datavorous/entrance-exam-dataset huggingface dataset:
If the number of integral terms in the expansion of 312+518n is exactly 33, then the least value of n is

Options:
A 264
B 128
C 256
D 248

The correct answer is:
256

Solution:
312+518n

General term Tr+1=Crn312n-r518r

Tr+1=Crn3n-r25r8.

n-r2, r8 must be integer.

n-r2∈I⇒n-r=0, 2, 4, 6, 8.......

r8∈I⇒r=0, 8, 16, 24, 32....

For rational term common powers=0, 8, 16, 24.......

Number of integral term=33

Ts=a+s-1d ⇒T33=0+33-18=256.

Value of n=256.

We can cross check to verify answer.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (03 Sep Shift 1)
If the number of integral terms in the expansion of \( \left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)^{n} \) is exactly \( 33 \), then the least value of \( n \) is

Options:
A \( 264 \)
B \( 128 \)
C \( 256 \)
D \( 248 \)

The correct answer is:
\( 256 \)

Solution:
312+518n

General term Tr+1=Crn312n-r518r

Tr+1=Crn3n-r25r8.

n-r2, r8 must be integer.

n-r2∈I⇒n-r=0, 2, 4, 6, 8.......

r8∈I⇒r=0, 8, 16, 24, 32....

For rational term common powers=0, 8, 16, 24.......

Number of integral term=33

Ts=a+s-1d ⇒T33=0+33-18=256.

Value of n=256.

We can cross check to verify answer.

Topic: Mathematics, Binomial Theorem, JEE Main
If the number of principal solutions of the equation tanâ¡7Ïcosâ¡x=cotâ¡7Ïsinâ¡x is k, then k5 is equal to

Options:

The correct answer is:
8

Solution:
tan⁡7πcos⁡x=tan⁡π2-7πsin⁡x
⇒7πcos⁡x=nπ+π2-7πsin⁡x
⇒7π(sin⁡x+cos⁡x)=n+12π
⇒sin⁡x+cos⁡x=2n+114
⇒sin⁡x+π4=2n+1142
-1≤sin⁡x+π4≤1
-1≤2n+1142⁡≤1
-142-12⁡≤n≤142-12
-10.4≤n≤9.4
⇒n=-10,-9,……,-1,0,1,….,9
Total 20 values of n and for each value of n their will be two solutions.

Hence, k=40⇒k5=8

Topic: Mathematics, Trigonometric Equations, JEE Main
If the number of protons and neutrons of an element is 13 and 14 respectively, then what's the atomic number(Z) and mass number(A)?

Options:
A $13,13$
B $13,27$
C $14,13$
D $27,14$

The correct answer is:
$13,27$

Solution:
For an element, Atomic number $(Z)=$ number of protons in that atoms $=$ numbers of electrons in that atom; Mass number $=$ number of protons + number of neutrons. So $Z=13$ and $A=13+14=27$. Hence that element is Aluminium.

Topic: Physics, Atomic Physics, JEE Main
If the number of solutions of the equation x+y+z=20, where 1≤x<y<z and x,y,z∈I is k, then k10 is equal to

Options:

The correct answer is:
2.4

Solution:
Let, y=x+h, where h≥1 and z=x+h+k, where k≥1.
Then, x+y+z=20
⇒x+x+h+x+h+k=20
⇒3x+2h+k=20.
When x=1⇒2h+k=17, then there are 8 ways.
When x=2⇒2h+k=14, then there are 6 ways.
When x=3⇒2h+k=11, then there are 5 ways.
When x=4⇒2h+k=8, then there are 3 ways.
When x=5⇒2h+k=5, then there are 2 ways.
When x>5⇒ not possible.
⇒ Total number of ways =24

Topic: Mathematics, Permutation Combination, JEE Main
If the number of terms free from radicals in the expansion of 713+11196561 is k, then the value of k100 is equal to

Options:

The correct answer is:
7.3

Solution:
Tr+1=6561Cr7136561-r1119r
=6561Cr76561-r311r/9
=6561Cr72187-r311r9
Tr+1 is rational if r9 and r3 are integers i.e. if r is a multiple of 9.
0≤r≤6561
⇒0≤r9≤729
⇒r9=0,1,2,.....,728,729
⇒ Total terms =730=k
⇒k100=7.3

