datavorous/entrance-exam-dataset huggingface dataset:
If for all real triplets a,b,c,fx=a+bx+cx2; then ∫01fxdx is equal to:

Options:
A 23f1+2f12
B 12f1+3f12
C 13f0+f12
D 16f0+f1+4f12

The correct answer is:
16f0+f1+4f12

Solution:
∫01a+bx+cx2dx=ax+bx22+cx3301=a+b2+c3

f1=a+b+c

f0=a

f12=a+b2+C4

Now

16f1+f0+4f12=16a+b+c+a+4a+b2+c4=166a+3b+2c=a+b2+c3

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If for an Arithmetic progression, 9 times nineth term is equal to 13 times thirteenth term, then value of twenty second term is

Options:
A 0
B 2
C 4
D 5

The correct answer is:
0

Solution:
Let the first term and common difference of given $\mathrm{AP}$ be a and d, respectively.
Given
$\begin{array}{l}
9 . a_{9}=13 . a_{13} \\
\Rightarrow 9[a+(9-1) d]=13[a+(13-1) d] \\
\Rightarrow 9[a+8 d]=13[a+12 d] \\
\Rightarrow 9 a+72 d=13 a+156 d \\
\Rightarrow 4 a+84 d=0 \\
\Rightarrow 4[a+21 d]=0 \\
\Rightarrow 4[a+(22-1) d=0 \\
\Rightarrow a+(22-1) d=0 \\
\Rightarrow a_{22}=0
\end{array}$

Topic: Mathematics, Sequences and Series, JEE Main
If for $\mathrm{n} \geq 1, \mathrm{P}_{\mathrm{n}}=\int_1^{\mathrm{e}}\left(\log \mathrm{x}^{\mathrm{n}}\right) \mathrm{dx}$, then $\mathrm{P}_{10}-90 \mathrm{P}_8$ is equal to:

Options:
A
$-9$

B
$10 \mathrm{e}$

C
$-9 \mathrm{e}$

D
10

The correct answer is:
$-9 \mathrm{e}$

Solution:
$\mathrm{P}_{\mathrm{n}}=\int_1^e(\log x)^n d x$ put $\log x=t$ then $x=e^t$ and $d x=e^t d t$
Also, when $x=1$, then $t=\log 1=0$ and when $x=e$, then $t=\log _e e=1$
$$
\begin{gathered}
\therefore \mathrm{P}_{\mathrm{n}}=\int_0^1 t^n \cdot e^t d t \\
\therefore \mathrm{P}_{10}=\int_0^1 t^{10} e^t d t \text { and } \mathrm{P}_8=\int_0^1 t^8 e^t d t \\
\text { Now, } \mathrm{P}_{10}-90 \mathrm{P}_8=\int_0^1 t^{10} e^t d t-90 \int_0^1 t^8 e^t d t \\
\mathrm{P}_{10}-90 \mathrm{P}_8 \\
\mathrm{P}_{10}-90 \mathrm{P}_8 \\
=\left[t^{10} e^t\right]_0^1-10 \int_0^1 t^9 e^t d t-90 \int_0^1 t^8 e^t d t \\
\mathrm{P}_{10}-90 \mathrm{P}_8= \\
e-10\left[t^9 \int_0^1 e^t d t-\int_0^1 \frac{d}{d t}\left(t^9\right) \int e^t d t\right]-90 \int_0^1 t^8 e^t d t
\end{gathered}
$$
$$
\begin{gathered}
\mathrm{P}_{10}-90 \mathrm{P}_8=e-10\left[e-9 \int_0^1 t^8 e^t d t\right]-90 \int_0^1 t^8 e^t d t \\
\mathrm{P}_{10}-90 \mathrm{P}_8=e-10 e+90 \int t^8 e^t d t-90 \int_0^1 t^8 e^t d t \\
\therefore P_{10}-90 P_8=-9 e
\end{gathered}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2014 (11 Apr Online)
If for p≠q≠0, then function fx=p729+x7-3729+qx3-9 is continuous at x=0, then

