datavorous/entrance-exam-dataset huggingface dataset:
If the sum of the distances of a point from two perpendicular lines in a plane is 1 , then its locus is

Options:
A Square
B Circle
C Straight line
D Two intersecting lines

The correct answer is:
Square

Solution:
Required locus of the point $(x, y)$ is the curve $|x|+|y|=1$. If the point lies in the first quadrant, then $x\gt0, y\gt0$ and so $|x|+|y|=1 \Rightarrow x+y=1$, which is straight line $A B$. If the point ${ }^{(x, y)}$ lies in second quadrant then $x \lt 0, y\gt0$ and so $|x|+|y|=1 \Rightarrow-x+y=1$
Similarly for third and fourth quadrant, the equations are $-x-y=1$ and $x-y=1$ Hence the required locus is the curve consisting of the sides of the square $A B C D$.

Topic: Mathematics, Straight Lines, JEE Main
If the sum of the first \(2 n\) terms of \(2,5,8, \ldots \ldots\) is equal to the sum of the first \(n\) terms of 57,59 , \(61 \ldots \ldots\). , then \(\mathrm{n}\) is equal to

Options:
A 10
B 12
C 11
D 13

The correct answer is:
11

Solution:
Given,
\(\frac{2 n}{2}\{2.2+(2 n-1) 3\}=\frac{n}{2}\{2.57+(n-1) 2\}\)
or \(2(6 n+1)=112+2 n\) or \(10 n=110, ~\therefore n=11\)

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the first 2n terms of the arithmetic progression 2,5,8,â¦â¦. is equal to the sum of the first n terms of the arithmetic progression 57,59,61,â¦â¦â¦â¦, then the value of n is equals to

Options:
A 10
B 12
C 11
D 13

The correct answer is:
11

Solution:
For the AP 2,5,8,…….a=2,d=3
S2n=2n22×2+2n-13=n6n+1
For the AP 57,59,61,……..a=57,d=2
Sn=n22×57+n-12=nn+56
Now, S2n=Sn
⇒n6n+1=nn+56
⇒5n=55⇒n=11

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the first 15 terms of the series 343+1123+2143+33+334 3+… is equal to 225K, then K is equal to :

Options:
A 9
B 27
C 54
D 108

The correct answer is:
27

Solution:
Given series, 343+643+943+1243+1543+…15 terms

=34313+23+33+…+153

=276415×1622 ∵∑n=1n=nn3=nn+122

=2764×225×641=27×225

Since, given 225K=27×225

⇒K=27

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If the sum of the first 20 terms of the series log71/2x+log71/3x+log71/4x+…is 460, then x is equal to:

Options:
A 72
B 71/2
C e2
D 746/21

The correct answer is:
72

Solution:
Given log71/2x+log71/3x+log71/4x+…20 times =460
⇒ (2+3+4+…..+21)log7x=460
⇒ 202(2+21)log7x=460
⇒ log7x=2
⇒ x=49

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the sum of the first 40 terms of the series, 3+4+8+9+13+14+18+19+.... is 102m, then m is equal to

Options:
A 20
B 25
C 5
D 10

The correct answer is:
20

Solution:
S=3+4â+8+9â+13+14â+18+19ââ¦40 terms
S=7+17+27+37+47+â¦20 terms
S40=2022Ã7+1910=1014+190=102040=10220
⇒m=20

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If the sum of the first 100 terms of an arithmetic progression is -1 and the sum of the even terms lying in the first 100 terms is 1, then the 100th term of the arithmetic progression is

Options:
A 4725
B 14950
C 7425
D -14950

The correct answer is:
7425

Solution:
Given, S100=-1

⇒10022a+99d=-1
⇒502a+99d=-1
Also, a2+a4+a6+......+a100=1
⇒5022a+d+492d=1
⇒50a+50d=1
On solving, we get,

d=350 and a=-14950
a100=a+99d=7425

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the first n terms of an arithmetic progression, whose first term is the sum of the first n positive integers and whose common difference is n, is 8n2-11n-20, then the sum of all the possible values of n is