Topic: Mathematics, Binomial Theorem, JEE Main
If the number of terms in the expansion of 1-2x+4y2n, x, y≠0, is 28, then the sum of the coefficients of all the terms in this expansion is

Options:
A 243
B 729
C 64
D 2187

The correct answer is:
729

Solution:
1-2x+4y2n

We know, if having three variables then total number of terms would be n+1n+22

⇒n+1n+22=28
∴ n+1n+2 =56
Which gives n=6 , and sum of coefficients would be

1-2+46=36=729 .

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2016 (03 Apr)
If the number of terms in the expansion of 1+x1011+x2-x100 is n, then the value of n25 is equal to

Options:

The correct answer is:
8.08

Solution:
1+x1011+x2-x100
=1+x1+x1-x+x2100
=1+x1+x3100
=1×1+x3100+x×1+x3100
So, the number of terms=( 101 terms of the form x3k ) + (101 terms of the form x3k+1 )
=202 terms

⇒n=202

Topic: Mathematics, Binomial Theorem, JEE Main
If the number of terms in the expansion of \( (1+x)^{101}\left(1+x^{2}-x\right)^{100} \) is \( n \), then the value of \( \frac{n}{25} \) is equal to

Options:

The correct answer is:
8.08

Solution:
1+x1011+x2-x100

=1+x1+x1-x+x2100

=1+x1+x3100

=1×1+x3100+x×1+x3100

So, the number of terms=( 101 terms of the form x3k ) + (101 terms of the form x3k+1 )
=202 terms

⇒n=202

∵n25=8.08

Topic: Mathematics, Binomial Theorem, JEE Main
If the number of turns in the coil of galvanometer are decreased then the resistance of galvanometer

Options:
A may increase Or decrease.
B increaseS.
C remains the same
D decreases.

The correct answer is:
decreases.

Solution:
Resistance of the wire is proportional to its length. If the number of turns is decreased, the length of the wire decreases, and hence the resistance decreases.

Topic: Physics, Current Electricity, MHT CET, MHT CET 2020 (14 Oct Shift 1)
If the number of turns per unit length of a solenoid is doubled, what happens to the magnetic field in the solenoid?

Options:
A It remains unchanged
B It becomes half
C It doubles
D It quadruples

The correct answer is:
It doubles

Solution:
The formula of the strength of magnetic field due to current carrying solenoid in terms of current and number of turns per unit length is,

B=μ0ni, here n is defined as, Nl, here N is the number of turns, if we double the number of turns of solenoid then the strength of the magnetic field also double.

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
If the number of ways in which four distinct balls can be put into two identical boxes such that no box remains empty is equal to k, then k is

Options:

The correct answer is:
7

Solution:
Every ball has two options
⇒ 4 balls can be put in 24 ways.
Arrangement among themselves is 2!
So, total ways=12!×24= 23
But, above count also includes the one case in which all the balls are put in one box,
⇒ required number of ways k= 23-1=7

Topic: Mathematics, Permutation Combination, JEE Main
If the number of ways of selecting 3 numbers out of 1, 2, 3,...., 2n+1 such that they are in arithmetic progression is 441, then the sum of the divisors of n is equal to

Options:
A 21
B 32
C 45
D 60

The correct answer is:
32

Solution:
If a,b,c are in A.P., then a+c=2b
i.e. the sum of two numbers is even
⇒ Both numbers are even or odd
Odd numbers →1,3,5,......,2n+1
(total n+1 )
Even numbers →2,4,6,......,2n
(total n )
⇒ Required number of ways =n+1C2+nC2
⇒n+1C2+nC2=441
⇒n+1n2+nn-12=441
⇒n2n+1+n-1=441
⇒n2=441⇒n=21=3×7
⇒ Sum of divisors =1+31+7=4×8=32

Topic: Mathematics, Permutation Combination, JEE Main
If the number of words, with or without meaning. which can be made using all the letters of the word MATHEMATICS in which C and S do not come together, is (6!)k then k is equal to

Options:
A 2835
B 5670
C 1890
D 945

The correct answer is:
5670

Solution:
The word 'MATHEMATICS' consists of 11 letters including two M's, two T's and two A's,

Now, first we will arrange letters _M_A_T_H_E_M_A_T_I_.

