-b^2}$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$

Topic: Mathematics, Sequences and Series, JEE Main
If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b}$ is

Options:
A $a+b$
B $a b$
C $\frac{1}{a}+\frac{1}{b}$
D $\frac{1}{a}-\frac{1}{b}$

The correct answer is:
$\frac{1}{a}+\frac{1}{b}$

Solution:
Putting $\mathrm{H}=\frac{2 a b}{a+b}$, we have

$\begin{array}{l}

\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b} \\

=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}

\end{array}$

$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$

Topic: Mathematics, Sequences and Series, JEE Main
If the height of a cone of greatest volume that can be inscribed in a sphere of radius $\mathrm{R}$ is $\mathrm{kR}$, then ratio of the volume of the cone to the volume of the sphere is

Options:
A $8: 27$
B 27:64
C 8:125
D $4: 5$

The correct answer is:
$8: 27$

Solution:
$\begin{aligned} & \text { In } \triangle O D C \text {, } \\ & r^2=R^2-(h-R)^2 \\ & =2 h R-h^2\end{aligned}$
Hence volume of cone $A B C=\frac{1}{3} \pi r^2 h$
$\Rightarrow \mathrm{V}=\frac{1}{3} \pi\left(2 h R-h^2\right) h=\frac{1}{3} \pi\left(2 R h^2-h^3\right)$
Now for volume to be maximum,
$$
\begin{aligned}
& \frac{d V}{d h}=0 \\
& \Rightarrow \frac{1}{3 \pi}\left[4 h R-3 h^2\right]=0 \\
& \Rightarrow h \neq 0 \text { Hence } h=\frac{4 R}{3} \\
& \text { at } h=\frac{4 R}{3}, \frac{d^2 V}{d h^2} < 0
\end{aligned}
$$
Hence at $h=\frac{4 R}{3}$, volume of cone is maximum.
$$
\begin{aligned}
& \frac{\text { Volume of cone }}{\text { Volume of sphere }}=\frac{\frac{1}{3} \pi r^2 h}{\frac{4}{3} \pi R^3}=\frac{\left(2 h R-h^2\right) h}{4 R^3} \\
& =\frac{\left(\frac{8 R^2}{3}-\frac{16 R^2}{9}\right) \frac{4 R}{3}}{4 R^3}=\frac{8}{27}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
If the height of TV tower is increased by 21% , then the transmission range is enhanced by

Options:
A 10%
B 5%
C 15%
D 25%

The correct answer is:
10%

Solution:
Range =2Rh⇒ Range ∝h
Here h increases to 1.21 h
∴ Enhancement range ∝1.2112=1.1
∴ Increase =1.1-1=0.10
∴% increase =0.10×100=10%

Topic: Physics, Communication System, JEE Main
If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are

Options:

The correct answer is:
9

Solution:
B. W. (Bandwidth) =2× maximum frequency at modulating signal

=2×5 kHz

=10 kHz

∴ No of stations accommodate =9010=9

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If the highest oxidation states shown by any Lanthanide and any Actinide are +X and +Y, the sum of X+Y is equal to?

Options:

The correct answer is:
11.00

Solution:
Lanthanide [Ce = +4]
Actinide [Np = +7]
Sum of X + Y = 4 + 7 = 11.

Topic: Chemistry, d and f Block Elements, JEE Main
If the I.F. of the differential equation $\frac{d y}{d x}+5 y$ $=\cos x$ is $\int \mathrm{e}^{\mathrm{Adx}}$, then $\mathrm{A}=$

Options:
A 0
B 1
C 3
D 5

The correct answer is:
5

Solution:
The I.F. of the differential equation
$\frac{d y}{d x}+P y=Q$ is $e^{\int P d x} .$ Here $P=5$ therefore
I.F. $={ }_{\mathrm{e}} \int 5 \mathrm{dx} .$ Hence $\mathrm{A}=5$