Options:

The correct answer is:
9

Solution:
a1=nn+12,d=n
âSn=n2nn+1+n-1n
=n2n2+n+n2-n=n3
But, given that Sn=8n2-11n-20.
â8n2-11n-20=n3
ân3-8n2+11n+20=0
ân+1n-4n-5=0
ân=4 or 5 nâ -1
â sum of all possible values =9

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the first n terms of the series 3+ 75+ 243+ 507+… is 4353, then n equals:

Options:
A 13
B 15
C 29
D 18

The correct answer is:
15

Solution:
3 1+25+ 81+ 169+â¦=4353
3 1+5+9+13+â¦Tn=4353
= 3Ãn22+n-14=4353
=2n+4n2-4n=870
=4n2-2n-870=0
=2n2-n-435=0
n=1Â± 1+4Ã2Ã4354
=1Â±594 âµn is +ve
â´n=1+594=15

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2017 (08 Apr Online)
If the sum of the first ten terms of the series 1352+2252+3152+42+4452+…., is 165m , then m is equal to

Options:
A 100
B 99
C 102
D 101

The correct answer is:
101

Solution:
852+1252+1652 ........ upto ten terms
Tn=4n+452
Tn=16n2+2n+125
Tn=1625n2+2n+1

Sn=1625∑n2+2n+1
Sn=1625 nn+1 2n+16+2nn+12+n
Put n=10
162510·11·216+2·10·112+10
=1625 385+110+10=1625· 505
=165·101
Hence m=101

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2016 (03 Apr)
If the sum of the first ten terms of the series 15+265+3325+41025+52501+…. is mn, where m and n are co-prime numbers, then m+n is equal to ______.

Options:

The correct answer is:
276

Solution:
For the series 15+265+3325+41025+52501+……

Tn=n4n4+1

=n2n2+12-(2n)2=n2n2+2n+12n2-2n+1

=1412n2-2n+1-12n2+2n+1

Now S10=∑n=110 Tn=∑n=1101412n2-2n+1-12n2+2n+1

=1411-15+15-113+113………+1181-1221

=141-1221=55221=mn

Hence, m+n=55+221=276

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is _______ km (Take radius of Earth =6400 km)

Options:

The correct answer is:
64

Solution:
hr+ht=160 (Given)
LOS=2Rhr+ht
LOS=2Rhr+160-hr
dLOSdhr=0=2R12hr-12160-hr=0
hr=ht=80 m
LOS=2R(80+80)=22×6400×103×80=64 km

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If the sum of the roots of the equation $a x^2+b x+c=0$ be equal to the sum of their squares, then

Options:
A $a(a+b)=2 b c$
B $c(a+c)=2 a b$
C $b(a+b)=2 a c$
D $b(a+b)=a c$

The correct answer is:
$b(a+b)=2 a c$

Solution:
$\alpha+\beta=\alpha^2+\beta^2$
$\frac{-b}{a}=(\alpha+\beta)^2-2 \alpha \beta$
$-\frac{b}{a}=\frac{b^2}{a^2}-\frac{2 c}{a}$
$\therefore \frac{b}{a}+\frac{2 c}{a}=\frac{b^2}{a^2}$
$\frac{-b+2 c}{a}=\frac{b^2}{a^2}$
$-a b+2 a c=b^2$
$2 a c=b^2+a b$
$2 a c=b(a+b)$
Option 3 is correct

Topic: Mathematics, Quadratic Equation, JEE Main
If the sum of the roots of the equation $x^2+p x+q=0$ is three times their difference, then which one of the following is true