Here, we have 10 gaps to fill C & S.

The number of words with or without meaning can be formed from the word MATHEMATICS when C, S does not come together is

=9!2!·2!·2!×C210×2!

=9×8×7×6!2!·2!×10×92

=5670

Hence this is the correct option.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (08 Apr Shift 2)
If the numbers 32a-1, 14, 34-2a 0<a<1 are the first three terms of an arithmetic progression, then its fifth term is equal to

Options:
A 33
B 43
C 53
D 63

The correct answer is:
53

Solution:
Given, 32a-1+34-2a=28
â9a3+819a=28
Put 9a=xâx3+81x=28
x2+243=84x
âx2-84x+243=0
âx=81 or 3
â9a=81 or 3
âa=2 or 12
Since, aâ0,1âa=12
So, the numbers are 1,14,27 which are in A.P. with d=13

Hence, t5=a+4d=1+4Ã13=53

Topic: Mathematics, Sequences and Series, JEE Main
If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x2+αx+β>0, for all x∈R, is

Options:
A 1736
B 49
C 12
D 1936

The correct answer is:
1736

Solution:
Total sample space of two dice will be equal to 6×6=36

Now for x2+αx+β>0 its discriminant D must be less than zero,

So D=b2-4ac<0 or α2-4β<0⇒α2<4β

Now, if α=1,β can take values 1,2,3,4,5,6

If α=2,β can take values 2,3,4,5,6

If α=3,β can take values 3,4,5,6

If α=4,β can take values 5,6

If α=5 and 6, no value of β possible

So number of favourable ways =17

Required probability=favourable waystotal sample space=1736

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (25 Jul Shift 1)
If the observations 1,2,3,….,n occur with frequency n,n-1,n-2,..,1 respectively such that the mean of observations is 133, then n is equal to

Options:
A 10
B 11
C 12
D 13

The correct answer is:
11

Solution:
Mean =∑fixi∑fi=∑r=1nrn+1-rnn+12=133 (given)
⇒nn+12·n+1-nn+12n+16nn+12=133
⇒n+1-2n+13=133⇒n3+23=133⇒n=11

Topic: Mathematics, Statistics, JEE Main
If the order of the differential equation of the family of circles touching the x-axis at the origin is k, then 2k is equal to

Options:

The correct answer is:
2

Solution:
The equation of the family is x2+y-r2=r2
i.e. only one arbitrary constant.
Hence, the order of its differential equation will be '1'
∴2k=2

Topic: Mathematics, Differential Equations, JEE Main
If the origin and the non-real roots of the equation 3z2+3z+λ=0, ∀λ∈R are the vertices of an equilateral triangle in the argand plane, then 3 times the length of the triangle is

Options:
A 2 units
B 1 unit
C 3 units
D 4 units

The correct answer is:
1 unit

Solution:
Let, z1 and z2 are roots of the equation 3z2+3z+λ=0
⇒z1+z2=-1 and z1z2=λ3
If 0, z1 and z2 form an equilateral triangle, then

z12+z22=z1z2
⇒z1+z22=3z1z2
⇒-12=3λ3⇒λ=1⇒z1z2=13

⇒3z1=1 unit

Topic: Mathematics, Complex Number, JEE Main
If the origin lies between zeroes of the polynomial fx=x24-ax+a2+a-2, then the number of integral value(s) of a is

Options:
A 1
B 2
C 3
D more than 3

The correct answer is:
2

Solution:
For zeroes on either side of the origin.
Opening upward parabola
⇒f0<0
⇒a2+a-2<0
⇒a+2a-1<0
⇒-2<a<1
⇒2 integers i.e. -1,0

Topic: Mathematics, Quadratic Equation, JEE Main
If the orthocentre of the triangle formed by the lines $2 x+3 y-1=0, x+2 y-1=0$ and $a x+b y-1=0$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3,4)$ and $(-6,-8)$, then the value of $|a-b|$ is_______

Options:

The correct answer is:
16

Solution:
$\begin{aligned} & 2 x+3 y-1=0 \\ & x+2 y-1=0 \\ & a x+b y-1=0\end{aligned}$
$\begin{aligned} & \left(\frac{6-6}{3}, \frac{8-8}{3}\right) \\ & =(0,0)\end{aligned}$
$\begin{aligned} & a x+b y-1=0 \\ & \left(\frac{1-0}{-1-0}\right)\left(\frac{-a}{b}\right)=-1 \\ & \Rightarrow-\mathrm{a}=\mathrm{b} \\ & \Rightarrow \quad \mathrm{ax}-\mathrm{ay}-1=0 \\ & a x-a\left(1-\frac{2 x}{3}\right)-1 \\ & x\left(a+\frac{2 a}{3}\right)=\frac{a}{3} \\ & \mathrm{x}=\frac{\mathrm{a}+3}{5 \mathrm{a}} \\ & 2\left(\frac{a+3}{5 a}\right)+3 y-1=0 \\ & y=\frac{1-\frac{2 a+6}{5 a}}{3}=\frac{3 a-6}{3 \times 5 a} \\ & y=\frac{a-2}{5 a} \\ & \frac{\left(\frac{a-2}{5 a}\right)}{\left(\frac{a+3}{5 a}\right)}=2 \Rightarrow a-2=2 a+6 \\ & a=-8 \\ & \mathrm{~b}=8 \\ & -8 \mathrm{x}+8 \mathrm{y}-1=0 \\ & |a-b|=16 \\ & \end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (08 Apr Shift 1)
If the oxygen $\left(\mathrm{O}_2\right)$ has root mean square velocity of $C \mathrm{~ms}^{-1}$, then root mean square velocity of the hydrogen $\left(H_2\right)$ will be

Options:
A $\mathrm{Cms}^{-1}$
B $\frac{1}{C} m s^{-1}$
C $4 \mathrm{Cms}^{-1}$
D $\frac{C}{4} m s^{-1}$

The correct answer is:
$4 \mathrm{Cms}^{-1}$

Solution:
$v_{r m s} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{v_{\mathrm{O}_2}}{v_{\mathrm{H}_2}}=\sqrt{\frac{M_{\mathrm{H}_2}}{M_{\mathrm{O}_2}}} \Rightarrow \frac{C}{v_{\mathrm{H}_2}}=\sqrt{\frac{2}{32}}=\frac{1}{4}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
If the $p^{\text {th }}$ term of an A.P. be $\frac{1}{q}$ and $q^{\text {th }}$ term be $p q^{t h}$ terms will be

Options:
A $\frac{p q-1}{2}$
B $\frac{1-p q}{2}$
C $\frac{p q+1}{2}$
D $-\frac{p q+1}{2}$

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

Solution:
Since $T_p=a+(p-1) d=\frac{1}{q}$\ldots(i)
and $T_q=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) and (ii), we get
$a=\frac{1}{p q}$ and $d=\frac{1}{p q}$
Now sum of $p q$ terms
$=\frac{p q}{2}\left[\frac{2}{p q}+(p q-1) \frac{1}{p q}\right]$
$=\frac{p q}{2} \cdot \frac{2}{p q}\left[1+\frac{1}{2}(p q-1)\right]$ $=\left[\frac{2+p q-1}{2}\right]=\frac{p q+1}{2}$
Note : Students should remember this question as a formula.

Topic: Mathematics, Sequences and Series, JEE Main
If the $p^{\text {th }}$ term of an A.P. be $q$ and $q^{\text {th }}$ term be $p$, then its $r^{\text {th }}$ term will be

Options:
A $p+q+r$
B $p+q-r$
C $p+r-q$
D $p-q-r$

The correct answer is:
$p+q-r$

Solution:
Given that, $\quad T_p=a+(p-1) d=q \ldots(1)$
and $T_q=a+(q-1) d=p \ldots(2)$
From (i) and (ii), we get $d=-\frac{(p-q)}{(p-q)}=-1$
Putting value of $d$ in equation (i), then $a=p+q-1$
Now, $r^{\text {th }}$ term is given by A.P.
$T_r=a+(r-1) d=(p+q-1)+(r-1)(-1)=p+q-r$
Note : Students should remember this question as a formula.