Options:
A 7pqf0-1=0
B 63qf0-p2=0
C 21qf0-p2=0
D 7pq f0-9=0

The correct answer is:
63qf0-p2=0

Solution:
For a function to be continuous fx to be x=0, we know

f0=limx→0fx

For fx=p729+x7-3729+qx3-9

limit should be 00 from

So, limx→0p729+x7-3

p.7297-3=0

⇒p·729=37 ⇒ p=3

Now, f0=limx→0336+x7-336+qx3-9

=limx→031+x3617-191+qx3613-1=39×17·36q3·36

⇒f0=13×37q=17q

⇒7qf0-1=0

⇒7·p2·qf0-p2=0 (multiplying the equation by p2)

⇒63qf0-p2=0

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (27 Jul Shift 2)
If for positive integers $r>1, n>2$, the coefficients of the $(3 r)^{\text {th }}$ and $(r+2)^{\text {th }}$ powers of $x$ in the expansion of $(1+x)^{2 n}$ are equal, then $n$ is equal to:

Options:
A
$2 r+1$

B
$2 r-1$

C
$3 r$

D
$r+1$

The correct answer is:
$2 r+1$

Solution:
Expansion of $(1+x)^{2 n}$ is $1+{ }^{2 n} \mathrm{C}_1 x+{ }^{2 n} \mathrm{C}_2 x^2$ $+\ldots \ldots .+{ }^{2 n} \mathrm{C}_r x^n+{ }^{2 n} \mathrm{C}_{r+1} x^{n+1}+\ldots \ldots+{ }^{2 n} \mathrm{C}_{2 n} x^{2 n}$ As given ${ }^{2 n} \mathrm{C}_{r+2}={ }^{2 n} \mathrm{C}_{3 r}$
$$
\begin{aligned}
&\Rightarrow \frac{(2 n) !}{(r+2) !(2 n-r-2) !}=\frac{(2 n) !}{(3 r) !(2 n-3 r) !} \\
&\Rightarrow(3 r) !(2 n-3 r) !=(r+2) !(2 n-r-2) !
\end{aligned}
$$
Now, put value of $n$ from the given choices.
Choice (a) put $n=2 r+1$ in (1)
LHS : $(3 r) !(4 r+2-3 r) !=(3 r) !(r+2) !$
RHS : $(r+2) !(3 r) !$
$$
\Rightarrow \text { LHS }=\text { RHS }
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2013 (25 Apr Online)
If for some α and β in R , the intersection of the following three planes
x+4y-2z=1
x+7y-5z=β
x+5y+αz=5
is a line in R3 , then α+β is equal to:

Options:
A 0
B 10
C 2
D -10

The correct answer is:
10

Solution:
â=0⇒14-217-515Î±=0
7Î±+25-4Î±+10+-20+14=0
3Î±+9=0⇒Î±=-3
Also Dz=0⇒14117Î²155=0
135-5Î²-15+14Î²-7=0
Î²=13

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If for some m, n; 6Cm+26Cm+1+6Cm+2>8C3 and n−1P3:nP4=1:8, then nPm+1+n+1Cm is equal to

Options:
A 380
B 376
C 384
D 372

The correct answer is:
372

Solution:
Given: 6Cm+26Cm+1+6Cm+2>8C3

⇒6Cm+6Cm+1+6Cm+1+6Cm+2>8C3

⇒7Cm+1+7Cm+2>8C3

⇒8Cm+2>8C3

⇒m=2

Also, n−1P3:nP4=1:8

⇒n−1n−2n−3nn−1n−2n−3=18

⇒n=8

⇒nPm+1+n+1Cm=8P3+9C2

⇒nPm+1+n+1Cm=8×7×6+9×82

⇒nPm+1+n+1Cm=372

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (31 Jan Shift 2)
If for some p,q,r∈R, all have positive sign, one of the roots of the equation p2+q2x2-2qp+rx+q2+r2=0 is also a root of the equation x2+2x-8=0, then q2+r2p2 is equal to-

Options:

The correct answer is:
272

Solution:
Given, p2+q2x2-2qp+rx+q2+r2=0

On simplifying we get, px-q2+qx-r2=0

⇒px-q=0 & qx-r=0

⇒x=qp=rq

⇒ x=qp=rq=4 [because roots of equation x2-2x-8=0 are 4 or -2]

As p,q,r are positive, so x must be 4.