Options:
A $9 p^2=2 q$
B $2 q^2=9 p$
C $2 p^2=9 q$
D $9 q^2=2 p$

The correct answer is:
$2 p^2=9 q$

Solution:
Let $\alpha, \beta$ are roots of $x^2+p x+q=0$
So $\alpha+\beta=-p$ and $\alpha \beta=q$
Given that $(\alpha+\beta)=3(\alpha-\beta)=-p \Rightarrow \alpha-\beta=\frac{-p}{3}$
Now $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta$
$\Rightarrow \frac{p^2}{9}=p^2-4 q$ or $2 p^2=9 q$

Topic: Mathematics, Quadratic Equation, JEE Main
If the sum of the roots of the quadratic equation $a x^2+b x+c=0$ is equal to the sum of the squares of their reciprocals, then $\frac{\mathrm{a}}{\mathrm{c}}, \frac{\mathrm{b}}{\mathrm{a}}$ and $\frac{\mathrm{c}}{\mathrm{b}}$ are in

Options:
A
Arithmetic - Geometric Progression

B
Arithmetic Progression

C
Geometric Progression

D
Harmonic Progression

The correct answer is:
Harmonic Progression

Solution:
$a x^2+b x+c=0, \alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
As for given condition, $\alpha+\beta=\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
$\alpha+\beta=-\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}-\frac{b}{a}=\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}$
On simplification $2 a^2 c=a b^2+b c^2$
$\Rightarrow \frac{2 \mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}} \quad \therefore \frac{\mathrm{a}}{\mathrm{b}}, \frac{\mathrm{b}}{\mathrm{a}}, \& \frac{\mathrm{c}}{\mathrm{b}}$ are in H.P.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2003
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

Options:
A 126349-1
B 126350-1
C 213350-1
D 113350-1

The correct answer is:
126350-1

Solution:
Let a, ar, ar…2….. be terms of G.P.

Given,T2+T3+T4=3 ⇒ ar 1+r+r2=3

T6+T7+T8=243 ⇒
ar51+r+r2=243

by (i) and (ii)

r4=81
⇒r=3
∴a=113

S50=ar50-1r-1=350-126

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the sum of the series $\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$ is equal to 5 , then $50 \mathrm{~d}$ is equal to :

Options:
A 10
B 5
C 15
D 20

The correct answer is:
5

Solution:
$\begin{aligned} & \frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots \ldots \ldots \ldots . . \\ & \frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}=5\end{aligned}$
$\begin{aligned} & \frac{1}{d}\left[\frac{(1+d)-1}{1 \cdot(1+d)}+\frac{(1+2 d)-(1-d)}{(1+d)(1+2 d)}\right]+\ldots \ldots \ldots \ldots . . . \\ & \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots \ldots \ldots \ldots . .\right. \\ & \left.\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{(1+10 d)}\right]=5 \\ & \frac{10 d}{1+10 d}=5 d \\ & 50 d=5\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (09 Apr Shift 1)
If the sum of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+$ ... $2.6^2+\ldots$ upto $\mathrm{n}$ terms, when $\mathrm{n}$ is even, is $\frac{n(n+1)^2}{2}$, then the sum of the series, when $\mathrm{n}$ is odd, is

Options:
A
$n^2(n+1)$

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

C
$\frac{n^2(n+1)}{2}$

D
$n^2(n-1)$

The correct answer is:
$\frac{n^2(n+1)}{2}$

Solution:
If $n$ is odd, the required sum is
$$
\begin{aligned}
& 1^2+2.2^2+3^2+2.4^2+\ldots . .+2(n-1)^2+n^2 \\
& =\frac{(n-1)(n-1+1)^2}{2}+n^2 \quad(\because n-1 \text { is even }) \\
& =\left(\frac{n-1}{2}+1\right) n^2=\frac{n^2(\mathrm{n}+1)}{2}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (26 May Online)
If the sum of the series 1+32+54+78+....+2n-12n-1 is fn, then the value of f8 is