Topic: Mathematics, Sequences and Series, JEE Main
If the pair of lines $a x^2+2(a+b) x y+b y^2=0$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

Options:
A
$3 a^2-10 a b+3 b^2=0$

B
$3 a^2-2 a b+3 b^2=0$

C
$3 a^2+10 a b+3 b^2=0$

D
$3 a^2+2 a b+3 b^2=0$

The correct answer is:
$3 a^2+2 a b+3 b^2=0$

Solution:
$$
\begin{aligned}
& \left|\frac{2 \sqrt{(a+b)^2-a b}}{a+b}\right|=1 \\
& \Rightarrow(a+b)^2=4\left(a^2+b^2+a b\right) \\
& \Rightarrow 3 a^2+3 b^2+2 a b=0 .
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2005
If the pair of lines $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ intersect on the $y$ - axis then

Options:
A
$2 f g h=b g^2+c h^2$

B
$\mathrm{bg}^2 \neq \mathrm{ch}^2$

C
$a b c=2 f g h$

D
none of these

The correct answer is:
$2 f g h=b g^2+c h^2$

Solution:
No Solution Available

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2002
If the pair of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 p x y-y^2=0$ be such that each pair bisects the angle between the other pair, then

Options:
A
$p q=-1$

B
$\mathrm{p}=\mathrm{q}$

C
$p=-q$

D
$\mathrm{pq}=1$

The correct answer is:
$p q=-1$

Solution:
Equation of bisector of both pair of straight lines,
$p x^2+2 x y-p y^2=0$
$q x^2+2 x y-q y^2=0$
From (1) and (2). $\frac{\mathrm{q}}{1}=\frac{2}{-2 \mathrm{p}}=\frac{-\mathrm{q}}{-1} \Rightarrow \mathrm{pq}=-1$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2003
If the parabola $y^2=4 a x$ passes through $(-3,2)$, then length of its latus rectum is

Options:
A $2 / 3$
B $1 / 3$
C $4 / 3$
D $4$

The correct answer is:
$4 / 3$

Solution:
The point $(-3,2)$ will satisfy the equation $y^2=4 a x$
$\Rightarrow 4=-12 a \Rightarrow 4 a=-\frac{4}{3}=\frac{4}{3}, \quad$ (Taking positive sign).

Topic: Mathematics, Parabola, JEE Main
If the parabola $y^2=4 a x$ passes through the point $(1,-2)$, then the tangent at this point is

Options:
A $x+y-1=0$
B $x-y-1=0$
C $x+y+1=0$
D $x-y+1=0$

The correct answer is:
$x+y+1=0$

Solution:
Parabola passes through the point $(1,-2)$, then $4=4 a \Rightarrow a=1$
Formula for tangent, $y y_1=2 a\left(x+x_1\right) \Rightarrow-2 y=2(x+1)$
Required tangent is, $x+y+1=0$

Topic: Mathematics, Parabola, JEE Main
If the percentage errors in measuring the length and the diameter of a wire are 0.1% each. The percentage error in measuring its resistance will be:

Options:
A 0.2%
B 0.3%
C 0.1%
D 0.144%

The correct answer is:
0.3%

Solution:
Resistance of the wire in terms of the diameter can be written as, R=ρLπd24

Hence, percentage error relation will be, ∆RR×100=∆LL×100+2∆dd×100

Given: ∆LL=0.1% and ∆dd=0.1%

Therefore, ∆RR=0.3%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If the perpendicular bisector of the line segment joining the points P(1,4) and Q(k,3) has y-intercept equal to -4, then a value of k is;

Options:
A -2
B -4
C 14
D 15

The correct answer is:
-4

Solution:
Mid point of PQ is k+12,72
And slope of PQ=11-k
So equation of perpendicular bisector of PQ

y-72=(k-1)x-k+12
It's y intercept =-4. So, this equation must satisfy 0,-4.

⇒-152=-k2-12
k2=16⇒k=±4

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If the phase difference between the two wave is $2 \pi$ during superposition, then the resultant amplitude is

Options:
A Maximum
B Minimum
C Maximum or minimum
D None of the above

The correct answer is:
Maximum

Solution:
Phase difference is $2 \pi$ means constrictive interference so resultant amplitude will be maximum.