Now, q=4p and r=4q=16p

So, q2+r2p2=4p2+4×4p2p2=16+256=272.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (26 Jul Shift 1)
If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of 1+xn+5 are in the ratio 5:10:14, then the largest coefficient in the expansion is :

Options:
A 462
B 330
C 792
D 252

The correct answer is:
462

Solution:
Let three consecutive terms are Tr, Tr+1, Tr+2
So, Cr-1n+5Crn+5=510 and Crn+5Cr+1n+5=1014

Using property CrnCr-1n=n-r+1r, we have

n+5-r+1r=2 and n+5-r+1+1r+1=75

⇒n-3r+6=0 and 5n-12r+18=0

Solving both equations, n=6, r=4
Hence, the greatest coefficient will be of middle term, which is =C5n+5=C511=462

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If for some positive $x \hat{I} R$, the frequency distribution of the marks obtained by 20 students in a certain test, is as follows
Then the mean of the marks is

Options:
A $3.0$
B $2.5$
C $2.8$
D $3.2$

The correct answer is:
$2.8$

Solution:
Here, total number of students $=20$
$\begin{aligned} & \Rightarrow(x+1)^2+(2 x-5)+\left(x^2-3 x\right)+x=20 \\ & \Rightarrow 2 x^2+2 x-4=20 \\ & \Rightarrow x^2+x-12=0 \\ & \Rightarrow(x+4)(x-3)=0 \\ & \text { But } x=3[\text { as } x>0]\end{aligned}$
Now required mean $=$
$\begin{aligned} & \frac{2 \times(3+1)^2+3 \times(2 \times 3-5)+5 \times\left(3^2-3 \times 3\right)+7 \times 3}{20} \\ & =\frac{2 \times 16+3 \times 1+5 \times 0+7 \times 3}{20} \\ & =2.8\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If for some,α∈R, the lines L1:x+12=y-2-1=z-11 and L2:x+2α=y+15-α=z+11 are coplanar, then the line L2 passes through the point :

Options:
A (10, 2, 2)
B (2, –10, –2)
C (10, –2, –2)
D (–2, 10, 2)

The correct answer is:
(2, –10, –2)

Solution:
Lines are coplanar
so α5-α12-11+1+32=0
⇒-5α+(α-5)3+7=0
⇒-2α=8⇒α=-4
⇒L2:x+2-4=y+19=z+11
Now by cross checking option (1) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If for some reason the parietal cells of the gut epithelium become partially non-functional, what is likely to happen?

Options:
A The pancreatic enzymes and specially the trypsin and lipase will not work efficiently
B The $\mathrm{pH}$ of stomach will fall abruptly
C Steapsin will be more effective
D Proteins will not be adequately hydrolysed by pepsin into proteoses and peptones

The correct answer is:
Proteins will not be adequately hydrolysed by pepsin into proteoses and peptones

Solution:
The parietal cells or oxyntic cells are large, rounded cells located upon the surface of gastric glands. These cells secrete $\mathrm{HCl}$ (which converts inactive pepsinogen into active pepsin) and Castle's Intrinsic Factor or CIF (which helps in the absorption of vitamin- $B_{12}$ or cyanocobalamine). If somehow, parietal cells become partially non-functional, the patient will suffer from achlorhydria (no $\mathrm{HCl}$ ) or hypochlorhydria (less $\mathrm{HCl}$ ). This will result in the non-activation of pepsinogen due to which proteins will not be adequately hydrolysed into proteoses and peptones. Lack of CIF will cause vitamin- $\mathrm{B}_{12}$ deficiency, which will result in pernicious anaemia.