Options:
A 4+1225
B 5+1327
C 6-1927
D 5-1327

The correct answer is:
6-1927

Solution:
Sn=1+32+54+78+.....+2n-12n-1
12Sn=12+34+58+.....+2n-32n-1+2n-12n
Subtracting, we get,
12Sn=1+22+24+28+.....+22n-1-2n-12n
12Sn=1+212+14+18+.....n-1terms-2n-12n
12Sn=1+2121-12n-11-12-2n-12n
12Sn=1+21-12n-1-2n-12n
=1+2-12n-2-2n-12n
fn=Sn=6-12n-3-2n-12n-1⇒f8=6-125-1527
f8=6-4+1527=6-1927

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the series 20+1935+1915+1845+.......... up to nth term is 488 and the nth term is negative, then :

Options:
A nth term is -425
B n=41
C nth term is -4
D n=60

The correct answer is:
nth term is -4

Solution:
The given series is an arithmetic progression whose first term is 20 and common difference is 1935-20=-25.

According to question,

488=n2220+n-1-25

⇒488=n240-2n5+25⇒n2-101n+2440=0

⇒n=61 or 40

For n=40⇒Tn=T40=20+40-1-25=225>0

For n=61⇒Tn=T61=20+61-1-25=-4<0

Thus, Tn=T61=-4

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If the sum of the series $54+51+48+$ ________ is 513, then the number of terms are

Options:
A $18$
B $20$
C $17$
D None of these

The correct answer is:
$18$

Solution:
According to condition
$\Rightarrow 513=\frac{n}{2}\{2 \times 54+(n-1)(-3)\}$
$\Rightarrow 1026=n(111-3 n)$ $\Rightarrow 3 n^2-111 n+1026=0$
$\Rightarrow \quad(3 n-57)(n-18)=0$
Hence $n=18$.

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of the slopes of the lines given by $x^2-2 c x y-7 y^2=0$ is four times their product, then $c$ has the value

Options:
A
1

B
$-1$

C
2

D
$-2$

The correct answer is:
2

Solution:
$m_1+m_2=-\frac{2 c}{7}$ and $m_1 m_2=-\frac{1}{7}$
$m_1+m_2=4 m_1 m_2 \quad$ (given)
$\Rightarrow c=2$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2004
If the sum of the square of the roots of the equation $x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0$ is least, then $\alpha$ is equal to

Options:
A
$\frac{\pi}{6}$

B
$\frac{\pi}{4}$

C
$\frac{\pi}{3}$

D
$\frac{\pi}{2}$

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

Solution:
Given equation is
$$
x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0
$$
Let $x_1$ and $x_2$ be two roots of quadratic equation.
$$
\begin{aligned}
& \therefore x_1+x_2=\sin \alpha-2 \text { and } x_1 x_2=-(1+\sin \alpha) \\
& \left(x_1+x_2\right)^2=(\sin \alpha-2)^2=\sin ^2 \alpha+4-4 \sin \alpha
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad x_1^2+x_2^2=\sin ^2 \alpha+4-4 \sin \alpha-2 x_1 x_2 \\
& =\sin ^2 \alpha+4-4 \sin \alpha+2(1+\sin \alpha) \\
& =\sin ^2 \alpha-2 \sin \alpha+6
\end{aligned}
$$
Now, By putting
$\alpha=\frac{\pi}{6}, \alpha=\frac{\pi}{4}, \alpha=\frac{\pi}{3}$ and $\alpha=\frac{\pi}{2}$ in (A) one by one
We get least value of $x_1^2+x_2^2$ at $\frac{\pi}{2}$
Hence, $\alpha=\frac{\pi}{2}$

Topic: Mathematics, Quadratic Equation, JEE Main
If the sum of the square of the roots of the equation $x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0$ is least, then $\alpha$ is equal to