Topic: Physics, Waves and Sound, JEE Main
If the pituitary gland is surgically removed, the blood level of sodium falls and that of potassium rises due to

Options:
A atrophy of adrenal cortex
B atrophy of adrenal medulla
C fact that ADH from pituitary is no longer available
D fact that TSH from pituitary is no longer available

The correct answer is:
atrophy of adrenal cortex

Solution:
Pituitary gland is the master endocrine gland. It is made up of Adenohypophysis and Neurohypophysis. Adenohypophysis secretes many hormones depending on the signals from hypothalamus in form of releasing and inhibiting hormones. One such example is ACTH i.e. Adreno corticotrophic hormone. This is released when hypothalamus secrets CRH i.e. Corticotrophic releasing hormone. ACTH acts on adrenal cortex and thus it causes release of Glucocoticoids, Mineralocorticoids and Sex corticoids as per body's demand. One such mineralocorticoid is Aldosterone which when released causes Sodium and Water reabsorption and release of Potassium and thus electrolyte and water balance is maintained. If pituitary gland is removed, ACTH production will stop and thus Aldosterone will not be released leading to atrophy of Adrenal cortex. Sodium and water reabsorption and Potassium excretion will be reversed.

Topic: Biology, Chemical Coordination and Integration, NEET
If the pK a of a weak acid HA is 4.80 and the pK b of a weak base BOH is 4.78. Then, the pH of an aqueous solution of the corresponding salt, BA will be

Options:
A 8.23
B 9.41
C 7.01
D 5.91

The correct answer is:
7.01

Solution:
BOH+HA→ B + + A − + H 2 O
pH= 1 2 pK w + 1 2 pK a − 1 2 pK b
= 1 2 14+4.80−4.78 =7.01

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the pK a of acetic acid and pK b of NH 4 OH are 4.76 and 4.75 respectively, what will be the pH of ammonium acetate solution?

Options:
A 9.51
B 7.005
C 7.00
D 6.9

The correct answer is:
7.005

Solution:
When a weak acid (acetic acid) and weak base NH 4 OH are mixed to form a solution. pH of the solution is then determined by following formula:
pHâ=â7â+â 1 2 pK a â pK b
Given, pK a  = 4.76,   pK b  = 4.75
pH = 7 +  1 2 4.76−4.75
pH = 7 + 0.005
pH = 7.005

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25° C is __________ × 10–1 (Nearest integer)

Options:

The correct answer is:
85

Solution:
Concentration of calcium lactate = 0.005 M

concentration of lactate ion = = (2 × 0.005) M

Calcium lactate is a salt of weak acid + strong base

∴ Salt hydrolysis will take place for a basic salt:

⇒pH=7+12pKa+12logC=7+12×5+12log(0.01)=85 × 10-1

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If the plane $2 a x-3 a y+4 a z+6=0$ passes through the midpoint of the line joining the centres of the spheres
$$
x^2+y^2+z^2+6 x-8 y-2 z=13 \text { and }
$$
$x^2+y^2+z^2-10 x+4 y-2 z=8$, then a equals

Options:
A
-1

B
1

C
-2

D
2

The correct answer is:
-2

Solution:
Plane
$2 a x-3 a y+4 a z+6=0$ passes through the mid point of the centre of spheres $x^2+y^2+z^2+6 x-8 y-2 z=13$ and $x^2+y^2+z^2-10 x+4 y-2 z=8$ respectively centre of spheres are $(-3,4,1) \&(5,-2,1)$
Mid point of centre is $(1,1,1)$
Satisfying this in the equation of plane, we get
$2 a-3 a+4 a+6=0 \Rightarrow a=-2$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
If the plane 2x-y+2z+3=0 has the distances 13 and 23 units from the planes 4x-2y+4z+λ=0 and 2x-y+2z+μ=0 , respectively, then the maximum value of λ+μ is equal to:

Options:
A 9
B 15
C 13
D 5

The correct answer is:
13

Solution:
The distance between the parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is d1-d2a2+b2+c2.