Topic: Biology, Digestion and Absorption, NEET, NEET 2010 (Mains)
If for the complex numbers z satisfying |z-2-2i|≤1, the maximum value of |3iz+6| is attained at a+ib, then a+b is equal to _____ .

Options:

The correct answer is:
5

Solution:
Given that: |z-2-2i|≤1 which represents a circle with centre at (2,2) and radius =1.

Now |3iz+6|=|3i(z-2i)|=|3i||z-2i|

As we know that, |z1z2|=|z1||z2|

From tha diagram we can see that, |3iz+6|=3|z-2i| is maximum at z=3+2i because a point on the circle, which has maximum distance from (0,2) is (3,2).

So, a=3 & b=2

⇒a+b=5

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (01 Sep Shift 2)
If for the matrix, A=1-ααβ, AAT=I2, then the value of α4+β4 is :

Options:
A 3
B 1
C 2
D 4

The correct answer is:
1

Solution:
A=1-ααβ AAT=I2

⇒1-ααβ1α-αβ=1001

⇒1+α2α-αβα-αβα2+β2=1001

⇒1+α2=1 & α2+β2=1

⇒α2=0 & β2=1

∴ α4=0 & β4=1

∴ α4+β4=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If for two vector $\vec{A}$ and $\vec{B}$, sum $(\vec{A}+\vec{B})$ is perpendicular to the difference $(\vec{A}-\vec{B})$. The ratio of their magnitude is

Options:
A 1
B 2
C 3
D None of these

The correct answer is:
1

Solution:
$(\vec{A}+\vec{B})$ is perpendicular to $(\vec{A}-\vec{B})$. Thus
$(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0$
or $A^2+\vec{B} \cdot \vec{A}-\vec{A} \cdot \vec{B}-B^2=0$
Because of commutative property of dot product $\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}$
$\therefore A^2-B^2=0$ or $A=B$
Thus the ratio of magnitudes $A / B=1$.

Topic: Physics, Mathematics in Physics, JEE Main
If for x∈0,14, the derivative of tan-1⁡6xx1-9x3 is x ⋅gx , then gx equals:

Options:
A 91+9x3
B 3xx1-9x3
C 3x1-9x3
D 31+9x3

The correct answer is:
91+9x3

Solution:
Let 3x32=y, then y∈0,38
tan-1⁡6xx1-9x3=tan-1⁡2y1-y2
⇒ 2tan-1⁡y=2tan-1⁡3x32
Now ⇒ddx2tan-1⁡3x32=21+9x3⋅3⋅32⋅x12
⇒ddx2tan-1⁡3x32=91+9x3 ⋅ x

⇒x.gx=91+9x3 ⋅ x

⇒gx=91+9x3

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2017 (02 Apr)
If for x≥0,y=yx is the solution of the differential equation, x+1dy=x+12+y-3dx,y2=0 then y3 is equal to ________

Options:

The correct answer is:
3

Solution:
dydx=1+x+y-31+x

dydx-11+xy=1+x-3(1+x)

I.F=e-∫1(1+x)dx=1(1+x)

∴ddxy1+x=x+31+x-1+c

⇒y1+x=x+31+x-1+c

⇒y=1+xx+3(1+x)+c , ∴ at x=2, y=0

⇒0=32+1+c ⇒c=-3

At x=3, y=3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If for x,y∈R, x>0, y=log10x+log10x1/3+log10x1/9+…upto ∞ terms and 2+4+6+…+2y3+6+9+…+3y=4log10x, then the ordered pair x, y is equal to

Options:
A 106, 6
B 106, 9
C 102, 3
D 104, 6

The correct answer is:
106, 9

Solution:
2+4+6+…+2y3+6+9+…+3y=4log10x

21+2+3+----y31+2+3+----y=4log10x

log10x=6

x=106

Now

y=log10x+log10x13+log10x19---∞

=1+13+19---∞log10x

=11-13log10x=9

So, x, y=106, 9

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If for z=α+iβ, z+2=z+4(1+i), then α+β and αβ are the roots of the equation