Options:
A
$\frac{\pi}{6}$

B
$\frac{\pi}{4}$

C
$\frac{\pi}{3}$

D
$\frac{\pi}{2}$

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

Solution:
Given equation is
$$
x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0
$$
Let $x_1$ and $x_2$ be two roots of quadratic equation.
$$
\begin{aligned}
& \therefore x_1+x_2=\sin \alpha-2 \text { and } x_1 x_2=-(1+\sin \alpha) \\
& \left(x_1+x_2\right)^2=(\sin \alpha-2)^2=\sin ^2 \alpha+4-4 \sin \alpha
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad x_1^2+x_2^2=\sin ^2 \alpha+4-4 \sin \alpha-2 x_1 x_2 \\
& =\sin ^2 \alpha+4-4 \sin \alpha+2(1+\sin \alpha) \\
& =\sin ^2 \alpha-2 \sin \alpha+6
\end{aligned}
$$
Now, By putting
$\alpha=\frac{\pi}{6}, \alpha=\frac{\pi}{4}, \alpha=\frac{\pi}{3}$ and $\alpha=\frac{\pi}{2}$ in (A) one by one
We get least value of $x_1^2+x_2^2$ at $\frac{\pi}{2}$
Hence, $\alpha=\frac{\pi}{2}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (12 May Online)
If the sum of the squares of the reciprocals of the roots α and β of the equation 3x2+λx-1=0 is 15, then 6α3+β32 is equal to

Options:
A 46
B 36
C 24
D 18

The correct answer is:
24

Solution:
Given 1α2+1β2=15

For equation 3x2+λx-1=0

α+β=-λ3, αβ=-13⇒α2β2=19

then α2+β2=α+β2-2αβ=λ29-2-13=λ29+23

⇒α2+β2=λ2+69

Now 1α2+1β2=15⇒α2+β2α2β2=15

⇒λ2+69×1α2β2=15 ⇒λ2+69×119=15

⇒λ2+6=15 ⇒λ=±3

Now 6α3+β32 =6α+βα2-αβ+β22

=6×1×159+132

=6×1×22=6×4=24

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If the sum of three terms of G.P. is 19 and product is 216 , then the common ratio of the series is

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

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

Solution:
Let three terms of G.P. are $a, a r, a r^2$. Then
$a+a r+a r^2=19 \Rightarrow a\left[1+r+r^2\right]=19$\ldots(i)
a. $a r \cdot a r^2=216 \Rightarrow a^3 r^3=216 \Rightarrow a r=6$
\ldots(ii)
Dividing (ii) by (i),
$\frac{6}{r}+\frac{6}{r} r+\frac{6}{r} r^2=19 \Rightarrow \frac{6}{r}+6+6 r=19$
$\Rightarrow r^2-\frac{13}{6} r+1=0$ Hence $\quad r=\frac{3}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
If the sum of two particular roots of the equation $x^4-4 x^3$ $-7 x^2+22 x+24=0$ is equal to the sum of the remaining two roots, then the sum of the cubes of all the roots of this equation is

Options:
A $100$
B $196$
C $0$
D $82$

The correct answer is:
$82$

Solution:
$x^4-4 x^3-7 x^2+22 x+24=0$
$\begin{aligned} & \Rightarrow \quad(x+1)\left(x^3-5 x^2-2 x+24\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)\left(x^2-7 x+12\right)=0 \\ & \Rightarrow \quad(x+1)(x+2)(x-3)(x-4)=0 \\ & \Rightarrow \quad x=-1,-2,3,4 \\ & \therefore \quad 3-1=4-2 \\ & \therefore \quad \alpha^3+\beta^3+\gamma^3+\delta^3=(3)^3+(4)^3+(-1)^3+(-2)^3=82 .\end{aligned}$

Topic: Mathematics, Quadratic Equation, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
If the sum to infinity of the series 1+4x+7x2+10x3+.â¦â¦. is 3516, where x<1, then x is equal to