Given, the distance between the planes 2x-y+2z+3=0 and 4x-2y+4z+λ=0, ⇒2x-y+2z+λ2=0 is 13 units

⇒λ2-34+1+4=13

⇒λ2-33=13

⇒λ2-3=1

⇒λ2-3=1 or λ2-3=-1

⇒λ=8 or λ=4.

Also, distance between the planes 2x-y+2z+3=0 and 2x-y+2z+μ=0 is 23

⇒µ-34+1+4=23

⇒µ-33=23

⇒µ-3=2

⇒µ-3=2 or µ-3=-2

⇒µ=5 or ⇒µ=1.

Hence, the maximum value of λ+μ=8+5=13.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If the plane 2x+y-5z=0 is rotated about its line of intersection with the plane 3x-y+4z-7=0 by an angle of π2, then the plane after the rotation passes through the point

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

The correct answer is:
1,0,2

Solution:
Given equation of plane, P1:2x+y-5z=0 and P2:3x-y+4z-7=0

We know that equation of family of planes P1 and P2 is given by,

P:P1+λP2

∴P:2+3λx+1-λy+-5+4λz-7λ=0

∵P⊥P1 ∴4+6λ+1-λ+25-20λ=0⇒λ=2

So, equation of plane P:8x-y+3z-14=0

It passes through the point 1,0,2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If the plane $3 x-2 y-z-18=0$ meets the coordinate axes in $A, B, C$ then the centroid of $\triangle A B C$ is

Options:
A $(2,3,-6)$
B $(2,-3,6)$
C $(-2,-3,6)$
D $(2,-3,-6)$

The correct answer is:
$(2,-3,-6)$

Solution:
Given equation $3 x-2 y-z=18$ can be rewritten as
$$
\begin{aligned}
\frac{x}{18 / 3}-\frac{y}{18 / 2}-\frac{z}{18} & =1 \\
\frac{x}{6}-\frac{y}{9}-\frac{z}{18} & =1
\end{aligned}
$$
$\therefore$ Points of a coordinates axes are $A(6,0,0), B(0,-9,0)$ and $C(0,0,-18)$.
$\therefore$ Centroid of a triangle
$$
\begin{aligned}
& =\left(\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}\right) \\
& =(2,-3,-6)
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the plane ax-by+cz=d contains the line x-aa=y-2db=z-cc, then the value of b4d is equal to b, d≠0

Options:

The correct answer is:
0.5

Solution:
Since the line lies on the given plane, therefore the point a,2d,c on the line also lies on the plane

⇒a2-2bd+c2=0

Also, the line will be perpendicular to the normal vector of the plane

⇒a2-b2+c2=0

Hence, b2=2bd⇒bd=2⇒b4d=12

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the plane P passes through the intersection of two mutually perpendicular planes 2x+ky-5z=1 and 3kx-ky+z=5,k<3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is

Options:
A 111
B 511
C 6
D 7

The correct answer is:
7

Solution:
Given the plane P passes through the intersection of two mutually perpendicular planes 2x+ky-5z=1 and 3kx-ky+z=5,k<3

Since planes are mutually perpendicular

Then, 23k+k-k+-51=0

⇒6k-k2-5=0

⇒k=1,5

But k<3

So k=1

Now plane passing through these planes will be,

2x+y-5z-1+λ3x-y+z-5=0

⇒x5λ+12+3λ+y5λ+11-λ+z5λ+1λ-5=1

Given 5λ+12+3λ=1⇒λ=12

So intercept on y-axis =5λ+11-λ=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (27 Jul Shift 1)
If the plane $x-3 y+5 z=d$ passes through the point $(1,2,4)$, then the lengths of intercepts cut by it on the axes of $x, y, z$ are respectively

Options:
A $15,-5,3$
B $1,-5,3$
C $-15,5,-3$
D $1,-6,20$

The correct answer is:
$-15,5,-3$

Solution:
If plane $x-3 y+5 z=d$ passes through point $(1,2,4)$. Then $1-6+20=d \Rightarrow d=15$
$\therefore$ Plane, $x-3 y+5 z=15 \Rightarrow \frac{x}{15}+\frac{y}{-5}+\frac{z}{3}=1$
Hence length of intercept cut by it on the axes $(x, y, z)$ are respectively $(15,-5,3)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main