Options:
A x2+3x-4=0
B x2+7x+12=0
C x2+x-12=0
D x2+2x-3=0

The correct answer is:
x2+7x+12=0

Solution:
Given,

z=α+iβ & z+2=z+41+i

Now putting the value of z=α+iβ in z+2=z+41+i we get,

z+2=z+41+i

⇒α+22+β2=α+4+iβ+4

Now on comparing real and imaginary part, we get

(α+2)2+β2=α+4...(i) and β+4=0⇒β=-4 ...(ii)

Now solving,

(α+2)2+16=α+4

⇒α2+4α+20=α2+8α+16

⇒α=1

So, α+β=-3, αβ=-4

We know that quadratic equation is given by,

x2-sum of rootsx+product of roots=0

So, equation with roots -3 and -4 will be,

x2+7x+12=0

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If force F→=3i^+4j^-2k^ acts on a particle having position vector 2i^+j^+2k^ then, the torque about the origin will be:

Options:
A -10i^+10j^+5k^
B 3i^+4j^-2k^
C 10i^+5j^-10k^
D 10i^+j^-5k^

The correct answer is:
-10i^+10j^+5k^

Solution:
Torque in vector form is expressed as vector multiplication of position vector and force.

τ→=r→×F→

⇒τ→=i^j^k^21234-2

⇒τ→=-2-8i^--4-6j^+8-3k^

⇒τ→=-10i^+10j^+5k^ N m

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (25 Jun Shift 1)
If force F is related with distance x and time t as F=Ax+Bt2 , the dimensions of AB is

Options:
A M0L-1/2T
B ML-1/2T-2
C M0L-1/2T2
D M0 LT–2

The correct answer is:
M0L-1/2T2

Solution:
All terms of an equation must have same dimensions
MLT–2=AL12 =BT2
A=MLT-2L
B=MLT-2T2

Topic: Physics, Units and Dimensions, JEE Main
If force $(F)$, length $(L)$ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

Options:
A $\left[\mathrm{FL}^{-1} \mathrm{~T}^{2}\right]$
B $\left[\mathrm{FL}^{-1} \mathrm{~T}^{-2}\right]$
C $\left[\mathrm{FL}^{-1} \mathrm{~T}^{-1}\right]$
D $\left[\mathrm{FL}^{2} \mathrm{~T}^{2}\right]$

The correct answer is:
$\left[\mathrm{FL}^{2} \mathrm{~T}^{2}\right]$

Solution:
Let $\mathrm{m}=\mathrm{KF}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}$

Substituting the dimensions of

$[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right],[\mathrm{L}]=[\mathrm{L}]$ and $[\mathrm{T}]=[\mathrm{T}]$ and comparing both side, we get $\mathrm{m}=\mathrm{FL}^{-1} \mathrm{~T}^{2}$

Topic: Physics, Units and Dimensions, JEE Main
If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density:

Options:
A FL-4 T2
B FL-3 T3
C FL-3 T2
D FL-5 T2

The correct answer is:
FL-4 T2

Solution:
d∝FaLbTc

ML-3=MLT-2aLbT c

=MaLa+bT-2a+c

comparing :

a=1 ......(i)

a+b=-3 ....(ii)

-2 a+c=0 .......(iii)

c=2 ; b=-4 ; ∴ [d]=FL-4 T2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (27 Aug Shift 2)
If force F, length L and time T be considered fundamental units, then the units of mass will be

Options:
A [F L T-2]
B [F L-2T-1]
C [F L-1T2]
D [F2L T-2]

The correct answer is:
[F L-1T2]

Solution:
Let M∝FaLbTc
Writing dimensions on both sides and using the principle of homogeneity of dimensions we have,
M1L0T0=kM L T-2aLbTc
On comparing the power on both sides we get a=1,a+b=0 and -2a+c=0
On solving we have b=-1,c=2,a=1
∴ units of mass is F L-1T2

Topic: Physics, Units and Dimensions, JEE Main
If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :

Options:
A F V4 T-6
B F V-4 T-2
C F2 V-2 T6
D F V-2 T2

The correct answer is:
F V-4 T-2

Solution:
The dimensional formula of force is

F=MLT-2

The dimensional formula of velocity is

V=LT-1

The formula for density is

Density =ML3=ML-3

Comparing the density dimensions with the assumed fundamental quantities in the question,
ML-3=FxVyTz⇒ML-3=MLT-2xLT-1y[T]z⇒ML-3=MxL(x+y)T-2x-y+z

Comparing the exponents,
x=1 ...(i)x+y=-3 ...(ii)-2x-y+z=0 ...(iii)

Solving equations (i), (ii) & (iii), we get

y=-4, z=-2.

The dimensions of density =FV-4T-2.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If four distinct points with position vectors a→,b→,c→ and d→ are coplanar, then [a→b→c→] is equal to

Options:
A d→b→a→+a→c→d→+d→b→c→
B a→d→b→+d→c→a→+d→b→c→
C d→c→a→+b→d→a→+c→d→b→
D b→c→d→+d→a→c→+d→b→a→

The correct answer is:
d→c→a→+b→d→a→+c→d→b→

Solution:
Given that a→,b→,c→,d→ are coplanar.

⇒b→-a→,c→-a→,d→-a→ are also coplanar.

⇒b→-a→c→-a→d→-a→=0

⇒b→-a→.c→-a→×d→-a→=0

⇒b→-a→.c→×d→-c→×a→-a→×d→=0 (Since a→×a→=0)

⇒b→.c→×d→-b→.c→×a→-b→.a→×d→-a→.c→×d→+a→.c→×a→+a→.a→×d→=0

⇒b→c→d→-b→c→a→-b→a→d→-a→c→d→=0

We know that x→y→z→=y→z→x→=z→x→y→ and x→y→z→=-x→z→y→

⇒b→c→d→+a→b→d→+a→d→c→=a→b→c→

⇒c→d→b→+b→d→a→+d→c→a→=a→b→c→

Hence, the required answer is d→c→a→+b→d→a→+c→d→b→

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If \( \frac{{ }^{\mathrm{n}} \mathrm{C}_{0}}{2}-\frac{{ }^{\mathrm{n}} \mathrm{C}_{1}}{3}+\frac{{ }^{\mathrm{n}} \mathrm{C}_{2}}{4}+\ldots . .+(-1)^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}=\frac{1}{2001 \times 2000} \), then \( \frac{(n+1)}{1000} \) is..

Options:

The correct answer is:
2

Solution:
(1 - x)n=C0n-C1nx+C2nx2-......(-1)n Cnnxn
multiplying both sides by x
x1-xn=C0nx-C1nx2+C2nx3-.......-1n Cnnxn+1
Integrating both sides
∫ 0 1 x 1 - x n = ∫ 0 1 n C 0 x - n C 1 x 2 + n C 2 x 3 + ..... + n C n x n + 1
1 n + 1 n + 2 = n C 0 2 - n C 1 3 + n C 2 4 .....
⇒ 1 n + 1 n + 2 = 1 2 0 0 0 × 2 0 0 1 i.e. n = 1 9 9 9

Topic: Mathematics, Binomial Theorem, JEE Main
If \( \frac{z-\alpha}{z+\alpha}(\alpha \in R) \) is a purely imaginary number and \( |z|=2 \), then a value of \( \alpha \) is

Options:
A \( 1 \)
B \( \frac{1}{2} \)
C \( \sqrt{2} \)
D \( 2 \)

The correct answer is:
\( 2 \)

Solution:
If a complex number is purely imaginary, then it must be equal to negative of its conjugate.