Options:
A 197
B 15
C 14
D 47

The correct answer is:
15

Solution:
S=1+4x+7x2+10x3+.â¦â¦.
xS=x+4x2+7x3+.â¦â¦..
On subtracting, we get,
S1-x=1+3x+3x2+3x3+.â¦â¦..
=1+3x11-x,x<1
1-xS=1-x+3x1-x=1+2x1-x
S=1+2x1-x2=3516 (given)
16+32x=35+35x2-70x
â35x2-102x+19=0
â5x-17x-19=0
x=15,197
But, x<1âx=15

Topic: Mathematics, Sequences and Series, JEE Main
If the surface area of a cube is increasing at a rate of 3.6cm2/sec, retaining its shape; then the rate of change of its volume (in cm3/sec), when the length of a side of the cube is 10cm, is:

Options:
A 20
B 10
C 18
D 9

The correct answer is:
9

Solution:
Surface area of cube s=6a2

⇒dsdt=12adadt=3.6

⇒1210dadt=3.6⇒dadt=0.03cm/sec

Now, volume of cube V=a3

⇒dvdt=3a2.dadt=3(10)20.03=9cm3/sec

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \mathrm{~cm}^2 / \mathrm{s}$, then the rate of change of its volume is:

Options:
A
constant

B
proportional to $\sqrt{r}$

C
proportional to $r^2$

D
proportional to $r$

The correct answer is:
proportional to $r$

Solution:
$$
\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi r^3 \Rightarrow \frac{d \mathrm{~V}}{d t}=4 \pi r^2 \cdot \frac{d r}{d t} \\
& \mathrm{~S}=4 \pi r^2 \Rightarrow \frac{d \mathrm{~S}}{d t}=8 \pi r \cdot \frac{d r}{d t} \\
& \Rightarrow 8=8 \pi r \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{\pi r}
\end{aligned}
$$
Putting the value of $\frac{d r}{d t}$ in (i), we get
$$
\frac{d \mathrm{~V}}{d t}=4 \pi r^2 \times \frac{1}{\pi r}=4 r
$$
$\Rightarrow \frac{d \mathrm{~V}}{d t}$ is proportional to $r$.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (09 Apr Online)
If the system of equation
$3 x-2 y+z=0, \quad \lambda x-14 y+15 z=0, \quad x+2 y+3 z=0$
have a non-trivial solution, then $\lambda=$

Options:
A 5
B -5
C -29
D 29

The correct answer is:
29

Solution:
If the given system of equations has a non-trivial solution, then $\left|\begin{array}{ccc}3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & 3\end{array}\right|=0 \Rightarrow \lambda=29$

Topic: Mathematics, Determinants, JEE Main
If the system of equations 2x-3y+5z=12, 3x+y+pz=q and x-7y+8z=17 is consistent, then which of the following is not true?

Options:
A p=2,q=7
B p≠2,q=7
C p≠2,q≠7
D p=2,q≠7

The correct answer is:
p=2,q=7

Solution:
Using Cramerâs rule:
â=2-3531p1-78=28+7p+324-p+5-21-1
=-22+11p=-11p-2
â1=12-35q1p17-78=128+7p+38q-17p+5-7q-17
=33p-11q+11=113p-q+1
â2=21253qp1178=28q+17p-1224-p+551-q
=11a-2p-3
â3=2-31231q1-717=217+7q+351-q+12-21-1
=11q-77=11q-7
For unique solution âpâ 2,qâR
For infinite solutions âp=2,q=7

Topic: Mathematics, Determinants, JEE Main
If the system of equations 2x+3y-z=0, x+ky-2z=0 and 2x-y+z=0 has a non-trivial solution x, y, z, then xy+yz+zx+k is equal to

Options:
A -14
B 12
C -4
D 34

The correct answer is:
12

Solution:
The system of equations a1x+b1y+c1z=0, a2x+b2y+c2z=0 and a3x+b3y+c3z=0 has a non-trivial solution, if

a1b1c1a2b2c2a3b3c3=0.