Topic: Mathematics, Differential Equations, JEE Main
If the image of point $\mathrm{P}(2,3)$ in a line $\mathrm{L}$ is $\mathrm{Q}(4,5)$, then the image of point $\mathrm{R}(0,0)$ in the same line is:

Options:
A
$(2,2)$

B
$(4,5)$

C
$(3,4)$

D
$(7,7)$

The correct answer is:
$(7,7)$

Solution:
Mid-point of $\mathrm{P}(2,3)$ and $\mathrm{Q}(4,5)=(3,4)$
Slope of PQ = 1
Slope of the line $\mathrm{L}=-1$
Mid-point $(3,4)$ lies on the line $L$.
Equation of line $\mathrm{L}$,
$$
y-4=-1(x-3) \Rightarrow x+y-7=0
$$
Let image of point $\mathrm{R}(0,0)$ be $\mathrm{S}\left(x_1, y_1\right)$
Mid-point of $\mathrm{RS}=\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$
Mid-point $\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$ lies on the line (i) $x_1+y_1=14$
Slope of RS $=\frac{y_1}{x_1}$
Since $\mathrm{RS} \perp$ line $\mathrm{L}$
$$
\begin{aligned}
&\frac{y_1}{x_1} \times(-1)=-1 \\
&x_1=y_1
\end{aligned}
$$
From (ii) and (iii),
$$
x_1=y_1=7
$$
Hence the image of $R=(7,7)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (25 Apr Online)
If the image of the point 1,-2,â3 in the plane 2x+3y-z=7 is the point Î±,âÎ²,âÎ³, then the value of Î±+Î²+Î³ is equal to

Options:
A -6
B 10
C 8
D -4

The correct answer is:
-6

Solution:
Since, α, β, γ is the image of the point 1,-2, 3 in the plane. So, we have,
∵Image of (x1,y1,z1) in the plane ax+by+cz+d=0 is given byx−x1a=y−y1b=z−z1c=−2(ax1+by1+cz1)a2+b2+c2
α-12=β+23=γ-3-1=-221+3-2-3-722+32+-12
⇒α-12=β+23=γ-3-1=-22-6-3-74+9+1
⇒α-12=β+23=γ-3-1=-2
∴ α-12=-2 ⇒α=-4+1=-3
β+23=-2⇒β=-6-2=-8
and γ-3-1=-2⇒γ=2+3=5
So, α+β+γ=-3-8+5=-6

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the image of the point $(-4,5)$ in the line $x+2 y=2$ lies on the circle $(x+4)^2+(y-3)^2=r^2$, then $\mathrm{r}$ is equal to:

Options:
A 2
B 3
C 1
D 4

The correct answer is:
2

Solution:
Image of point $(-4,5)$
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-2\left(\frac{a x_1+b y_1+c}{a^2+b^2}\right)$
Line: $x+2 y-2=0$
$\frac{x+4}{1}=\frac{y-5}{2}=-2\left(\frac{-4+10-2}{1^2+2^2}\right)$
$=\frac{-8}{5}$
$x=-4-\frac{8}{5}=-\frac{28}{5}$
$y=-\frac{16}{5}+5=\frac{9}{5}$
Point lies on circle $(x+4)^2+(y-3)^2=r^2$
$\begin{aligned} & \frac{64}{25}+\left(\frac{9}{5}-3\right)^2=r^2 \\ & \frac{100}{25}=r^2, r=2\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the image of the point MÎ»,Î»2 on the line x+y=Î»2 is N0,2, then the sum of the square of all the possible values of Î» is equal to

Options:
A 5
B 2
C 1
D 4

The correct answer is:
5

Solution:
Since the mid-point of the line segment MN lie on the given line,

⇒λ2+λ2+22=λ2…..i

Also, the line segment MN is perpendicular to the given line
⇒λ2-2λ-0×-1=-1…..ii
As i and ii must be satisfied simultaneously,
∴ On solving, we get,

λ=-1,2⇒λ12+λ22=1+4=5

Topic: Mathematics, Straight Lines, JEE Main