⇒z-αz+α=-z-αz+α⇒z-αz+α=-z--αz-+α

⇒z-αz-+α=z+αz--α

⇒ zz-+αz-αz--α2=-zz--αz+αz--α2

⇒z2=α2

⇒α2=4 ∵z=2

⇒α=±2

Topic: Mathematics, Complex Number, JEE Main
If frequency of electromagnetic wave is 60 MHz and it travels in air along z direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave in m is :

Options:
A 2.5
B 10
C 5
D 2

The correct answer is:
5

Solution:
As speed of electromagnetic wave is given by, c=fλ. Therefore,

λ=cf=3×10860×106=5 m.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (01 Feb Shift 2)
If from a point $\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})$, perpendiculars $\mathrm{PA}$ and $\mathrm{PB}$ are drawn to $\mathrm{YZ}$ and $\mathrm{ZX}$ planes respectively, then the equation of the plane $\mathrm{OAB}$ is

Options:
A $b c x+$ cay $+a b z=0$
B bcx $+$ cay $-a b z=0$
C $\mathrm{bcx}-\mathrm{cay}+\mathrm{abz}=0$
D bcx - cay - abz = 0

The correct answer is:
bcx $+$ cay $-a b z=0$

Solution:
$\mathrm{A}(\mathrm{o}, \mathrm{b}, \mathrm{c}) \mathrm{B}(\mathrm{a}, \mathrm{o}, \mathrm{c}), \mathrm{O}(\mathrm{o}, \mathrm{o}, \mathrm{o})$ Perpendicular vector is
$\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}$
$\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c\end{array}\right|=b c \hat{i}+a c \hat{j}-a b \hat{k}$
$\therefore$ Equation of plane is $b c x+a c y-a b z=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If from a wire of length 36 meter a rectangle of greatest area is made, then its two adjacent sides in meter are

Options:
A 6,12
B 9, 9
C 10, 8
D 13, 5

The correct answer is:
9, 9

Solution:
$\begin{aligned} & \text { Given } 2(a+b)=36, \quad a+b=18 \\ & \text { Area of rectangle }=a b=a(18-a) \\ & A=18 a-a^2, \therefore \frac{d A}{d a}=18-2 a \\ & \text { Put } \frac{d A}{d a}=0 \text { i.e., } 18-2 a=0 \Rightarrow a=9 ; b=9\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
If from mean value theorem, $f^{\prime}\left(x_1\right)=\frac{f(b)-f(a)}{b-a}$, then

Options:
A $a \lt x_1 \leq b$
B $a \leq x_1 \lt b$
C $a \lt x_1 \lt b$
D $a \leq x_1 \leq b$

The correct answer is:
$a \lt x_1 \lt b$

Solution:
According to mean value theorem, In interval $[a, b]$ for $f(x)$
$\begin{aligned}& \frac{f(b)-f(a)}{b-a}=f^{\prime}(c) \\& \therefore a \lt x_1 \lt b .\end{aligned}, \text { where } a \lt c \lt b$

Topic: Mathematics, Application of Derivatives, JEE Main
If from $R \rightarrow R, f(x)=(x+1)^2, g(x)=x^2+1$, then $(f \circ g)(-3)$ equals

Options:
A $121$
B $112$
C $211$
D None of these

The correct answer is:
$121$

Solution:
$\begin{aligned} & f(x)=(1+x)^2 \text { and } g(x)=x^2+1 \\ & \text { and } f \circ g(-3)=f[g(-3)]=f[9+1]=f[10] \\ & \Rightarrow f \circ g(-3)=f[10]=[11]^2=121 .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If from the top of a tower, 60 metres high, the angles of depression of the top and floor of a house are 30o and 60o respectively, then the height (in metres) of the house is

Options:

The correct answer is:
40

Solution:
Let d be the distance between the tower and house
∴d=60cot⁡60o
⇒d=60-hcot⁡30o
⇒60cot⁡60o=60-hcot⁡30o
⇒h=603-133=40m

Topic: Mathematics, Heights and Distances, JEE Main
If from the top of a tower 80 meters high the angles of depression of the top and bottom of a house are 30o and 45o respectively, then the height of the house is

Options:
A 403 meters
B 403-13 meters
C 803-13 meters
D 403+13 meters

The correct answer is:
803-13 meters

Solution:
Let, OA is tower and BC is house
Given, OA=80 meters
∠ACD=30o
∠ABO=45o
Let, BC=h meters
In ΔABO,