Hence, for the given lines 2x+3y-z=0, x+ky-2z=0 and 2x-y+z=0, we have

23-11k-22-11=0

⇒2k-2-31+4-1-1-2k=0

⇒2k-4-15+1+2k=0

⇒k=92

Thus, the lines are

2x+3y-z=0 ...1

x+92y-2z=0 ...2

2x-y+z=0 ...3

1-3⇒4y-2z=0

2y=z ...4

yz=12 ...5

put z from eq 4 into 1

2x+3y-2y=0

2x+y=0

xy=-12 ...6

From eq 4 and 6

zx=-4

Hence, xy+yz+zx+k=-12+12-4+92=12.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If the system of equations
$\begin{aligned}
& 2 x+7 y+\lambda z=3 \\
& 3 x+2 y+5 z=4 \\
& x+\mu y+32 z=-1
\end{aligned}$
has infinitely many solutions, then $(\lambda-\mu)$ is equal to________

Options:

The correct answer is:
38

Solution:
$\begin{aligned} & \mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0 \\ & \mathrm{D}_3=\left|\begin{array}{ccc}2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & \mu & -1\end{array}\right|=0 \Rightarrow \mu=-39 \\ & \mathrm{D}=\left|\begin{array}{ccc}2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & -39 & 32\end{array}\right|=0 \Rightarrow \lambda=-1 \\ & \lambda-\mu=38\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the system of equations
2x+y-z=5

2x-5y+λz=μ

x+2y-5z=7
has infinitely many solutions, then(λ+μ)2+(λ-μ)2 is equal to

Options:
A 904
B 916
C 912
D 920

The correct answer is:
916

Solution:
Given system of equations.

2x+y-z=5

2x-5y+λz=μ

x+2y-5z=7

For infinite solution Δ=0=Δ1=Δ2=Δ3

⇒Δ=21-12-5λ12-5=0

⇒51-3λ=0

⇒λ=17

⇒Δ3=521μ2-5712=0

⇒-3(μ+13)=0

⇒μ=-13

Now (λ+μ)2+(λ-μ)2

=(17+13)2+(17-13)2

=900+16

=916

Hence this is the correct option.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (13 Apr Shift 2)
If the system of equations 3x+y+z=1, 6x+3y+2z=1 and Î¼x+Î»y+3z=1 is inconsistent, then

Options:
A Î¼â 9, Î»â 5
B Î¼â 9, Î»=5
C Î¼=9, Î»=5
D Î¼=9, Î»â 5

The correct answer is:
Î¼=9, Î»â 5

Solution:
Δ=311632μλ3=39-2λ-118-2μ+16λ-3μ=9-μ
Δ1=1111321λ3=19-2λ-11+1λ-3=5-λ
Δ2=311612μ13=33-2-118-2μ+16-μ=μ-9
Δ3=311631μλ1=33-λ-16-μ+16λ-3μ=3+3λ-2μ
For the system of equations to be inconsistent, Δ=0⇒μ=9
and atleast one of Δ1, Δ2, Δ3 is non zero ⇒λ≠5

Topic: Mathematics, Determinants, JEE Main
If the system of equations
$\begin{array}{r}
11 x+y+\lambda z=-5 \\
2 x+3 y+5 z=3 \\
8 x-19 y-39 z=\mu
\end{array}$
has infinitely many solutions, then $\lambda^4-\mu$ is equal to :

Options:
A 51
B 45
C 47
D 49

The correct answer is:
47

Solution:
$\begin{aligned}
& 11 x+y+\lambda z=-5 \\
& 2 x+3 y+5 z=3 \\
& 8 x-19 y-39 z=\mu
\end{aligned}$
for infinite sol.
$\begin{aligned}
& \mathrm{D}=\left|\begin{array}{ccc}
11 & 1 & \lambda \\
2 & 3 & 5 \\
8 & -19 & -39
\end{array}\right|=0 \\
& \Rightarrow 11(-117+95)-1(-78-40)+\lambda(-38-24) \\
& \Rightarrow 11(-22)+118-\lambda(62)=0 \\
& \Rightarrow 62 \lambda=118-242 \\
& \Rightarrow \lambda=\frac{-124}{62}=-2
\end{aligned}$
$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc}-5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39\end{array}\right|=0 \\ & \Rightarrow-5(-117+95)-1(-117-5 \mu)-2(-57-3 \mu)=0 \\ & \Rightarrow-5(-22)+117+5 \mu+114+6 \mu=0 \\ & \Rightarrow 11 \mu=-110-231=-341 \\ & \Rightarrow \mu=-31 \\ & \lambda^4-\mu=(-2)^4-(-31)=16+31=47\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If the system of equations 14x-3y+z=12, x-2y=0 and x+2z=0 has a solution x0,y0,z0, then the value of x02+y02+z02 is equal to

Options:
A 32
B 34
C 92
D 94

The correct answer is:
32

Solution:
Using cramerâs rule, we get,
â=14-311-20102=14-4+32+12=-48
â1=12-310-20002=12-4=-48
â2=14121100102=-122=-24
â3=14-3121-20100=122=24
âx=â1â=1, y=â2â=12, z=â3â=-12
âx02+y02+z02=1+14+14=32

Topic: Mathematics, Determinants, JEE Main
If the system of equations C3nx+C4ny+35z=0, C4nx+35y+C3nz=0 and 35x+C3ny+C4nz=0 has a non-trivial solution, then the value of n is equal to ânâN, nâ¥4

Options:
A 6
B 7
C 8
D 9

The correct answer is:
7

Solution:
For non trival solution, ∆=0
C3nC4n35C4n35C3n35C3nC4n=0

⇒C3n+C4n+35=0 (not possible)
or C n 3 = C n 4 =35⇒n=7

Topic: Mathematics, Determinants, JEE Main
If the system of equations
$(k+1)^3 x+(k+2)^3 y =(k+3)^3$
$(k+1) x+(k+2) y =k+3$
$x+y =1$
is consistent, then the value of $k$ is

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

The correct answer is:
-2

Solution:
System of equations
$(k+1)^3 x+(k+2)^3 y=(k+3)^3$
$(k+1) x+(k+2) y=(k+3)$
$x+y=1$ is consistent.
Since, the given system of equations are consistent. Then $D=0$ and Also, $\left(D_1=D_2=D_3=0\right)$ have infinitely many solutions. By Crammer Rule
$D=\left|\begin{array}{ccc}(k+1)^3 & (k+2)^3 & (k+3)^3 \\(k+1) & (k+2) & (k+3) \\1 & 1 & 1\end{array}\right|=0$
$\because \quad C_2 \rightarrow C_2-C_1$
$C_3 \rightarrow C_3-C_1$
$D=\left|\begin{array}{ccc}(k+1)^3 & (k+2)^3-(k+1)^3 & (k+3)^3-(k+1)^3 \\ (k+1) & (k+2)-(k+1) & (k+3)-(k+1) \\ 1 & 0 & 0\end{array}\right|$
$=0$
$D=\left|\begin{array}{ccc}(k+1)^3 & \left(3 k^2+9 k+7\right) & \left(6 k^2+24 k+26\right) \\ (k+1) & 1 & 2 \\ 1 & 0 & 0\end{array}\right|$
$=0$
Expand with $r$ to $R_3$
$2\left(3 k^2+9 k+7\right)-2\left(3 k^2+12 k+13\right)=0$
$\Rightarrow 3 k^2+9 k+7-3 k^2-12 k-13=0$
$\Rightarrow \quad-3 k-6=0 \Rightarrow 3 k=-6$
$\Rightarrow \quad k=-2$

Topic: Mathematics, Determinants, JEE Main