OAOB=tan⁡45o⇒OB=80tan⁡45o=80 meters
In ΔADC,

ADDC=tan⁡30o
OA-ODOB=13∵OB=DC
80-h80=13
803-3h=80⇒h=803-13 meters

Topic: Mathematics, Heights and Distances, JEE Main
If function fx=ax+7;0≤x<2bx+1;x≥2 is differentiable ∀x≥0, then 2a+4b is equal to

Options:
A 1
B 5
C 4
D 9

The correct answer is:
1

Solution:
fx must be continuous and differentiable at x=2
For continuity at x=2
f2-=f2+=f2
a2+7=b2+1⇒3a=2b+1…i
For differentiability at x=2
f ′ x = a 2 x+7 ; 0≤x<2 b ; x≥2
f ′ 2 − = f ′ 2 +
a22+7=b⇒a=6b…ii
From i & (i), we get,

18b=2b+1⇒b=116
2a+4b=12b+4b=16b=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If f(x) = Î»2x+3, 0â¤xâ¤3Î¼x+12, 3<xâ¤9 is differentiable at x=3, then the value of Î»+Î¼ is equal to

Options:

The correct answer is:
8

Solution:
∵f(x) is continuous at x=3
∴LHL=RHL=f(3)
λ2(3)+3=μ(3)+12 ⇒λ=μ+4 ......(1)
Also f(x) is differentiable at x=3
∴ at x=3 LHD=RHD
λ22(3)+3·2=μ⇒λ=3μ ......(2)
By (1) and (2)

μ=2,λ=6

Topic: Mathematics, Continuity and Differentiability, JEE Main
If \(f(x), f^{\prime}(x) f^{\prime \prime}(x)\), are positive functions and \(f(0)=1, f^{\prime}(0)=2\), then the solution of the differential equation \(\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0\) is

Options:
A \(e^{2 x}\)
B \(2 \sin x+1\)
C \(\sin ^2 x+2 x+1\)
D \(e^{4 x}\)

The correct answer is:
\(e^{2 x}\)

Solution:
\(\begin{aligned}
& \text {Since, }\left|\begin{array}{ll}
f(x) & f^{\prime}(x) \\
f^{\prime}(x) & f^{\prime \prime}(x)
\end{array}\right|=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^2 \\
& \Rightarrow \quad \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\
\end{aligned}\)
On, integrating both sides, we get
\(\begin{aligned}
& \int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} d x=\int \frac{f^{\prime}(x)}{f(x)} d x \\
& \Rightarrow \quad \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+c \\
& \because \quad f(0)=1 \text { and } f^{\prime}(0)=2 \\
& \therefore \quad c=\log _e 2 \\
& \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+\log _e 2 \\
& \Rightarrow \quad f^{\prime}(x)=2 f(x) \\
& \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2 \\
& \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\int 2 d x \Rightarrow \log _e f(x)=2 x+c \\
& \because \quad f(0)=1 \text {, so } c=0 \\
& \because \quad \log _e f(x)=2 x \Rightarrow f(x)=e^{2 x} \\
\end{aligned}\)
Hence, option (1) is correct.

Topic: Mathematics, Determinants, JEE Main
If \( f(x) \) is a differentiable function such that \( \int f(x) d x=2[f(x)]^{2}+C \), (where, \( C \) is the constant of integration) and \( f(1)=1 / 4 \), then \( f(\pi) \) equals

Options:
A \( \frac{3}{4} \)
B \( \frac{\pi}{4} \)
C \( \frac{2}{3} \)
D \( \frac{7}{6} \)

The correct answer is:
\( \frac{\pi}{4} \)

Solution:
Differentiating w.r.t. ‘x’ we get f x =4f x · f ′ x
f ′ x =0 or 14=fx
Integrating, we have
∫ 1 4 dx=∫ f ′ x dx
⇒x4=fx+C
As f1=14→C=0
⇒fx=x4
⇒fπ=π4

Topic: Mathematics, Indefinite Integration, JEE Main